10113_9789813143746_tp.indd 1 25/1/18 5:05 PM Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Mathematical Olympiad Series ISSN: 1793-8570 Series Editors: Lee Peng Yee (Nanyang Technological University, Singapore) Xiong Bin (East China Normal University, China) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Published Vol. 15 Mathematical Olympiad in China (2011–2014): Problems and Solutions edited by Bin Xiong (East China Normal University, China) & Peng Yee Lee (Nanyang Technological University, Singapore) Vol. 14 Probability and Expectation by Zun Shan (Nanjing Normal University, China) translated by: Shanping Wang (East China Normal University, China) Vol. 13 Combinatorial Extremization by Yuefeng Feng (Shenzhen Senior High School, China) Vol. 12 Geometric Inequalities by Gangsong Leng (Shanghai University, China) translated by: Yongming Liu (East China Normal University, China) Vol. 11 Methods and Techniques for Proving Inequalities by Yong Su (Stanford University, USA) & Bin Xiong (East China Normal University, China) Vol. 10 Solving Problems in Geometry: Insights and Strategies for Mathematical Olympiad and Competitions by Kim Hoo Hang (Nanyang Technological University, Singapore) & Haibin Wang (NUS High School of Mathematics and Science, Singapore) Vol. 9 Mathematical Olympiad in China (2009–2010) edited by Bin Xiong (East China Normal University, China) & Peng Yee Lee (Nanyang Technological University, Singapore) Vol. 8 Lecture Notes on Mathematical Olympiad Courses: For Senior Section (In 2 Volumes) by Xu Jiagu (Fudan University, China) The complete list of the published volumes in the series can be found at http://www.worldscientific.com/series/mos Suqi - 10113 - Mathematical Olympiad in China.indd 1 19-01-18 10:39:14 AM 10113_9789813143746_tp.indd 2 25/1/18 5:05 PM Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Published by East China Normal University Press 3663 North Zhongshan Road Shanghai 200062 China and World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE Library of Congress Cataloging-in-Publication Data Names: Xiong, Bin. | Lee, P. Y. (Peng Yee), 1938– Title: Mathematical Olympiad in China (2011–2014) : problems and solutions / editors, Bin Xiong, East China Normal University, China, Peng Yee Lee, Nanyang Technological University, Singapore. Description: [Shanghai, China] : East China Normal University Press ; [Hackensack, New Jersey] : World Scientific, 2017. | Series: Mathematical olympiad series ; vol. 15 Identifiers: LCCN 2017002034| ISBN 9789813143746 (hardcover : alk. paper) | ISBN 9813143746 (hardcover : alk. paper) | ISBN 9789813142930 (pbk. : alk. paper) | ISBN 9813142936 (pbk. : alk. paper) Subjects: LCSH: International Mathematical Olympiad. | Mathematics--Problems, exercises, etc. | Mathematics--Competitions--China. Classification: LCC QA43 .M31453 2017 | DDC 510.76--dc23 LC record available at https://lccn.loc.gov/2017002034 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. Copyright © 2018 by East China Normal University Press and World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher. For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher. For any available supplementary material, please visit http://www.worldscientific.com/worldscibooks/10.1142/10113#t=suppl Typeset by Stallion Press Email: [email protected] Printed in Singapore Suqi - 10113 - Mathematical Olympiad in China.indd 2 19-01-18 10:39:14 AM Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Pr e f ac e Thef i r s tt ime Ch i na pa r t i c i t edi nIMO wa si n1985,when pa twos t uden t swe r es en tt ot he26 t hIMO.S i nce1986,Ch i naha s at eamo fs i xs t uden t sa teve r ti n1998wheni twa s yIMOexcep he l di n Ta iwan.Sof a r,upt o2014,Ch i naha sach i evedt he numbe rone r ank i ng f or 20 t ime si nt eam e f for t. A g r ea t ma or i t fs t uden t sr e c e i vedgo l d meda l s.Thef ac tt ha tCh i na j yo ob t a i neds uchencou r ag i ngr e s u l ti sduet o,ont heonehand, Ch i ne s es t uden t s ha rdworkandpe r s eve r ance,andont heo t he r hand,t hee f f or to ft eache r si ns choo l sandt het r a i n i ngo f f e r ed by na t i ona lcoache s. We be l i eve i ti sa l s o ar e s u l to ft he educa t i on s s t em i n Ch i na,i n pa r t i cu l a r,t he empha s i s on y t r a i n i ngo fba s i cs k i l l si ns c i enc eeduca t i on. Thema t e r i a l so ft h i sbookcomef r omas e r i e so ffou rbooks ( i nCh i ne s e)onFo rwa rdt oIMO :ACo l l e c t i ono thema t i ca l fMa Ol i adPr ob l ems ( 2011 2014).I ti saco l l ec t i onofpr ob l ems ymp ands o l u t i on so ft hema orma t hema t i ca lcompe t i t i onsi nCh i na. j I tpr ov i de sag l imp s eo fhowt heCh i nana t i ona lt eami ss e l ec t ed and f ormed. F i r s t, t he r e i s t he Ch i na Ma t hema t i ca l Compe t i t i on,ana t i ona leven t.I ti she l dont hes econdSunday o f Oc t obe reve r r.Thr ought hecompe t i t i on,abou t300 y yea s t uden t sa r es e l e c t edt oj o i nt heCh i na Ma t hema t i ca lOl i ad ymp ( common l st hewi n t e rcamp),ori nshor tCMO,i n yknowna ⅴ Ma t hema t i c a lOl i adi nCh i na ymp t hef o l l owi ngJ anua r a s t sf orf i veday s.Bo t ht het y.CMOl ype and t he d i f f i cu l t ft he pr ob l ems ma t ch t ho s e of IMO. yo S imi l a r l t uden t sa r eg i vent hr ee prob l emst os o l vei n4 .5 y,s hour seach day.Fr om CMO,abou t 50 t o 60 s t uden t sa r e Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. s e l e c t edt of orm ana t i ona lt r a i n i ngt eam.Thet r a i n i ngt ake s l ac ef ortwo week si nt he mon t hof Ma r ch.Af t e rf ou rt os i x p t e s t s,p l u stwoqua l i f i ngexami na t i ons,s i xs t uden t sa r ef i na l l y y eam,t ak i ngpa r ti nIMOi nJ u l s e l e c t edt oformt hena t i ona lt y o ft ha tyea r. In v i ew o ft he d i f f e r enc e si n educa t i on, cu l t u r e and e conomy o f we s t e rn pa r to f Ch i nai n compa r i s on wi t ht he coa s t a lpa r ti nEa s tCh i na,ma t hema t i ca lcompe t i t i onsi n We s t Ch i nad i dno tdeve l opa sf a s ta si nt her e s toft hecoun t r y.In orde rt opr omo t et heac t i v i t fma t hema t i ca lcompe t i t i onand yo t oenhanc et hel eve lo fma t hema t i ca lcompe t i t i on,s t a r t i ngf rom 2001,Ch i na Ma t hema t i ca lOl i ad Commi t t eeor i ze st he ymp gan Ch i na We s t e rn Ma t hema t i ca lOl i ad.Thet optwo wi nne r s ymp wi l lbeadmi t t edt ot he na t i ona lt r a i n i ngt eam.Throught he CWMO,t he r e have beentwos t uden t sen t e r i ng t he na t i ona l t eamandr e c e i v i ngGo l d Meda l sf ort he i rpe r f ormancea tIMO. S i nc e1995,f ora qu i t el ongpe r i odt he r e wa snof ema l e s t uden ti nt he Ch i ne s e na t i ona lt eam.In orde rt oencour age mor e f ema l e s t uden t s pa r t i c i t i ng i n t he ma t hema t i ca l pa compe t i t i on, s t a r t i ng f r om 2002, Ch i na Ma t hema t i ca l Ol i adCommi t t eeconduc t edt heCh i na Gi r l s Ma t hema t i ca l ymp Ol i ad.Aga i n,t het optwowi nne r swi l lbeadmi t t edd i r ec t l ymp y i n t ot hena t i ona lt r a i n i ngt eam. Theau t hor so ft h i sbooka r ecoache soft heCh i nana t i ona l t eam.Theya r eXi ongBi n,LiShenghong,LengGang s ong,Wu J i anp i ng, Chen Yonggao, Li We i i ng, Zhu gu, Yu Hongb ⅵ Pr e a c e f Huawe i,Feng Zh i u Sh i x i ong, Qu Zhenhua, Wang gang, Li We i i hu i. Tho s e who t ook pa r ti nt he ye, and Zhang S t r ans l a t i on work a r e Xi ong Bi n, Wang Shanp i ng, Li u Yongmi ng. Wea r eg r a t e f u lt o Qi u Zonghu, Wang J i e, Wu Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. J i anp i ng,andPanChengb i aofort he i rgu i danceanda s s i s t anc e t oau t hor s.Wea r eg r a t e f u lt o NiMi ngofEa s tCh i na Norma l Un i ve r s i t e s s.The i re f for tha she l obea s i e r. yPr ped makeourj Wea r ea l s og r a t e f u lt oZhangJ iof Wor l dSc i en t i f i cPub l i sh i ng f orhe rha rdworkl ead i ngt ot hef i na lpub l i ca t i onoft hebook. Xi ongBi n Oc t obe r2017 ⅶ b2530 International Strategic Relations and China’s National Security: World at the Crossroads Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. This page intentionally left blank b2530_FM.indd 6 01-Sep-16 11:03:06 AM Int roduc t i on Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Ea r l s yday The In t e rna t i ona l Ma t hema t i ca l Ol i ad (IMO ), ymp f oundedi n 1959,i sone o ft he mo s tcompe t i t i veand h i l gh y i n t e l l e c t ua lac t i v i t i e si nt hewor l df orh i choo ls t uden t s. ghs Even be f or e IMO,t he r e we r ea l r eady many coun t r i e s wh i ch had ma t hema t i c scompe t i t i on.They we r e ma i n l he yt coun t r i e si n Ea s t e rn Eu r opeandi n As i a.In add i t i ont ot he l a r i za t i on o f ma t hema t i c s and t he conve r n popu gence i educa t i ona ls s t emsamong d i f f e r en tcoun t r i e s,t hes uc c e s sof y ma t hema t i ca lcompe t i t i on sa tt he na t i ona ll eve l prov i ded a f ounda t i on f or t he s e t t i ng up o f IMO. The coun t r i e st ha t a s s e r t edg r ea ti nf l uenc ea r e Hunga r hef orme rSov i e tUn i on y,t andt heUn i t edS t a t e s.He r ei sabr i e fr ev i ew o ft heIMO and ma t hema t i ca lcompe t i t i oni nCh i na. In1894,t heDepa r tmen to fEduca t i oni nHunga r s s eda ypa mo t i onandde c i dedt oconduc ta ma t hema t i ca lcompe t i t i onfor t he s e conda r choo l s. The we l l -known s c i en t i s t, J. von ys Et övö s,wa st heMi n i s t e ro fEduca t i ona tt ha tt ime.Hi ss uppor t i nt he even t had made i tas uc c e s s and t hu si t wa s we l l l i c i z ed.Inadd i t i on,t hes uc c e s so fh i ss on,R .vonEt övö s, pub whowa sa l s oaphy s i c i s t, i nprov i ngt hepr i nc i l eofequ i va l ence p n s t e i nt hrough o ft he gene r a lt heor fr e l a t i v i t yo y by A .Ei ⅸ Ma t hema t i c a lOl i adi nCh i na ymp expe r imen t,hadbrough tHunga r ot hewor l ds t agei ns c i ence. yt The r ea f t e r,t hepr i z ef orma t hema t i c scompe t i t i oni n Hunga r y övö s pr wa s named “Et i ze”. Th i s wa st he f i r s t forma l l y or i z ed ma t hema t i ca lcompe t i t i on i nt he wor l d.In wha t gan Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. fo l l ows,Hunga r ndeed pr oduced al o tof we l l -known y hadi s c i en t i s t si nc l ud i ng L .Fe é r,G .S z e r j gö,T.Radó,A .Haa and M .Ri e s z( i nr ea lana l s i s),D .Kön i i ncomb i na t or i c s), g( y T.von Kármán ( i nae rodynami c s),andJ.C .Ha r s any i( i n heor l s owont heNobe lPr i zeforEconomi c s gamet y),whohada i n1994.Theya l lwe r et he wi nne r sof Hunga r t hema t i ca l y ma compe t i t i on.Thet op s c i en t i f i c gen i u so f Hunga r y,J.von soneo ft hel ead i ng ma t hema t i c i an si nt he20 t h Neumann,wa c en t u r sove r s ea s wh i l et he compe t i t i on t ook y.Neumann wa l ac e.La t e rhed i di th ims e l fandi tt ookh im ha l fanhourt o p comp l e t e. Ano t he r ma t hema t i c i an wor t h men t i on i ng i st he s.Hewa sapup i lof h i l oduc t i venumbe rt heor i s tP.Erdö gh ypr Fe é r and a l s o a wi nne roft he Wo l f Pr i ze.Erdö s wa sve r j y s s i ona t e abou t ma t hema t i ca l compe t i t i on and s e t t i ng pa compe t i t i onque s t i on s.Hi scon t r i bu t i ont od i s c r e t ema t hema t i c s wa s un i r ea t l i i f i can t.The r ap i d pr og r e s s and que and g ys gn deve l opmen to fd i s c r e t e ma t hema t i c s ove rt he s ub s equen t de cade shadi nd i r e c t l nf l uencedt het sofque s t i onss e ti n yi ype IMO.Ani n t e rna t i ona l l e cogn i zed pr i ze nameda f t e r Erdö s yr wa st ohonourt ho s e whohadcon t r i bu t edt ot heeduca t i onof ma t hema t i ca lcompe t i t i on.Pr o f e s s or Qi u Zonghu f rom Ch i na hadwont hepr i zei n1993. In1934,af amou s ma t hema t i c i an B.De l one conduc t eda ma t hema t i ca lcompe t i t i onf orh i choo ls t uden t si nLen i ng r ad ghs ( nowS t.Pe t e r s bu r s cowa l s os t a r t edor i z i ng g).In1935,Mo gan s ucheven t.Ot he rt hanbe i ngi n t e r r up t eddur i ngt he Wor l d Wa r ⅹ In t r oduc t i on I I,t he s eeven t shad beenca r r i edonun t i lt oday.Asfort he Ru s s i an Ma t hema t i ca lCompe t i t i on ( l a t e rr enameda st heSov i e t Ma t hema t i ca lCompe t i t i on),i t wa s no ts t a r t ed un t i l1961. Thu s,t hef orme rSov i e tUn i onand Ru s s i abecamet hel ead i ng Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. r so f Ma t hema t i ca lOl i ad.A l o tofg r andma s t e r si n powe ymp ma t hema t i c si nc l ud i ngt he g r ea t A .N .Ko lmogo r ov we r ea l l ve r t hu s i a s t i c abou tt he ma t hema t i ca lcompe t i t i on.They y en s t i ons for t he wou l d pe r s ona l l nvo l ve i ns e t t i ng t he que yi compe t i t i on. The f orme r Sov i e t Un i on even ca l l ed i tt he Ma t hema t i ca l Ol i ad, be l i ev i ng t ha t ma t hema t i c si st he ymp “ s t i c so ft h i nk i ng”.The s e po i n t sofv i ew gavea g r ea t gymna impac tont heeduca t i ona lcommun i t nne roft heFi e l d s y.Thewi Meda li n1998,M .Kon t s e v i ch,wa sonc et hef i r s tr unne r upof he t he Ru s s i an Ma t hema t i ca l Compe t i t i on. G . Ka s r ov,t pa i n t e rna t i ona lche s sg r andma s t e r,wa soncet hes econdr unne r up.Gr i r iPe r e lman,t e l d s Meda li n2006 hewi nne ro ft heF i go ( bu thede c l i ned),whos o l vedt hePo i nca r é sCon ec t ur e,wa sa j l dmeda l i s to fIMOi n1982. go In t he Un i t ed S t a t e so f Ame r i ca, due t ot he ac t i ve omo t i onbyt her enowned ma t hema t i c i an G . D. B i r kho f f and pr h i ss on,t oge t he r wi t h G .Pó l he Pu t nam ma t hema t i c s ya,t compe t i t i on wa sor i z edi n1938f orj un i orunde r r adua t e s. gan g Manyo ft he que s t i on s we r e wi t h i nt hes cope o fh i choo l ghs s t uden t s.Thet opf i vecon t e s t an t so ft hePu t nam ma t hema t i ca l compe t i t i on wou l dbeen t i t l edt ot he membe r sh i t nam. p ofPu Many o ft he s e we r e even t ua l l t s t and i ng ma t hema t i c i ans. y ou The r ewe r et hef amou sR .Fe wi nne ro ft heNobe lPr i ze ynman ( forPhy s i c s,1965),K .Wi l s on ( wi nne roft heNobe lPr i z efor Phy s i c s,1982),J.Mi l no r( wi nne ro ft heFi e l d sMeda l,1962), D .Mumf o rd (wi nne ro ft he F i e l d s Meda l,1974),and D . ⅺ Ma t hema t i c a lOl i adi nCh i na ymp i l l en ( wi nne ro ft heF i e l d sMeda l,1978). Qu S i nce1972,i norde rt o pr epa r efort heIMO,t he Un i t ed S t a t e s of Ame r i ca Ma t hema t i ca l Ol i ad (USAMO ) wa s ymp or i z ed.Thes t anda rd of que s t i on s po s ed wa s ve r i gan yh gh, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. r a l l e lt ot ha to ft he Wi n t e rCampi nCh i na.Pr i ort ot h i s,t he pa Un i t ed S t a t e s had or i z ed gan Ame r i can Hi l gh Schoo Ma t hema t i c s Exami na t i on (AHSME ) for t he h i choo l gh s s t uden t ss i nc e1950.Th i swa sa tt hej un i orl eve lye tt he mo s t t hema t i c scompe t i t i oni nAme r i ca.Or i i na l l i twa s l a rma g y, popu l anned t os e l e c t abou t 100 con t e s t an t sf rom AHSME t o p r t i c i t ei nUSAMO.Howeve r,duet ot hed i s c r epancyi nt he pa pa l eve lo fd i f f i cu l t tween t he two compe t i t i on s and o t he r y be r e s t r i c t i on s,f r om 1983 onwa rd s, an i n t e rmed i a t el eve l of compe t i t i on, name l r i can Inv i t a t i ona l Ma t hema t i c s y, Ame Exami na t i on (AIME ), wa si n t r oduced. Hence for t h bo t h AHSMEandAIMEbe camei n t e rna t i ona l l l l -known.Af ew y we c i t i e si n Ch i na had pa r t i c i t edi nt he compe t i t i on and t he pa r e s u l t swe r eencou r ag i ng. S imi l a r l si nt hef orme rSov i e tUn i on,t he Ma t hema t i ca l ya Ol i adeduca t i on wa s wi de l e cogn i zedi n Ame r i ca.The ymp yr book “How t oSo l vei t”wr i t t en by Geo r ePo l l ong wi t h g ya a many o t he rt i t l e s had been t r ans l a t ed i n t o many d i f f e r en t l anguage s.Ge o r e Po l i ded a who l es e r i e s of gene r a l g ya prov heur i s t i c sf ors o l v i ngpr ob l emsofa l lk i nd s.Hi si nf l uencei nt he educa t i ona lcommun i t nCh i nas hou l dno tbeunde r e s t ima t ed. yi I n t e r na t i ona lMa t hema t i c a lOl i ad ymp In1956,t heEa s tEu r opeancoun t r i e sandt heSov i e tUn i on t ookt hei n i t i a t i vet o or i z et heIMO forma l l i r s t gan y.Thef In t e rna t i ona l Ma t hema t i ca l Ol i ad ( IMO ) wa s he l di n ymp ⅻ In t r oduc t i on Br a s ov,Roman i a,i n1959.Att het ime,t he r ewe r eon l even ys r t i c i t i ngcoun t r i e s,name l i a,Bu l r i a,Po l and, pa pa y,Roman ga Hunga r echo s l ovak i a, Ea s t Ge rmany and t he Sov i e t y, Cz Un i on.Sub s equen t l he Un i t ed S t a t e sof Ame r i ca, Un i t ed y,t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Ki ngdom,Fr anc e,Ge rmanyanda l s oo t he rcoun t r i e si nc l ud i ng t ho s ef r om As i aj o i ned.Today,t heIMOhadmanagedt or each a lmo s ta l lt hedeve l opedanddeve l op i ngcoun t r i e s.Exc ep ti n e sf aced by t he ho s t t heyea r1980 duet of i nanc i a ld i f f i cu l t i coun t r l i a,t he r ewe r ea l r eady49 Ol i ad she l dand y,Mongo ymp 97coun t r i e spa r t i c i t i ng. pa The ma t hema t i ca lt op i c si nt he IMO i nc l ude Al a, gebr Comb i na t or i c s,Geome t r rt heor s ea r ea s had y, Numbe y.The ov i dedgu i danc ef ors e t t i ngque s t i onsfort hecompe t i t i ons. pr , Ot he rt hant hef i r s tf ew Ol i ad s eachIMOi snorma l l l d ymp yhe i nmi d J u l r randt het e s tpape rcons i s t sof6que s t i ons yeve yyea i na l l.Theac t ua lcompe t i t i onl a s t sf or2day sforat o t a lof9 hou r swhe r e pa r t i c i t sa r er equ i r edt ocomp l e t e3 que s t i ons pan eachday.Eachque s t i oni s7po i n t swi t ht o t a lupt o42po i n t s. Thef u l ls cor ef orat eam i s252 ma rks.Abou tha l f oft he r t i c i t s wi l l be awa rded a meda l, whe r e 1/12 wi l l be pa pan awa rdedago l dmeda l.Thenumbe r sofgo l d,s i l ve randbronze meda l sawa rdeda r ei nt her a t i oo f1: 2: 3appr ox ima t e l he y.Int ca s e when a pa r t i c i t pr ov i de sa be t t e rs o l u t i on t han t he pan o f f i c i a lan swe r,as c i a lawa rdi sg i ven. pe Eachpa r t i c i t i ngcoun t r l lt aket u rnt oho s tt heIMO. pa y wi Theco s ti sbornebyt heho s tcoun t r i nahads uc c e s s f u l l y.Ch y ho s t edt he31 s tIMOi n Be i i ng.Theeven thad madea g r ea t j impac tont hema t hema t i ca lcommun i t nCh i na.Ac cord i ngt o yi t her u l e sandr egu l a t i on so ft heIMO,a l lpa r t i c i t i ngcoun t r i e s pa a r er equ i r ed t os end a de l ega t i on cons i s t i ng o f al eade r,a Ma t hema t i c a lOl i adi nCh i na ymp depu t eade rand6con t e s t an t s.Thepr ob l emsa r econ t r i bu t ed yl byt hepa r t i c i t i ngcoun t r i e sanda r el a t e rs e l ec t edca r e f u l l pa yby t heho s tcoun t r ors ubmi s s i ont ot hei n t e rna t i ona lj ur e tup yf ys by t he ho s t coun t r t ua l l l ob l ems wi l l be y.Even y,on y 6 pr Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. ac c ep t edf oru s ei nt hecompe t i t i on.Theho s tcoun t r sno t ydoe i de any que s t i on. The s hor t l i s t ed prob l ems a r e prov s ub s equen t l r an s l a t ed,i f nec e s s a r n Eng l i s h, Fr ench, yt y,i t he rwork i ngl anguage s.Af t e rt ha t,t he Ge rman,Ru s s i anando t eam l eade r s wi l lt r an s l a t et he prob l ems i n t ot he i r own l anguage s. The an swe rs c r i t so f each pa r t i c i t i ng t eam wi l l be p pa ma rkedbyt het eam l eade randt hedepu t eade r.Thet eam yl l eade rwi l ll a t e rpr e s en tt hes c r i t so ft he i rcon t e s t an t st ot he p coord i na t or sf ora s s e s smen t.I ft he r ei sanyd i s t e,t hema t t e r pu wi l lbes e t t l edbyt hej u r u r sf ormedbyt heva r i ou s y.Thej yi t eaml eade r sandanappo i n t edcha i rmanbyt heho s tcoun t r y. Thej u r sr e s s i b l ef orde c i d i ngt hef i na l6prob l emsfort he yi pon compe t i t i on.The i r du t i e sa l s oi nc l udef i na l i z i ng t he g r ad i ng s t anda rd,en s u r i ng t he ac cu r acy o ft he t r an s l a t i on of t he ob l ems,s t anda rd i z i ng r ep l i e st o wr i t t en que r i e sr a i s ed by pr r t i c i t sdu r i ngt hecompe t i t i on,s i z i ng d i f f e r enc e s pa pan ynchron i ng r ad i ngbe tweent het eaml eade r sandt hecoord i na t or sand a l s ode c i d i ngont hecu t o f fpo i n t sf ort hemeda l sdepend i ngon t hecon t e s t an t s r e s u l t sa st hed i f f i cu l t i e so fprob l emseachyea r a r ed i f f e r en t. Ch i nahadpa r t i c i t edi nf orma l l nt he26 t hIMOi n1985. pa yi On l t uden t s we r es en t.S t a r t i ng f rom 1986,excep ti n ytwos 1998whent heIMO wa she l di nTa iwan,Ch i nahada lway ss en t 6 o f f i c i a l con t e s t an t st ot he IMO. Today, t he Ch i ne s e con t e s t an t sno ton l r f ormedou t s t and i ng l nt heIMO,bu t ype yi In t r oduc t i on a l s oi nt heIn t e rna t i ona lPhy s i c s,Chemi s t r orma t i c s,and y,Inf Bi o l ogy Ol i ad s.Th i scanber ega rdeda sani nd i ca t i ont ha t ymp Ch i na pay sg r ea ta t t en t i ont ot het r a i n i ng of ba s i csk i l l si n Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. ma t hema t i c sands c i enc eeduca t i on. Wi nne r so ft heIMO Amonga l lt heIMO meda l i s t s,t he r e we r e many oft hem whoeven t ua l l r ea t ma t hema t i c i an s.They we r ea l s o y becameg awa rdedt heF i e l d sMeda l,Wo l fPr i zeandNevan l i nnaPr i ze ( a nen tma t hema t i c spr i zeforcompu t i ngandi nforma t i c s). promi Inwha tf o l l ows,wenames omeo ft hewi nne r s. i l ve r meda l i s to f IMO i n 1959, wa s G . Ma r l i s,a s gu awa rdedt heF i e l d s Meda li n1978.L .Lova s z,who wont he Wo l fPr i zei n1999,wa sawa rdedt heSpec i a lAwa rdi nIMO i n e l d,ago cons e cu t i ve l n1965and1966.V .Dr l dmeda l i s tof f yi IMOi n1969,wa sawa rdedt heF i e l d s Meda li n1990.J.-C . Yoc coz and T.Gowe r s,who we r e bo t h awa rdedt he Fi e l d s Meda li n1998,we r ego l d meda l i s t si nIMOi n1974and1981 r e s c t i ve l i l ve rmeda l i s to fIMOi n1985,L .La o r f f gue, pe y.As wont heFi e l d sMeda li n2002.Ago l dmeda l i s tofIMOi n1982, Gr i r iPe r e lmanf r om Ru s s i a,wa sawa rdedt heF i e l d sMeda li n go 2006f ors o l v i ngt hef i na ls t ep oft hePo i nca r écon ec t ur e.In j 1986,1987,and1988,Te r enc eTao wonabronze,s i l ve r,and l d meda lr e s c t i ve l st he younge s t pa r t i c i tt o go pe y.He wa pan da t ei nt heIMO,f i r s tcompe t i nga tt heageoft en.Hewa sa l s o awa rdedt heF i e l d sMeda li n2006.Go l d meda l i s tofIMO1988 and1989, Ngo Bau Chao, won t he Fi e l d s Meda li n 2010, t oge t he r wi t h t he br onz e meda l i s t of IMO 1988, E. Li nden s t r aus s.Go l d meda l i s to fIMO 1994and1995,Ma r yam Mi r z akhan iwont heF i e l d s Meda li n2014.A go l d meda l i s tof Ma t hema t i c a lOl i adi nCh i na ymp IMOi n1995,Ar t u rAv i l awont heFi e l d sMeda li n2014. r,wa As i l ve rmeda l i s to fIMOi n1977,P.Sho sawa rded t he Nevan l i nna Pr i z e.A go l d meda l i s to fIMO i n1979,A . Razbo r ov,wa sawa rdedt he Nevan l i nna Pr i z e.Ano t he r go l d Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. rnov,wa sawa rdedt heCl ay meda l i s tofIMOi n1986,S.Smi l d meda l i s tofIMO i n Re s ea r ch Awa rd.V .La o r f f gue,a go 1990,wa sawa rdedt heEu r opean Ma t hema t i ca lSoc i e t i ze. ypr Hei sL .La o r syounge rbr o t he r. f f gue Al s o,a f amou s ma t hema t i c i an i n numbe rt heor y, N . El k i e s,whoi sa l s o a pro f e s s ora t Ha r va rd Un i ve r s i t s y, wa awa rdedago l d meda lofIMOi n1982.Ot he rwi nne r si nc l ude P.Kr onhe ime rawa rdedas i l ve rmeda li n1981andR . Tay l o ra con t e s t an to fIMOi n1980. Ma t hema t i c a lc ompe t i t i oni nCh i na Duet ova r i ou sr ea s ons,ma t hema t i ca lcompe t i t i oni nCh i na s t a r t edr e l a t i ve l a t ebu ti spr og r e s s i ngv i s l yl gorou y. “Wea r ego i ngt ohaveou rown ma t hema t i ca lcompe t i t i on t oo!”s a i d HuaLuog eng.Huai sahou s e -ho l dnamei nCh i na. Thef i r s t ma t hema t i ca lcompe t i t i on wa she l dconcur r en t l n yi Be i i ng,Ti an i n,Shangha iand Wuhani n 1956.Duet ot he j j l i t i ca ls i t ua t i ona tt het ime,t h i seven twa si n t e r r up t edaf ew po , t ime s.Un t i l1962 whent he po l i t i ca lenv i ronmen ts t a r t edt o impr ove, Be i i ng and o t he rc i t i e ss t a r t ed or i z i ng t he j gan compe t i t i on t hough no tr egu l a r l he e r a of Cu l t ur a l y.In t Revo l u t i on,t he who l e educa t i ona ls s t em i n Ch i na wa si n y chao s.Thema t hema t i ca lcompe t i t i oncamet oacomp l e t eha l t. Incon t r a s t,t hema t hema t i ca lcompe t i t i oni nt hef orme rSov i e t Un i onwa ss t i l lon i ngdu r i ngt hewa randa tat imeunde rt he go d i f f i cu l tpo l i t i ca ls i t ua t i on.Thecompe t i t i onsi n Mo s cow we r e In t r oduc t i on i n t e r r up t edon l ime sbe tween1942and1944.I twa si ndeed y3t commendab l e. eng In 1978,i t wa st he s i ng o fs c i ence. Hua Luog pr conduc t edt he Mi dd l eSchoo l Ma t hema t i ca lCompe t i t i onfor8 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. i nc e si n Ch i na.The ma t hema t i ca lcompe t i t i oni n Ch i na prov wa st henmak i ngaf r e s hs t a r tandemba rkedonaroado fr ap i d s s ed awayi n 1985.In commemor a t i ng deve l opmen t.Hua pa h im,acompe t i t i onnamedHuaLuog eng Go l dCupwa ss e tupi n 1986f ors t uden t si nGr ade6and7andi tha sag r ea timpac t. The ma t hema t i ca lcompe t i t i on si n Ch i nabe for e1980can becon s i de r ed a st hei n i t i a l pe r i od.The pr ob l ems we r es e t wi t h i nt hes copeo f mi dd l es choo lt ex t book s.Af t e r1980,t he compe t i t i on swe r eg r adua l l i ngt owa rd st hes en i or mi dd l e y mov s choo ll eve l.In 1981, t he Ch i ne s e Ma t hema t i ca l Soc i e t y de c i dedt o conduc tt he Ch i na Ma t hema t i ca l Compe t i t i on,a na t i ona leven tf orh i choo l s. ghs In1981,t heUn i t edS t a t e so fAme r i ca,t heho s tcoun t r yof IMO,i s s uedani nv i t a t i ont oCh i nat opa r t i c i t ei nt heeven t. pa On l n 1985, Ch i na s en t two con t e s t an t st o pa r t i c i t e yi pa i nf orma l l nt heIMO.Ther e s u l t swe r eno tencour ag i ng.In yi v i ew o ft h i s,ano t he r ac t i v i t l l ed t he Wi n t e r Camp wa s y ca conduc t ed a f t e rt he Ch i na Ma t hema t i ca l Compe t i t i on. The Wi n t e rCamp wa sl a t e rr enamed a st he Ch i na Ma t hema t i ca l Ol i adorCMO.The wi nn i ngt eam wou l dbeawa rdedt he ymp Che rnSh i i ng Shen Cup.Ba s edont heou t comea tt he Wi n t e r Camp,a s e l e c t i on wou l d be made t o form t he 6 -membe r na t i ona lt eam f orIMO.Fr om 1986 onwa rd s,o t he rt hant he r when IMO wa s or i z edi n Ta iwan, Ch i na had been yea gan s end i nga6 -membe rt eamt oIMO.Upt o2011,Ch i nahadbeen awa rdedt heove r a l lt eamchamp i onf or17t ime s. ⅹ ⅴ ⅱ Ma t hema t i c a lOl i adi nCh i na ymp In1990,Ch i nahads uc c e s s f u l l s t edt he31s tIMO.I t y ho s howedt ha tt hes t anda rdo fma t hema t i ca lcompe t i t i oni nCh i na ha sl eve l edt ha to fo t he rl ead i ngcoun t r i e s.F i r s t,t hef ac tt ha t Ch i naach i eve st heh i s tma rk sa tt he31s tIMOfort het eam ghe Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. i sanev i denc eo ft hee f f e c t i vene s so ft hepy r ami dappr oachi n s e l e c t i ng t he con t e s t an t si n Ch i na.Second l he Ch i ne s e y,t ma t hema t i c i an shads imp l i f i edand mod i f i edove r100pr ob l ems ot het eaml eade r so ft he35coun t r i e sfor ands ubmi t t edt hemt t he i rpe r u s a l.Even t ua l l ob l emswe r er ecommended.At y,28pr t heend,5pr ob l emswe r echo s en ( IMOr equ i r e s6 pr ob l ems). Th i si sano t he rev i denc et os howt ha tCh i naha sach i evedt he h i s tqua l i t ns e t t i ngpr ob l ems.Th i rd l heanswe rs c r i t s ghe yi y,t p o ft hepa r t i c i t swe r ema rkedbyt heva r i ou st eaml eade r sand pan a s s e s s edbyt hecoord i na t or swho we r enomi na t edbyt heho s t coun t r i e s.Ch i na had f ormed a g r oup 50 ma t hema t i c i anst o scoord i na t or swho wou l den s ur et heh i cur acyand s e r vea ghac f a i rne s si nma rk i ng.Thema rk i ngpr oc e s swa scomp l e t edha l fa dayea r l i e rt hani twa ss chedu l ed.Four t h l ha twa st hef i r s t y,t eve rIMO or i z edi n As i a.Theou t s t and i ngpe r formanc eby gan Ch i nahadencou r agedt heo t he rdeve l op i ngcoun t r i e s,e s i a l l pec y t ho s ei n As i a.The or i z i ng andcoord i na t i ng work oft he gan IMObyt heho s tcoun t r sa l s or ea s onab l y wa ygood. In Ch i na,t he ou t s t and i ng pe r f ormancei n ma t hema t i ca l compe t i t i oni sar e s u l to f many con t r i bu t i on sf r om t he a l l r t e r so f ma t hema t i ca l commun i t r ea r et he o l de r qua y. The r a t i ono fma t hema t i c i an s,mi dd l e agedma t hema t i c i ansand gene a l s ot he mi dd l eande l emen t a r choo lt eache r s.The r ei sone ys r s onwhode s e r ve sas c i a lmen t i onandhei sHuaLuog eng. pe pe Hei n i t i a t edandpr omo t edt hema t hema t i ca lcompe t i t i on.Hei s a l s ot he au t horo ft hef o l l owi ng book s:Beyond Yang hu i s ⅹ ⅴⅲ In t r oduc t i on Tr i ang l e,Beyondt he p i of Zu Chong zh i,Beyondt he Mag i c z i, Ma Compu t a t i on of Sun t hema t i ca l Induc t i on, and Ma t hema t i ca lPr ob l ems of Bee Hi ve.The s e we r eh i s books de r i vedf r om ma t hema t i c scompe t i t i on s.When Ch i nar e s umed Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. ma t hema t i ca lcompe t i t i oni n 1978,he pa r t i c i t edi ns e t t i ng pa l emsandg i v i ngc r i t i os o l u t i on soft heprob l ems.Ot he r prob quet ou t s t and i ng books de r i ved f r om t he Ch i ne s e ma t hema t i c s t t i ceandAr ea compe t i t i on sa r e:Symme t r fu,La yby DuanXue by Mi nS i he,OneS t r oke Dr awi ng and Po s tman Prob l em by J i angBo u. j Af t e r1980,t he younge r ma t hema t i c i ansi n Ch i na had t aken ove rf rom t he o l de r gene r a t i on o f ma t hema t i c i ansi n r unn i ng t he ma t hema t i ca l compe t i t i on. They worked and s t r i vedha rdt obr i ngt hel eve lo f ma t hema t i ca lcompe t i t i oni n sones uch ou t s t and i ng Ch i nat oa new he i t.Qi u Zonghu i gh r epr e s en t a t i ve.Fromt het r a i n i ngo fcon t e s t an t sandl ead i ngt he t eam 3t ime st oIMO t ot he or i z i ng o ft he31 t hIMO i n gan Ch i na,hehadcon t r i bu t edpr omi nen t l sawa rdedt heP. yandwa i z e. Erdö spr Pr e r a t i onf o rIMO pa Cur r en t l hes e l e c t i onpr oc e s sofpa r t i c i t sf orIMOi n y,t pan Ch i nai sa sf o l l ows. Fi r s t,t he Ch i na Ma t hema t i ca l Compe t i t i on,a na t i ona l compe t i t i onforh i l s, i sor i z edont hes econdSunday ghSchoo gan i n Oc t obe r eve r r.The ob e c t i ve sa r e:t oi nc r ea s et he y yea j i n t e r e s tofs t uden t si nl ea rn i ng ma t hema t i c s,t o pr omo t et he deve l opmen to fco cur r i cu l a rac t i v i t i e si n ma t hema t i c s,t ohe l p impr ove t he t each i ng o f ma t hema t i c si nh i choo l s,t o gh s d i s cove randcu l t i va t et het a l en t sanda l s ot o pr epa r ef ort he ⅹ ⅰ ⅹ Ma t hema t i c a lOl i adi nCh i na ymp IMO.Th i s happens s i nce 1981. Cur r en t l he r ea r e abou t yt 200, 000pa r t i c i t st ak i ngpa r t. pan Thr ought heCh i na Ma t hema t i ca lCompe t i t i on,a round350 o fs t uden t sa r es e l e c t edt ot akepa r ti nt heCh i na Ma t hema t i ca l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Ol i adorCMO,t ha ti s,t he Wi n t e rCamp.TheCMOl a s t s ymp f or5 day sandi she l di nJ anua r r r.Thet sand y eve yyea ype d i f f i cu l t i e so ft he prob l emsi n CMO a r es imi l a rt ot heIMO. The r ea r ea l s o3prob l emst obecomp l e t edwi t h i n4. 5hour seach r,t hes cor ef oreach pr ob l emi s21 ma rkswh i ch day.Howeve addupt o126 ma rk si nt o t a l.S t a r t i ngf rom 1990,t he Wi n t e r Camp i n s t i t u t ed t he Che rn Sh i i ng Shen Cup for t eam champ i on sh i he Wi n t e r Camp wa s of f i c i a l l p.In 1991, t y r enameda st he Ch i na Ma t hema t i ca lOl i ad (CMO).I ti s ymp s imi l a rt ot heh i s tna t i ona lma t hema t i ca lcompe t i t i oni nt he ghe f orme rSov i e tUn i onandt heUn i t edS t a t e s. The CMO awa rd st he f i r s t,s econd and t h i rd pr i ze s. Amongt hepa r t i c i t so fCMO,abou t60s t uden t sa r es e l ec t ed pan t opa r t i c i t ei nt het r a i n i ngf orIMO.Thet r a i n i ngt ake sp l ace pa i n Ma r cheve r r.Af t e r6t o8t e s t sandano t he r2round sof yyea l i f i ngexami na t i on s,on l t e s t an t sa r eshor t l i s t edt o qua y y6con f ormt heCh i naIMO na t i ona lt eamt ot akepa r ti nt heIMOi n J u l y. Be s i de st he Ch i na Ma t hema t i ca l Compe t i t i on ( for h i gh s choo l s),t heJ un i or Mi dd l eSchoo l Ma t hema t i ca lCompe t i t i on i sa l s odeve l op i ng we l l.S t a r t i ngf r om1984,t hecompe t i t i oni s or i z edi nApr i leve r rbyt hePopu l a r i za t i onCommi t t ee gan yyea o ft he Ch i ne s e Ma t hema t i ca lSoc i e t r i ou sprov i nce s, y.Theva c i t i e sandau t onomou sr eg i on swou l dro t a t et oho s tt heeven t. Ano t he rma t hema t i ca lcompe t i t i onf ort hej un i ormi dd l es choo l s i sa l s o conduc t edi n Apr i leve r r by t he Mi dd l e Schoo l y yea ⅹ ⅹ In t r oduc t i on Ma t hema t i c s Educa t i on Soc i e t ft he Ch i ne s e Educa t i ona l yo Soc i e t i nce1998t i l lnow. ys eng Go l dCup,acompe t i t i onbyi nv i t a t i on, The HuaLuog had a l s o been s uc c e s s f u l l t ed s i nc e 1986. The y conduc Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. r t i c i t i ngs t uden t scompr i s ee l emen t a r i xandj un i ormi dd l e pa pa ys ones t uden t s.The f orma t oft he compe t i t i on cons i s t s of a e l imi na r ound,s emi f i na l si nva r i ou sprov i nce s,c i t i e sand pr yr au t onomou sr eg i on s,t hent hef i na l s. Ma t hema t i ca lcompe t i t i oni nCh i naprov i de sap l a t formf or s t uden t st os howca s et he i rt a l en t si nma t hema t i c s.I tencour age s l ea rn i ng o f ma t hema t i c s among s t uden t s.I t he l den t i f ps i y t a l en t ed s t uden t s and t o pr ov i de t hem wi t h d i f f e r en t i a t ed l ea rn i ng oppor t un i t t deve l op s co cur r i cu l a r ac t i v i t i e si n y.I ma t hema t i c s.F i na l l i tbr i ng sabou tchange si nt het each i ngof y, ma t hema t i c s. ⅹ ⅹ ⅰ b2530 International Strategic Relations and China’s National Security: World at the Crossroads Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. This page intentionally left blank b2530_FM.indd 6 01-Sep-16 11:03:06 AM Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Con t en t s Pr e f ace I n t r oduc t i on ⅲ Ch i naMa t hema t i c a lCompe t i t i on 2010 ( Fu i an) j 1 1 2011 ( Hube i) 13 2013 ( J i l i n) 35 2012 ( Shaanx i) 23 Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r y Te s t) 44 2010 ( Fu i an) j 44 2012 ( Shaanx i) 56 2011 ( Hube i) 51 2013 ( J i l i n) 62 Ch i naMa t hema t i c a lOl i ad ymp 68 2011 ( Changchun,J i l i n) 68 2013 ( Shenyang,Li aon i ng) 86 2012 ( Xi an,Shaanx i) 77 2013 ( Nan i ng,J i ang s u) j 99 ⅹ ⅹⅲ Ma t hema t i c a lOl i adi nCh i na ymp Ch i naNa t i ona lTe amS e l e c t i onTe s t 2011 ( Fuzhou,Fu i an) j 109 2013 ( J i angy i n,J i ang s u) 131 2012 ( Nanchang,J i anx i) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 109 2014 ( Nan i ng,J i ang s u) j Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 119 145 160 2010 ( Sh i i azhuang,Hebe i) j 2011 ( Shenzhen,Guangdong) 160 2013 ( Zhenha i,Zhe i ang) j 199 2012 ( Guangzhou,Guangdong) Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 174 190 211 2010 ( Ta i i) yuan,Shanx 2011 ( Yu s han,J i angx i) 211 2013 ( Lanzhou,Gan s u) 240 2012 ( Hohho t,Inne rMongo l i a) Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 221 230 252 2010 ( Lukang,Changhua,Ta iwan) 252 2012 ( Pu t i an,Fu i an) j 278 2011 ( Ni ngbo,Zhe i ang) j 2013 ( Yi ng t an,J i angx i) 265 286 I n t e rna t i ona lMa t hema t i c a lOl i ad ymp 299 2011 ( Ams t e rdam,Ho l l and) 299 2013 ( San t a Ma r t a,Co l omb i a) 321 2012 (Ma rDe lP l a t a,Ar t i na) gen 2014 ( CapeTown,TheRepub l i co f Sou t hAf r i ca) ⅹ ⅹ ⅰ ⅴ 311 336 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Ch i naMa thema t i ca l Compe t i t i on 2010 Commi s s i oned by Ch i nese ( F u i a n) j Ma t hema t i ca l Soc i e t i an y, Fu j Ma t hema t i ca l Soc i e t r i zed t he 2010 Ch i na Ma t hema t i ca l yo gan Compe t i t i onhe l donOc t obe r17,2010. Compa r edt ot hecompe t i t i on si nt hep r ev i ousyea r s,wh i l et he t es tt imeandp r ob l emt ema i nunchangedi nt h i scompe t i t i on, ypesr t hea l l oca t i ono fma r kst oeachp r ob l emi sad us t eds l i t l omake j gh yt i tmo r er easonab l e. Thet imef o rt hef i r s tr oundt es ti s80mi nu t es,andt ha tf o rt he supp l emen t a r es ti s150mi nu t es. yt Ma t hema t i c a lOl i adi nCh i na ymp 2 Pa r tI Sho r t -An swe r Qu e s t i on s( e s t i on s1 8,e i tma rk s Qu gh e a ch) 1 Ther angeo ff( x)= x -5 - 24 -3x i s . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. So l u t i on.I ti sea s os eet ha tf( i si nc r ea s i ngoni t sdoma i n x) yt [ 5,8].The r e f or e,i t sr angei s [-3, 3]. 2 Themi n imumo fy = ( acos2x -3) s i nxi s -3.Thent he r angeo fr ea lnumbe rai s So l u t i on.Le ts i nx . = t.Th eexpr e s s i oni st henchangedt o t)= (-a t +a -3) t,or g( 2 t)= -a t3 + ( a -3) t. g( Fr om -a t3 + ( a -3) t ≥ -3,wege t -a t( t -1)-3( t -1)≥ 0, 2 ( t -1)(-a t( t +1)-3)≥ 0. S i ncet -1 ≤ 0,wehave -a t( t +1)-3 ≤ 0,or a( t2 +t)≥ -3. ① Whent =0,-1, expr e s s i on ① a lway sho l d s;when0 <t ≤ 1 2 ≤t + 1,wehave0 <t2 +t ≤ 2;andwhen -1 <t < 0,4 3 ≤a ≤ 1 t < 0.The r e f or e,2. 2 3 Thenumbe ro fi n t eg r a lpo i n t s( i. e.,t hepo i n t swho s ex - andy i na t e sa r ebo t hi n t ege r s)wi t h i nt hea r ea ( no t -coord i nc l ud i ngt hebounda r l o s edbyt her i anchof y)enc ghtbr hype rbo l ax2 -y2 = 1andl i nex = 100i s . So l u t i on.Bysymme t r l ocons i de rt hepa r tof y,weon yneedt -ax i s.Suppo s el i ney = k i n t e r cep t st he t hea r eaabovet hex Ch i naMa t hema t i c a lCompe t i t i on 3 r i tbr ancho ft hehype rbo l aandl i nex = 100a tpo i n t sAk and gh Bk ( k = 1,2, ...,99),r e s c t i ve l he numbe ro f pe y.Thent i n t eg r a lpo i n t swi t h i nt hes egmen tAkBk i s99 -k.The r e for e, t henumbe ro fi n t eg r a lpo i n t swi t h i nt hea r eaabovet hex -ax i si s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 99 ∑ (99 -k)= 99 ×49 = 4851. k=1 Fi na l l t a i nt he t o t a l numbe ro fi n t eg r a l po i n t s y,we ob wi t h i nt hewho l ea r eaa s2 ×4851 +98 = 9800. 4 I ti sknownt ha t{ an }i sana r i t hme t i cs equencewi t hnon - z e r ocommond i f f e r enc eand { t r i cs equence, bn }ageome s a t i s f i nga1 =3, b1 =1,a2 =b2 ,3a5 =b3 ; f ur t he rmor e, y t he r ea r econs t an t sα andβ s ucht ha tforeve r s i t i ve y po i n t ege rn,we havean = l ogαbn + β.Thenα + β = . So l u t i on.Le an }bed andt he tt hecommon d i f f e r enceof { commonr a t i oo f{ bn }beq.Then 3 +d =q, 3( 3 +4d)=q2 . ① ② Sub s t i t u t i ng ① i n t o ② ,wehave9 +12d = d2 +6d +9. Thenwege td = 6andq = 9. The r e for e,3 +6( n -1)= l ogα9n-1 +βor6 n -3 = ( n- 1) l ogα9 +βho l d sf oreve r s i t i vei n t ege rn.Le t t i ngn =1and ypo nt u rn,wef i ndt ha tα = 3andβ = 3. n = 2i 3 Con s equen t l α +β = 3 +3. y, 3 5 Func t i onf( x)=a2x +3ax -2 ( a >0,a ≠1)r eache st he max imumva l ue8oni n t e r va l[-1,1].Theni t smi n imum Ma t hema t i c a lOl i adi nCh i na ymp 4 va l ueont h i si n t e r va li s So l u t i on.Le ta x . =y.Th eor i i na lf unc t i oni st henchangedt o g æ 3 ö 2 i chi si nc r ea s i ngove r ç - ,+ ∞ ÷ . g( y)= y + 3y - 2,wh è 2 ø Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. a,a-1]and When0 < a < 1,wehavey ∈ [ 1 -2 -1 -1 g( y)max = a +3a -2 = 8⇒a = 2⇒a = . 2 Then æ 1 ö2 1 1 -2 = . g( y)min = ç ÷ +3 × è2ø 2 4 Whena > 1,wehavey ∈ [ a-1 ,a]and 2 g( y)max = a +3a -2 = 8⇒a = 2. Then 1 -2 -1 g( y)min = 2 +3 ×2 -2 = - . 4 1 I ns umma r i s- . hemi n imumv a l u eo ff( x)onx ∈ [-1,1] y,t 4 6 Twope r s on sr o l ltwod i c ei nt u rn.Whoeve rge t st hes um numbe rg r ea t e rt han 6 f i r s t wi l l wi nt he game.The obab i l i t ort hepe r s onr o l l i ngf i r s tt owi ni s pr yf . So l u t i on.Theprobab i l i t orr o l l i ngtwod i cet oge tt hes um yf 7 21 = .The r e f or e,t her equ i r ed numbe rg r ea t e rt han 6i s 12 36 obab i l i t s pr yi æ 5 ÷ö2 7 æ 5 ÷ö4 7 … 7 7 × × +ç +ç + = × 12 è12ø 12 12 è12ø 12 1 12 = . 25 17 1144 Ch i naMa t hema t i c a lCompe t i t i on 7 5 Thel eng t hso ft hen i needge so fr egu l a rt r i angu l a rpr i sm ABC A1B1C1 a st hemi dpo i n to fCC1,andt r eequ a l,Pi he i nα = d i hed r a lang l eB A1P B1 =α.Thens . So l u t i on 1. Le t t he l i ne Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. t hr oughs egmen tAB bex -ax i s i ng t he wi t ht he or i i n O be g mi dpo i n to fAB andl e tt hel i ne t hr oughs egmen tOC bey -ax i st o e s t ab l i s h as er ec t angu l a r pac coord i na t e s s t em s hown i n y F i .7 .1.As s umi ng t hel eng t h g , of an edge i s 2 we have B( 1, 0, 2), 1,0,0), B1( ,P( -1, A1( 0,2) 0,3,1) .Th e n F i .7 .1 g → = (-2,0,2),B→ BA P = (-1, 3,1), 1 → = (-2,0,0),B → -1, 3,-1) B1A . 1 1P = ( → → Le tve c t or sm = ( x1 ,y1 ,z1)andn = ( x2 ,y2 ,z2)be rpend i cu l a rt oBA1P andB1A1P ,r e s t i ve l pe pec y.Wehave → → = -2x +2 m ·BA z1 = 0, 1 1 { { → m ·B→ P = -x1 + 3y1 +z1 = 0, → → = -2x = 0, n·B A 1 1 2 n·B1→ P = -x2 + 3y2 -z2 = 0. → → → Wecant hena s s umet ha tm = ( 1,0,1),n = ( 0,1, 3). Fr om|m ·n|=|m |·|n||c o sα|,wehave → → → → 6 3 = 2·2|c o sα|⇒|cosα|= . 4 Ma t hema t i c a lOl i adi nCh i na ymp 6 The r e f or e,s i nα = 10 . 4 So l u t i on2.Ass eeni nF i .7 .2,wehave g PC =PC1 ,PA1 =PB .Suppo s eA1Band Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. AB1i n t e r s e c ta tO .Then wege tOA1 = OB ,OA = OB1 ,A1B ⊥ AB1 . henPO ⊥ AB1 . S i ncePA = PB1 ,t The r e f or eAB1 ⊥ p l anePA1B . Onp l anePA1B t hr oughO dr awl i ne OE ⊥ A1P wi t hf oo tpo i n tE . F i .7 .2 g st hent he p l aneang l eoft he Conne c t i ngB1E , ∠B1EO i d i hedr a lang l eB A1P B1 .As s umi ngAA1 =2, i ti sea s of i nd yt t ha tPB = PA1 = 5,A1O = B1O = 2,PO = 3. Inr i tt r i ang l e△PA1O ,wehaveA1O·PO = A1P·OE , gh 6 or 2· 3 = 5·OE .SoOE = . 5 AsB1O = 2,wehave B1E = B1O2 +OE2 = 2+ F i na l l y, i n∠B1EO = s i nα = s 8 6 45 = . 5 5 B1O 2 10 = = . 4 B1E 45 5 Thenumbe ro fpo s i t i vei n t ege rs o l u t i onsofequa t i onx + s t hx ≤ y ≤zi y +z =2010wi . So l u t i on.I ti sea s of i ndt ha tt henumbe rofpo s i t i vei n t ege r yt sC2 009 ×1004. s o l u t i on so fx +y +z = 2010i 2009 = 2 Wenowc l a s s i f he s es o l u t i on si n t ot hr eeca t egor i e s: yt Ch i naMa t hema t i c a lCompe t i t i on 7 ( 1)x =y =z, t henumbe ri nt h i sca t egor sobv i ou s l yi y1; ( 2)t he r ea r eexac t l ha ta r eequa lamongx,y,z — ytwot t henumbe ri nt h i sca t egor s1003; yi ( 3)x,y,z a r ed i f f e r en tf r om eacho t he r—s uppo s et he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. numbe ri nt h i sca t egor sk. yi Fr om k = 2009 ×1004, 1 +3 ×1003 +6 wehave k = 2009 ×1004 -3 ×1003 -1 6 =2 006 ×1005 -2009 +3 ×2 -1 =2 006 ×1005 -2004. Wege t k = 1003 ×335 -334 = 335671. The r e f or e, t he numbe r o f po s i t i ve i n t ege r s o l u t i ons s a t i s f i ngx ≤ y ≤zi s y 1 +1003 +335671 = 336675. Pa r tI I Wo r dPr ob l ems ( 16 ma rk sf o r Qu e s t i on9,20 ma rk s e a chf o r Qu e s t i on s10and11,andt hen56 ma rk si n t o t a l) 9 I ti sknownt ha tf( x)=ax3 +bx2 +cx +d ( a ≠0),and '( x)|≤ 1for0 ≤ x ≤ 1.P l ea s ef i ndt he max imum |f va l ueofa. So l u t i on1.f'( x)= 3ax2 +2 bx +c.Wehave '( 0)=c, ìïf ï æ1ö 3 'ç ÷ = a +b +c, íf ï è2ø 4 ï () îf ' 1 = 3a +2 b +c. Ma t hema t i c a lOl i adi nCh i na ymp 8 Then Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Wege t æ1ö a = 2f 3 '( 0)+2f '( 1)-4f 'ç ÷ . è2ø æ1ö 3|a| = 2f '( 0)+2f '( 1)-4f 'ç ÷ è2ø æ 1 ö÷ ≤ 8. è2ø ≤ 2|f '( 0)|+2|f '( 1)|+4 f 'ç The r e f or e,a ≤ 8 .Fu r t he rmor e,i ti sea s of i ndt ha t yt 3 8 x)= x3 -4x2 +x +m ( sanycon s t an t)s a t i s f i e s whe r emi f( 3 8 t heg i vencond i t i on.The r e for e,t hemax imumva l ueofai s . 3 So l u t i on2.Le tg( x)=f '( x)+1.Then0 ≤g( x)≤2for0 ≤ x ≤ 1.Le tz = 2x -1.Thenx = z +1 and -1 ≤z ≤ 1.Le t 2 æz +1ö÷ 3 a 2 3a +2 b 3 a = +b +c +1. h( z)= g ç z + z+ è 2 ø 4 2 4 I ti sea s oche ckt ha t0 ≤h( z)≤2and0 ≤h(-z)≤2for yt -1 ≤z ≤ 1. The r e for e,0 ≤ t ha ti s 0≤ h( z)+h(-z) ≤ 2f or -1 ≤ z ≤ 1.And 2 3 a 2 3a +b +c +1 ≤ 2. z + 4 4 3 a 3 a +b +c +1 ≥ 0a Thenwehave nd z2 ≤ 2.Fr om0 ≤ 4 4 z2 ≤ 1wege ta ≤ 8 . 3 8 Asf( x)= x3 -4x2 +x + m ( whe r em i sanycons t an t) 3 Ch i naMa t hema t i c a lCompe t i t i on 9 s a t i s f i e st heg i vencond i t i on.Weob t a i nt ha tt hemax imumva l ue 8 s . o fai 3 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 10 Gi ventwo mov i ngpo i n t sA ( x1 ,y1)andB ( x2 ,y2)on r abo l acu r vey2 =6x wi t hx1 +x2 =4andx1 ≠x2 ,and pa n t e r s ec t sx -ax i s t hepe rpend i cu l a rb i s e c t oro fs egmen tABi a tpo i n tC .F i ndt hemax imuma r eaof△ABC . So l u t i on1.Le tt hemi dpo i n tofAB beM ( x0 ,y0).Thenx0 = y1 +y2 x1 +x2 = 2a .Wehave ndy0 = 2 2 kAB = y2 -y1 y2 -y1 6 3 = = . 2 = + x2 -x1 y2 2 1 0 y y y 2 y1 6 6 Theequa t i ono ft hepe rpend i cu l a rb i s e c t orofABi s y0 x -2). y -y0 = - ( 3 ① I ti sea s of i ndt ha tones o l u t i ono fi ti sx = 5,y = 0. yt The r e f or e,t hei n t e r s ec t i onC i saf i xed po i n twi t hcoord i na t e ( 5,0). From ① ,weknowt heequa t i onofl i neABi sy -y0 = ( x -2),or x = or y0 ( y -y0)+2. 3 3 y0 ② Sub s t i t u t i ng ②i ny2 = 6x,wege ty2 = 2y0( y -y0)+12, 2 2 y -2y0y +2y0 -12 = 0. ③ Asy1 andy2 a r etwor ea lr oo t so f③ andy1 ≠y2 ,wehave Ma t hema t i c a lOl i adi nCh i na ymp 10 Δ = 4y2 2y2 2)= -4y2 8 > 0. 0 -4( 0 -1 0 +4 The r e for e,-2 3 < y0 < 2 3.Thenwehave 2 2 ( +( x1 -x2) y1 -y2) |AB | = æç æy0 ö2 ö 2 1+ ç ÷ ÷ ( y1 -y2) è è3ø ø Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. = 2 æç y0 ö 2 1 + ÷ [( y1 +y2) -4 y1y2] è 9ø = = = 2 3 2 æç y0 ö 2 2 4 2 1+ ÷ ( y0 -4( y0 -12)) è 9ø ( ( 9 +y2 12 -y2 . 0) 0) F i .10 .1 g s Thed i s t anc ef r om po i n tC( 5,0)t os egmen tABi h =|CM |= The r e f or e, S△ABC = = ≤ = 2 2 ( +( = 5 -2) 0 -y0) 1 1 |AB |·h = 2 3 1 3 1 3 9 +y2 0. ( ( · 9 +y2 9 +y2 12 -y2 0) 0) 0 1( ( ( 9 +y2 24 -2y2 9 +y2 0) 0) 0) 2 2 2 2 3 1 æç9 +y0 +24 -2y0 +9 +y0 ö÷ 2è ø 3 14 7. 3 Theequa l i t l d si fandon l f9 +y2 4 - 2y2 i. e. 0 =2 0, yho yi t y0 = ± 5.Thenwege æ ö æ ö A ç6 + 35, 5 + 7 ÷ ,B ç6 - 35, 5 - 7 ÷ è ø è ø 3 3 and æ ö æ ö A ç6 + 35,- (5 + 7)÷ ,B ç6 - 35,- 5 + 7÷ . è ø è ø 3 3 Ch i naMa t hema t i c a lCompe t i t i on 11 14 Con s equen t l hemax imuma r eaof△ABCi s 7. y,t 3 So l u t i on 2. S he imi l a rt o So l u t i on 1, we ge tt ha t C ,t i n t e r s e c t i onoft hepe rpend i cu l a rb i s e c t orofAB andt hex -ax i s, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. i saf i xedpo i n twi t hcoord i na t e( 5,0). Le tx1 =t2 x2 =t2 t1 >t2 , t2 t2 st he ThenS△ABCi 1, 2, 1 + 2 =4. ab s o l u t eva l ueo f 5 1 t2 1 2 t2 2 0 1 6t1 1 , 6t2 1 s o S2△ABC = = = ≤ æç 1 ( ö2 5 6t1 + 6t2 t2 - 6t1t2 t2)÷ 1 2 -5 6 è2 ø 3( 2 2 ( t1 -t2) t1t2 +5) 2 3( 4 -2 t1t2)( t1t2 +5)( t1t2 +5) 2 3 çæ14÷ö . 2è3ø 3 The r e f or e,S△ABC ≤ 14 7andt heequa l i t l d si fandon l yho y 3 7+ 5 2 =t1t2 +5a i f( ndt2 t1 -t2) t2 henge tt1 = 1 + 2 =4.Wet 6 andt2 = - or 7 - 5, wh i chimp l i e se i t he r 6 ö ö æ æ A ç6 + 35, 5 + 7 ÷ ,B ç6 - 35, 5 - 7 ÷ è ø è ø 3 3 ö æ ö æ A ç6 + 35,- (5 + 7)÷ ,B ç6 - 35,- 5 + 7 ÷ . ø è è ø 3 3 Ma t hema t i c a lOl i adi nCh i na ymp 12 14 F i na l l hemax imuma r eaof△ABCi s 7. y,t 3 11 Pr ovet ha tequa t i on2x3 +5x -2 = 0ha sexac t l ea l yoner Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. r oo t (deno t ed a s r),and t he r ei s a un i t r i c t l que s y 2 i nc r e a s i ngs equ en c e{ an } s u cht ha t =ra1 +ra2 +ra3 + … . 5 So l u t i on.Le tf( x)= 2x3 +5 x -2.Then wehavef x)= '( i ch mean s f( ss t r i c t l nc r ea s i ng. 6x2 +5 > 0, wh x) i y i æ1ö 3 > 0.Th Fu r t he rmor e,f( 0) = - 2 < 0,f ç ÷ = e r e for e, è2ø 4 1ö æ saun i ea lr oo tr ∈ ç0, ÷ .From2 x)ha r3 +5 r -2 =0, f( quer 2ø è wehave r 2 4 7 10 = =r +r +r +r + … . 5 1 -r3 The r e for e,s equencean = 3 n -2 ( n = 1,2,...)s a t i s f i e st he r equ i r edcond i t i on. As s umet he r ea r etwod i f f e r en tpo s i t i vei n t ege rs equence s a1 < a2 < … < an < … andb1 <b2 < … <bn < … s a t i s f i ng y ra1 +ra2 +ra3 + … =rb1 +rb2 +rb3 + … = 2 . 5 De l e t i ngt het e rmst ha tappea ra tt he bo t hs i de soft he expr e s s i on,wehave rs1 +rs2 +rs3 + … =rt1 +rt2 +rt3 + …, whe r es1 <s2 <s3 < …, t1 <t2 <t3 < … wi t ha l lt hesiandtj d i f f e r en tf r omeacho t he r. Wemaya swe l la s s umet ha ts1 <t1 .Then Ch i naMa t hema t i c a lCompe t i t i on 13 rs1 <rs1 +rs2 + … =rt1 +rt2 + …, 1 <rt1-s1 +rt2-s1 + … ≤r +r2 + … Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. = 1 -1 < 1 -r 1 -1 = 1. 1 12 I ti sacon t r ad i c t i on.Th i spr ove st ha t{ an } i sun i que. 2011 ( Hu b e i) Comm i s s i o n edb i n e s e Ma t h ema t i c a lSo c i e t iMa t h ema t i ca l y Ch y,Hube Soc i e t r i zedt he2011Ch i naMa t hema t i ca lCompe t i t i onhe l don yo gan Oc t obe r16,2011. Pa r tI Sho r t -An swe r Qu e s t i on s( e s t i on s1 8,e i tma rk s Qu gh e a ch) 1 Le tA = { a1 ,a2 ,a3 ,a4}.Suppo s et hes e tofs umsofa l l t hee l emen t si neve r e rna r ub s e tofAi sB = {-1,3, yt ys 5,8}.ThenA = . So l u t i on.Obv i ou s l r l emen tofA appea r st hr eet ime s y,eve ye i na l lt het e rna r ub s e t s.Thenwehave ys 3( a1 +a2 +a3 +a4)= (-1)+3 +5 +8 = 15, ora1 +a2 +a3 +a4 = 5.The r e f or e,t hefoure l emen t sofA a r e 5 - (-1)= 6,5 -3 = 2,5 -5 = 0,5 -8 = -3,r e s t i ve l pec y. Theanswe ri sA = {-3,0,2,6}. 2 Theva l uef i e l do ff( x)= x2 +1 i s x -1 . 14 Ma t hema t i c a lOl i adi nCh i na ymp So l u t i on.Le tx =t anθ,- Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. x)= f( π π π <θ < andθ ≠ .Wehave 2 2 4 1 co sθ 1 = = t anθ -1 s osθ i nθ -c 1 . πö æç i nθ - ÷ 2s 4ø è πö æ Le tu = 2s i nçθ - ÷ .Then - 2 ≤ u < 1andu ≠ 0. 4ø è The r e for e,f( x)= 3 æ 1 2 ùú ∪ ( ∈ ç - ∞ ,1,+ ∞ ). ú u è 2û 1 1 Suppo s eaandba r epo s i t i ver ea lnumbe r ss a t i s f i ng + y a b ≤ 2 2a nd ( a -b) = 4( ab).Thenl ogab = 2 So l u t i on.From 1 + 1 a o t he rhand, b 3 . ≤22,weh avea +b ≤22ab.Ont he 2 2 3 ( =4 =4 a +b) ab + ( a -b) ab +4( ab) ≥ 4·2 a b·( ab) = 8( ab), 3 2 andt ha tmeans The r e f or e, a +b ≥ 2 2ab. ① a +b = 2 2ab. ② Theequa l i t n ① ho l d son l s oc i a t i ngi t yi y whenab = 1.As wi t h ② ,wef i nd a = 2 -1, a = 2 +1, and b = 2 +1, b = 2 -1. { { Sot hean swe ri sl ogab = -1. Ch i naMa t hema t i c a lCompe t i t i on 4 15 Suppo s ec o s5θ - s i n5θ < 7( s i n3θ - c os3θ),θ ∈ [ 0,2π). Thent her angeo fθi s . So l u t i on.Fromtheinequa l i t y c o s5θ -s i n5θ < 7( s i n3θ -co s3θ), Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. wehave s i n3θ + 1 5 1 5 s i nθ > c sθ. o s3θ + co 7 7 S i ncef( x)= x3 + 1 5 si nc r ea s i ngove r (- ∞ ,+ ∞ ), x i 7 t hens i nθ > c o sθ,andt ha tmean s kπ + 2 π 5π <θ < 2 kπ + ( k ∈ Z). 4 4 æ π 5πö Bu tθ ∈ [ sç , ÷ . 0,2π), s ot her angeofθi è4 4 ø 5 Sevens t uden t sa r ea r r angedt oa t t endf i ves t i ngeven t s. por I ti sr equ i r edt ha ts t uden t sA andBcanno ta t t endt hes ame even t,eve r ti sa t t endedbya tl ea s tones t uden t,and yeven eachs t uden t mu s ta t t endoneandon l t.Then y oneeven t henumbe ro ft hea r r angemen tp l ansmee t i ngt her equ i r ed cond i t i oni s nume r i ca lva l ue) .( t heanswe rshou l d be g i veni n So l u t i on.Therearetwopos s i b l eca s e st ha ts a t i s f her equ i r ed yt cond i t i on s: ( 1)t he r ei saneven ta t t endedbyt hr ees t uden t s— t h i sca s e 1 ha sC3 600p l an s; 7 ·5! - C 5 ·5! = 3 ( 2)t he r ea r etwoeven t seacha t t endedbytwos t uden t s— 1 2 · 2)· ! t h i sca s eha s ( 1400p l ans. C7 C5 5 - C2 5 ·5! = 1 2 Ma t hema t i c a lOl i adi nCh i na ymp 16 The r e f or e,t he r ea r e3600+11400 =15000p l anst ha tmee t t her equ i r edcond i t i on. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 6 Gi venat e t r ahedr onABCD ,i ti sknownt ha t ∠ADB = ∠BDC = ∠CDA =6 0 °,AD =BD =3andCD =2.Then t he r ad i u s of t he s r e c i r cums c r i b i ng ABCD i s phe . So l u t i on.Le tt he c en t e ro ft he s r ec i r cums c r i b i ngABCD beO . phe Then O i s on t he ve r t i ca ll i ne of hr ough po i n tN t he l ane ABD t p I ti sknown c i r cumc en t e rof△ABD . t ha t△ABD i sr egu l a r,s oN i st he cen t e ro fi t.Le tP and M bet he mi dpo i n t s AB of and CD , r e s c t i ve l sonDP wi t h pe y.ThenN i F i .6 .1 g ON ⊥ DP andOM ⊥ CD . Le tθdeno t et heang l ebe tweenCD andp l aneABD .From ∠CDA = ∠CDB = ∠ADB = 6 0 °, wef i ndc osθ = 1, 2 s i nθ = . 3 3 S i nc eDM = 1 2 2 3 CD =1,DN = ·DP = · ·3 = 3, 2 3 3 2 byco s i net heor em wehave,i n△DMN , MN2 = DM 2 + DN2 -2·DM ·DN ·cosθ = 1 + (3) -2·1· 3· 2 2 1 = 2, 3 t ha ti sMN = 2.Ther ad i u so ft hes r ec i r c ums c r i b i ngABCD phe i st hen Ch i naMa t hema t i c a lCompe t i t i on OD = 17 MN 2 = = 3. s i nθ 2 3 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Thean swe ri sR = 3. 7 Li nex - 2y - 1 = 0and pa r abo l ay2 = 4x i n t e r s ec ta t sont hepa r abo l a,and ∠ACB = i n t sA ,B ,po i n tC i po 90 °.Thent hecoord i na t eo fCi s . So l u t i on.Le x1 ,y1),B ( x2 ,y2),C( t ,2 t).From tA ( 2 x -2y -1 = 0, { 2 y = 4x, weg e ty2 -8 i chme an sy1 +y2 = 8,y1·y2 =-4. y -4 = 0,wh S i nc ex1 = 2y1 +1,x2 = 2y2 +1,wehave x1 +x2 = 2( y1 +y2)+2 = 18, x1 ·x2 = 4y1 ·y2 +2( y1 +y2)+1 = 1. °,wehaveC→ A ·C→ B = 0, Fur t he rmor e,by ∠ACB = 90 wh i ch mean s ( t2 -x1)( t2 -x2)+ ( 2 t -y1)( 2 t -y2)= 0, t ha ti s t4 - ( x1 +x2) t2 +x1 ·x2 +4 t2 -2( t +y1 ·y2 = 0. y1 +y2) Then t4 -14 t2 -16 t -3 = 0, or ( t2 +4 t +3)( t2 -4 t -1)= 0. he rwi s e,wehavet2 - 2· Obv i ou s l t2 -4 t -1 ≠ 0;ot y, 2 t -1 = 0,wh i ch meansC i sonx - 2y - 1 = 0,i. e.,C Ma t hema t i c a lOl i adi nCh i na ymp 18 co i nc i de swi t he i t he rA orB .Sot2 + 4 t + 3 = 0.Thent1 = - 1, t2 = -3. The r e f or e,t hecoord i na t eofCi s( 1,-2)or ( 9,-6). Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 8 Le tan = Cn 200·(6) 3 200-n æ 1 ön ·ç ÷ ( n =1,2,...,95).Then è 2ø t henumbe ro ft e rmst ha ta r ei n t ege r si n{ an } i s So l u t i on.Wehavean = C200·3 n 200-n 3 ·2 400-5n 6 . .Whenan ( 1 ≤n ≤ 400 -5 n 200 -n and mu s tbei n t ege r s.Then 95) i sani n t ege r, 3 6 6|n +4. Whenn = 2,8,14,20,26,32,38,44,50,56,62,68, 400 -5 200 -n n and a r ea l lnon -nega t i vei n t ege r s.So 74,80, 3 6 t hecor r e s i ngan , t o t a l l r ei n t ege r s. pond y14,a 6 38 -5 Thenumbe rof Whenn =86,wehavea86 = C8 200·3 ·2 . t hef ac t or so f2i n200! i s 200 200 200 200 200 200 0 + + + + + + =1 97. [20 2 ] [2 ] [2 ] [2 ] [2 ] [2 ] [2 ] 2 3 4 5 6 7 Byt hes amer ea s on,t henumbe r soft hef ac t or sof2i n86! and114! a r e82and110,r e s c t i ve l r e for e,t henumbe r pe y.The 6 o ft hef ac t or so f2i nC8 200 = a86i sani n t ege r. 200! i s197 -82 -110 =5.So 86!·114! 2 36 -10 .Int hes ame Whenn = 92,wehavea92 = C9 200 ·3 ·2 way,wef i ndt henumbe r soft hef ac t or sof2i n92! and108! 6 a r e88and105,r e s c t i ve l i ch meanst ha ti nC8 s197 200i pe y,wh 88 -105 = 4.The r e f or e, a92i sno tani n t ege r. Ove r a l l,t her equ i r ednumbe ri s14 +1 = 15. Ch i naMa t hema t i c a lCompe t i t i on 19 Pa r tI I Wo r dPr ob l ems ( 16 ma r k sf o r Qu e s t i on9,20 ma rk s hen56 ma rk si n e a chf o r Qu e s t i on s10and11,andt t o t a l) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 9 Suppo s ef( x)=|l x +1) b( a <b) ea lnumbe r sa, |andr g( æ b +1÷ö , ( s a t i s f a)= f ç b + 21) = 4 l g2. f 10a + 6 yf( è b +2ø F i ndt heva l ue so fa,b. æ b +1ö÷ , wehave è b +2ø So l u t i on.Asf( a)= f ç - æ 1 ö÷ æ b +1 ö +1÷ = l =|l a +1)|= l b +2)|. |l gç gç g( g( èb +2ø è b +2 ø Thene i t he ra +1 =b +2or( a +1)( b +2)=1.S i nc ea < oa +1 ≠b +2.The b,s a +1)( b +2)= 1. r e for e,( a)=|l a +1)|weknow0 < a +1.Then Fromf( g( 0 < a +1 <b +1 <b +2, wh i chimp l i e s The r e f or e, 0 < a +1 < 1 <b +2. ( 10 a +6 b +21)+1 = 10( a +1)+6( b +2) 10 > 1. b +2 = 6( b +2)+ Then 10 b +2)+ 10 a +6 b +21)= l g 6( f( b +2 [ ] 10 . b +2 b +2)+ =l g 6( [ Ont heo t he rhand, a +6 b +21)= 4 l 10 g2. f( ] Ma t hema t i c a lOl i adi nCh i na ymp 20 So 10 b +2)+ =4 l l g 6( g2, b +2 [ ] Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 10 1 =1 wh i ch mean s6( b +2)+ 6.Thene i t he rb = - orb = b +2 3 -1 ( d i s ca rded). Sub s t i t u t i ngb = - 2 . 5 The r e f or ea = - 1 i n t o( a +1)( b +2)= 1,wef i nda = 3 2, 1 b =- . 5 3 10 Suppo s es e en c e{ an } t-3( t ∈ Randt ≠±1), s a t i s f i e sa1 =2 qu an+1 = ( tn+1 -3) an +2( t -1) tn -1( 2 n ∈ N* ). an +2 tn -1 ( 1)Fi ndt heformu l aofgene r a lt e rmabou t{ an }. ( 2)I ft > 0, f i ndou twh i chi sl a r rbe tweenan+1andan . ge So l u t i on.( 1)Theg i venexpr e s s i oncanber ewr i t t ena s an+1 = 2( tn+1 -1)( an +1) -1. an +2 tn -1 Then 2( an +1) ( ) an+1 +1 2 an +1 tn -1 = = . n+1 n t -1 an +2 t -1 an +1 + 2 tn -1 an +1 2 bn , a1 +1 =bn .Th = Le tn enbn+1 = wi t hb1 = bn +2 t -1 t -1 2 t -2 = 2. t -1 Ch i naMa t hema t i c a lCompe t i t i on 21 1 1 1 ,1 1 = + = Fu r t he rmor e, .Then bn+1 bn 2 b1 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 1 ( 1 ·1 = n . = + n -1) 2 bn b1 2 an +1 2 , 2( tn -1) = -1. The r e f or e, n wh i chmeansan = n t -1 n ( 2)Wehave an+1 -an = = 2( tn+1 -1) 2( tn -1) n +1 n 2( t -1)[ ( n 1 +t + … +tn-1 +tn )n( n +1) ( n +1)( 1 +t + … +tn-1)] = = = t -1)[ n ( 2( n t - 1 +t + … +tn-1)] n( n +1) t -1)[(n 2( t -1)+ ( tn -t)+ … + ( tn -tn-1)] n( n +1) 2 2( t -1) [( tn-1 +tn-2 + … +1)+ ( ) n n +1 t( tn-2 +tn-3 + … +1)+ … +tn-1]. I ti sobv i ou st ha tan+1 -an >0fort >0( t ≠1).The r e for e, an+1 > an . 11 S t r a i tl i nel wi t hs l ope gh 1 3 2 x2 y i n t e r c ep t se l l i s eC : + p 36 4 = 1a t po i n t s A ,B ,and i n tP ( 3 2, 2)i si nt he po ( t op l e f to fl a sshowni n F i .11 .1 g Fi .11 .1). g ( 1)Pr ovet ha tt hecen t e roft hei ns c r i bedc i r c l eof△PAB Ma t hema t i c a lOl i adi nCh i na ymp 22 i sonag i venl i ne; ( 2)When ∠APB = 60 °,f i ndt hea r eaof△PAB . So l u t i on.( 1)Le tlbeas t r a i tl i nes ucht ha ty gh Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. andA ( x1 ,y1),B ( x2 ,y2). Sub s t i t u t i ngy = i t,wehave = 1 x +m, 3 1 x2 y + = 1, x +mi n t o ands imp l i f i ng y 36 4 3 2 2x2 +6mx +9m2 -36 = 0. Thenx1 +x2 = -3m ,x1x2 = kPB = y2 - 2 y1 - 2 , 9m2 -36, kPA = 2 x1 -3 2 .The r e f or e, x2 -3 2 kPA +kPB = = y1 - 2 x1 -3 2 + y2 - 2 x2 -3 2 ( x2 -3 2)+ ( x1 -3 2) y1 - 2)( y2 - 2)( . ( x1 -3 2)( x2 -3 2) Int heexpr e s s i onabove,t henume r a t ori sequa lt o æç 1 ö æ1 ö x1 + m - 2 ÷ ( x2 -3 2)+ ç x2 + m - 2 ÷ ( x1 -3 2) è3 ø è3 ø = = 2 x1x2 + ( m -2 2)( x1 +x2)-6 2( m - 2) 3 2 ·9m2 -36 ( (-3m )-6 2( + m -2 2) m - 2) 2 3 = 3m -1 2 -3m +6 2m -6 2m +12 = 0. 2 2 The r e for e, i nc ePi kPA +kPB =0.S si nt het op l e f to fl,we s pa r a l l e lt ot he y -ax i s. know t ha tt he b i s ec t or o f ∠APB i The r e for e,t hec en t e ro ft hei n s c r i bedc i r c l eo f△PABi sonl i ne x = 3 2. Ch i naMa t hema t i c a lCompe t i t i on 23 ( 2)When ∠APB = 60 °,byt her e s u l ti n( 1),wehav ekPA = 3,kPB = - 3.Thent heequa t i onforl i nePA i sy - 2 = 2 x2 y + = 1,a x -3 2).Sub 3( nde l imi na t i ng s t i t u t i ngi ti n t o 36 4 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. t y,wege 1 -3 3) x +18( 13 -3 3)= 0, 14x2 +9 6( wh i chha sr oo t sx1 and3 2.Sox1 ·3 2 = x1 = 3 2( 13 -3 3) .Thenwef i nd 14 2 ·|x1 -3 2|= 1 + (3) |PA |= 18( 13 -3 3), i. e. 14 3 2( 3 3 +1) . 7 3 2( 3 3 -1) In t he s ame way, we have | PB |= . 7 The r e f or e, S△PAB = = = 1· i n60 ° |PA |·|PB |·s 2 1 ·3 2( 3 3 +1)·3 2( 3 3 -1)· 3 2 7 7 2 117 3 . 49 2012 ( Sh a a n x i) Commi s s i oned by Ch i nese Ma t hema t i ca l Soc i e t i y, Shaanx Ma t hema t i ca l Soc i e t r i zed t he 2012 Ch i na Ma t hema t i ca l yo gan Ma t hema t i c a lOl i adi nCh i na ymp 24 Compe t i t i onhe l donOc t obe r14,2012. i tma rk s Pa r tI Sho r t -An swe r Qu e s t i on s( e s t i on s1 8,e gh Qu e a ch) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 1 Le tP beapo i n tont heimageo fy = x + 2( x > 0). x awl i ne spe rpend i cu l a rt oy = x andy -ax i s Thr oughP dr wi t hf oo tpo i n t sA ,B ,r e s c t i ve l heva l ueof pe y.Thent → → PA ·PBi s . 2 ö÷ .Theexpr e s s i onforl i nePA x0 ø So l u t i on1.Le tP çx0 ,x0 + æ è i st hen 2 ö÷ æ =- ( x -x0), y - çx0 + x0 ø è 2 y = -x +2x0 + . x0 or Fr om ìïy = x, ï í 2 ïy = -x +2x0 + , x0 î 1 1 ö÷ æ wege tA çx0 + ,x0 + . x0 x0 ø è 2 ÷ö æ Ont heo t he rhand,wehaveB ç0,x0 + .ThenP→ A = x0 ø è æç 1 , 1 ö÷ B = (-x0 ,0).The r e f or e, andP→ x0 ø èx0 1 ·( -x0)= -1. P→ A ·P→ B = x0 Thean swe ri s -1. Ch i naMa t hema t i c a lCompe t i t i on 25 So l u t i on2.Ass eeni nF i .1 .1, g 2 ÷ö æ t hed i s t anc e sf r om P çx0 ,x0 + t o x0 ø è i s,r e s c t i ve l l i ne sy = x andy -ax pe y, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. a r e 2 ö÷ æ x0 - çx0 + x0 ø è |PA |= 2 = 2 x0 F i .1 .1 g and |PB |= x0 . S i nc eO ,A ,P andB a r econcyc l i cpo i n t s,t hen ∠APB = π - ∠AOB = 3π . 4 3π = -1. The r e f or e,P→ A ·P→ B =|P→ A |·|P→ B |·cos 4 2 Suppo s e △ABC wi t h ang l e s A , B and C ,and t he cor r e s i ngs i de sa,bandcs a t i s f i e sequa t i onaco sB pond bcosA = t anA 3 c.Thent heva l ueo f i s 5 t anB . So l u t i on1.Byt h eg i v e nc ond i t i onandt heLawo fCo s i n e s,weha v e 3 c2 +a2 -b2 b2 +c2 -a2 -b· = c, a· 2 ca 2 b c 5 ora2 -b2 = 3 2 c .The r e f or e, 5 c2 +a2 -b2 a· t anA s i nAc o sB 2 ca = = 2 2 2 +c -a t anB s i nBc osA b b· b c 2 Ma t hema t i c a lOl i adi nCh i na ymp 26 8 2 c c2 +a2 -b2 5 = 2 = = 4. 2 2 b +c2 -a2 c 5 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Thean swe ri s4. So l u t i on 2. As s een i n F i .2 .1, g aw CD ⊥ AB wi t hfoo t t hr oughC dr i n tD .Wehaveac o sB = DB and po bc o sA = AD .Byt heg i v enc ond i t i on, wehaveDB -AD = 3 c.Comb i n i ng 5 i twi t hDB + AD = c,wege tAD = F i .2 .1 g 4 1 candDB = c.The r e f or e, 5 5 CD t anA BD AD = = = 4. AD t anB CD BD So l u t i on3.Byt hep r o e c t i ont he o r em,wehav eac o sB +bc o sA j = 3 4 c.Comb i n i ngi twi t hac o sB -bc o sA = c,weg e tac o sB = c 5 5 andbc osA = 1 c.The r e f or e, 5 4 c t anA s i nAc osB a·c osB 5 = = = = 4. t anB s i nBc osA b·co sA 1 c 5 3 Le tx,y,z ∈ [ 0,1].Thent hemax imumva l ueo fM = s |x -y| + |y -z| + |z -x|i So l u t i on.Wemayas s ume0 ≤ x M = ≤ y ≤z ≤ 1.Th en y -x + z -y + z -x . . Ch i naMa t hema t i c a lCompe t i t i on 27 S i nce z -y)] = 2[( y -x)+ ( y -x + z -y ≤ wehave Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. M ≤ z -x), 2( 2( z -x)+ z -x = (2 +1) z -x ≤ 2 +1. Theequa l i t l d si fandon l fy -x =z -y,x =0,z = yho yi 1 æç ö . e.x = 0,y = ,z = 1÷ . 1i 2 è ø The r e for e,t hean swe ri sM max = 2 +1. 4 Le tt hef ocu sandd i r e c t r i xo fpa r abo l ay2 = 2px( p > 0) beFandl, r emov i ngpo i n t sont he r e s c t i ve l pe y.A andBa π .Le tt hepro ec t i ono fM j 3 r abo l as a t i s f i ng ∠AFB = pa y t he mi dpo i n to fs egmen t AB on l be N .Then t he MN | i s max imumva l ueof| |AB | So l u t i on1.Suppos e ∠ABF Lawo fS i ne,wehave |AF | = s i nθ Andt hen So . = θç0 <θ < æ è 2πö÷ .Thenbyt he 3ø |BF | |AB |. = π æ2π ö ç ÷ -θ s i n s i n 3 è3 ø |AF |+|BF | = |AB |. π æ2π ö -θ÷ s i n s i nθ +s i nç 3 è3 ø |AF |+|BF | = |AB | æ2π ö -θ÷ s i nθ +s i nç πö æ è3 ø =2 c osçθ - ÷ . 3ø è π s i n 3 Ma t hema t i c a lOl i adi nCh i na ymp 28 Ass eeni nF i .4 .1,byu s i ngt he g de f i n i t i on o f a pa r abo l a and t he ope r t fat r ape zo i d, we have pr yo Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. |AF |+|BF |.Then |MN |= 2 πö æ |MN | = c o sçθ - ÷ . 3ø è |AB | MN | The r e f or e,| r eache st he |AB | max imumva l ue1whenθ = Thean swe ri s1. F i .4 .1 g π . 3 So l u t i on2.By us i ng t he de f i n i t i on ofa pa r abo l a andt he |AF |+|BF |.In ope r t fat r ape zo i d,wehave|MN |= pr yo 2 △AFB ,byu s i ngt heLawo fCo s i ne swehave π 2 2 2 |AB | =|AF | +|BF | -2|AF |·|BF |cos 3 2 =( |AF |+|BF |) -3|AF |·|BF | æ AF |+|BF |÷ö2 2 è ø | 2 ≥( |AF |+|BF |) -3ç = æç|AF |+|BF |ö÷2 2 =|MN | . 2 è ø The equa l i t l d si f and on l f | AF |=| BF |. y ho yi MN | The r e f or e,t hemax imumva l ueof| i s1. |AB | 5 Suppo s etwor egu l a rt r i angu l a rpyr ami d sP ABC andQ ABCsha r i ngt hes ameba s ea r ei n s c r i bedi nt hes ames r e. phe I ft heang l eb e twe ent hes i de f a c eandt heba s eo fP ABCi s 45 °, t hent het ang en tv a l u eo ft heang l eb e twe ent hes i d e f a c e Ch i naMa t hema t i c a lCompe t i t i on andt heba s eo fQ ABCi s 29 . So l u t i on. As s e en i n F i .5 .1, g t henPQi spe r i c u l a r c onne c t i ngPQ, pend t op l aneABC wi t ht hef oo tpo i n tH Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. be i ngt hec en t e ro f△ABC .Thec en t e r sa l s oonPQ .Conne c t o ft hes r eOi phe andex t endCH t ol e ti ti n t e r s ec twi t h ABa tpo i n tM .M i st hent hemi dpo i n t o fAB ,andCM ⊥ AB . I ti sea s os ee yt t ha t ∠PMH and ∠QMH a r et hep l ane ang l e sformed by t hes i de s f ac e sand F i .5 .1 g t heba s e so ft her egu l a rt r i ang u l a rpy r ami d sP ABCandQ ABC, °, s oPH = MH = r e s c t i v e l pe y.Then ∠PMH = 45 1 AH . 2 S i nce ∠PAQ =90 °,AH ⊥PQ , t henAH2 =PH ·QH .We t henhaveAH2 = 1 AH ·QH . 2 The r e for e,QH = 2AH = 4MH . QH = 4. F i na l l t an∠QMH = y, MH Thean swe ri s4. 6 Le tf( x)beanoddf unc t i ononR,andf( x)=x2forx ≥ 0.Suppo s eforanyx ∈ [ a,a +2],f( x +a)≥ 2f( x). Thent her angeo fr ea lnumbe rai s . So l u t i on.Accord i ngt ot heg i vencond i t i on,wehave x)= f( x2 { 2 -x ( x ≥ 0), ( x < 0). So2f( x)= f(2x).The r e for e,t heor i i na li nequa l i t s g yi equ i va l en tt of( x +a)≥ f(2x). Ma t hema t i c a lOl i adi nCh i na ymp 30 Asf( x) i si nc r ea s i ngove rR,t henx +a ≥ 2x, i. e., a ≥ (2 -1) x. a,a + 2],(2 - 1)x r eache s Fu r t he rmor e,s i nc ex ∈ [ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. (2 -1)( a +2)t hemax imumva l uewhenx = a +2. a ≥ (2 -1)( a +2), The r e for e, f rom wh i chweob t a i na ≥ a ∈ [2,+ ∞ ). 2, i. e., Thean swe ri st hen [2,+ ∞ ). 7 Thes umo fa l lt hepo s i t i vei n t ege r sns a t i s f i ng y < 1 i s 3 1 π <s i n n 4 . So l u t i on.Ass i nxi saconvexf unc t i onf orx π ö÷ , we 6ø ∈ ç0, æ è 3 have x < s i nx < x.Then π π π 1, π 3 π 1, < < > × = s i n s i n 13 13 4 12 π 12 4 π 1, 3 π 1, π π < < > × = s i n s i n π 9 3 10 10 3 9 t ha ti s 1 π π π 1 π π < <s <s <s < <s i n i n i n i n . s i n 12 11 10 3 9 13 4 r e The r e f or e,a l lt hepo s s i b l eva l ue sofpo s i t i vei n t ege r sna 10,11,12,andt he i rs umi s33. Thean swe ri s33. 8 Ani nf orma t i ons t a t i onemp l oy sfou rd i f f e r en tcode s,A , B ,C andD , f orcommun i ca t i on,bu teachweeku s e son l y Ch i naMa t hema t i c a lCompe t i t i on 31 one of t hem. The code u s ed i n a de f i n i t e week i s r andoml e l ec t edwi t hequa lchanceamongt het hr eeone s ys t ha thave no tbeen u s edi nt hel a s t week.Suppo s et he codeu s edi nt hef i r s tweeki sA .Thent hepr obab i l i t ha t yt Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Ai sa l s ou s edi nt hes even t hweeki s a sani r r educ i b l ef r ac t i on) .( expr e s s ed So l u t i on.Le t et hepr obab i l i t ha tcodeA i su s edi n tPk deno yt t hekt hweek.Thent heprobab i l i t ha tAi sno tu s edi nt hekt h yt weeki s1 -Pk .The r e f or e,wehave Pk+1 = 1( 1 -Pk ). 3 Or Pk+1 AsP1 =1, Pk - { a st hef i r s tt e rmand - 1 1 çæP - 1 ÷ö =. k 4ø 4 3è 1 3 i st henageome t r i cs equencewi t h 4 4 } 1 a st hecommonr a t i o.Sowehave 3 Pk - k-1 1 3 æç - 1 ö÷ = . 4 4è 3ø Or Pk = The r e f or e,P7 = k-1 1 3 æç - 1 ö÷ + . 4è 3ø 4 61 . 243 61 Thean swe ri s . 243 Ma t hema t i c a lOl i adi nCh i na ymp 32 Pa r tI I Wo r dPr ob l ems ( 56ma rk si nt o t a lf o rt hr e equ e s t i on s) 1 3 + 9 ( 16ma rk s)Suppo s ef( x)= as i nx - c os2x +a 2 a Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 1, a ∈ R,a ≠ 0. 2 ( 1)I ff( x)≤ 0foranyx ∈ R,f i ndt her angeofa. ( 2)I fa ≥2andt he r eex i s t sx ∈ Rs ucht ha tf( x)≤0, f i nd t her angeofa. So l u t i on.( 1)Wehavef( x)= s i n2x t =s i nx(-1 ≤t ≤ 1).Then +as i nx +a - 3 .Le t a 3 t)=t2 +a t +a - . g( a Thes u f f i c i en tandne c e s s a r i t i onf orf( x)≤ 0,∀x ycond ∈ Ri s ìï (-1)= 1 - 3 ≤ 0, g a ï í ï () 3 ≤ 0. aïg 1 = 1 +2 î a The r e f or e,weob t a i nt her angeaofi s( 0,1]. a ( ≤ -1.Weh 2)Asa ≥ 2, t hen ave 2 3 t)min = g(-1)= 1 - . g( a 3 Thenf( x)min = 1 .The r e for e,t he s u f f i c i en t and a s1ne c e s s a r i t i onf orf( x)≤0,∃x ∈ Ri ycond a ≤ 3. 3 ≤0, or0 < a F i na l l t a i nt ha tt her angeofai s[ 2,3]. y,weob Ch i naMa t hema t i c a lCompe t i t i on 33 10 ( 20ma rk s)I ti sknownt ha teacht e rmofs equence { an }i s anon z e r or ea lnumbe r,andforany po s i t i vei n t ege rn ho l d st heequa t i on Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 3 3 3 2 ( = a1 +a2 + … +an . a1 +a2 + … +an ) ( 1)Whenn = 3,f i ndou ta l lt hes equence scons i s t i ngof t hr eet e rmsa1 ,a2 ,a3 . ( an }s ucht ha t 2)Doe st he r eex i s tani nf i n i t es equence { a2013 = -2012?I fi tex i s t s,wr i t eou tt heformu l aofgene r a l t e rm;i fno t,g i veyourr ea s on. So l u t i on.( 1)Whenn =1,wehavea2 1 ta1 = 1. ge =a1 . S i ncea1 ≠0,we 3 2 3 = 1 +a2 . 1 +a2) S i ncea2 ≠0,we Whenn =2,wehave( ta2 = 2ora2 = -1. ge 2 3 3 =1+a2 +a3 .F Whenn = 3,weha v e( 1+a2 +a3) o ra2 = 2,wege ta3 = 3ora3 = -2;f ora2 = -1,wege ta3 = 1. Ins umma r tt hr ees equence scons i s t i ng o ft hr ee y,we ge t e rmst ha ts a t i s f heg i vencond i t i on: yt { 1,2,3},{ 1,2,-2},and { 1,-1,1}. ( 2)Le tSn = a1 +a2 + … +an .Thenwehave 2 3 3 3 = a1 +a2 + … +an ( Sn n ∈ N), 2 3 3 3 3 ( = a1 +a2 + … +an +an+1 . Sn +an+1) F i nd i ngou tt hed i f f e r enc eo ft hetwoexpr e s s i on saboveand 2 byan+1 ≠ 0,wehave 2Sn = an +1 -an+1 . r om ( 1)t ha ta1 = 1. Whenn = 1,weknowf Whenn ≥ 2,wehave 2 2 -an ) 2 an = 2( Sn -Sn-1)= ( an an . +1 -an+1 )- ( Ma t hema t i c a lOl i adi nCh i na ymp 34 Andt ha ti s ( an+1 +an )( an+1 -an -1)= 0. Thenwege tan+1 = -an oran+1 = an +1. i ndt heformu l a F i na l l roma1 =1anda2013 = -2012,wef y,f Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. ofgene r a lt e rmf orar equ i r eds equenc ea s an = n, { 1 ≤ n ≤ 2012, 2012(-1), n ≥ 2013. n 11 ( 20ma rk s)Ass eeni nF i .11 .1, g i n t he r e c t angu l a r coord i na t e s s t em XOY ,t he s i de o ft he y , r hombu sABCDi s4 and|OB |= |OD |= 6. ( sa 1)Pr ovet ha t|OA|·|OC|i con s t an t. ( 2) When po i n tA i smov i ngon 2 2 +y = t heha l fc i r c l e( x - 2) F i .11 .1 g 4( 2 ≤ x ≤ 4),f i ndt het r ac eo f C. So l u t i on.( 1)S i nc e|OB |=|OD | and|AB|=|AD |=|BC|=|CD |, t henO ,A ,C a r eco l l i nea r. Ass eeni nF i .11 .2,connec t i ng g BD ,t henBD i spe rpend i cu l a rt oAC andt hr oughi t s mi dpo i n t K .So we have F i .11 .2 g Ch i naMa t hema t i c a lCompe t i t i on 35 |OK |-|AK |)( |OK |+|AK |) |OA |·|OC | = ( 2 2 =|OK | -|AK | 2 2 2 2 =( |OB | -|BK | )- ( |AB | -|BK | ) =|OB | -|AB | = 6 -4 = 2 0( acons t an t). 2 2 2 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. ( 2)Le tC( x,y),A ( 2 +2c o sα,2s i nα),whe r e π ÷ö æ π ≤α ≤ . α = ∠XMA ç 2ø è 2 Then ∠XOC = α .As 2 α 2 2 2 +( = 8( 2 +2c osα) 2s i nα) 1 +c osα)= 16cos2 , |OA | = ( 2 α i n i ngi twi t ht her e s u l ti n( 1),we t hen|OA |= 4co s .Comb 2 α t|OC |co s = 5. ge 2 α α os =5,andy =|OC|s i n = Thenwehavex =|OC|c 2 2 α ∈[ -5,5] . 5 t an 2 The r e f or e,t het r ac eo fpo i n tCi sas egmen twi t ht heend s ( 5,5)and ( 5,-5). 2013 ( J i l i n) Comm i s s i o n ed b i n e s e Ma t h ema t i c a lSo c i e t i l i n Ma t h ema t i ca l y Ch y,J Soc i e t r i zedt he2013Ch i naMa t hema t i ca lCompe t i t i onhe l don yo gan Oc t obe r13,2013. Ma t hema t i c a lOl i adi nCh i na ymp 36 Pa r tI Sho r t -An swe r Qu e s t i on s( e s t i on s1 8,e i tma rk s Qu gh e a ch) 1 Gi venA = { 2,0,1,3}, l e tB = { x|-x ∈ A ,2 -x2 ∉ A }.Thent hes umofe l emen t si nBi s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. So l u t i on.I ti sea s of i ndt ha tB yt . ⊆ {-2,0,-1,-3} .We have2 -x2 ∉ A whenx = -2,-3,and2 -x2 ∈ A whenx = t hes umofwho s ee l emen t s 0,-1.The r e for e,B = {-2,-3}, i s -5. Thean swe ri s -5. 2 Inap l aner e c t angu l a rcoord i na t es s t emxOy,po i n t sA , y Ba r eont hepa r abo l ay2 =4x, A ·O→ B = -4, s a t i s f i ngO→ y st hef ocu soft he pa r abo l a.ThenS△OFA · andpo i n tF i S△OFB = . So l u t i on.Le tF ( 1,0),A ( x1 ,y1),B ( x2 ,y2).Thenx1 2 1 = 2 2 y , y , x2 = and 4 4 → → -4 = OA ·OB = x1x2 +y1y2 = f r om wh i ch we have The r e for e, S△OFA ·S△OFB = = 1( 2 y1y2) +y1y2 , 16 1( 2 y1y2 + 8) = 0,ory1y2 = - 8. 16 æç 1 ö æ1 ö |OF |·|y1 |÷· ç |OF |·|y2 |÷ è2 ø è2 ø 1· 2 |OF | ·|y1y2 |= 2. 4 Thean swe ri s2. 3 i nA = 10s i nBs i nC ,cosA = Suppo s ei n△ABC wehaves 10c osBc o sC .Thent anA = . Ch i naMa t hema t i c a lCompe t i t i on So l u t i on.Ass i nA 37 -c i nC -c sC)= o sA = 10( s i nBs osBco -1 0co s( B +C)=10cosA ,wehaves r e f or e i nA =11c osA .The t anA = 11. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. Thean swe ri s11. 4 Suppo s et hes i deo ft heba s eandt hehe i fr egu l a r ghto t r i angu l a rpy r ami dP ABC a r e1and 2,r e s t i ve l pec y. Thent her ad i u so ft hei n s c r i beds r eo ft hepyr ami di s phe . So l u t i on.Ass eeni nF i .4 .1,s uppo s et he g o e c t i on so ft hei n s c r i beds r e scen t e rO pr j phe on f ace s ABC and ABP a r e H , K, sM ,and r e s c t i ve l he mi dpo i n to fAB i pe y,t t her ad i u soft hes r ei sr.ThenP ,K ,M phe π a r eco l l i nea r,∠PHM = ∠PKO = ,and 2 OH = OK =r,PO = PH -OH = 2 -r, MH = 3 3, AB = PM = 6 6 MH2 +PH2 = F i .4 .1 g 1 53 +2 = . 12 6 Thenwehave r OK MH 1 = =s = . i n∠KPO = PM 5 2 -r PO 2 The r e f or e, r= . 6 Thean swe ri s 5 2 . 6 Le ta,b ber ea lnumbe r s,andf( x) = ax +bs a t i s f i e s x)| ≤1foranyx ∈ [ 0,1].Thent hemax imumo fab |f( Ma t hema t i c a lOl i adi nCh i na ymp 38 i s . So l u t i on.I ti sea s of i nd t ha ta yt 0).Then f( = f( 1) - f( 0),b = Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. 1 æ ö2 2 0)- f( 1)÷ + 1 ( ab = f( 0)·( 1)-f( 0))= - çf( 1)) f( f( 2 è ø 4 ≤ 1 ( ( )) 1 2 ≤ . f1 4 4 1 1 a =b = ± ,wege When2f( 0)=f( 1)= ±1, i. e., tab = . 2 4 1 The r e f or e,t hemax imumo fabi s . 4 1 Thean swe ri s . 4 6 Taker andoml i ved i f f e r en tnumbe r sf rom 1,2,..., yf 20.Then t he pr obab i l i t ha tt he r ea r ea tl ea s ttwo yt ad ac en tnumbe r samongt hemi s j So l u t i on.Suppos ea1 . < a2 < a3 < a4 < a5 t akenf rom 1, 2,...,20.I fa1 ,a2 ,a3 ,a4 ,a5 a r e no tad acen tt o each j o t he r,t henwehave 1 ≤ a1 < a2 -1 < a3 -2 < a4 -3 < a5 -4 ≤ 16, f r om wh i ch weknow t ha tt henumbe rof way st os e l e c tf i ve numbe r sno tad ac en tt oeacho t he rf r om 1,2,...,20i st he j s amea ss e l ec t i ngf i ved i f f e r en tnumbe r sf rom 1,2, ...,16, i. e.,C5 The r e for e,t her equ i r edpr obab i l i t s 16 . yi 5 C5 C5 232 20 - C 16 16 =1- 5 = . 5 C20 C20 323 232 Thean swe ri s . 323 Ch i naMa t hema t i c a lCompe t i t i on 39 Suppo s er ea lnumbe r sx,ys a t i s f yx -4 y = 2 x -y . 7 Thent her angeo fxi s So l u t i on.Le ty . =a, x -y =b( a,b ≥0).Thenx =y + ( x -y)= a2 +b2 .Theequa t i oni nt heque s t i onbecome sa2 + Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. b2 -4a =2 b,wh i chi sequ i va l en tt o 2 2 ( +( =5 ( a -2) b -1) a,b ≥ 0). Ass eeni nF i .7 .1,t het r ac eo f g i n t( a,b)i np l aneaObi st hepa r t po ( , ) oft hec i r c l ewi t hc en t e r 2 1 and a t i s f i nga,b ≥ 0,i. r ad i u s 5s e., y r c t he un i on of po i n t O and a ︵ ACB .Then F i .7 .1 g a2 +b2 ∈ { 0}∪ [ 2,2 5]. 0}∪ The r e f or e,x =a2 +b2 ∈ { [ 4,20]. Thean swe ri s{ 0}∪ [ 4,20]. 8 Suppo s es e en c e{ an }c on s i s t so fn i net e rms,wh i chs a t i s f qu y 1 ai+1 ∈ 2,1,a1 =a9 =1and f o ranyi ∈ { 1,2,..., 2 ai { } 8}.Thent henumb e ro fs equ en c e sl i ket h i si s So l u t i on.Le tbi = . ai+1 ( 1 ≤ i ≤ 8).Then foreach { an } ai s a t i s f i ngt heg i vencond i t i on,wehave y 8 ∏b i=1 i = 8 ai+1 a9 1 ( = = 1,w i t hbi ∈ 2,1,1 ≤i ≤ 8). 2 a a i 1 i=1 ∏ { } ① Conve r s e l equenceo fe i e rms{ bn } s a t i s f i ng ① can y,as ghtt y Ma t hema t i c a lOl i adi nCh i na ymp 40 un i l t e rmi neas equenc e{ an } i nt heque s t i on. que yde 1 bn },t Ineach { he r ea r eobv i ou s l rof - and yevennumbe 2 t hes amenumbe ro f2,wi t ht her ema i nde rbe i ng1.Or, i no t he r Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. word s,t he numbe r so f- 1 i l et he and 2 a r e bo t h2 k,wh 2 numbe rof1i s8 -4 k.He r e,i ti sea s oche ckt ha tkcanon l yt y k 2k be0,1,2.Onceki sg i ven,t he r ea r eC2 ay st ocons t r uc t 8 C 8-2k w { bn }.The bn }s r e for e,t het o t a lnumbe ro f{ a t i s f i ng ①i s y 2 4 4 N = 1 + C2 8 ×15 +70 ×1 = 491. 8C 6 +C 8C 4 = 1 +2 Thean swe ri s491. Pa r tI I Wo r dPr ob l ems ( 56ma rk si nt o t a lf o rt hr e equ e s t i on s) 9 ( xn } 16 ma rk s) Suppo s e po s i t i ve numbe rs equenc e { s a t i s f i e sSn ≥2Sn-1 ,n =2,3,...,whe r eSn =x1 + … + xn .Provet ha tt he r eex i s t sacons t an tC > 0,s ucht ha t xn ≥ C ·2n ,n = 1,2,.... So l u t i on.Whenn ≥ 2,Sn ≥ 2 Sn-1i sequ i va l en tt o xn ≥ x1 + … +xn-1 . Le tC = 1 l lpr ove x1 .Wewi 4 xn ≥ C ·2n ,n = 1,2,.... ① ② byi nduc t i on. Whenn = 1, i ti sobv i ou s l r ue.Whenn = 2,wehavex2 yt ≥ x1 = C ·2 . 2 Whenn ≥ 3,a s s umexk ≥ C ·2k ,k = 1,2,...,n -1. Thenf rom ① ,wehave Ch i naMa t hema t i c a lCompe t i t i on 41 xn ≥ x1 + ( x2 + … +xn-1) ≥ x1 + ( C ·22 + … +C ·2n-1) = C( 22 +22 +23 + … +2n-1)= C ·2n . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. The r e f or e,② ho l d sf oreve r yn. 2 x2 y 10 ( 20ma rk s) Gi venane l l i s eequa t i on 2 + 2 = 1( a >b p a b > 0) l e tA1 , i nap l aner e c t angu l a rcoord i na t es s t emxOy, y A2 ,F1 ,F2 bei t sl e f tandr i -po i n t s,l e f tandr i ghtend ght focu s e s,r e s t i ve l i n tont hee l l i s e pec y,andP beanypo p s et he r ea r e po i n t sQ ,R d i f f e r en tf r om A1 ,A2 .Suppo s a t i s f i ngQA1 ⊥ PA1 ,QA2 ⊥ PA2 ,RF1 ⊥ PF1 ,RF2 ⊥ y ndandpr ovet her e l a t i onsh i tweent hel eng t h PF2 .Fi pbe o fs egmen tQR andb. So l u t i on.Le tc = a2 -b2 .ThenA1(- a,0),A2( a,0), F1(-c,0),F2( c,0). x2 0 Deno t eP ( x0 ,y0),Q ( x1 ,y1),R ( x2 ,y2),whe r e 2 + a 2 y0 = 1,y0 ≠ 0. b2 Fr omQA1 ⊥ PA1 ,QA2 ⊥ PA2 ,wehave A1→ Q ·A1→ P =( x1 +a)( x0 +a)+y1y0 = 0, A2→ Q ·A2→ P =( x1 -a)( x0 -a)+y1y0 = 0. ① ② Sub t r ac t i ng ① and ② ,wehave2 a( x1 +x0)=0, i. e. x1 = -x0 . Sub s t i t u t i ngi ti n t o ① ,wege t-x0 +a +y1y0 = 0,or 2 2 2 2 x2 æ x2 0 -a ö 0 -a ÷ .ThenQ ç -x0 , . è y0 ø y0 Fr omRF1 ⊥ PF1 ,RF2 ⊥ PF2 , i nt hes amewayweob t a i n y1 = 2 x2 æ 0 -c ö ÷ .The r e f or e, R ç -x0 , è y0 ø Ma t hema t i c a lOl i adi nCh i na ymp 42 |QR |= 2 2 x2 x2 0 -a 0 -c b2 = . y0 y0 |y0 | S i nc e|y0 |∈ ( 0,b],t hen|QR |≥ b,whe r et heequa l i t y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. ho l d si fandon l f|y0 |=b ( i. e.,P ( 0,±b)). yi 11 ( 20ma rk s)F i nda l lt hepo s i t i ver ea lnumbe rpa i r s( a,b), s ucht ha tf( x)= ax2 +bs a t i s f i e s xy)+f( x +y)≥ f( x) f o ranyr e a lnumbe r sx,y). f( f( y)( So l u t i on.Theg i vencond i t i oni sequ i va l en tt o 2 ( +b)≥ ( ax2y2 +b)+ ( a( x +y) ax2 +b)( ay2 +b). ① ·b,or ax2 +b)≥ ( ax2 +b) In① ,l e ty =0.Wehaveb + ( ( 1 -b) ax2 +b( 2 -b)≥ 0. Asa > 0andax2canbes u f f i c i en t l a r hen1 -b ≥ 0, yl ge,t i. e.,0 <b ≤ 1. 2 ,or In ① ,l e ty = -x.Wehave( ax4 +b)+b ≥ ( ax2 +b) ( a -a2) x4 -2abx2 + ( 2 b -b2)≥ 0. ② sg( x).I ti sobv i ou st ha t Deno t et hel e f t -hands i deo f② a o t he rwi s e,f r oma >0weknowa =1.Theng( a -a2 ≠0( x)= -2 bx2 +( 2 b - b2)wi t hb > 0,wh i ch meansg( x)can be nega t i ve.A con t r ad i c t i on).Then 2 ( ab ö÷2 æ 2 ab) +( x)= ( a -a2)çx b -b2) 2 g( 2 è a -a ø a -a2 b ö÷2 b ( + 2 -2 a -b) 1 -a ø 1 -a 2 =( a -a2)çx - æ è ≥0 ho l d sf oranyr ea lnumbe rx.Sowehavea -a2 > 0, i. e.,0 < a < 1. Ch i naMa t hema t i c a lCompe t i t i on 43 b > 0a Fu r t he rmor e,f rom nd 1 -a æ b ö÷ = b ( 2 -2 a -b)≥ 0, gç è 1 -a ø 1 -a Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by 91.124.253.25 on 05/31/18. For personal use only. wehave2 a +b ≤ 2. Sof a r,wege tt hene c e s s a r i t i ont ha ta, b mu s ts a t i s f ycond y a sf o l l ows: 0 <b ≤ 1,0 < a < 1,2 a +b ≤ 2. ③ a,b) Wea r ego i ngt opr ovet ha tf oranypa i r( s a t i s f i ng ③ y andanyr ea lnumbe r sx,y,① ho l d s,orequ i va l en t l y, h( x,y)= ( a -a2) x2y2 +a( x2 +y2)+ 1 -b)( 2 axy + ( b -b2)≥ 0. 2 Asama t t e ro ff ac t,when ③ ho l d s,wet henhave b ( a( 1 -b)≥ 0,a -a2 > 0and 2 -2 a -b)≥ 0. 1 -a Comb i n i ngi twi t hx2 +y2 ≥ -2xy,wege t h( x,y)≥ ( a -a2) x2y2 +a( 1 -b)(-2xy)+2 axy + ( 2 b -b2) =( a -a2) x2y2 +2abxy + ( 2 b -b2) b ö÷2 b ( + 2 -2 a -b)≥ 0. 1 -a ø 1 -a =( a -a2)çxy + æ è The r e for e,t hes e tofa l lt hepa i r s( a,b)mee t i ngt heg i ven cond i t i oni s {( a,b)|0 <b ≤ 1,0 < a < 1,2a +b ≤ 2}. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naMa thema t i ca l Compe t i t i on ( Comp l ement aryTe s t) 1 2010 ( F u i a n) j (40 ma rk s) As s een i n F i .1 .1,t he c i r cumc en t e r g o facu t et r i ang l eABCi sO , K i s a po i n t (no tt he mi dpo i n t)ont hes i deBC , Di sapo i n tont heex t ended l i ne o fs egmen t AK ,l i ne s BD and AC i n t e r s ec t a t i n tN ,andl i ne sCD and po F i .1 .1 g ABi n t e r s e c ta tpo i n tM .Pr ovei fOK ⊥ MN , t henA ,B , Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 45 D ,C a r econcyc l i c. So l u t i on. By reduc t i on t o ab s u rd i t s s umet ha tA ,B ,D , y,a Ca r eno tconcyc l i c.Le tt hec i r cu - Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. mc i r c l e (wi t hr ad i u sr)of ABC i n t e r s e c tAD a tpo i n tE .J o i nBE andex t endi tt oi n t e r s e c tl i neAN a tpo i n tQ ;j t endi t o i nCE andex s t oi n t e r s ec tAM a tP .J o i nPQ ,a F i .1 .2 g s eeni nF i .1 .2. g Wehave PK2 =t hepowe rofP wi t hr e s tt o ☉O + pec t hr e s tt o ☉O t hepowe rofQ wi pec =( PO2 -r2)+ ( KO2 -r2). (Wewi l lpr ovei ti nt heappend i x) Int hes ameway, QK2 = ( QO2 -r2)+ ( KO2 -r2). Thenwehave PO2 -PK2 = QO2 -QK2 . The r e f or e,OK ⊥ PQ .Byt heg i vencond i t i onOK ⊥ MN , wege tt ha tPQ ‖MN .Thenwehave AQ AP = . QN PM ① By Mene l au s Theor em,weob t a i n NB ·DE ·AQ = 1, BD EA QN ② MC ·DE ·AP = 1. CD EA PM ③ Ma t hema t i c a lOl i adi nCh i na ymp 46 NB MC , ND MD = = or Fr om ① ,② ,③ ,wege t .Then BD CD BD DC i chimp l i e s ∠DMN = ∠DCB .Then △DMN ∽ △DCB ,wh ha t means K i BC ‖MN .The r e f or e,OK ⊥ BC ,andt st he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i chi sacon t r ad i c t i on.Th i scomp l e t e st he mi dpo i n tofBC ,wh r econcyc l i c. ft ha tA ,B ,D ,C a proo Append i x.Wearego i ngt opr ove PK2 =t hepowe rofP wi t hr e s tt o ☉O + pec t hepowe rofQ wi t hr e s tt o ☉O . pec Ex t endPK t opo i n tF , s ucht ha t PK ·KF = AK ·KE ④ ( s eeF i .1 .3).ThenP ,E ,F ,A g a r econcyc l i c,andwehave ∠PFE = ∠PAE = ∠BCE . r e Then E , C , F , K a concyc l i c,andwehave PK ·PF = PE ·PC . ⑤ Fr om ⑤ and④ ,weg e tPK2 = PE ·PC -AK ·KE =t hepowe r F i .1 .3 g o fP wi t hr e s c tt o ☉O +t hepowe ro fQ wi t hr e s c tt o ☉O . pe pe Remark.I fEi sont heex t endedl i neofAD ,t hent hepr oofi s s imi l a r. 1 2 ( 40ma rk s)Gi venpo s i t i vei n t ege rk, l e tr =k + .De f i ne 2 ( () ( 1) l-1) ( r)= f( r)=rr ,f l ( r)= f( r)),x ∈ R+ ,l≥ 2. f ( f Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 47 ( He r e x deno t e st he mi n imumi n t ege rno tl e s st hanx;e. g., 1 )Pr = 1,1 = 1. ovet ha tt he r eex i s t sapo s i t i vei n t ege rm 2 ( ) s ucht ha tf m ( r) i sani n t ege r. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on.Def i nev2( n)a st heexponen tof2i npo s i t i vei n t ege r ( ) n.Wewi l lpr ovet ha tf m ( r) i sani n t ege rform =v2( k)+1. Weu s ema t hema t i ca li nduc t i ononv2( k)=v. Whenv = 0,ki soddandk +1i seven.Then 1ö 1 ö÷ ( æ æ () 1 = çk + r)= f 1 ( r)= çk + ÷ k + k +1) f( 2ø 2ø è è 2 i sani n t ege r. v ≥1).Thenforv As s umet hepropo s i t i oni st r uef orv -1( ≥ 1, l e t k = 2v +αv+1 ·2v+1 +αv+2 ·2v+2 + …, whe r eαi ∈ { 0,1}f ori =v +1,v +2,....Wehave 1ö 1 ö÷ ( æ æ 1 = çk + r)= çk + ÷ k + k +1) f( 2ø 2ø è è 2 = = 1 k 2 + +k +k 2 2 1 v 1 +2 - + ( αv+1 +1)·2v + ( αv+1 +αv+2)· 2 2v+1 + … +22v + … =k '+ 1, 2 ① wi t h k ' = 2v-1 + ( αv+1 +1)·2v + ( αv+1 +αv+2)·2v+1 + … +22v + … . Obv i ou s l k ') = v - 1.Le tr ' = k ' + y,v2( 1 .By 2 () r ')i a s s ump t i on,weknowf v ( sani n t ege r,wh i chi sequa lt o Ma t hema t i c a lOl i adi nCh i na ymp 48 ( r)by ①.Theproofi st hencomp l e t ed. ( v+1) f 3 ( 50 ma rk s)Gi veni n t ege rn > 2,s uppo s e po s i t i ver ea l numbe r sa1 ,a2 ,...,ans a t i s f yak ≤1,k =1,2,...,n. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Le tAk = Prove a1 +a2 + … +ak , k = 1,2,...,n. k n n k=1 k=1 ∑ ak - ∑ Ak < n -1 . 2 So l u t i on.For1 ≤k ≤n -1,wehave0 < ∑i=1ai ∑ k ≤ka nd0 < a ≤n -k.Byu s i ngt hef ac tt ha t|x -y|< max{ x,y} n i=k+1 i f orx,y > 0,wege t k n æç 1 1 ö÷ 1 ai + ∑ ai ∑ èn k ø i n i=k+1 =1 |An - Ak | = n = < ma x ≤ k æ1 1 ö 1 ai - ç - ÷ ∑ ai è k n ø i=1 ni∑ =k+1 n { 1 n max { ( n =1- k æ1 1 ö 1 ai ,ç - ÷ ∑ ai è k n ø i=1 ni∑ =k+1 } æ 1 1 ÷ö k èk n ø ,ç -k) } k . n The r e f or e, n n k=1 k=1 n ∑ ak - ∑Ak = nAn - ∑ Ak k=1 = < n-1 ∑ (An -Ak ) ≤ k=1 n-1 k ö÷ ∑ è1 - n ø = æç k=1 Th i scomp l e t e st hepr oo f. n-1 ∑ |A k=1 n -1 . 2 n - Ak | Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 49 4 ( 50ma rk s)Thecodes e t t i ngo fac i rl ocki se s t ab l i shed phe -r egu l a r l t hve r t i ce sA1 ,A2 , ...,An : onann po ygon wi eachve r t exi sa s s i r( 0or1)andaco l or ( r ed gnedanumbe orb l ue),s ucht ha te i t he rt henumbe r sort heco l or son Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. eachpa i rofad acen tve r t i ce sa r et hes ame.Wea sk:How j manycode s e t scanber ea l i z edfort h i sl ock? So l u t i on.Gi ven an a rb i t r a r s e tf ort hel ock,i ftwo y code ad ac en tve r t i c e shave d i f f e r en tnumbe r s,wel abe lt hes i de s j i ft heyhaved l i nk i ngt hembyl e t t e ra; i f f e r en tco l or s,wel abe l i fbo t ht henumbe r sandco l or sa r et hes ame,wel abe li t i tbyb; byc.Onc et henumbe randco l oronve r t exA1a r es e t( t he r ea r e ) , , , , fou rd i f f e r en ts e t sf ori t wecant hens e tA2 A3 ... An one byoneac cord i ngt ot hel e t t e r sl abe l l edoneachs i de.Inorde rt o l e ti tr e t urnt ot hei n i t i a ls e tofA1f i na l l henumbe r sofs i de s y,t s tbebo t heven.Sot henumbe rofcode s e t s l abe l l edaandb mu f ort he l ock i s fou rt ime so ft he numbe r of l abe l l ed s i de t i s f hecond i t i ont ha tt henumbe r sofs i de s s equenc e swh i chs a yt l abe l l edbyaandba r ebo t heven. n ö÷ æ Suppos et he r ea r e2 iç0 ≤i ≤ s i de sl abe l l edbya, 2 ø è [ ] n -2 i ö÷ æ and2j ç0 ≤ j ≤ s i de sl abe l l edbyb.Thent he r ear e 2 è ø [ ] 2i 2j Cn way st ol abe l2 is i de sbyaf r omn one s,Cn ay st ol abe l -2i w 2js i de sbyb f r om n - 2 i one s,andt her ema i n i ngs i de sar e l abe l l ed byc.The r e f or e, by t he Mu l t i l i ca t i on Pr i nc i l e, p p 2i 2j t he r e ar e Cn Cn-2i way st ol abe la l lt he s i de s. So t he r e a r et ot a l l y [ n2 ] æ 4∑ çç C i=0 è 2 i n [n-22i ] ∑ j=0 ö 2 j Cn i÷ ÷ -2 ø ① Ma t hema t i c a lOl i adi nCh i na ymp 50 code s e t sf ort hel ock.He r ewes t i l a t eC0 0 = 1. pu sodd,wehaven -2 i > 0,andt Whenni hen [n-22i ] ∑ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. j=0 2 n-2 i-1 j Cn . i =2 -2 ② Sub s t i t u t i ngi ti n t o ① ,wege t [ n2 ] æ 4∑ çç C i=0 è 2 i n [n-22i ] ∑ j=0 n C 2 j n-2 i n [2 ] [2 ] ö 2 2 i n-2 i-1 i n-2 ÷÷ = 4 )= 2 ( ( C 2 Cn 2 i) n ∑ ∑ i i =0 =0 ø n = n k n-k k n-k k ∑Cn2 + ∑Cn2 (-1) k=0 k=0 n =( 2 +1) + ( 2 -1) n = 3 +1. n seven,i fi < Whenni n, n t hen ② r ema i nst r ue;i fi = , 2 2 ha t t hena l lt hes i de so ft he po l r el abe l l ed bya,andt ygona meanst he r ei son l ol abe lt hes i de s.The r e for e,t he r e yonewayt a r et o t a l l y [ n2 ] æ 4∑ çç C i=0 è 2 i n [n-22i ] ∑ j=0 C 2 j n-2 i [ n2 ] -1 ö æ ö 2 i n-2 ÷÷ = 4 × çç1 + ( Cn 2 i-1)÷÷ ∑ i=0 ø è ø [ n2 ] ∑ (C 2 = 2 +4 i n-2 i-1 2 n i=0 code s e t sfort hel ock. Ins umma r y, t henumbe rofcode s e t sf ort hel ock = )= 3n +3 sodd, 3n +1whenni { seven. 3n +3whenni Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 1 2011 51 ( Hu b e i) ( r e, 40 ma rk s) As s een i nF i .1 .1, po i n t s P,Q a g r e s c t i ve l he mi dpo i n t s of AC , BD — t he two pe y,t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. d i agona l sofcyc l i c quadr i l a t e r a lABCD .Le t ∠BPA = ∠DPA .P r ove ∠AQB = ∠CQB . F i .1 .1 g F i .1 .2 g So l u t i on.Asshownin Fi .1 .2,weex t ends egmen tDP t o g i n t e r c ep twi t ht hec i r c l ea tpo i n tE .Then ∠CPE = ∠DPA = ︵ ︵ ∠BPA .S i nc ePi st hemi dpo i n to fAC ,wege tAB =CE ,wh i ch mean s ∠CDP = ∠BDA .Fur t he rmor e, ∠ABD = ∠PCD . AB PC , = The r e f or e,△ABD ∽ △PCB .Then i. e.,AB · BD CD CD =PC ·BD . Thenwehave AB ·CD = æ1 ö 1 AC ·BD = AC · ç BD ÷ = AC ·BQ , è2 ø 2 AB BQ = or .Comb i n i ngi twi t h ∠ABQ = ∠ACD ,wede r i ve AC CD t ha t△ABQ ∽ △ACD .So ∠QAB = ∠DAC . Ex t end i ngs egmen tAQ t oi n t e r c ep twi t ht hec i r c l ea tpo i n t F ,wet henhave Ma t hema t i c a lOl i adi nCh i na ymp 52 ∠CAB = ∠QAB - ∠QAC = ∠DAC - ∠QAC = ∠DAF , ︵ ︵ wh i ch meansBC = DF .Fu r t he rmor e,a sQ i st he mi dpo i n tof BD , t hen ∠CQB = ∠DQF . S i nc e ∠AQB = ∠DQF ,wet henhave ∠AQB = ∠CQB . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Thepr oo fi scomp l e t ed. 2 ( 40 ma rk s)Pr ovef oranyi n t ege rn ≥ 4,t he r eex i s t sa l a lofdeg r een, po ynomi x)= xn +an-1xn-1 + … +a1x +a0 f( wi t ht hef o l l owi ngprope r t i e s. ( 1)a0 ,a1 ,...,an-1 a r ea l lpo s i t i vei n t ege r s; ( 2)Foranypo s i t i vei n t ege rm anda rb i t r a r k ≥ 2) yk( s i t i vei n t ege r sr1 ,r2 , ...,rk t ha ta r ed i f f e r en tf rom po eacho t he r,wehave m )≠ f( r1) r2)…f( rk ). f( f( So l u t i on.Le t x)= ( x +1)( x +2)…( x +n)+2. f( ① Obv i ou s l x)i sa mon i cpo l a lofdeg r ee n wi t h y,f( ynomi s i t i vei n t ege rcoe f f i c i en t s.Wea r ego i ngt oprovet ha tf( x) po ha spr ope r t 2). y( Foranyi n t ege rt,s i nc en ≥ 4,weknowt ha tt he r eex i s t s de f i n i t e l l t i l eo f4i nanyn con s ecu t i venumbe r st +1, t yamu p +2,. .., t +n.Thenf r om ① ,wehavef( t)≡ 2( mod4). s i t i vei n t ege r sr1 , Thenf oranyk ( k ≥2)po r2 ,..., rk ,we have mod4). r1) r2)…f( rk )≡ 2k ≡ 0( f( f( Ont heo t he rhand,f oranypo s i t i vei n t ege r m ,wehave Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 53 m )≡ 2( mod4).The r e f or e, f( mod4), m )≢ f( r1) r2)…f( rk )( f( f( wh i chimp l i e st ha tf( m )≠ f( r1) r2)…f( rk ).Wet henf i nd f( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t her equ i r edf( x)andcomp l e t et heproof. 3 ( s i t i ver ea l 50 ma rk s)Le ta1 ,a2 ,...,an ( n ≥ 4)bepo numbe r swi t ha1 < a2 < … < an .Foranypo s i t i ver ea l numbe r r,t he numbe ro ft e rna r r oups ( i,j,k) yg aj -ai =r ( 1 ≤i <j < k ≤ n)i sdeno t eda s s a t i s f i ng y ak -aj r).Provefn ( r)< fn ( So l u t i on.Gi 1 <j venj( n2 . 4 ,t < n) henumbe roft e rna r r oups yg ( i,j,k)s a t i s f i ng1 ≤i <j <k ≤ nand y aj -ai =r ak -aj ① i sdeno t eda sgj ( r).Forf i xedi,j wi t hi <j,t he r ei sa tmo s t ; , oneks st ochoo s ei wh a t i s f i ng ① s ot he r ea r ej -1way i ch y meansgj ( r)≤j -1. t hk > Inas imi l a rway,forf i xedj,k wi he r ei sa tmo s toneis a t i s f i ng ① ;s ot he r ea r en -j way st o j,t y choo s ek,wh i ch meansgj ( r)≤ n -j.The r e for e, r)≤ mi n{ gj ( j -1,n -j}. Then,whenni seven ( i. e., n = 2m ),wehave r)= fn ( ≤ n-1 r)= ∑gj ( j=2 m m-1 2m-1 j=2 j=m r)+ ∑gj ( r) ∑gj ( 2m-1 2m -j)= j -1)+ ∑ ( ∑( j=m+1 j=2 2 2 = m -m < m = n2 . 4 m( m -1) m ( m -1) + 2 2 Ma t hema t i c a lOl i adi nCh i na ymp 54 Whenni sodd ( i. e. n = 2m +1),wehave r)= fn ( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ≤ n-1 r)= ∑gj ( j=2 m r)+ ∑gj ( j=2 m j -1)+ ∑( 2 r) ∑g ( j=m+1 j ∑ (2m +1 -j) j=m+1 j=2 =m < 2m 2m n2 . 4 Thepr oo fi scomp l e t ed. 4 ( 50ma rk s)Gi vena3×9a r r ayA wi t heachce l lcon t a i n i ng apo s i t i vei n t ege r,wes ayam ×n ( 1 ≤ m ≤3,1 ≤n ≤9) sa “ e c t ang l e”i ft hes um oft he s uba r r ayo fA i goodr numbe r si ni t sc e l l si samu l t i l eof10,andca l la1×1ce l l p o fA “ bad”i fi ti sno tcon t a i nedi nany “ ec t ang l e”. goodr F i ndt hemax imumnumbe ro f“ badce l l s”i nA . So l u t i on.Wef i r s tc l a imt ha tt henumbe rof “ badce l l s”i nAi s no mor et han25.Ot he rwi s e,t he r ewi l lbea tmo s tonec e l li nA t ha ti sno t“ bad”.Wi t hou tl o s so fgene r a l i t s s umet he y,wea c e l l si nt hef i r s tr owo fA a r ea l l“ bad”.Thenl e tt henumbe r s f r omt op t o bo t t om i nt heit hco l umn beai ,bi ,ci( i = 1, 2,...,9) i nt u rn,andde f i ne Sk = k ∑ai ,Tk = i=1 k b ∑( i=1 i ,k = 1,2,...,9, +ci ) wi t hS0 = T0 = 0.Wea r ego i ngt o pr ovet ha tt hr eenumbe r r oup sS0 ,S1 ,...,S9 ,T0 ,T1 ,...,T9 ,andS0 +T0 ,S1 + g T1 , ...,S9 + T9 each form a comp l e t es e to fr e s i due s modu l o10: I ft he r eex i s tm ,n,0 ≤ m <n ≤9s ucht ha tSm ≡Sn ( mod 10),t hen Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe n ∑a i=m+1 i 55 = Sn -Sm ≡ 0( mod10), wh i ch meanst ha tt hec e l l si nt hef i r s tr ow andf rom co l umns m +1t onf orma “ e c t ang l e”.Bu ti ti sacon t r ad i c t i ont o goodr Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t hea s s ump t i ont ha tt hec e l l si nt hef i r s trowa r ea l l“ bad”. I ft he r eex i s tm ,n,0 ≤ m <n ≤9s ucht ha tT m ≡Tn ( mod 10),t hen n b ∑( i=m+1 i +ci )= Tn -T m ≡ 0( mod10). Sot hec e l l sr ang i ngf r omrows2t o3andco l umnsm +1t o n forma “ ec t ang l e”,wh i ch meanst he r ea r ea tl ea s ttwo goodr c e l l st ha ta r eno t“ bad”.Bu ti ti sa l s oacon t r ad i c t i on. Inas imi l a rway,wecana l s opr ovet ha tt he r ea r enom ,n, 0 ≤ m < n ≤ 9s ucht ha t Sm +T m ≡ Sn +Tn ( mod10). The r e f or e,wehave 9 ∑Sk ≡ k=0 9 ∑Tk ≡ k=0 9 ∑ (S k=0 k +Tk )≡ 0 +1 +2 + … +9 ≡ 5( mod10). Then 9 ∑ (Sk +Tk )≡ k=0 9 9 k=0 k=0 ∑ Sk + ∑Tk ≡ 5 +5 ≡ 0(mod10). I ti saga i nacon t r ad i c t i on! The r e for e,t henumbe rof “ bad c e l l s”i nAi snomor et han25. Ont heo t he rhand,wecancons t r uc ta3 × 9a r r ayi nt he fo l l owi ngandche ckt ha teachc e l li ni tt ha tdoe sno tcon t a i n numbe r10i s“ bad”. Ma t hema t i c a lOl i adi nCh i na ymp 56 1 1 1 2 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 1 10 1 1 2 1 1 1 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. The r e f or e,wef i ndou tt ha tt hemax imum numbe rof “ bad c e l l s”i nAi s25. 2012 ( Sh a a n x i) 1 ( 40ma rk s)Ass eeni nF i .1 .1,i nacu t et r i ang l e△ABC , g AB > AC ,M ,N a r etwod i f f e r en tpo i n t sonBCs ucht ha t ∠BAM = ∠CAN ,a ndO1 ,O2 a r et hec i r cumcen t e r sof △ABC , △AMN ,r e s c t i ve l ove O1 ,O2 , A a r e pe y.Pr co l l i nea r. F i .1 .1 g F i .1 .2 g So l u t i on.AsshowninFi .1 .2,weconne c tAO1 ,AO2 ,and g t hr oughA dr awal i nepe rpend i cu l a rt oAO1 andi n t e r s ec twi t h t heex t endedl i neo fBCa tpo i n tP .ThenAPi sat angen tl i neof ☉O1 ,wh i ch mean s ∠B = ∠PAC . As ∠BAM = ∠CAN ,wehave ∠AMP = ∠B + ∠BAM = ∠PAC + ∠CAN = ∠PAN . ThenAPi sa l s oat angen tl i neo f☉O2 t hec i r cumc i r c l eof Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 57 △AMN .The r e f or e,AP ⊥AO2 , andt ha tmeansO1 ,O2 ,A a r e co l l i nea r. Thepr oo fi scomp l e t e. 2 ( 40ma rk s)Le tA = { 2,22 ,...,2n ,...}.Prove: Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 1)Foranya ∈ A ,b ∈ N* , i fb < 2 a -1, t henb( b +1) wi l lno tbeamu l t i l eof2 a. p ( a t i s f i nga ≠ 1,t 2)Foranya ∈ A (= N* - A )s he r e y ex i s t sb ∈ N* s a t i s f i ngb < 2 a -1, s ucht ha tb( b +1) i sa y a. mu l t i l eof2 p So l u t i on.( 1)Foranya k ∈ A ,a = 2 ( k ∈ N* ).Then2a = s i t i vei n t ege rs t r i c t l e s st han2 a -1.Then 2k+1 .Le tbbeanypo yl ( b +1)≤ 2a -1. Be tweenbandb +1,onei sanoddnumbe rt ha tcon t a i nsno imef ac t or2,andt heo t he ri sanevennumbe rt ha tcon t a i nsa t pr mo s tt hekt hpowe ro f2.The r e f or e, b( b +1) i sde f i n i t e l ta yno mu l t i l eo f2 a. p ( 2)Fora ∈ Aanda ≠1, s uppo s ea =2km whe r eki sanon - nega t i vei n t ege randm i sanoddnumbe rg r ea t e rt han1.Then 2 a = 2k+1m . We wi l l pr e s en tt hr ee d i f f e r en t proof si nt he f o l l owi ng. Proo f1.Le tb = mx, b +1 =2k+1y.El imi na t i ngb,wehave 2k+1y - mx = 1.S i nc e( 2k+1 ,m )= 1, t heequa t i onha si n t eg r a l s o l u t i onst ha tcanbeexpr e s s eda s x = x0 +2k+1t, ( whe r et ∈ Z,and ( x0 ,y0)i sas i a l pec y = y0 + mt { s o l u t i ono ft heequa t i on). Deno t et hesma l l e s ts o l u t i on amongt hem a s( x* ,y* ). Thenx* < 2k+1 . The r e f o r e, i samu l t i l eo f2 b = mx* <2 a -1andb( b +1) a. p Pr oo f2.S i nc e( 2k+1 ,m ) = 1,byt heCh i ne s e Rema i nde r 58 Ma t hema t i c a lOl i adi nCh i na ymp Theor em,t hecong r uenc eequa t i on x ≡ 0( mod2k+1), { x ≡ m -1( modm ) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ha sas o l u t i onx =b wi 0,2k+1m ). t hb ∈ ( I ti sea s os eet ha tb yt <2 a -1andb( b +1) i samu l t i l eof2 a. p t hent he r eex i s t sr ∈ N* , Pr oo f3.S i nc e( 2k+1 ,m )=1, r≤ m -1,s ucht ha t2r ≡ 1( modm ). Taket ∈ N* s ucht ha tt modm ). r >k +1.Then2tr ≡ 1( I t i sea s os eet ha tt he r eex i s t s yt 2tr -1)-q·2k+1m > 0( b=( q ∈ N), s ucht ha t0 < b < 2 a - 1.Then wehavem |b,2k+1 |b + 1. The r e for e, b( b +1) i samu l t i l eof2a. p 3 ( 50ma rk s)Le tP0 ,P1 ,P2 ,...,Pn ben +1po i n t sona l ane,andt hemi n imumd i s t ancebe tweeneachtwopo i n t s p oft hemi sd( d > 0).Prove æ d ön |P0P1 |·|P0P2 |·…·|P0Pn |> ç ÷ è3 ø ( n +1)! . So l u t i on1.Wemaya s s umet h a t|P0P1|≤|P0P2|≤ … ≤|P0Pn |. Atf i r s t,wewi l lpr ovet ha t|P0Pk |> s i t i vei n t ege rn. po Obv i ou s l y,|P0Pk|≥d ≥ d k +1forany 3 d k +1fork =1,2,...,8, 3 andt hes e condequa l i t l d son l l yho y whenk = 8.Then weon y needt opr ovet ha t|P0Pk |≥ d ≥ d k +1fork ≥ 9. 3 TakeeachPi( i = 0,1,2,...,k)a st hecen t e rt odr awa c i r c l ewi t hr ad i u s d .Thent he s ec i r c l e sa r ee i t he rex t e rna l l y 2 Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 59 t angen tt oorapa r tf rom eacho t he r.TakeP0 a st hec en t e rt o dr awac i r c l ewi t hr ad i u s|P0Pk |+ d .Thent hepr ev i ou sk +1 2 sma l l e rc i r c l e sa r ea l ll oca t edi nt h i sl a r rone. ge Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. d ö2 æ æ d ö2 k +1) Thenπç|P0Pk |+ ÷ > ( πç ÷ ,f r om wh i ch we 2ø è è2 ø have|P0Pk |> d( k +1 -1). 2 I ti sea s oche ckt ha t yt Then|P0Pk |> k +1 -1 > 2 d k +1fork ≥ 9. 3 k +1 fork ≥ 9. 3 d Ove ra l l,wehave|P0Pk |> k +1fork ≥ 9. 3 The r e for e, |P0P1 |·|P0P2 |·…·|P0Pn |> æç d ö÷n è3 ø ( n +1)! . So l u t i on2.Wemaya s s ume|P0P1|≤|P0P2|≤ … ≤|P0Pn|. i = 0,1,2,...,k)a TakeeachPi( st hecen t e rt odr awa c i r c l ewi t hr ad i u s d .Thent he s ec i r c l e sa r ee i t he rex t e rna l l y 2 t angen tt oorapa r tf r omeacho t he r. Le tQ beanypo i n ton☉Pi .S i nc e |P0Q | ≤|P0Pi |+|PiQ |=|P0Pi |+ ≤|P0Pk |+ d 2 1 3 |P0Pk |= |P0Pk |, 2 2 3 weg e tt ha tt hec i r c l ewi t hc en t e rP0andr ad i u s |P0Pk|c ov e rt he 2 ö2 æ3 r e v i ou sk + 1sma l l e rc i r c l e s.Then wehav eπç |P0Pk |÷ > p è2 ø æ d ö2 ( k +1) πç ÷ ,andt ha ti s è2 ø Ma t hema t i c a lOl i adi nCh i na ymp 60 |P0Pk |> The r e for e, d k +1( i = 0,1,2,...,k). 3 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. æ d ön |P0P1 |·|P0P2 |·…·|P0Pn |> ç ÷ è3 ø ( n +1)! . 1 1 4 ( 50ma rk s)Le tSn =1+ + … + ,whe r eni sapo s i t i ve 2 n i n t ege r.Pr ovet ha tforanyr ea lnumbe r sa, b wi t h0 ≤a < b ≤1, t he r ea r ei nf i n i t emanyt e rmsi nt hes equenc e{ Sn - [ Sn ]}t a,b). (He x]deno ha ta r e wi t h i n( r e[ t e st he l a r s ti n t ege rno tg r ea t e rt hanr ea lnumbe rx.) ge So l u t i on1.Foranyn S2n = 1 + * ∈ N ,weh ave 1 ö÷ æ 1 1 1 … 1 1 + + + + n =1+ +ç 1 + è2 +1 22 ø 2 3 2 2 1ö æç 1 +… + n÷ è2n-1 +1 2 ø >1+ =1+ Le tN0 = 1ö æ1 1 ö æ1 1 + ç 2 + 2 ÷+ … + ç n + … + n ÷ è2 2 ø è2 2 ø 2 1 1 … 1 1 + + + > n. 2 2 2 2 [b 1-a ] +1,m = [S N0 ]+1.Then 1 < N0 , b -a 1 <b -a, andSN0 < m ≤ m +a. N0 Le tN1 = 22 m+1 .ThenSN1 = S22(m+1) > m +1 ≥ m +b. ( ) u cht ha t Wec l a imt ha tt he r ee x i s tn ∈ N* wi t hN0 <n < N1s m +a < Sn < m +b ( o r,i no t he rwo r d s,Sn - [ Sn ]∈ ( a,b)). Ot he rwi s e,a s s umi ngt hec l a im i sf a l s e,t hent he r e mu s t ex i s tk > N0s ucht ha tSk-1 ≤ m +aandSk ≥ m +b. ThenSk - Sk-1 ≥ b - a.Bu ti tcon t r ad i c t st hef ac tt ha t Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe Sk -Sk-1 = 61 1 1 < <b -a.Th e r e f or e,t hec l a imi st r ue. k N0 Fu r t he rmor e,a s s umet he r ea r e on l i n i t e numbe r of y af a t i s f i ng s i t i vei n t ege r sn1 ,...,nk s y po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Snj - [Snj ]∈ ( a,b)( 1 ≤j ≤k). De f i nec = mi Snj - [Snj ]}.Thent n{ he r ee x i s t snon ∈ N* 1≤j≤k Sn ]∈ ( a,c). s u cht ha tSn - [ I tc on t r ad i c t st heabov ec l a im. Sn The r e for e,t he r ea r ei nf i n i t et e rmsi nt hes equenc e{ [ Sn ]}t a,b). ha ta r ewi t h i n( Thepr oofi scomp l e t e. So l u t i on2.Foranyn S2n = 1 + * ∈ N ,weh ave 1 ö÷ æ 1 1 1 … 1 1 + + + + n =1+ +ç 1 + è2 +1 22 ø 2 3 2 2 1ö æç 1 +… + n÷ è2n-1 +1 2 ø >1+ =1+ 1ö æ1 1 ö æ1 1 + ç 2 + 2 ÷+ … + ç n + … + n ÷ è2 2 ø è2 2 ø 2 1 1 … 1 1 + + + > n. 2 2 2 2 The r e f or e,Sn canbel a r rt hananypo s i t i venumbe ra s ge l onga snbe come ss u f f i c i en t l a r yl ge. Le tN0 = [b 1-a ] +1.ThenN1 <b -a,andwhenk > N , 0 wehave Sk -Sk-1 = 0 1 1 < <b -a. k N0 Sof oranypo s i t i vei n t ege rm > SN0 ,t he r eex i s t sn > N0 a,b),or,i s ucht ha tSn - m ∈ ( no t he rword s,m +a < Sn < m +b.Ot he rwi s e,t he r emu s tbek > N0s ucht ha tSk-1 ≤ m +a andSk ≥ m +b, i. e.,Sk -Sk-1 ≥b -a.Bu ti tcon t r ad i c t st he Ma t hema t i c a lOl i adi nCh i na ymp 62 f ac tt ha tSk -Sk-1 = 1 1 < <b -a. k N0 SN0 ]+i( i = 1,2,3,...).Thent Nowl e tmi = [ he r e ucht ha tmi +a < Sni < mi +b,i. e.,Sni ex i s t sni > N0s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. [ Sni ]∈ ( a,b). The r e for e,t he r ea r ei nf i n i t et e rmsi nt hes equenc e{ Sn - [ Sn ]}t a,b). ha ta r ewi t h i n( 2013 ( J i l i n) 1 ( 40ma rk s)Ass eeni nF i .1 .1,ABi sachordo fc i r c l eω, g Pi sa po i n tona r cAB ,andE ,F a r e2 po i n t son AB s a t i s f i ngAE = EF = FB .Connec tPE ,PF andex t end y t hemt oi n t e r s e c twi t hωa tC ,D ,r e s t i ve l pec y.Prove EF ·CD = AC ·BD . F i .1 .1 g F i .1 .2 g So l u t i on.Asshownin Fi .1 .2,weconnec tAD ,BC ,CF , g DE .S i nc eAE = EF = FB ,wehave Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 63 BC ·s i n∠BCE d i s t anc ebe tweenB andCP BE = = = 2. AC ·s i n∠ACE d i s t ancebe tweenA andCP AE ① Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Int hes ameway, AD ·s AF i n∠ADF d i s t anc ebe tweenA andPD = = = 2. BD ·s BF i n∠BDF d i s t ancebe tweenB andPD Ont heo t he rhand,s i nc e ② ∠BCE = ∠BCP = ∠BDP = ∠BDF , ∠ACE = ∠ACP = ∠ADP = ∠ADF , BC ·AD = 4,o mu l t i l i ng ① by ② ,wehave r p y AC ·BD BC ·AD = 4AC ·BD . ③ AD ·BC = AC ·BD + AB ·CD . ④ ByPt o l emy sTheor em,wehave Comb i n i ng ③ and ④ ,wege tAB ·CD = 3AC ·BD ,and t ha ti s EF ·CD = AC ·BD . Thepr oofi scomp l e t e. 2 ( t hes equenc e{ an } 40ma rk s)Gi venpo s i t i vei n t ege r su,v, i sde f i neda s: a1 = u +v,andform ≥ 1, a2m = am +u, { a2m+1 = am +v. Deno t eSm =a1 +a2 + … +am ( m =1,2,...).Prove t ha tt he r ea r ei nf i n i t et e rmsi ns equence { Sn }t ha ta r e s r enumbe r s. qua Ma t hema t i c a lOl i adi nCh i na ymp 64 So l u t i on.Forpos i t i vei n t ege rn,wehave S2n+1-1 = a1 + ( a2 +a3)+ ( a4 +a5)+ … + ( a2n+1-2 +a2n+1-1) = u +v + ( a1 +u +a1 +v)+ ( a2 +u +a2 +v)+ … + Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( a2n-1 +u +a2n-1 +v) n =2 ( u +v)+2S2n-1 . Then S2n-1 = 2n-1( u +v)+2S2n-1-1 n 1 =2- ( u +v)+2( 2n-2( u +v)+2S2n-2-1) n 1 = 2·2 - ( u +v)+22S2n-2-1 = … =( n -1)·2n-1( u +v)+2n-1( u +v) =( u +v)n·2n-1 . Suppo s eu +v = 2k ·q,whe r eki sanon -nega t i vei n t ege r, andqi sanoddnumbe r.Taken =q·l2 ,whe r eli sanypo s i t i ve 2 2 2 k-1+q· l n , i n t ege rs a t i s f i ngl ≡k -1( mod2).ThenS2 -1 =ql ·2 y and 2 k -1 +q·l2 ≡k -1 +l2 ≡k -1 + ( k -1) =k( mod2). k -1)≡ 0( The r e for e,S2n-1 i sas r e numbe r.S i nce t he r ea r e qua i nf i n i t el s,t he r ea r ei nf i n i t et e rmsi n{ Sn }t ha ta r es r e qua numbe r s.Theproo fi scomp l e t e. 3 (50 ma s t i ons i n an rk s) Suppo s et he r e a r e m que exami na t i ona t t endedbyns t uden t s,whe r em ,n ≥ 2a r e i ven na t ur a l numbe r s. The ma rk i ng r u l e for each g s t i oni sa sfo l l ows:i ft he r ea r e exac t l t uden t s que yx s f a i l i ngt oanswe rt he que s t i oncor r e c t l hent hey wi l l y,t eachge t0ma rk s,andt ho s ewhoan swe ri tcor r ec t l l l y wi eachge tx ma rk s.Thet o t a lma rksofas t uden ta r et he Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 65 s um o f ma rk she/s he ge t sf r om t he m que s t i on s.Now t uden t sa sp1 ≥ p2 ≥ … ≥ r ankt het o t a lma rk soft hens ndt hemax imum po s s i b l eva l ueo fp1 +pn . pn .Fi = 1,2,. ..,m ,a s s umi ngt he r ea r exk So l u t i on.Foranyk Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s t uden t sf a i l i ngt oan swe rt hekt hque s t i onco r r e c t l hent he r e y,t swhoanswe ri tco r r e c t l a chge t sxk ma rks a r en -xk one yande f r omi ta c co rd i ng l et hesum o ft hen s t uden t s t o t a l y.Suppos ma rksi sS.Thenwehave n ∑ pi = S = i=1 m m m k=1 k=1 k=1 2 ∑xk (n -xk )=n∑xk - ∑xk . As e a ch s t uden t ge t sa t mos t xk ma rks f r om t he kt h s t i on,wehave que p1 ≤ m ∑x k k=1 p2 S i nc ep2 ≥ … ≥ pn ,t henpn ≤ The r e f o r e, p1 +pn ≤ p1 + ≤ . +p3 + … +pn n -1 S -p1 n -2 S = p1 + n -1 n -1 n -1 m m m S -p1 . n -1 n -2· 1 · xk + (nk∑ xk - k∑ xk2 ) ∑ n -1 k n -1 =1 =1 =1 m ∑x =2 k=1 k m 1 · 2 xk . ∑ n -1 k =1 - Byt heCauchyI nequa l i t y,wehave m ∑x k=1 Then = 2 k ≥ m 2 1 ( ∑ xk ) . m k =1 Ma t hema t i c a lOl i adi nCh i na ymp 66 m ∑ xk - p1 +pn ≤ 2 k=1 =- m 2 1 · ( ∑ xk ) m( n -1) k =1 m 2 1 · ( ∑ xk - m ( n -1)) + m ( n -1) m( n -1) k =1 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ≤ m( n -1). Ont heo t he rhand,i ft he r ei sas t uden twhoanswe r sa l lt he s t i onsc o r r e c t l i l et heo t he rn -1s t uden t sf a i lt oanswe r que y,wh anyque s t i ons,t henwehave p1 +pn = p1 = m ∑ (n -1)= m (n -1). k=1 The r e f o r e,t he max imum pos s i b l eva l ueo fp1 + pn i sm ( n -1). 4 ( 50ma rks)Le tn,kbei n t ege r sg r e a t e rt han1ands a t i s f yn < 2 .P r ovet ha tt he r ea r e2 ki n t ege r sno td i v i s i b l ebyn, k sucht ha ti f wed i v i det hem i n t otwo g r oups,t hent he r e mus tex i s tag r oupi nwh i cht hesumo fsomei n t ege r sc anbe d i v i dedbyn. So l u t i on.Atf i r s t,wecon s i de rt heca s et ha tn = 2 ,r ≥ 1. r Obv i ou s l tt h i st imer <k.Wet aket hr ee“ 2r-1”sand2 k -3 y,a “ 1” s — each o ft hem canno tbe d i v i ded byn.I ft he s e2 k numbe r sa r ed i v i dedi n t otwog r oup s, t hent he r emu s te x i s tag r oup t ha tc on t a i n stwo“ 2r-1” s,who s es umi s2r — d i v i s i b l ebyn. Nex t,wecon s i de rt heca s et ha tni sno tapowe rof2.At t h i st ime,t he2 ki n t ege r swet akea r e 2 -1,-1,-2,-2 ,. ..,-2k-2 ,1,2,22 ,...,2k-1 . Thent heyeachcanno tbed i v i dedbyn. As s umet he s enumbe r scanbed i v i dedi n t otwog roups,s uch t ha tanypa r t i a ls umo fnumbe r si noneg roupi sno td i v i s i b l eby Ch i naMa t hema t i c a lCompe t i t i on ( Comp l emen t a r s t) yTe 67 n.Wemays ayt ha t“ 1”i si nt hef i r s tg roup.S i nce(-1)+1 = 0i sd i v i s i b l ebyn, t hetwo “-1” smu s tbei nt hes econdg roup; s i nc e(-1)+ (-1)+2 = 0, t he “ 2”i si nt hef i r s tg roup;t hen t he “-2”i si nt hes e condg r oup. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Nowbyi nduc t i on,a s s umi ng1,2,...,2l a r ei nt hef i r s t nt hes econdone( 1 ≤l < roupand -1,-1,-2,...,-2li g k -2),s i nce (-1)+ (-1)+ (-2)+ … + (-2l )+2l+1 = 0 i sd i v i s i b l ebyn,wege tt ha t2l+1i si nt hef i r s tg roup,andt hen 2l+1i nt hes e cond. The r e f or e,1,2,22 ,...,2k-2i si nt hef i r s tg r oupand -1, -1,-2,-2 ,. ..,-2 - i nt hes e cond.F i na l l i nc e y,s 2 k 2 (-1)+ (-1)+ (-2)+ … + (-2k-2)+2k-1 = 0, t hen2k-1i si nt hef i r s tg roup.The r e for e,1,2,22 ,...,2k-1a r e a l li nt hef i r s tg r oup. Ont heo t he rhand,t heknowl edgeabou tt heb i na r r ynumbe s s t emt e l l su st ha teve r s i t i vei n t ege rwh i chi sno tg r ea t e r y ypo k t han2 - 1can ber epr e s en t ed a st he pa r t i a ls um of1,2, 22 ,...,2k-1 .S i ncen ≤2k -1, t heni ti st hepa r t i a ls umof1,2, 22 ,...,2k-1 t ha ti so fcou r s ed i v i s i b l ebyn i t s e l f.Th i si sa con t r ad i c t i ont ot hea s s ump t i on. The r e f or e,wehavefound ou t2 ki n t ege r st ha t mee tt he r equ i r emen ti nt heque s t i on.Theproo fi st hencomp l e t e. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naMa thema t i ca l Ol i ad ymp The Ch i na Ma t hema t i ca lO l i ad, o r i zed by t he Ch i na ymp gan Ma t hema t i ca lO l i adCommi t t ee,i she l di nJanua r r r. ymp yeve yyea Abou t150wi n ne r so ft heCh i naMa t hema t i ca lCompe t i t i ont akepa r t i ni t.The compe t i t i onl as t sf o rtwo days,andt he r ea r et h r ee r ob l emst obecomp l e t edwi t h i n4 .5hou r seachday. p 2011 ( Ch a n c h u n,J i l i n) g F i r s tDay 8: 00~12: 30Janua r y15,2011 1 Le ta1 ,a2 ,...,an ( n ≥ 3)ber ea lnumbe r s.Pr ovet ha t Ch i naMa t hema t i c a lOl i ad ymp n ∑a 2 i i=1 n - ∑ aa ≤ i i+1 i=1 69 [ n2 ] (M -m ), 2 whe r ean+1 =a1 ,M = max1≤i≤nai ,m = mi n1≤i≤nai .[ x] i st he l a r s ti n t ege rno texc eed i ngx. ge =2 k (ki sapo s i t i vei n t ege r),t hen Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on.I fn n n i=1 i=1 2( ∑ ai2 - ∑ aiai+1 ) = n ∑ (a i=1 t he r e f or e, n n i=1 i=1 2 ∑ ai - ∑ aiai+1 ≤ i -ai+1) ≤ n( M - m ), 2 2 n ( n( 2 2 = M -m ) M -m ) . 2 2 [ ] I fn = 2 k + 1( ki sa po s i t i vei n t ege r),t henfor2 k +1 numbe r sa r r angedi nacyc l i c way,onecana lway sf i ndt hr ee con s e cu t i vei nc r ea s i ng or dec r ea s i ng t e rms ( a s ai-1)( ai+1 -ai)= ∏ 2k+1 i=1 ∏ 2k+1 i=1 ( ai - 2 ( ≥ 0, ai -ai-1) s oi ti sno tpo s s i b l e t ha tf oreve r i ngoppo s i t es i s). yi,ai -ai-1 andai+1 -ai hav gn Wi t hou tl o s so f gene r a l i t s s ume t ha ta1 ,a2 ,a3 a r e y, we a mono t on i c,t hen 2 2 2 ( +( ≤( a1 -a2) a2 -a3) a1 -a3) . Henc e, n 2( ∑ i=1 2 i a - n n ∑ aa ) ∑ (a i=1 i i+1 = i=1 i -ai+1) 2 2 + ≤( a1 -a3) n ∑ (a i=3 i -ai+1), 2 knumbe r s.We wh i cht r an s formedt heque s t i oni n t ot heca s eof2 have n n i=1 i=1 n 2 2 + ∑( 2( ∑ ai2 - ∑ aiai+1 ) ≤ ( a1 -a3) ai -ai+1) i=3 ≤2 k( M - m ), 2 Ma t hema t i c a lOl i adi nCh i na ymp 70 i. e., n ( ∑a Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i=1 2 2 i n - ∑ aa ) ≤k(M i i+1 i=1 -m ) = 2 [ n2 ] (M -m ). 2 Ass howni nF i .2 .1,D i st he g mi dpo i n to fa r c BC o ft he c i r cumc i r c l eo ft r i ang l eABC , Xl i e s on a r c BD ,E i st he i e son mi dpo i n tofa r cAX ,Sl a r cAC ,SD i n t e r s e c t sBC a t R ,SE i n t e r s e c t s AX a t T. Pr ovet ha ti fRT ‖DE ,t hen F i .2 .1 g t hei nc en t e ro ft r i ang l e ABC l i e sonl i neRT . Proo f.Connec tAD ,deno t et hei n t e r s e c t i onofAD andRT by I, t henAIi st heb i s e c t oro f ∠BAC .Connec tAS,SI,t henby RT ‖DE ,wehave ∠STI = ∠SED = ∠SAI, s oA ,T , IandSa r econcyc l i c,andwedeno t et h i sc i r c l ebyω1 . Conne c tCE ,deno t et hei n t e r s e c t i onofCE andRT byJ, t hen conne c tSC , ∠SRJ = ∠SDE = ∠SCE , r econcyc l i c,andwedeno t et h i sc i r c l ebyω2 . s oS,J,RandCa Deno t ebyK t hei n t e r s ec t i onpo i n tofω1 andω2 o t he rt han S,weprovenex tt ha tK i st hei n t e r s ec t i ono fAJ andCI. Deno t eby K1 t hei n t e r s ec t i on ofω1 and AJ o t he rt han A, t hen Ch i naMa t hema t i c a lOl i ad ymp ∠SK1A = ∠STA = 71 1 1( SA + XE )= ( SA + AE ) 2 2 = ∠SDE = ∠SRT = ∠SRJ, s oS,K1 ,J and R a r econcyc l i c,i.e.,K1 be l ong st oω2 . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t he rt han S imi l a r l t ebyK2t hei n t e r s e c t i onofω2andCIo y,deno C ,t s henK2 be l ong st oω1 .Henc e,K1andK2co i nc i de,andK i t hei n t e r s ec t i onofAJ andCI. As ∠CAD = ∠CAI,and ∠TJE = ∠CJR = ∠CED = ∠CAD , s oA ,I,J andC a r econcyc l i c,t he r e f or e,∠ACI = ∠AJI. Ont heo t he rhand,byt heconcyc l i c i t yofC ,K ,J andR , s oIi wehave ∠BCI = ∠ICR = ∠AJI,and ∠ACI = ∠BCI, s t hei nc en t e ro ft het r i ang l eABC . 3 Le tA1 ,A2 ,...,An bennon emp t ub s e t so faf i n i t es e t ys A ofr ea lnumbe r ss a t i s f i ngt hefo l l owi ngcond i t i on s: y ( sequa lt o0; 1)Thes umofe l emen t so fAi ( 2)P i cka rb i t r a r i l rf romeachAi ,andt he i rs um yanumbe i ss t r i c t l s i t i ve. ypo Pr ovet ha tt he r eex i s ts e t sAi1 ,Ai2 ,...,Aik ,1 ≤i1 <i2 < … <ik ≤ n, s ucht ha t k |Ai1 ∪ Ai2 ∪ … ∪ Aik |< |A |. n t e st henumbe rofe l emen t so faf i n i t es e tX . |X |deno So l u t i on.Le tA ={ a1 ,...,am }wi t ha1 > … >am .By ( 1)we s i de rt hesma l l e s te l emen tofeach havea1 + … +am = 0.Con Ai ,t hes um o ft he s enumbe r si sg r ea t e rt han0.As s umet ha t t he r ea r eexac t l e t samong A1 , ...,An who s e mi n ima l yki s e l emen ti sai , i = 1,2,...,m .Then,oneha s Ma t hema t i c a lOl i adi nCh i na ymp 72 By ( 2),wehave k1 + … +km = n. k1a1 + … +kmam > 0. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. e t s, Fors =1,2,...,m -1, t he r ea r ei nt o t a lk1 + … +kss who s e mi n ima le l emen t sa r eg r ea t e rt han or equa lt o as . The r e f or e,t heun i ono ft he s es e t si scon t a i nedi n{ a1 ,...,as }, whenc et henumbe rofe l emen t sdoe sno texceeds. 1,2,...,m - 1} Nex t,wepr ovet ha tt he r eex i s t ss ∈ { s n s ucht ha tk = k1 + … + ks > .We provet h i sc l a im by m con t r ad i c t i on.Suppo s et ha t s n k1 + … +ks ≤ ,s = 1,2,...,m -1. m Wi t ht hehe l heAbe lt r an s f ormandt hef ac tt ha tas poft as+1 > 0,1 ≤s ≤ m -1,weknowt ha t 0< = ≤ m ∑ka j j j=1 m-1 ∑ (a s s=1 m-1 ∑ (a s s=1 ( -as+1) k1 + … +ks )+am ( k1 + … +km ) s n +amn m -as+1) m = n aj = 0. ∑ mj =1 Wet hen ge tacon t r ad i c t i on.Fors uchans,wet aket hes e t s amongA1 ,...,An ,who s e mi n ima le l emen t sa r eg r ea t e rt han as ,s ay Ai1 ,Ai2 , ...,Aik .Then,byt heabover e s u l t s,we s n knowt ha tt het o t a lnumbe rofs uchs e t si sk =k1 + … +ks > , m andt henumbe ro fe l emen t so ft he i run i ondoe sno texceeds, Ch i naMa t hema t i c a lOl i ad ymp 73 km k = i. e.,|Ai1 ∪ Ai2 ∪ … ∪ Aik |≤s < |A |. n n S e c ondDay Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 8: 00~12: 30,Janua r y16,2011 4 Gi venpo s i t i vei n t ege rn, l e tS = { ndt he 1,2,...,n}.Fi mi n imum of|AΔS |+|BΔS |+|CΔS |fornonemp t y fr ea lnumbe r s,whe r eC = { a +b|a f i n i t es e t sA andB o ∈ A, b ∈ B },XΔY = { x|xbe l ong st oexac t l yoneofX andY},|X |deno t e st henumbe rofe l emen t so faf i n i t e s e tX . So l u t i on.Theminimumi sn +1. F i r s t,byt ak i ngA = B = S,wehave |AΔS|+|BΔS|+|CΔS|= n +1. Se cond,wecanpr ovet ha tl =|AΔS|+|BΔS|+|CΔS|≥ n +1.Le Y ={ x|x ∈ X ,x ∉Y}.Wehave tX\ l =|A\S|+|B\S|+|C\S|+|S\A |+|S\B |+|S\C |. Al lweneedt opr ovea r et hef o l l owi ng: ( i)|A\S|+|B\S|+|S\C |≥ 1, ( i i)|C\S|+|S\A |+|S\B |≥ n. Fo r( i).I nf a c t,i f|A\S|=|B\S|= 0, t henA,B ⊆S. So 1c anno tb eane l eme n to fC, t h e r e f o r e( i) i sv a l i d. h en c e|S\C|≥1, For ( i i).I fA ∩ S = ⌀ ,t hen|S\A |≥ n,t hec l a imi s s s umet ha tt he max ima l a l r eadyva l i d.I fA ∩ S ≠ ⌀ ,wea e l emen to fA ∩ Si t hen sn -k,0 ≤k ≤ n -1, |S\A |≥k. ① Ont heo t he rhand,fori =k +1,k +2,...,n,e i t he ri ∉ Ma t hema t i c a lOl i adi nCh i na ymp 74 B( t heni ∈ S\B )ori ∈ B ( t henn -k +i ∈ C , i. e., n -k + i ∈ C\S ),hence |C\S|+|S\B |≥ n -k. ② Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. From ① and ② ,weob t a i n( i i). Inconc l u s i on,( i)and ( i i)a r eva l i d,s ol ≥n +1.Hence, t hemi n imumi sn +1. 5 Gi ven i n t ege r n ∑ ∑ n ≥ 4. F i nd t he max imum of a( ai +bi) f or non -nega t i ve r ea l numbe r s a1 , b( ai +bi) i=1 i n i=1 i a2 ,...,an ,b1 ,b2 ,...,bn s a t i s f i ng y a1 +a2 + … +an =b1 +b2 + … +bn > 0. So l u t i on.The max imum i sn - 1.By homo i t gene y,wecan a s s umewi t hou tl o s sofgene r a l i t ha t∑i=1ai = yt n ∑ n i=1 i b = 1. Fi r s t,i ti sc l ea rt ha ti fa1 = 1,a2 =a3 = … =an = 0and b1 =0,b2 =b3 = … =bn = ∑ b( ai +bi)= n i=1 i n 1 , ai +bi)=1, t hen ∑i=1ai( n -1 1 , henc e n -1 n ∑a (a i=1 n i i ∑b (a i=1 i i +bi ) +bi ) = n -1. Now wepr ovet ha tf oranyr ea lnumbe r sa1 ,a2 ,...,an , b1 ,b2 ,...,bn s a t i s f i ng ∑i=1ai = y n n ∑a (a i=1 n i i +bi ) ∑bi(ai +bi) i=1 ∑ b = 1,wehave n i=1 i ≤ n -1. Ch i naMa t hema t i c a lOl i ad ymp 75 No t et ha tt hedenomi na t ori spo s i t i ve,i ti sequ i va l en tt o s howt ha t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i. e., n n i=1 i=1 ∑ ai(ai +bi)≤ (n -1)∑bi(ai +bi), n n i=1 i=1 ( n -1)∑bi2 + ( n -2)∑ aibi ≥ n ∑a . i=1 2 i Bys t r s s umet ha tb1 i st hesma l l e s tone ymme y,wecana amongb1 ,b2 ,...,bn .Then n n ( n -1)∑bi2 + ( n -2)∑ aibi i=1 i=1 n n i=2 i=1 2 ∑bi + (n -2)∑ aib1 ≥( n -1) b2 n -1) 1 +( ≥( n -1) b2 1 + n ( ∑b ) i=2 i 2 +( n -2) b1 2 =( +( n -1) b2 1 -b1) n -2) b1 1 +( =n b1 + ( n -4) b1 +1 2 ≥1 = 6 n ∑ ai ≥ i=1 n ∑a . i=1 2 i Pr ovet ha tf oranyg i ven po s i t i vei n t ege r sm ,n,t he r e ex i s ti nf i n i t e l i r so fcopr imepo s i t i vei n t ege r sa, y manypa b,s ucht ha ta +b|ama +bnb . So l u t i on.I fmn =1, t hent hec l a imi sva l i d.Formn ≥2, s i nce a a b , -n + ) na ( ama +bnb )= ( a +b) na+b +a (( mn) i ti ss u f f i c i en tt oprovet heex i s t enceofi nf i n i t e l ime y manycopr numbe rpa i r sa,b, s ucht ha t a a b ( -n + , a +b| ( mn) a +b,n)= 1. Ma t hema t i c a lOl i adi nCh i na ymp 76 Le tp = a + b,we on l o provet ha tt he r ea r e y needt s i t i vei n t ege r1 ≤ a ≤ i nf i n i t e l imenumbe r sp andpo y manypr ucht ha t p -1s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a -np . mn) p| ( ByFe rma t st heor em,i. e.,whena1 ≡ a2( modp -1),a1 ≥ 1,a2 ≥ 1,( mn)1 ≡ ( mn)2 ( modp). a a So weon l o pr ovet ha tt he r ea r ei nf i n i t e l y needt y many imenumbe r sp andpo s i t i vei n t ege ras ucht ha t pr a -n. mn) p| ( ① I ft he r ea r e on l i n i t e l uch pr ime s,s ay p1 , yf y many s a s mn ≥ 2,t he ex i s t ence of s uch pr ime si s p2 ,...,pr ( s).Suppo s et ha t obv i ou 2 α α α ( -n = p11p22 . .. mn) r enon -nega t i vei n t ege r s prr ,αisa ( 1 ≤i≤ r). ② α2 αr 1 Le ta = pα 1 p2 …pr ( p1 - 1)…( pr - 1) + 2,and s uppo s et ha t a βr 1 β2 ( -n = pβ mn) r enon -nega t i vei n t ege r s 1 p2 …pr , βisa ③ ) ( 1 ≤i ≤r . I fpi |n,t hen,by ③ anda ≥ 2,weknowt ha tpiβi |n, 2 -n, henc epiβi | ( mn) andby ② wehaveβi ≤αi . I fpi n,t henpi α +1 m ,s o( l e r s pii ,mn) = 1.By Eu α +1 α Theor em ( a sφ( i saf ac t orofa -2 ) pii )= pii ( pi -1) a 2 ( -n ≡ ( -n( mn) mn) modpiαi+1). Becau s epiαi+1 2 ( -n,t mn) hecong r uencer e l a t i onabove a -n. imp l i e st ha tpiαi+1 ( mn) Soβi ≤αi .Hence, Ch i naMa t hema t i c a lOl i ad ymp 77 α α α a 2 βr 1 β2 ( -n = pβ ≤ p11p22 …prr = ( -n, mn) mn) 1 p2 …pr he r ea r ei nf i n i t e l wh i chi si ncon t r ad i c t i on wi t ha > 2.Sot y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a -n. manypr ime sp andpo s i t i vei n t ege r sas ucht ha tp| ( mn) 2012 ( X i a n,Sh a a n x i) F i r s tDay 8: 00~12: 30Janua r y15,2012 1 Ass howni nF i .1 .1,∠A i st heb i s tang l ei nt r i ang l e g gge ABC .Ont hec i r cumc i r c l eo f△ABC ,t hepo i n t sD andE a r et hemi dpo i n t so fABC andACB ,r e s t i ve l t e pec y.Deno by ☉O1t hec i r c l epa s s i ngt hr oughA andB , andt angen tt o l i neAC ,by ☉O2t hec i r c l epa s s i ngt hroughA andE ,and t angen tt ol i neAD .☉O1i n t e r s e c t s☉O2 a tpo i n t sA and P .Provet ha tAPi st heb i s e c t oro f ∠BAC . So l u t i on.Jo i nr e s c t i ve l he pe yt i r so f po i n t sEP ,AE ,BE , pa BP , CD . For t he s ake of conven i ence,we deno t e by A , B, C t he ang l e s ∠BAC , ∠ABC , ∠ACB ,t henA + B + C = 180 °. Take an a rb i t r a r y i n tX ont heex t en s i onofCA , po F i .1 .1 g apo i n tY ont heex t en s i ono fDA .I ti sea s os eet ha tAD = yt DC ,AE = EB .By t hef ac tt ha t A ,B ,C ,D and E a r e Ma t hema t i c a lOl i adi nCh i na ymp 78 concyc l i c,wege t ∠BAE = 9 0 °- Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. =9 0 °- 1∠ C AEB = 90 °- ,∠CAD 2 2 B 1∠ ADC = 90 °- . 2 2 Asl i neAC and ☉O1 a r et angen ta tpo i n tA ,l i neAD and ☉O2 a r et angen ta tpo i n tA ,wege t ∠APB = ∠BAX = 1 80 °- A ,∠ABP = ∠CAP , and ∠APE = ∠EAY = 1 8 0 °- ∠DAE = 1 8 0 °- (∠BAE + ∠CAD -A) °=1 80 °- ç90 æ è C ö÷ æç Bö A 0 °- ÷ + A = 90 - 9 °+ . 2ø è 2ø 2 Bycompu t a t i on,weob t a i n ∠BPE = 3 60 °- ∠APB - ∠APE = 90 °+ A = ∠APE . 2 In △APE and △BPE ,weapp l i ne,andt ake y Law ofS i n t oac coun tt ha tAE = BE ,weob t a i n s i n∠PAE PE PE s i n∠PBE = = = . ∠ BE s i n APE AE s i n∠BPE The r e f or e,s i n∠PAE = s i n∠PBE .Ont heo t he rhand, ∠APE a nd ∠BPE a r ebo t hob t u s e,s o ∠PAE and ∠PBE a r e bo t hacu t e,t hu s ∠PAE = ∠PBE . Henc e,∠BAP = ∠BAE - ∠PAE = ∠ABE - ∠PBE = ∠ABP = ∠CAP . 2 Gi venapr imenumbe rp, l e tA beap ×p ma t r i xs ucht ha t 2 i t sen t r i e sa r eexac t l ns omeorde r.The y1,2,...,p i fo l l owi ng ope r a t i oni sa l l owedfora ma t r i x:addonet o Ch i naMa t hema t i c a lOl i ad ymp 79 eachnumbe ri nar ow oraco l umn,ors ub t r ac tonef rom eachnumbe ri naroworaco l umn.Thema t r i xAi sca l l ed “ fonecant akeaf i n i t es e r i e so fs uchope r a t i ons good”i r e s u l t i ngi nama t r i xwi t ha l len t r i e sze r o.F i ndt henumbe r Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ofgoodma t r i ce sA . So l u t i on.Wemaycomb i net heope r a t i onsont hes amerowor co l umn,t hu st hef i na lr e s u l to fas e r i e sofope r a t i onscanbe r ea l i z eda ss ub s t r ac t i ngi n t ege rxif r omeachnumbe rofi -t hrow ands ub s t r ac t i ngi n t ege ryj f r om eachnumbe rofj -t hco l umn. sgoodi fandon l ft he r eex i s ti n t ege r sxi Thu s,t hema t r i xAi yi ucht ha taij = xi +yj f ora l l1 ≤i,j ≤ p. yj ,s S i ncet heen t r i e sofA a r ed i s t i nc t,x1 ,x2 , ...,xp a r e i rwi s ed i s t i nc t,ands oa r ey1 ,y2 ,...,yp .Wemaycons i de r pa i nceswapp i ngt heva l ueof on l heca s et ha tx1 <x2 < … <xps yt xi andxjr e s u l t si nswapp i ngt hei -t hr owandj -t hr ow,wh i chi s aga i nagood ma t r i x.S imi l a r l i de ron l heca s e y,we maycons yt t ha ty1 < y2 < … < yp , t hu st hema t r i xi si nc r ea s i ngf roml e f t t or i t,a l s of r omt opt obo t t om. gh Fr omt hea s s ump t i on sabove,wehavea11 = 1,a12 ora21 i ncet he equa l s2.Wemaycon s i de ron l heca s et ha ta12 = 2s yt t r ans s e of t he ma t r i xi s aga i n good. Now we a r po gue by con t r ad i c t i ont ha tt hef i r s tr owi s1,2,...,p.As s umeont he sont hef i r s trow,bu tk +1i con t r a r ha t1,2,...,ki sno t,2 yt ≤k <p , t he r e f or ea21 =k +1.Weca l lkcon s ecu t i vei n t ege r sa “ b l ock”,andwes ha l lprovet ha tt hef i r s tr owcons i s t so fs eve r a l b l ock s,t ha ti s,t hef i r s tk numbe r si sa b l ock,t he nex tk numbe r si saga i nab l ock,ands oon. I fi ti sno ts o,a s s umet hef i r s tn g roupsofk numbe r sa r e “ r si sno ta “ b l ock”( ort he r ea r e b l ock s”,bu tt henex tk numbe noknumbe r sr ema i n i ng).I tfo l l owst ha tforj = 1,2,...,n, 80 Ma t hema t i c a lOl i adi nCh i na ymp sa “ b l ock”,t hef i r s tnkco l umn sof y(j-1)k+1 ,y(j-1)k+2 ,...,yjki ubma t r i c e sai,(j-1)k+1 , t hema t r i xcanbed i v i dedi n t opn 1 ×ks ai,(j-1)k+2 ,...,ai,jk , i =1,2,...,p,j =1,2,...,n,each s ubma t r i xi sa “ b l ock”.Now a s s umea1,nk+1 = a,l e tb bet he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. sno tont hef i r s trow, sma l l e s tpo s i t i vei n t ege r ss ucht ha ta +bi t henb ≤k -1.S i nc ea2,nk+1 -a1,nk+1 =x2 -x1 =a21 -a11 =k, wehavea2,nk+1 = a + k,t he r e f or ea + bl i e si nt hef i r s tnk co l umn s. The r e f or e,a + b i scon t a i nedi n one oft he1 × k s ubma t r i c e smen t i onedabove,wh i chi sa “ b l ock”,howeve ra, a +ka r eno ti nt h i s“ b l ock”,wh i chi sacon t r ad i c t i on. Wes howed t ha tt he f i r s trow i sformed by b l ocks,i n r t i cu l a rk|p,howeve r, 1 <k <p,andpi sapr ime,wh i chi s pa impo s s i b l e. So weconc l udet ha tt hef i r s trowi s1,2,...,p, t hek -t hr ow mu s tbe( k -1) k -1) s p +1,( p +2,...,kp.Thu upt oi n t e r chang i ng rows,co l umn sandt r an s s e,t he good po 2 ma t r i xi sun i hean swe ri st he r e f or e2( p!). que,t 3 Pr ovet ha tf orany r ea lnumbe r M > 2,t he r eex i s t sa s t r i c t l nc r ea s i ngi nf i n i t es equenceofpo s i t i vei n t ege r sa1 , yi a2 ,...s a t i s f i ngbo t ht hef o l l owi ngtwocond i t i ons: y ( 1)ai > M if oranypo s i t i vei n t ege ri. ( 2)Ani n t ege rni snon z e roi fandon l ft he r eex i s t sa yi s i t i vei n t ege rm andb1 ,b2 ,...,bm ∈ {-1,1},wi t hn po =b1a1 +b2a2 + … +bmam . So l u t i on.For g i ven M > 2,we c ons t r uc t by i nduc t i on a s equence{ an }t ha t ha ts a t i s f i e st her equ i r emen t s.Takea1 ,a2 t 2 s a t i s f uppo s ea1 ,a2 ,...,a2k ya2 -a1 = 1anda1 > M .Nows a r ea l r eadycho s en,s ucht ha tai > M i ,i = 1,2,...,2 kand s ucht ha tt hes e tAk = { b1a1 + … +bmam |b1 ,...,bm = ±1,1 ≤ m ≤ 2 k}doe s no t con t a i n 0.I ti s obv i ou st ha t Ak i s Ch i naMa t hema t i c a lOl i ad ymp 81 s t r i c,i. e.,Ak = -Ak .A1 = { a1 ,-a1 ,1,-1}.Le tn be ymme t hesma l l e s tpo s i t i vei n t ege rno ti n Ak ,N = ∑ 2k i=1 ai ,now choo s epo s i t i vei n t ege r sa2k+1 ,a2k+2s a t i s f i nga2k+2 -a2k+1 = N + y n,a2k+1 > M 2k+2 ,a2k+1 > ∑i=1ai .Wenowshowt ha tAk+1doe s 2k i r s t, no tcon t a i n0andn ∈ Ak+1 .F n = - ∑i=1ai -a2k+1 +a2k+2 . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 2k Ont heo t he rhand,i f∑i=1biai = 0,m ≤ 2 k +2,a s0 ∉ m Ak ,wemu k +1or2 k +2. s thavem = 2 I fm = 2 k +1, t hen| ∑i=1biai |≥a2k+1 - ∑i=1ai > 0. 2k+1 2k I fm = 2 k +2andb2k+1 andb2k+2 a r eoft hes ames i hen gn,t i fb2k+1andb2k+2a r e | ∑i=1biai|≥a2k+1 +a2k+2 - ∑i=1ai >0; 2k+2 2k o fd i f f e r en ts i s,t hen gn k+2 2 ∑biai = i=1 k 2 2 k i=1 i=1 a2k+1 -a2k+2) ≥|a2k+1 -a2k+2|- ∑ ai ∑biai ± ( = N +n - N = n > 0. The{ an }t hu scon s t r uc t eds a t i s f i e st her equ i r emen t ss i nce0 i sno tcon t a i nedi nanyAk ,andanynon ze r oi n t ege rbe tween -ka ndki scon t a i nedi nAk . S e c ondDay 8: 00~12: 30Janua r y16,2012 4 Le tf( x)= ( x +a)( x +b)whe r eg i venpo s i t i ve r ea,ba r ea lnumbe r s, n ≥ 2beag i veni n t ege r.Fornon -nega t i ve r ea lnumbe r sx1 ,x2 ,...,xn t ha ts a t i s f yx1 +x2 + … + xn = 1,f i ndt he max imum o fF = xj )}. f( So l u t i on1.As ∑ 1≤i<j≤n mi n{ xi), f( Ma t hema t i c a lOl i adi nCh i na ymp 82 mi n{ xi),f( xj )}= mi n{( xi +a)( xi +b),( xj +a)( xj +b)} f( ( xi +a)( xi +b)( xj +a)( xj +b) ≤ 1 (( xi +a)( xj +b)+ ( xi +b)( xj +a)) 2 ≤ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. = xixj + 1( xi +xj )( a +b)+ab, 2 s o F ≤ = = ≤ = = a +b 2 ( ·ab xixj + xi +xj )+ Cn ∑ 2 1≤i<j≤n 1≤i<j≤n ∑ 1 [ 2 n ( ∑x ) i i=1 2 n - n n a +b( 2 ·ab xi2 ] + n -1)∑ xi + Cn ∑ 2 i=1 i=1 -1( 1 2 ·ab a +b)+ Cn (1 - i∑ xi2 ) +n 2 2 =1 n 2ö 1 æç - 1 2 ·ab 1 xi ) ÷ +n 1( a +b)+ Cn ( ∑ n i=1 ø 2è 2 1 çæ1 - 1 ö÷ n -1( n( n -1) + a +b)+ ab n è ø 2 2 2 n -1æç 1 +a +b +nab ö÷ . ø 2 èn 1 Theequa l i t l d swhenx1 = x2 = … = xn = .Sot he yho n n -1æç 1 +a +b +nab ö÷ . max imumo fFi s ø 2 èn So l u t i on2.Weshowthatthemax imumva l uei sa t t a i nedwhen x1 =x2 = … =xn = 1, n -1æç 1 +a +b +nab ö÷ . andFmax = ø n 2 èn Wei nduc tonnt oshowamor egene r a ls t a t emen t:fornon - nega t i ver ea lnumbe r sx1 ,x2 ,...,xns a t i s f i ngx1 +x2 + … + y xn = s (whe r es i saf i xed non -nega t i ver ea lnumbe r),t he max imum va l ue of F = ∑ 1≤i<j≤n mi n{ xi), f( xj )}i s f( Ch i naMa t hema t i c a lOl i ad ymp 83 s a t t a i nedwhenx1 = x2 = … = xn = . n S i nc eFi ss t r i c,wemaya s s umex1 ≤ x2 ≤ … ≤ xn . ymme No t et ha t f( x)i ss t r i c t l nc r ea s i ng on non -nega t i ve r ea l yi Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. numbe r s,wehave F =( n -1) x1)+ ( n -2) x2)+ … +f( xn-1). f( f( æs ö l i t l d swhenx1 = Whenn =2,F =f( x1)≤f ç ÷ ,equa yho è2ø s umet ha tt hes t a t emen tho l d sforn,cons x2 .As i de rt heca s eof n +1.App l i ngi nduc t i vehypo t he s i sonx2 +x3 + … +xn+1 = y s -x1 ,wehave æs -x1 ÷ö 1 = g( F ≤ nf( x1)+ n( n -1) x1), fç è n ø 2 whe r e g( s a quadr a t i cf unc t i on o f x,t he l ead i ng x)i coe f f i c i en ti s1 + n -1, andt hecoe f f i c i en to fx i sa + b 2 n2 n -1æça +b + s ö÷ , t he r e f or e,t heax i sofs t r s ymme yi 2 nø 2 n è n -1æça +b + s ö÷ -a -b 2 nø 2 n è s ≤ . n -1 2( n +1) 2+ 2 n ( Theabovei nequa l i t sequ i va l en tt o[( n -1) s -2n( n +1)( a+ yi b)]( n + 1) ≤ 2 s( 2 n2 + n - 1);obv i ou s l e f t -hands i de < y,l æ s ÷ö ( n2 -1) s < r i t -hand s i de.) The r e for e,g ç i st he gh èn +1ø s max imumo fg( .Thu s,F a x)on 0, t t a i nsi t smax imum n +1 [ ] s -x1 s = = x1 , comp l e t i ng whenx2 = x3 = … = xn+1 = n n +1 t hes o l u t i on. Ma t hema t i c a lOl i adi nCh i na ymp 84 5 Le tn be as r e f r ee po s i t i ve even numbe r,k be an qua i n t ege r,p beapr imenumbe r,s a t i s f i ngp < 2 n ,p n, y ha tncanbewr i t t ena sn =ab +b c +ca, p|n +k .Provet whe r ea,b,ca r ed i s t i nc t i vepo s i t i vei n t ege r s. 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on.S i nc eni seven,wehavep ≠ 2.Asp n,weh ave s s umewi t hou tl o s so fgene r a l i t t p k.Wemaya y0 <k < p.Se 2 n -k( p -k) n +k = -k. t henc = a =k,b = p -k, p p By a s s ump t i on,c i s an i n t ege r,and a,b a r ed i s t i nc t s i t i vei n t ege r s.I tr ema i n st obes hownt ha tc >0,andc ≠a, po n t hu s b.Byt heAM GMi nequa l i t y,wehave +k ≥2 n >p, k n +k2 -k =k, n +k2 > pk,hencec > 0.I fc = a, t hen t hu s p n =k( 2p -k).S i nc eni seven, ki sa l s oeven,a sacons equence ni sd i v i s i b l eby4,wh i chcon t r ad i c t st hef ac tt ha tni ss r e qua seven,ki sodd, f r ee.I fc = b,t henn = p2 - k2 .S i nceni imp l i ngt ha tni saga i nd i v i s i b l eby4,wh i chi sacon t r ad i c t i on. y a t i s f l lt he r equ i r emen t s, Weconc l udet ha ta,b,c s ya comp l e t i ngt heproo f. 6 F i ndt hesma l l e s t po s i t i vei n t ege rk wi t ht hefo l l owi ng ope r t oranyk e l emen ts ub s e tA oft hes e tS = { 1, pr y:f 2,...,2012}, t he r eex i s tt hr eepa i rwi s ed i s t i nc te l emen t s a,b,cofSs ucht ha ta +b,b +c,c +aa l lbe l ongt oA . So l u t i on.Wi t hou tl o s so fgene r a l i t s s umea y,wemaya <b < c.Wr i t ex = a +b,z = b +c,y = a +c,t henx < y < z, x +y >z,andx +y +zi seven.Ont heo t he rhand,i ft he r e ex i s tx,y,z ∈ A s ucht ha tx < y < z,x +y > z,andt ha t x +y + zi seven,s e ta = x +y -z x +z -y ,b = ,c = 2 2 Ch i naMa t hema t i c a lOl i ad ymp 85 y +z -x , i ti sc l ea rt ha ta,b,ca r epa i rwi s ed i s t i nc te l emen t s 2 ofS,andx = a +b,y = a +c,z =b +c. Ther equ i r ed pr ope r t sequ i va l en tt ot hefo l l owi ng:f or yi Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. anyk he r eex i s tt hr eee l emen t sx,y, -e l emen ts ub s e tA o fS,t z ∈ As ucht ha t x < y <z,x +y >z,andx +y +zi seven. (* ) 1,2,3,5,7,...,2011},|A|=1007,andA doe s I fA = { no tcon t a i nt hr eee l emen t ss a t i s f i ngprope r t r e for e, y y (* ).The k ≥ 1008. Wenex tprovet ha tany1008 e l emen ts ub s e to fS con t a i ns t hr eee l emen t ss a t i s f i ngpr ope r t y y (* ). oveagene r a ls t a t emen t:Foranyi n t ege rn ≥ 4,any Wepr ( n + 2) -e l emen ts ub s e t of { 1,2, ...,2 n}con t a i nst hr ee e l emen t ss a t i s f i ng (* ).Wei nduc tonn. y e l emen ts ub s e tof{ 1,2,...,8}, Whenn =4, l e tA bea6 t henA ∩ { 3,4,5,6,7,8} con t a i nsa tl ea s tfoure l emen t s.I fA ∩{ 3,4,5,6,7,8}con t a i nst hr eeevennumbe r s,t hen4,6, 8 ∈ As a t i s f i ng (* ).I fA ∩ { 3,4,5,6,7,8}con t a i ns y exac t l r s,t heni tcon t a i n stwooddnumbe r s. ytwoevennumbe Foranytwooddnumbe r sx,y o f{ 3,5,7}, twoo f( 4,x,y), ( 6,x,y),( 8,x,y)wou l ds a t i s f r t hu soneof yprope y (* ),t t hemi scon t a i nedi nA . I fA ∩ { 3,4,5,6,7,8} c on t a i n se x a c t l y onee v ennumbe rx, t heni tc on t a i n sa l lt h r e eoddnumbe r s, t hen( x, 5,7) s a t i s f i e s(*).Ther e s u l tho l d sf o rn = 4. As s umi ngt her e s u l tho l d sf orn( n ≥4), cons i de rt heca s eof n +1.Le tA be( n +3) -e l emen t sof { 1,2,...,2 n +2}, i f|A ∩{ 1,2,...,2 n} nduc t i vehypo t he s i s,t her e s u l t |≥n +2.Byi fo l l ows.I tr ema i n st ocon s i de r|A ∩ { 1,2,...,2 n} |=n +1, Ma t hema t i c a lOl i adi nCh i na ymp 86 and2 n +1,2n +2 ∈ A .I fA con t a i n sanoddnumbe rxi n{ 1, n},t n + 1,2n + 2s 2,...,2 henx,2 a t i s f i fnoodd y (* ); n}g r ea t e rt han1i scon t a i nedi nA , numbe rof { 1,2,...,2 t henA = { 1,2,4,6,...,2 n,2n +1,2n +2},and4,6,8 ∈ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. As a t i s f y (* ). t ht her equ i r edprope r t s1008. Henc e,t hesma l l e s tk wi yi 2013 ( S h e n a n i a o n i n y g,L g) F i r s tDay8: 00 12: 30 Janua r y12,2013 1 Twoc i r c l e sK1 andK2 o fd i f f e r en tr ad i ii n t e r s ec ta ttwo i n t sonK1 andK2 , i n t sA andB , l e tC andD betwopo po r e s c t i ve l ucht ha tA i st he mi dpo i n toft hes egmen t pe y,s tano t he rpo i n tE , CD .Theex t en s i onofDB mee t sK1 a andt heex t en s i ono fCB mee t sK2 a tano t he rpo i n tF .Le t l1 andl2 bet he pe rpend i cu l a rb i s ec t or so fCD andEF , r e s c t i ve l pe y. ( 1)Show t ha tl1 andl2 ha v eaun i ec ommon qu i n t( d eno t e db . po yP ) ( 2)Pr ov et ha tt hel eng t h s o fCA ,APandPEa r e t hes i d el eng t h so fa r i tt r i ang l e. gh F i .1 .1 g Ch i naMa t hema t i c a lOl i ad ymp 87 So l u t i on ( 1)S i nceC ,A ,B ,E a r econcyc l i c,andD ,A ,B , het heor em o fpowe rofa Fa r econcyc l i c,CA = AD ,andbyt i n t,wehave po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. CB ·CF = CA ·CD = DA ·DC = DB ·DE . ① Suppo s eont hecon t r a r ha tl1andl2dono ti n t e r s e c t,t hen yt CF DE = CD ‖EF ,hence .P l ugg i ngi n t o ① ,wege tCB2 = CB DB DB2 , tfo l l owst ha tCB and t hu sCB = DB ,henceBA ⊥ CD .I DB a r et hed i ame t e r so fK1andK2 , r e s c t i ve l pe y,henceK1and K2 haves amer ad i i,wh i chcon t r ad i c t swi t ha s s ump t i on.Thu s, l1 andl2 haveaun i i n t. quecommonpo ( 2)J o i nAE ,AF andPF ,wehave ∠CAE = ∠CBE = ∠DBF = ∠DAF . st heb i s e c t orof ∠EAF .S i ncePi son S i nc eAP ⊥CD ,APi t he pe rpend i cu l a rb i s e c t or o ft hes egmen tEF ,P i son t he c i r cumc i r c l eo f△AEF .Wehave ∠EPF = 1 80 °- ∠EAF = ∠CAE + ∠DAF = 2∠CAE = 2∠CBE . Henc e,B i sont hec i r c l e wi t hcen t e rP andr ad i u sPE , deno t i ngt h i sc i r c l e byΓ.Le tR bet her ad i u so fΓ.By t he t heor emofpowe rofapo i n t,wehave 2CA2 = CA ·CD = CB ·CF = CP2 -R2 , t hu s AP2 = CP2 -CA2 = ( 2CA2 +R2)-CA2 = CA2 +PE2 . I tfo l l owst ha tCA ,AP ,PE f orm t hes i del eng t hso fa r i tt r i ang l e. gh Ma t hema t i c a lOl i adi nCh i na ymp 88 F i nda l lnon emp t e t sS o fi n t ege r ss ucht ha t3m -2 n∈ ys 2 Sfora l l( no tnec e s s a r i l i s t i nc t)m ,n ∈ S. yd So l u t i on Ca l las e tS “ fi ts a t i s f i e st heprope r t ss t a t ed good”i ya i nt hepr ob l em. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 1)I fS ha son l l emen t,Si s“ yonee good”. ( 2)Now we can a s s ume t ha t S con t a i n sa tl ea s t two e l emen t s.Le t n{ d = mi |m -n|∶m ,n ∈ S,m ≠ n}. Thent he r ei sani n t ege ra,s ucht ha ta +d,a +2d ∈ S. No t et ha t a +4d = 3( a +2d)-2( a +d)∈ S, a -d = 3( a +d)-2( a +2d)∈ S, a +5d = 3( a +d)-2( a -d)∈ S, a -2d = 3( a +2d)-2( a +4d)∈ S. Sowehavepr ovedt ha ti f a +d,a +2d ∈ S, t hen a -2d,a -d,a +4d,a +5d ∈ S. Con t i nu i ngt h i spr oc edur e,wecandeduc et ha t { a +kd|k ∈ Z,3 k}⊆ S. Le tS0 = { a +kd|k ∈ Z,3 k}, i ti sea s ove r i f ha tS0 yt yt i s“ good”. ( 3)Now wehaveprovedt ha tS0 ⊆ S.I fS ≠ S0 ,p i cka numbe rb ∈ S\S0 , t hent he r eex i s t sani n t ege rl,s ucht ha ta + ld ≤ b < a + ( l + 1) d.S i nc ea tl ea s toneofl andl + 1i s l+ i nd i v i s i b l eby3,weknowt ha ta tl ea s toneofa +ld,a + ( 1) di scon t a i nedi nS0 .I fa +ld ∈ S0 ,no t et ha t0 ≤|b - ( a+ Ch i naMa t hema t i c a lOl i ad ymp 89 ld)|< d,byt hede f i n i t i ono fd,wemu s thaveb = a +ld.I f a +( l +1) d ∈ S0 ,no t et ha t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 0 <|a + ( l +1) d -b|≤ d, byt hede f i n i t i ono fd,wea l s ohaveb = a +ld. Inbo t hca s e s,wehaveprovedt ha tt he r ei sanumbe rb ∈ S\S0oft hef ormb =a +ld.Th i slmu s tbed i v i s i b l eby3,hence a +ld,a + ( l +1) d,a + ( l +2) d ∈ S, a+( l -2) d,a + ( l -1) d ∈ S. So a+( l +3) d = 3( a+( l +1) d)-2( a +ld)∈ S, a+( l -3) d = 3( a+( l -1) d)-2( a +ld)∈ S. Con t i nu i ngt h i spr oc edur e,wehavea + ( l +3 d ∈S,j ∈ j) l i e st ha t{ a +kd|d ∈ Z}⊆S.Wec l a imt ha tS = Z,wh i chimp { a +kd ∣k ∈ Z}, a +kd|d ∈ Z}, s i nc ef oranynumbe rx ∉ { t he r ei sane l emen ti ny ∈ { a +kd|k ∈ Z}s ucht ha t0 <|x hede f i n i t i ono fd,we mu s thavex ∉ S.SoS = y|< d.Byt { a +kd ∣k ∈ Z},andi ti sea s ove r i f ha ts uchSi s“ yt yt good”. Weconc l udet ha tt he r ea r et hr eec l a s s e sof “ e t s: good”s ( 1)S = { a}; ( 2)S = { a +kd ∣k ∈ Z,3 k}; ( 3)S = { a +kd ∣k ∈ Z},he r ea,d ∈ Z,d > 0. 3 F i nd a l l po s i t i ve r ea l numbe r st wi t ht he fo l l owi ng r t he r eex i s t sani nf i n i t es e tX o fr ea lnumbe r s prope y:t s ucht ha tt hei nequa l i t y max{ a -d)|,|y -a|,|z - ( a +d)|}>td |x - ( ho l d sf ora l l( no tnec e s s a r i l i s t i nc t)x,y,z ∈ X ,a l l yd Ma t hema t i c a lOl i adi nCh i na ymp 90 r ea lnumbe r saanda l lpo s i t i ver ea lnumbe r sd. 1 So l u t i onTheansweri s0 <t < . 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 -2 t ö÷ , æ 1 F i r s t l or0 <t < ,choo l e txi = s eλ ∈ ç0, ( y,f è 2 1 +t)ø 2 λi ,X = { x1 ,x2 ,...}.Wec l a imt ha tfora l l( no tnece s s a r i l y l lr ea lnumbe r sa anda d i s t i nc t)x,y,z ∈ X ,a l lpo s i t i ver ea l hef o l l owi ngi nequa l i t numbe r sd,wehavet y: max{ a -d)|,|y -a|,|z - ( a +d)|}>td. |x - ( Suppo s eont hecon t r a r ha tt he r eex i s ta ∈ R,d ∈ R+ and yt xi ,xj ,xk ,s ucht ha t max{ a -d)|,|xj -a|,|xk - ( a +d)|}≤td. |xi - ( Henc e, d ≤ xi - ( a -d)≤td, ìï -t ï d ≤ xj -a ≤td, í -t ïï î -t d ≤ xk - ( a +d)≤td, i. e., 1 -t) d ≤ a ≤ xi + ( 1 +t) d, ìïxi + ( ï d ≤ a ≤ xj +td, íxj -t ïï îxk - ( 1 +t) d ≤ a ≤ xk - ( 1 -t) d, wh i chimp l i e st ha t 1 +t) d ≤ a ≤ xi + ( 1 +t) d, ìïxk - ( ï 1 -t) d ≤ a ≤ xj +td, íxi + ( ïï îxj -t d ≤ a ≤ xk - ( d, 1 -t) no t et ha t0 <t < 1 .I tf o l l owst ha t 2 (* ) Ch i naMa t hema t i c a lOl i ad ymp Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. xk -xi , ìï d ≥ ( 2 1 +t) ï ï ï xj -xi , íd ≤ 1 -2 t ï ï ïïd ≤ xk xj . î 1 -2 t 91 ① ② ③ By ② ,③ andd > 0,wege txi < xj < xk ,henc ei >j > i+1 k+1 i j k,λ +λ ≤λ +λ ,wege t xj -xi λj -λi = ≤λ. xk -xi λk -λi ④ xj -xi xk -xi , ≥ By ① and ② ,wege t hence 1 -2 t 2( 1 +t) xj -xi 1 -2 t ≥ >λ, xk -xi 2( 1 +t) wh i chcon t r ad i c t s④ ! Thu sourea r l i e rc l a imabou tXi sproved. Se cond l y,fort ≥ 1, wes howt ha tf oranyi nf i n i t es e tX , 2 f oranyx < y <zi nX ,wecanchoo s ea ∈ Randd ∈ R+ s uch t ha t max{ a -d)|,|y -a|,|z - ( a +d)|}≤td. |x - ( Inf ac t,l e td = z -x , henc ex + ( 1 -t) d =z - ( 1 +t) d. 2 1 Le ta = max{ x +( 1 -t) d,y -td}.S i ncet ≥ ,weob t a i n 2 1 +2 t) d, y -x < 2d ≤ ( 2 x -y < 0 ≤ ( t -1) d, { i. e., 1 +t) d, y -td ≤ x + ( x +( 1 -t) d ≤ y +td, { Ma t hema t i c a lOl i adi nCh i na ymp 92 henc e 1 -t) d ≤a ≤ x + ( 1 +t) d, ìïx + ( ï d ≤ a ≤ y +td, íy -t ïï îz - ( 1 +t) 1 -t) d ≤ a ≤z - ( d, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. f r om wh i chweconc l udet ha t max{ a -d)|,|y -a|,|z - ( a +d)|}≤td. |x - ( Soeve r yt ≥ ob l em. pr 1 doe sno ts a t i s f her equ i r emen toft he yt 2 1ö æ s ç0, ÷ . Inconc l u s i on,t hes e to fa l lr equ i r edti 2ø è S e c ondDay8: 00 12: 30 Janua r y13,2013 4 Gi venani n t ege rn ≥ 2,s uppo s eA1 ,A2 ,...,An a r en nonemp t i n i t es e t ss a t i s f i ng|AiΔAj|=|i -j|fora l l yf y i,j ∈ { 1,2,...,n}. F i ndt hemi n imum va l ueo f|A1 |+|A2 |+ … +|An |. (He t e st henumbe rofe l emen t sofaf i n i t es e t r e|X |deno X andXΔY = { a|a ∈ X ,a ∉Y}∪ { a|a ∈Y ,a ∉ X } f oranys e t sX andY .) So l u t i on For each pos i t i ve i n t ege r k, we prove t ha tt he sk2 + 2;t s mi n imumva l ueo fS2k i he mi n imum va l ueofS2k+1 i k( k +1)+2. F i r s t l f i net hes e t sA1 ,A2 ,...,A2k ,A2k+1a sf o l l ows: y,de Ai = { i, i +1,...,k}, i = 1,2,...,k;Ak+1 = { k,k +1}; Ak+j = { k +1,k +2,...,k +j -1},j = 2,3,...,k +1. Ch i naMa t hema t i c a lOl i ad ymp 93 Fort h i sf ami l fs e t s,i ti sea s ove r i f ha t|AiΔAj|= yo yt yt l d si nt hef o l l owi ngca s e s: j -i =|i -j|ho ( 1)1 ≤i <j ≤k; ( 2)1 ≤i <j =k +1; Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 3)1 ≤i <k +1 <j ≤ 2 k +1; ( 4)k +1 =i <j ≤ 2 k +1; ( 5)k +2 ≤i <j ≤ 2 k +1. i =ji st r i v i a l,andt hec a s eo f i >jc an Mor eove r,t hec a s eo f b er edu c edt ot hec a s eo fi < j.Thu s,f o ra l li,j ∈ { 1,2,..., 2 k +1},wehaveve r i f i edt ha t |AiΔAj |=j -i =|i -j|. Fort heabove( 2 k +1)s e t s,wecanea s i l l cu l a t et ha t yca S2k+1 = k( k +1) k( k +1) +2 + =k( k +1)+2; 2 2 i fwechoo s et hef i r s t2 ks e t s,wege tt ha t S2k = S2k+1 -k =k2 +2. Se cond l ha tS2k ≥k2 +2andS2k+1 ≥k( k +1)+ y,weshowt 2.No t et hef o l l owi ngf ac t s: Fa c t1.Foranytwof i n i t es e t sX ,Y ,wehave|X |+|Y | ≥|XΔ Y |. i f|XΔY | Fa c t2.Foranytwonon emp t i n i t es e t sX ,Y , yf = 1, t hen|X |+|Y |≥ 3. Whenn = 2 k, i tf o l l owsf r om Fac t1t ha t k +1 -2 i, i |Ai |+|A2k+1-i | ≥|AiΔA2k+1-i |= 2 = 1,2,. ..,k -1. By|AkΔAk+1|= 1andFa c t2,weha v e|Ak |+|Ak+1|≥ 3.So Ma t hema t i c a lOl i adi nCh i na ymp 94 k-1 S2k =|Ak |+|Ak+1 |+ ∑ ( |Ai |+|A2k+1-i |) i=1 ≥3+ k-1 ∑ (2k +1 -2i)=k 2 i=1 +2. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. tt ha t S imi l a r l k +1,wege y,whenn = 2 k +2 -2 i, i |Ai |+|A2k+2-i | ≥|AiΔA2k+2-i |= 2 = 1,2,. ..,k -1. S i nce|AkΔAk+1 |= 1,wehave ( |Ak |+|Ak+1 |)+|Ak+2 |≥ 3 +1 = 4, s o k-1 S2k+1 =|Ak |+|Ak+1 |+|Ak+2 |+ ∑ ( |Ai |+|A2k+2-i |) i=1 ≥4+ k-1 k +1)+2. ∑ (2k +2 -2i)=k( i=1 Inconc l u s i on,t hemi n imumva l ueofS2k i sk2 +2,andt he mi n imumva l ueofS2k+1i sk( k +1)+2.Equ i va l en t l y,foranyn ≥ 2, t hemi n imumva l ueo fSni s 5 2 [n4 ] +2. Foreachpo s i t i vei n t ege rnandeachi n t ege ri( 0 ≤i ≤n), l e tCin ≡ c( n,i) ( mod2),whe r ec( n,i) ∈ { 0,1}, andde f i ne n,q)= f( n q. ∑c(n,i) i i=0 tapowe r Le tm ,nandqbepo s i t i vei n t ege r swi t hq +1no o f2.Suppo s et ha tf( m ,q)|f( n,q).Provet m, ha tf( r)|f( n,r)foreve r s i t i vei n t ege rr. ypo So l u t i on Foreach pos i t i vei n t ege rn,we wr i t en i nb i na r y Ch i naMa t hema t i c a lOl i ad ymp 95 r epr e s en t a t i ona sn =2a1 +2a2 + … +2ak ,whe r e0 ≤a1 <a2 < … < ak .De f i neas e tT ( n)= { 2a1 ,...,2ak },T ( 0) i scons i de r ed anemp t e t. ys ByLuca s Theor em,Ci soddi fandon l fT ( i)⊆ T ( n), ni yi Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. henc e n,q)= f( ∑q A) σ( A ⊆T ( n) = ∏ ( 1 +qa ), a∈T ( n) whe r eσ( t e st hes umofa l le l emen t sofA . A )deno Form ,nandqa sg i venbya s s ump t i on,weshowt ha ti f m ,q)= f( ∏ ( 1 +qa ) a∈T ( m) ∏ ( 1 +qa )= f( n,q), a∈T ( n) t henT ( m )⊆ T ( n),andcons equen t l m ,r)|f( n,r)for y,f( eve r yr. Foranyi n t ege r si,j,0 ≤ i < j,wehavet hefo l l owi ng f ac t or i za t i on: 2 2 q -1 = ( q j-1 j t he r e f or e, …( +1) q i 2 ( +1) q i 2 , -1) 2 2 2 ( q +1,q +1)= ( q +1,2)|2. j i i k)bet Le ts( hel a r s toddd i v i s oro fapo s i t i vei n t ege rk, ge 2 2 r ecopr ime.Cl ea r l fi t hens( q +1)ands( q +1)a q > 1.I y, i > 0, q i 2 j +1 ≡1o r2( mod4),andq i 2 +1 >2, t hu ss( q i 2 +1)> 1.I fi = 0, s i nc eq +1i sno tapowe rof2,wehaves( q +1)> a b m ),s( i nce 1.Foranya ∈ T ( q + 1)| ∏b∈T(n)s( q + 1).S s( 1 +qa )> 1,wehavea ∈ T ( n),henceT ( m )⊆ T ( n),wh i ch comp l e t e st hepr oof. 6 Gi venpo s i t i vei n t ege r sm andn,f i ndt hesma l l e s ti n t ege r N (≥ m )wi l emen ts e t t ht hef o l l owi ngpr ope r t fanN-e y:i Ma t hema t i c a lOl i adi nCh i na ymp 96 o fi n t ege r scon t a i n sacomp l e t er e s i dues s t em modu l om , y t heni tha sanon emp t ub s e ts ucht ha tt hes um o fi t s ys e l emen t si sd i v i s i b l ebyn. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i onTheansweri s N = max m ,m +n - { 1 [( , ) ] m m n +1 . 2 } F i r s t,wes howt ha tN ≥ ma x m,m +n - { 1 [ m ( m,n)+1] . 2 } Le td = ( m ,n),andwr I fn > i t em =dm 1 ,n =dn1 . 1 ( m d+ 2 1),t he r e ex i s t s a comp l e t er e s i due s s t em modu l o m ,x1 , y x2 ,...,xm , s ucht ha tt he i rr e s i duemodu l oncons i s t sexac t l yof r oup so f1,2, ...,d.Forexamp m1 g l e,t hefo l l owi ng m numbe r shavet her equ i r edprope r t y: i +dn1j, i = 1,2,...,d,j = 1,2,...,m 1 . F i nd i ngano t he rs e to fk = n - 1 ( m d + 1)- 1numbe r sy1 , 2 ha ta r econg r uen tt o1modu l on, t hes e t y2 ,...,yk t A ={ x1 ,x2 ,...,xm ,y1 ,...,yk } r,noneof con t a i n sacomp l e t er e s i dues s t em modu l om ,howeve y i t snon emp t ub s e t sha si t ss um o fe l emen t sd i v i s i b l ebyn.In ys f ac t,t hes umo ft he ( sma l l e s tnon -nega t i ve)r e s i duemodu l onof a l le l emen t sofA i sg r ea t e rt hanz e r oandl e s st hanorequa lt o m 1( 1 +2 + … +d)+k = n -1.Thu s, N ≥ m +n - i. e., 1 ( m d +1), 2 N ≥ max m ,m +n - { 1 [( , ) ] m m n +1 . 2 } Ch i naMa t hema t i c a lOl i ad ymp 97 1 (, ) ] m n +1 Ne x t,wes howt ha tN = ma x m,m +n - m[ 2 { } ha st her equ i r edpr ope r t y. Thef o l l owi ng key f ac ti sf r equen t l s edi nt he pr oof: yu Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. amonganyki n t ege r s,onecanf i nda ( non emp t ub s e twho s e y)s ta1 ,a2 ,...,ak bei n t ege r s,Si =a1 + s umi sd i v i s i b l ebyk.Le a2 + … +ai . I fs omeSii sd i v i s i b l ebyk, t hent her e s u l ti st r ue. Ot he rwi s e,t he r eex i s t1 ≤i <j ≤k, s ucht ha tSi ≡Sj ( modk), t henSj -Si =ai+1 + … +aji sd i v i s i b l ebyk, t her e s u l ti saga i n t r ue.Thef o l l owi ngf ac ti sanea s l l a r he pr ev i ou s y coro y oft r e s u l t:amonganyki n t ege r s,eacho fwh i chi samu l t i l eofa, p onecanf i nda ( non emp t ub s e twho s es umi sd i v i s i b l ebyka. y)s Re t u rn i ngt ot heprob l em,wesha l ld i s cu s stwoca s e s. 1 Ca s e1: n ≤ m( d +1),andN = m . 2 -s e ti ft hes um ofa l li t s Weca l laf i n i t es e to fi n t ege r sak e l emen t si sd i v i s i b l ebyk.Le tx1 ,x2 ,...,xm beacomp l e t e r e s i dues s t em modu l om .Cl ea r l i v i det he s enumbe r s y,wecand y i n t om 1 g r oup s,each g r oup con s i s t i ng ofacomp l e t er e s i due ty1 ,y2 , ...,yd beacomp l e t er e s i due s s t em modu l od.Le y s s t em modu l od,andyi ≡i ( modd).I fdi sodd,wec and i v i de y d +1 e a chg r oupi n t o d -s e t, f o re x amp l e,{ y1,yd-1},...,{ yd2-1 , 2 1 1 t m 1( d +1)d -s e t s.S i ncen1 ≤ m 1( d+ yd2+1 },{ yd }.Wege 2 2 -s e t ss ucht ha tt hes um of 1),wecanchoo s es omeoft he s ed t he i re l emen t si sd i v i s i b l ebyn1d(=n). seven,s imi l a r l I fdi y, acomp l e t er e s i dues s t em modu l o d can be d i v i dedi n t o y d 2 d -s e t s,wi t hyd2 r ema i n i ng.Twor ema i n i ng numbe r sc anf o rm 1 -s e t. I nt heend,wed i v i d ex1,x2,...,xmi n t o m1d + ano t he rd 2 Ma t hema t i c a lOl i adi nCh i na ymp 98 [m2 ] d-sets(possiblywithanumberleftifm isodd). 1 1 S i nc en1 ≤ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. m 1d + m1 , 1 1 1 wehaven1 ≤ m 1( d +1)= m 1d + 2 2 2 2 [m2 ] ,againwecanfindsomeofthesed-setssuchthat 1 t hes umo fa l lt he i re l emen t si sd i v i s i b l ebyn1d = n. 1 1 Ca s e2: n > m( d +1),N = m +n - m ( d +1). 2 2 Le tA beanN-e l emen ts e t,con t a i n i ngacomp l e t er e s i due t hs omeo t he rn s s t em modu l om ,x1 ,x2 ,...,xm ,wi y 1 m 2 ( d +1)numbe sodd,a ss howni n ca s e1,we may r s.I fd i d i v i dex1 ,x2 , ...,xm i n t o r ema i n i ngn - 1 m 1( d + 1)d -s e t s.Di v i det he 2 1 ( 1 m d +1)numbe r sa rb i t r a r i l n t on1 - m 1 yi 2 2 ( d +1)g roup s,eachwi t hd numbe r s.Amongeachg roupo fd numbe r s,one may f i ndad -s e t,t he r e for e,we haveano t he r n1 - 1 m 1( d +1)d -s e t s,andt o t a l l -s e t s.I fdi seven,a s yn1 d 2 d i s cu s s edi nca s e1,we mayd i v i dex1 ,x2 ,...,xm i n t o d+ 1 m1 2 [m2 ]d-sets.Ifm isodd,weareleftwithanumberx with 1 d|xi - 1 i d 1 .Di v i det heo t he rn - m ( d +1)numbe r sa rb i t r a r i l y 2 2 d i n t o2 n1 -m 1( d +1)g r oup s,eachwi t h numbe r s.Fromeach 2 r oup,wecanf i nda g d - ; d e t s.wecan s e t f r om anytwo -s 2 2 f i ndad -s e t.I fm 1 i seven,t hen wecanf i nd ano t he rn1 - 1 sodd,{ i sa m 1( d +1)d -s e t s,andt o t a l l -s e t s. I fm 1i xi} yn1d 2 Ch i naMa t hema t i c a lOl i ad ymp 99 d d -s e t,wehave2 n1 - m 1( d + 1)+ 1 -s e t s,anda l s on1 2 2 1 1 m 1( d +1)+ d -s e t s.Aga i n,wecanf i ndn1d -s e t s.Fi na l l y, 2 2 wecanchoo s es omeoft he s en1 d -s e t s,s ucht ha tt hes um ofa l l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t he i re l emen t si sd i v i s i b l ebyn1d(= n). 2013 ( Na n i n i a n s u) j g,J g F i r s tDay 30,Decembe r21,2013 8: 00 12: 1 Inanacu t et r i ang l eABC ,AB > AC ,t he b i s e c t or ofang l e BAC ands i deBCi n t e r s e c ta tpo i n tD , two po i n t sE and F a r ei ns i de s AB and AC ,r e s t i ve l uch pec y,s t ha tB ,C ,F ,E a r econcyc l i c. Prove t ha tt he c i r cumcen t e ro f t r i ang l e DEF co i nc i de s wi t ht he f i nne r c en t e ro ft r i ang l e ABC i F i .1 .1 g andon l fBE +CF = BC . yi So l u t i on.Le tIbet hei nne r c en t e rof△ABC . ( Suf i c i enc s eBC = BE +CF .Le tK bet hepo i n t f y)Suppo onBC s ucht ha tBK = BE ,t hu sCK = CF .S i nceBI b i s ec t s ∠ABC ,CI b i s e c t s ∠ACB , △BIK and △BIE a r er e f l e c t i on wi t hr e s c tt o BI, △CIK and △CIF a r er e f l ec t i on wi t h pe r e s c tt oCI,wehave ∠BEI = ∠BKI = π - ∠CKI = π pe Ma t hema t i c a lOl i adi nCh i na ymp 100 ∠CFI = ∠AFI.Th e r e f or e,A ,E ,I, Fa r econcyc l i c. S i nceB ,E ,F ,C a r e concyc l i c,wehave ∠AIE = ∠AFE = ∠ABC ,a nd hence B ,E ,I,D a r e Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. concyc l i c. S i nc et he b i s e c t oro f ∠EAF and t hec i r cumc i r c l eo f △AEF mee ta tI, IE =IF .S i nc et heb i s e c t oro f ∠EBD and t he c i r cumc i r c l e of △BED a l s o F i .1 .2 g mee ta tI,IE =ID .So, ID = IE = IF ,t ha ti s,Ii sa l s ot he c i r cumc en t e ro f△DEF . Q. E. D. ( Ne c e s s i t st hec i r c umc en t e ro f△DEF.S i nc eB, s eIi y)Suppo E,F,Ca r ec on c c l i c,AE·AB = AF·AC,AB > AC,weha v e y AE < AF.The r e f o r e,t heb i s e c t o ro f ∠EAFandt hepe r i c u l a r pend b i s e c t o ro fEF me e ta tI,wh i chl i e sont hec i r c umc i r c l eo f△AEF. S i nceBIb i s e c t s ∠ABC , l e tK bet hes t r i cpo i n tofE ymme wi t hr e s c tt oBI, t henwehave ∠BKI = ∠BEI = ∠AFI > pe ∠ACI = ∠BCI.Th e r e f or e,K l i e sonBC , ∠IKC = ∠IFC , ∠ICK = ∠ICF , and△IKC ≌ △IFC . Henc eBC = BK +CK = BE +CF . 2 Foranyi n t ege rn wi t hn > 1, l e t D( n)= { a -b|a,ba r epo s i t i vei n t ege r swi t hn =a banda >b}. Provet ha tf oranyi n t ege rk wi he r eex i s tk t hk > 1,t i rwi s ed i s t i nc ti n t ege r sn1 ,n2 ,...,nk wi t hni >1( 1≤ pa i ≤k),s n1)∩ D ( n2)∩ … ∩ D ( nk )ha ucht ha tD ( sa t l ea s ttwoe l emen t s. Ch i naMa t hema t i c a lOl i ad ymp Proo f.Le ta1 ,a2 ,...,ak+1 bek 101 + 1d i s t i nc tpo s i t i veodd s, whe r e each o ft hem i ssma l l e rt hant he pr oduc tofo t he rk numbe r s.Wr i t eN =a1a2 …ak+1 .Foreachi =1,2,...,k +1, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. l e txi = æN ö 1 æç N + ö÷ , ai yi = 1 ç -ai ÷ ,t henxi2 - yi2 = N . ø ø 2 èai 2 èai N >ai ,( S i nc eaiaj < N and xi ,yi)( 1 ≤i ≤k +1)a r ek +1 ai s i t i vei n t ege rs o l u t i on so fequa t i onx2 -y2 = N .Wi t hou tl o s s po , { , , , } ofgene r a l i t uppo s exk+1 = mi n x1 x2 ... xk+1 .Foreach ys 2 2 i∈{ 1,2,...,k}, s i nc exi2 -yi2 = xk wehave +1 -yk+1 , 2 2 2 ( xi +xk+1)( xi -xk+1)= xi2 -xk +1 = yi -yk+1 =( yi +yk+1)( yi -yk+1). Le tni = ( xi +xk+1)( xi -xk+1)= ( t hen yi +yk+1)( yi -yk+1), xi +xk+1)- ( xi -xk+1)∈ D ( ni), 2xk+1 = ( 2yk+1 = ( ni). yi +yk+1)- ( yi -yk+1)∈ D ( Soxk+1 > yk+1 ,2xk+1 and2yk+1 a r etwod i f f e r en tmembe r sof D( n1)∩ D ( n2)∩ … ∩ D ( nk ). 3 Le tN* bet hes e to fa l lpo s i t i vei n t ege r s.Pr ovet ha tt he r e ex i s t saun i unc t i onf:N* → N* s a t i s f i ngf( 1) = quef y 2)= 1and f( n)= f( n -1))+f( n -f( n -1)),n = 3,4,.... f( f( Fors uchf,f i ndt heva l ueoff( 2m )fori n t ege rm ≥ 2. 1 So l u t i on.S i nc ef( 1)= 1,wehave 2 ≤ f( 1)≤ 1. Weshowbyi nduc t i ont ha tf oranyi n t ege rn > 1,f( n)i s un i l t e rmi nedbyt heva l ueo ff( 1),f( 2),...,f( nque yde n ≤ f( 1),and n)≤ n. 2 Ma t hema t i c a lOl i adi nCh i na ymp 102 Forn = 2,f( 2)= 1, t hec l a imi st r ue. n ≥3),f( k) As s umet ha tf oranyk,1 ≤k <n( i sun i l que y k n -1 ≤ f( ≤ f( k)≤k, n -1)≤ de t e rmi ned,and t hen1 ≤ 2 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. n - 1,and1 ≤ n - f( n - 1) ≤ n - 1,hencebyi nduc t i on hypo t he s i s,t heva l ueo ff( n - 1))andf( n - f( n - 1))i s f( n),byde f i n i t i on,i s de t e rmi ned,andt heva l ueo ff( n)= f( n -1))+f( n -f( n -1)), f( f( wh i chi sun i l t e rmi ned.Fu r t he rmor e,wehave que yde ① 1 ( n -1))≤ f( n -1), f n -1)≤ f( f( 2 1( n -f( n -1))≤ f( n -f( n -1))≤ n -f( n -1). 2 n n)≤ f( n -1)+ ( n -f( n -1))= Equa l i t l i e s ≤ f( y ①imp 2 n.Thec l a imi sa l s ot r uef orn.Byi nduc t i on,weprovedt ha t t he r eex i s t s a un i unc t i on f: N* → N* s a t i s f i ng t he que f y n ≤ f( r equ i r edprope r t i e s,and n)≤ n. 2 Nex t,wes howbyi nduc t i ont ha tforanypo s i t i vei n t ege rn, wehave n +1)-f( n)∈ { 0,1}. f( Whenn = 1,②i st r ue. ② As s umet ha t②i st r uef orn ≤k.By ① ,wehave k +2)-f( k +1) f( =( k +1))+f( k +2 -f( k +1)))- ( k))+ f( f( f( f( k +1 -f( k))) f( =( k +1))-f( k)))+ ( k +2 -f( k +1))f( f( f( f( k +1 -f( k))). f( ③ Ch i naMa t hema t i c a lOl i ad ymp 103 Byi nduc t i onhypo t he s i s,f( k +1)-f( k)∈ { 0,1}. k +1)= f( k)+1,s k)≤ k,i I ff( i nc e1 ≤ f( tf o l l ows, f r om ③ andi nduc t i onhypo t he s i s,t ha t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 0,1}. k +2)-f( k +1)= f( k)+1)-f( k))∈ { f( f( f( k +1)=f( k), k)≤k, I ff( s i nce1 ≤k +1 -f( i tf o l l ows f r om ③ andt hei nduc t i onhypo t he s i st ha t k +2)-f( k +1)= f( k +2 -f( k))f( k +1 -f( k))∈ { 0,1}. f( Thu s,②i st r uef orn =k +1.Byi nduc t i on,② i st r uefor anypo s i t i vei n t ege rn. Fi na l l nduc t i ont ha tf oranypo s i t i vei n t ege r y,weshowbyi m ,wehavef( 2m )= 2m-1 . Form = 1, t her e s u l ti sc l ea r. e.,f( 2k ) = As s umet ha tt her e s u l ti st r uef orm = k,i. 2k-1 ,con s i de rt heca s ef orm =k +1. 2k+1)≠ 2k , s i ncef( 2k+1)≥ As s umeont hecon t r a r ha tf( yt 2k ,andf( i sani n t ege r,wehavef( i nce 2k+1) 2k+1)≥ 2k +1.S 1)=1,by ② ,l e tn bet hesma l l e s ti n t ege rs ucht ha tf( n)= f( 2k +1,wehaven ≤2k+1 ,byt hemi n ima l i t fn,f( n -1)=2k . yo No t i c et ha tn -2k ≤ 2k ,wehave 2k +1 = f( n)= f( n -1))+f( n -f( n -1)) f( = f( 2k )+f( n -2k )≤ 2f( 2k )= 2k , wh i chi sacon t r ad i c t i on.I tf o l l owst ha tf( 2k+1)=2k , t her e s u l t i sa l s ot r ueform =k +1.Byi nduc t i on,f( 2m )= 2m-1 forany s i t i vei n t ege rm . po 104 Ma t hema t i c a lOl i adi nCh i na ymp S e c ondDay 8: 00 12: 30Decembe r22,2013 αt 1 4 Foranyi e tn = pα n t ege rn wi t hn > 1,l ei t s 1 …pt b Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s t anda rdf ac t or i za t i on,wr i t e ω( n)=t,Ω( n)=α1 + … +αt . Pr ove or d i s ove t he fo l l owi ng s t a t emen t: Gi ven any pr s i t i vei n t ege rk andanypo s i t i ver ea lnumbe r sα andβ, po t he r eex i s t sapo s i t i vei n t ege rn wi t hn > 1s ucht ha t ω( n +k) Ω( n +k) >αa <β. nd ω( n) n) Ω( So l u t i on.Theansweri sYES. Fromt hede f i n i t i onofωandΩ,wehave ω( ab)≤ ω( a)+ω( b), Ω( ab)= Ω( a)+ Ω( b), ① ② f oranypo s i t i vei n t ege r sa,b.Gi venaf i xedpo s i t i vei n t ege rk andpo s i t i ver ea lnumbe r sα,β,wet akeapo s i t i vei n t ege rm > ( ω( k)+1) α.Ast he r ea r ei nf i n i t e l imenumbe r s,we y manypr Ω( k)+1 +l ogp2 cant akeas u f f i c i en t l a r imeps ucht ha t yl gepr m p <β,a ndt akem pa i rwi s ed i s t i nc tpr imenumbe r sq1 ,q2 ,..., ha ta r ea l lg r ea t e rt hanp.Wewi l lshowt ha tn = 2q1q2 qm t …q m k ha st hede s i r edprope r t y. … ω( n +k) n +k >α.Le =2q1q2 qm + F i r s t,weprove tn1 = ω( n) k r ea l loddpr imenumbe r s,2qi +1|n1 1.Asq1 ,q2 ,...,qm a when1 ≤i ≤ m , t hendi = 2qi +1 i sani n t ege rg r ea t e rt han1. 3 Ch i naMa t hema t i c a lOl i ad ymp 105 No t et ha t (, ) ( l lpo s i t i vei n t ege r sr,s,③ 2r -1,2s -1)= 2r s -1fora Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. and ( i ≠j),wehave qi ,qj )= 1( 1 qi 1 2qi ( di ,dj )= ( 2 +1,2qj +1)≤ ( 2 -1,22qj -1) 3 3 = 22qi ( ,2 qj ) 3 -1 = 22 -1 = 1. 3 d1 ,d2 , ...,dm a r et he pa i rwi s ecopr imef ac t or sofn1 ,and eachoft hemi sg r ea t e rt han1.Henc e,ω( n1) ≥ m .From ① andt hecho i ceofm ,wehave n1) ω( n1) ω( n +k) ω( m ≥ ≥ ≥ >α. ( ) ( ) ( ωn ωn ω k)+1 ω( k)+1 Ω( n +k) < β.A Ne x t,wep r ov e sq1q2…qm i saoddnumbe r Ω( n) … andc anno tbed i v i d edby3,weha v en1 =2q1q2 qm +1 ≡±3( mo d9), n1 t ha ti s, 3‖n1.S uppo s eqi sap r imef a c t o ro f andq ≤ p, t he n 3 22q1q2 …q m … -1 = ( 2q1q2 qm -1)·n1 ≡ 0( modq). Fr omt heFe rma t sLi t t l eTheor em,2q-1 ≡ 1( modq).From ③ , ( … , -1) 2 q|2 q1q2 qm q -1.From ( q -1,2 q1q2 …qm )= ( q -1,2)≤ 2,q -1 <p <qi( i = 1,2,...,m ),henceq|22 -1,q = 3. n1 Th i scon t r ad i c t st ha t i sno tamu l t i l eo f3.The r e for e,each p 3 n1 n1 Ω( n /3) > p 1 .F imef ac t oro f i sl a r rt hanp.So r om ② pr ge 3 3 andt hecho i c eofpr ime sp andq1 ,q2 ,...,qm ,wehave æn1 ÷ö æn1 ÷ö < Ω( Ω( n +k)= Ω( k)+ Ω( 3)+ Ωç k)+1 +l ogp ç è3ø è3ø < Ω( k)+1 +l n1 -1) ogp ( = Ω( k)+1 +q1q2 …qml ogp2, Ma t hema t i c a lOl i adi nCh i na ymp 106 k)+1 +q1q2 …qml ogp2 Ω( Ω( n +k) Ω( k)+1 < < +l ogp2 <β. m … n) Ω( q1q2 qm p Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 5 Gi venX = { 1,2,...,100},cons i de rf unc t i onf:X → X s a t i s f i ngbo t ht hefo l l owi ngcond i t i ons: y ( 1)f( x)≠ xfora l lx ∈ X ; ( 2)A ∩ f( A )≠ ⌀fora l lA ⊆ X wi t h|A |= 40. Fi ndt hesma l l e s tpo s i t i vei n t ege rk, s ucht ha tforanys uch he r eex i s t sas e tB ⊆ X s a t i s f i ng|B|=kand f unc t i onft y B ∪ f( B )= X . Remark.Forasubs e tT o fX ,wede f i nef( T)= { x |t he r e ex i s t st ∈ T s ucht ha tx = f( t)}. So l u t i on.Fi r s t,wede f i neaf unc t i onf:X → X wi t h i -2)= 3 i -1,f( i -1)= 3 i,f( i)= 3 i -2, 3 3 3 f( i = 1,2,...,30, 100)= 99. f( j)= 100,91 ≤j ≤ 99,f( Obv i ou s l a t i s f i e scond i t i on ( 1).ForanyA ⊆ X wi t h|A | y,fs =4 0, i f ( ucht ha t|A ∩ i)t he r eex i s t sani n t ege riwi t h1 ≤i ≤30s { 3 i -2,3 i -1,3 i}|≥ 2, t henA ∩ f( A )≠ ⌀ ;or ( i i)91,92,...,100 ∈ A , t henA ∩f( A )≠ ⌀a l s oho l d s. Inbo t hca s e s,fs a t i s f i e scond i t i on ( 2).I fas ub s e tB o fX 3 s a t i s f i e sf( B)∪B = X , t henweha v e|B ∩ { i -2,3 i -1,3 i}| ≥ 2f ora l l1 ≤i ≤ 30,{ 91,92,...,98}⊂ B ,andB ∩ { 99, 100}≠ ⌀.Hence,|B |≥ 69. Nex t,wewi l ls howt ha t,f oranyf unc t i onfs a t i s f i ngt he y de s c r i bedcond i t i on s,t he r eex i s t sas ub s e tB ⊆ X wi t h|B|≤69 B )∪ B = X . s ucht ha tf( Ch i naMa t hema t i c a lOl i ad ymp 107 Amonga l lt hes ub s e t sU ⊆ X wi t hU ∩ f( U )= ⌀ ,choo s e smax ima l.I ft he r ea r e manyU ⊆ X wi t h ones ucht ha t|U |i i ng max ima l,choo s eones ucht ha t|f( U) smax ima l. |U |be |i Theex i s t enc eo fU i sgua r an t eedby cond i t i on ( 1).Le tV = Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. U ),W = X\( U ∪ V ).No r e pa i rwi s e t et ha tU ,V ,W a f( d i s o i n tandX = U ∪ V ∪ W .Fromcond i t i on ( 2),|U |≤ 39, j hefo l l owi nga s s e r t i on s: |V |≤ 39,|W |≥ 22.Wemaket ( i)f( w )∈ U ,fora l lw ∈ W .Ot he rwi s e,l e tU' = U ∪ { i nc ef( w },s U )= V ,f( w )∉ U ,f( w )≠ w ,wehaveU' ∩ ti sacon t r ad i c t i ont o|U |be i ng max ima l. U')= ⌀.I f( ( w 1) ≠ f( w 2)fora i i)f( l lw 1 ,w 2 ∈ W ,w 1 ≠ w 2 . t henbycond i t i on ( 1), Ot he rwi s e,l e tu =f( w 1)=f( w 2) u∈ s i nc ef( U .Le tU' = ( U\{ u})∪ { w 1 ,w 2}, U')⊆V ∪ { u},U' ∩ ( V ∪ { u}) = ⌀ ,we have U' ∩ f( U') = ⌀.I ti sa con t r ad i c t i ont o|U |be i ng max ima l. w 1 ,w 2 ,...,w m },ui =f( wi),1 ≤i ≤ mt hen Le tW = { by ( i)and ( i i),u1 ,u2 ,...,um a r ed i s t i nc te l emen t so fU . ( i i i)f( ui)≠ f( uj )fora l l1 ≤i <j ≤ m .Ot he rwi s e,l e t v = f( ui)= f( uj )∈V ,U' = ( U\{ ui})∪ { wi}, U') t henf( =V ∪ { ui},U' ∩f( U')= ⌀.Howeve U') U) r, |f( |>|f( |. I ti sacon t r ad i c t i ont o|f( U )|be i ng max ima l. The r e for e, f( u1), f( u2), ..., f( um ) a r e d i s t i nc t e l emen t sofV .Inpa r t i cu l a r,|V |≥|W |.As|U |≤ 39,we tB =U ∪ W , have|V|+|W |≥61and|V|≥31.Le t hen|B|≤ 69andf( B )∪ B ⊇ V ∪ B = X .Ove r a l l,t hede s i r edsma l l e s t s69. i n t ege rki 6 Fornon emp t e t sS,T o fnumbe r s,wede f i ne ys S +T = { s +t|s ∈ S, t ∈ T },2S = { 2 s|s ∈ S}. Ma t hema t i c a lOl i adi nCh i na ymp 108 Le tnbeapo s i t i vei n t ege r,andA ,B benon emp t ub s e t s ys o f{ 1,2,...,n}.Pr ovet ha tt he r eex i s t sas ub s e tD ofA + Bs ucht ha t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. A ·B D + D ⊆ 2( A +B ),and|D |≥ | | | |, 2 n whe r e|X |deno t e st henumb e ro fe l emen t so faf i n i t es e tX . So l u t i on.Le tSy n-1 ∑ |S y=1-n y ( ={ i nce a,b) b ∈ B }.S |a -b =y,a ∈ A , t he r eex i s t sani n t ege ry0s ucht ha t1|=|A|·|B|, A ·B A ·B n ≤ y0 ≤ n -1and|Sy0 |≥ | | | | > | | | |. 2 2 n -1 n Le tD = { 2 b +y0 | ( a,b)∈ Sy0 }, t hen |A |·|B |. |D |=|Sy0 |> 2 n Fr omt hede f i n i t i ono fSy0 ,foreachd ∈ D ,t he r eex i s t s ( ucht ha td = 2 a,b)∈ Sy0 s b +y0 =a +b ∈ A +B .SoD ⊆ e td1 = 2 A +B .Foranyd1 ,d2 ∈ D ,l b1 +y0 = 2a1 -y0 , d2 =2 b2 +y0( b1 ,b2 ∈ B ,a1 ∈ A ), t hen d1 +d2 = 2a1 -y0 +2 b2 +y0 = 2( a1 +b2)⊆ 2( A +B ). The r e f or e,D s a t i s f i e st hecond i t i on. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naNa t i ona l TeamSe l e c t i onTe s t 2011 ( F u z h o u,F u i a n) j F i r s tDay 8: 00 12: 30,Ma r ch27,2011 1 Gi venani n t ege rn ≥3, f i ndt hemax imumr ea lnumbe rM , s ucht ha tf oranypo s i t i venumbe r sx1 ,x2 ,...,xn ,t he r e ex i s t sape rmu t a t i ony1 ,y2 ,...,yn o fx1 ,x2 ,...,xn t ha ts a t i s f i e s n ∑y i=1 2 i+1 2 yi ≥ M, 2 -yi+1yi+2 +yi+2 Ma t hema t i c a lOl i adi nCh i na ymp 110 whe r eyn+1 = y1 ,yn+2 = y2 .( s edby QuZhenhua) po So l u t i on.Le t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. F( x1 ,...,xn )= n ∑x i=1 2 i+1 xi2 2 . -xi+1xi+2 +xi+2 F i r s t,t akex1 = x2 = … = xn-1 = 1,xn = ε,t hena l l rmu t a t i on sa r et hes amei nt hes ens eofc i r cu l a t i on.Int h i s pe ca s e,wehave 2 2 +ε . 1 -ε +ε2 F( x1 ,...,xn )= n -3 + Le tε →0+ ,F → n -1, s oM ≤ n -1. Nex t,wes howt ha tforanypo s i t i venumbe r sx1 ,...,xn , a t i s f i ngF ( t he r eex i s t sape rmu t a t i ony1 ,...,yns y1 ,...,yn ) y ≥ n -1. Inf ac t,t aket hepe rmu t a t i ony1 ,...,yn wi t hy1 ≥y2 2 2 ≥ … ≥yn a ndbyt hei nequa l i t a2 ,b2),we ya -ab +b ≤ max( s eet ha t 2 2 2 y1 y2 yn-1 F( y1 ,...,yn )≥ 2 + 2 + … + 2 ≥ n -1, y2 y3 y1 whe r et he l a s ti nequ a l i t s ob t a i ned by AM-GM i nequa l i t yi y. Summi ngup,M = n -1. 2 Le tn > 1beani n t ege r, kbet henumbe rofd i s t i nc tpr ime f ac t or sofn.Pr ovet ha tt he r eex i s t sani n t ege ra,1 <a < n +1, s ucht ha tn|a2 -a.( s edby YuHongb i ng) po k αk 1 So l u t i on.Le tn =pα et hes t anda rdf ac t or i za t i onofn. 1 …pk b αk 1 S i nc epα ..,pk a r epa i rwi s ecopr ime,byt heCh i ne s e 1 ,. Rema i nde r Theor em,f or each i,1 ≤ i ≤ k,cong r uence equa t i on s Ch i naNa t i ona lTe amS e l e c t i onTe s t 111 x ≡ 1( modpiαi ) { αj ),j ≠i x ≡ 0( modpj haves o l u t i onxi . Foranys o l u t i onofx2 modn),wes eet ha tx0( x0 -1) 0 =x0 ( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ≡0( mod modn).Thenf oreachi =1,2,...,k,e i t he rx0 ≡0( α modpiαi ).Fu r t he r,l e tS( A )bet hes um of pii )orx0 ≡ 1( { , , , }( , e l emen t sofs ub s e tA o f x1 x2 ... xk pa r t i cu l a r l y S(⌀ ) = 0) .Obv i ou s l y,wehave S( A )( S( A )-1)≡ 0( modn). ( Th i si s be cau s eo ft he s e l ec t i on of xi ,s uch t ha t S( A) α mod( A )≢ i se i t he r0or1.)Mor eove ri fA ≠ A',t henS( pii ) S( A')( x1,x2,..., modn).The r e f o r e,t hes umo fa l ls ub s e t so f{ xn } i se x a c t l l ls o l u t i on so fx( x -1)≡ 0( modn). ya Le tS0 =n,Sr b et hel e a s tnon -n e a t i v er ema i nd e ro fx1 +x2 + g … +xr modu l l1 ≤r l en,r = 1,2,...,k.Thu sSk = 1.Fora ≤k -1,Sr ≠0. r ei n[ 1, S i ncek +1numbe r sS0 ,S1 ,...,Ska n],byDi r i ch l e t sDr awe rPr i nc i l e,t he r eex i s t0 ≤l < m ≤k, p n ,( æjn ( j +1) s ucht ha tSl ,Smi nt hes amei n t e r va lç , 0 ≤j ≤ k èk ] k -1),whe r el = 0andm =kdono tho l ds imu l t aneou s l y. Thu s,|Sl -Sm |< n .Deno t ey1 = S1 ,yr = Sr - Sr-1 k ( r = 2,3, ...,k).Soanys um o fyr ≡ xr ( modn)( r = 1, 2,...,k)mee t st her equ i r emen t. I fSm -Sl >1, t hena =yl+1 +yl+2 + … +ym =Sm -Sl ∈ æç ,n ö÷ 1 i st hes o l u t i ono ft heequa t i onx2 -x ≡ 0( modn). è kø henn | ( I fSm - Sl = 1,t y1 + y2 + … + yl )+ ( ym+1 + x1 + x2 + … + xl )+ ( xm+1 + ha ti s,n | ( ym+2 + … +yk ),t xm+2 + … +xk ).No t i c et ha tm > l,wh i chcon t r ad i c t st ot he Ma t hema t i c a lOl i adi nCh i na ymp 112 de f i n i t i ono fxi . ha ti s,n I fSm -Sl = 0,t henn|yl+1 +yl+2 + … +ym ,t i chcon t r ad i c t st hede f i n i t i ono fxi . |xl+1 +xl+2 + … +xm ,wh I fSm -Sl < 0, t hen Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a=( y1 +y2 + … +yl )+ ( ym+1 + … +yk ) = Sk - ( Sm -Sl )= 1 - ( Sm -Sl ) i st hes o l u t i ono fe t i onx2 -x ≡0( modn),and1 <a <1+ qua Summi ngup,t he r eex i s t sas a t i s f i ngt hecond i t i on. y n . k 3 Le t3 n2 bet heve r t exnumbe ro fas imp l eg r aphG ( i n t ege r I ft hedeg r eeo feachve r t exi sno tg r ea t e rt han4 n ≥2). n, t he r eex i s t sa tl ea s toneve r t ex wi t hdeg r ee1,andt he r e ex i s t sarou t ewi t hl eng t hno tg r ea t e rt han3be tweenany twove r t i ce s.Pr ovet ha tt hemi n imum numbe rofedge sof 3 7 Gi s n2 - n. 2 2 Remark.A rout ebe tweentwod i s t i nc tve r t i ce su andv wi t h l eng t hki sas equenc eo fve r t i c e su = v0 ,v1 ,...,vk = v, whe r eviandvi+1 , i =0,1,...,k -1,a r ead ac en t.( s edby j po Leng Gang s ong) So l u t i on.Foranytwod i s t i nc tve r t i c e suandv,wes ayt ha tt he d i s t ancebe tweenu andv i st hes hor t e s tl eng t h oft herou t e be tweenu andv.Cons i de ra g r aph G * wi x1 , t h ve r t exs e t{ x2 ,...,x3n2-n ,y1 ,y2 ,...,yn },whe r eyiandyja r ead ac en t j ( 1 ≤i <j ≤n),xiandxj a r eno tad acen t( 1 ≤i <j ≤3 n2 j n),xiandyja r ead ac en ti fandon l fi ≡j( modn).Thu s,t he j yi deg r eeo feachxii s1,andt hedeg r eeofyi doe sno texceed 3 n2 -n =4 n -2. n n -1 + Ch i naNa t i ona lTe amS e l e c t i onTe s t 113 I ti sea s os eet ha tt hed i s t anc ebe tweenxi andxj i sno t yt r ea t e rt han 3.So g r aph G * s a t i s f i e st he cond i t i on o ft he g 2 = sN = 3 l em.G * ha n2 -n + Cn prob 7 2 3 n - nedge s. 2 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Int hefo l l owi ng,wes howt ha tanyg r aphG = G ( V ,E ) s a t i s f i ngt hecond i t i ono ft heprob l emha sa tl ea s tN edge s.Le t y X ⊆ V bet hes e tofve r t i c e swi t hdeg r ee1,Y ⊆ ( V\X )bet he s e to fr ema i n i ngve r t i c e sad acen tt oX ,andZ ⊆V\( X ∪Y)be j t hes e to fr ema i n i ngve r t i ce sad ac en tt oY .Le tW =V\( X ∪Y j ∪ Z) .Wewi l lpo i n tou tt hefo l l owi ngf ac t s. Proper ty1.Anytwover t i ce si nYa r ead acen t.Th i si sbecau s e j oft hef ac tt ha ti fy1 ,y2 ∈Ya r etwov e r t i c e s, t he r ee x i s tx1,x2 ∈ Xt ha ta r ead ac en tt oy1 andy2 ,r e s t i ve l j pec y;hencey1 andy2 a r ead ac en ts i nc et hed i s t ancebe tweenx1 andx2i sno tg r ea t e r j t han3. Proper ty2.Thed r t exi nY i s t ancebe tweenve r t exi nW andve i s2.Th i si sb e c au s eo ft hef a c tt ha ti ft hed i s t anc ebe twe enw0 ∈W andy0 ∈Yi sg r ea t e rt han2 ( obv i ou s l i s t ance >1),s uppo s e y,d t ha tx0 ∈ Xi sad ac en tt oy0 , t hent hed i s t anc ebe tweenw 0and j x0i sg r ea t e rt han3,wh i chi sacon t r ad i c t i on.Fur t he rmor e,we knowt h i sPr ope r t seachve r t exi nW i sad acen tt os ome y2mean j ve r t exi nZ . Deno t ebyx,y,zandwt henumbe r sofe l emen ti ns e t sX , Y ,Z and W ,r e s c t i ve l tt henumbe ro fedge s: pe y.Now coun 2 t he r ea r eCy edge sbe tweenpo i n t si nY ,xedge sf rompo i n t sofX t oY ,a tl ea s tzedge sf r om po i n t so fZt oY ,anda tl ea s tw edge s f r om po i n t so fW t oZ .So,i fy ≥ n, t hen 2 2 2 -y ≥ 3 -n = N , n2 + Cy n2 + Cn |E |≥ Cy +x +z + w = 3 i nc eeachdeg r eeofve r t exi sa t mo s t4 n, andi fy ≤ n - 1,s Ma t hema t i c a lOl i adi nCh i na ymp 114 wehave 4 4 x +z ≤ y( n-( n +1 -y) y -1))= y( ≤( n -1)( n +2)= 3n2 -n -2, 3 w ≥ 3n2 -y -y( n +1 -y)≥ 3. 4 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. nW s ucht ha tP i sad acen ta sl e s sa s Se l e c tave r t exP i j s s i b l et ove r t i c e si nZ .Suppo s et hel ea s tnumbe ri sa,a > 0 po ( byPr ope r t t et hes e to ft he s eave r t i ce sbyNP ⊆ Z . y2).Deno 2 Coun t i ngt he numbe ro ft heedge saga i n,t he r ea r eCy edge s sf rom po i n t sofX t oY ,a tl ea s ty be tweenpo i n t si nY ,x edge edge sf r om po i n t sofNP t oY ( byPrope r t hed i s t ancef rom y2,t Pt ove r t exi nYi s2),a tl ea s tz -aedge sf rom po i n t sofZ\NP tl ea s taw edge t oY ,anda sf r om po i n t sofW t oZ .Thu s, 2 |E | ≥ Cy +x +y +z -a +aw =3 n -1 + Cy + ( a -1)( w -1). 2 2 I fa > 1, t hen 2 +( n2 -1 + Cy w -1) |E | ≥ 3 ≥3 n -2 + Cy +3n -y -y( 4 n +1 -y)> N . 2 2 2 I fa = 1, s i nc et hedeg r eeo feachve r t exi nW i sa tl ea s t2, whenwecoun tt heedge sf r om po i n t sofW t oZ ,wes hou l dadd a tl ea s tw/2edge s,s o 2 + n2 -1 + Cy |E | ≥ 3 ≥3 n -1 + Cy + 2 2 1 w 2 1( 2 3 n -y -y( 4 n +1 -y))> N . 2 Summi ngup,t hel ea s tnumbe ro fedge si sN = 7 2 3 n - n. 2 2 Ch i naNa t i ona lTe amS e l e c t i onTe s t 115 S e c ondDay 8: 00 12: 30,Ma r ch28,2011 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 4 Le t H bet heor t hoc en t e rofa cub i t ang l ed △ABC , P be a ︵ i n tonBC oft hec i r cumc i r c l e po ︵ n t e r s e c t sAC a t o f△ABC .PH i M .The r e ex i s t sa po i n t K on ︵ AB s uch t ha tt hel i ne i s KM r a l l e lt ot heS ims onl i neo fP pa wi t h r e s c tt o △ABC , t he pe chordQP ‖BC ,t hechord KQ i n t e r s ec t sBC a tpo i n tJ. Pr ovet ha t△KMJi si s o s ce l e s.( s edby Xi ongBi n) po So l u t i on.WeshowthatJK = JM . Dr awl i nef rom P andl e ti tbe rpend i cu l a rt oBC andi n t e r s e c tt he pe c i r cumc i r c l eandBCa tpo i n tSandL , r e s c t i ve l t N be t he pr o e c t pe y.Le j i n to fP onAB .S i nc eB ,P ,L and po Na r econcyc l i c, ∠SLN = ∠NBP = ∠ABP = ∠ASP . Thu s,NL ‖AS.S i nc eNL ‖KM , t henKM ‖SA . Le tT bet hei n t e r s e c t i onpo i n tofBCandPH .S i ncepo i n t s K ,Q ,P andM a r econcyc l i candBC ‖PQ , s oa r eK ,J,T and M .Suppo s et ha tt heex t en s i ono fAH i n t e r s ec t st hec i r cumc i r c l e a tpo i n tD .Thenwehave ∠JKM = ∠MTC , ∠KMJ = ∠KTJ. I ts u f f i c e st oshowt ha t ∠MTC = ∠KTJ. Ma t hema t i c a lOl i adi nCh i na ymp 116 I ti sea s os eet ha tD and H a r es t r i c ove rl i ne yt ymme BC .Then ∠SPM = ∠SPH = ∠THD = ∠HDT . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Fur t he rmor e KS Con s equen t l y, = o ∠ADM AM , s = ∠KPS. ∠TDM = ∠HDT + ∠ADM = ∠SPM + ∠KPS = ∠KPM = ∠KDM , wh i ch mean st ha tpo i n t sK ,T andD a r eco l l i nea r.Thu s ∠KTJ = ∠DTC = ∠MTC . So ∠JKM = ∠KMJ.Con s equen t l yJK = JM . 5 Le rmu t a t i onofa l lpo s i t i vei n t ege r s. ta1 ,a2 ,...beape Provet ha tt he r eex i s ti nf i n i t e po s i t i vei n t ege r si s,s uch 3 t ha t( ai ,ai+1)≤ i.( s edbyChenYonggao) po 4 So l u t i on. We provethi s pr ob l em by con t r ad i c t i on.I ft he conc l u s i ono ft hepr ob l emi sno tt r ue,t hent he r eex i s t si0 ,and 3 wehave( ai ,ai+1)> ifori ≥i0 . 4 oi fi ≥ 4M ,t ai , Takeapo s i t i venumbe rM > i0 ,s hen ( 3 ai+1)> i ≥ 3M . 4 So,i fi ≥4M ,ai ≥ ( ai ,ai+1)>3M , t hen{ 1,2,...,3M } ⊆{ a1 ,a2 ,...,a4M-1}. Henc e 1,2,...,3M }∩ { a2M ,a2M+1 ,...,a4M-1}| |{ ≥ 3M - ( 2M -1)= M +1. ByDi r i ch l e t sDr awe rPr i nc i l e,t he r ee x i s t s2M ≤j0 < 4M p Ch i naNa t i ona lTe amS e l e c t i onTe s t 117 1s ucht ha taj0 ,aj0+1 ≤ 3M .Thu s, 1 3M 3· 3 ( = aj0 ,aj0+1)≤ max{ aj0 ,aj0+1}≤ 2M ≤ j0 , 2 2 4 4 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. wh i chi sacon t r ad i c t i on. 6 Weca l lapo i n ts equenc e( A0 ,A1 ,...,An ) i n t e r e s t i ng, i f t heab s c i s s aandord i na t ea r eequa lforeachAi ,andt he t r i c t l nc r ea s e s l ope so fs egmen tOA0 ,OA1 ,...,OAn s yi (Oi st heor i i n),andt hea r eaofeach△OAiAi+1( 0 ≤i ≤ g 1 n -1) i s . 2 A0 ,A1 ,...,An ), Forapo i n ts equenc e( i ns e r tapo i n tA → → ad a c en tt otwopo i n t sAi ,Ai+1s a t i s f i ngOA = OAi +OAi→ +1 , j y t henwec a l lt hu sob t a i nednewpo i n ts equ en c e( A0,...,Ai , A ,Ai+1 ,...,An )anexpans i onof( A0 ,A1 ,...,An ). A0 ,A1 ,...,An )and( B0 ,B1 ,...,Bm )beanytwo Le t( i n t e r e s t i ngpo i n ts equence s.Pr ovet ha ti fA0 = B0 andAn = Bm , t hen wecanexpandbo t hpo i n ts equence st os ome s amepo i n ts equence ( C0 ,C1 , ...,Ck ).( s edby Qu po Zhenhua) So l u t i on.Wes eet ha tbyt hecond i t i on oft he prob l em,an expans i onofani n t e r e s t i ngs equenc ei ss t i l li n t e r e s t i ng. Fi r s t,wecons t r uc tt hei n t e r e s t i ngs equence( C0 ,C1 ,..., Ck )con t a i n i nga l lpo i n t sofs equenc e s( A0 ,A1 ,...,An )and ( B0 ,B1 ,...,Bm ),andC0 = A0 = B0 ,Ck = An = Bm . Byt heP i ck Theor em,weknowt ha tt hea r eaoft r i ang l e equa l s1/2i fandon l ft he r ei snog r i dpo i n tont het r i ang l e yi excep tt r i ang l e ve r t i ce s. Henc et he r ei s no g r i d po i n t on △OAiAi+1 exc ep ti t sve r t i ce s.The r e for e,i ft hes l ope sofOAi andOBj a r eequa l,t henAi = Bj . 118 Ma t hema t i c a lOl i adi nCh i na ymp Deno t et hes l ope so fs egmen t s f r om po i n t so f{ Ai}and { Bj }t ot he or i i ni ns t r i c t l nc r ea s i ngorde rby g yi D0 ,D1 ,...,Dl ,whe r eD0 =C0 = Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. A0 =B0andDl =Ck = An =Bm . I f Di , Di+1) i s no t a s equence ( i n t e r e s t i ng, t hen we can i ns e r t ucht ha t s eve r a lpo i n t sE1 ,...,Ess t he s equence ( Di , E1 , ..., Es , F i .6 .1 g i si n t e r e s t i ng.Inf ac t,con s i de rt heconvexhu l lP o Di+1) ft he ep tt he or i i n.P i sa convex r i d po i n t son △ODiDi+1 ,exc g g l egmen tDiDi+1 ( adegene r a t edpo l he po ygonors ygon).Thent s equenc eo f ve r t i c e so fP i si n t e r e s t i ng. Thu s, we have C0 ,C1 ,...,Ck ). cons t r uc t edt hei n t e r e s t i ngs equenc e( F i na l l ts u f f i c e st os how t ha tt hei n t e r e s t i ngs equence y,i ( A0 ,A1 ,...,An )canbeexpandedt o( C0 ,C1 ,...,Ck ),and t hes amei st r uef or( B0 ,B1 ,...,Bm ).Weon l oprove yneedt t h i sf ort heca s eo fn =1, s i nc ewecanapp l heconc l u s i onfor yt n =1t Ai ,Ai+1), i =0,1,...,n -1s o( uc ce s s i ve l tC0 = y.Le A0 andCk = A1 .Byi nduc t i ononk,fork = 1,weneedno expans i on.Suppo s et ha tt heconc l u s i oni st r uefora l lpo s i t i ve i n t ege r sl e s st hank. Then deno t et he g r i d po i n tA → → → s a t i s f i ngOA = OA0 + OA1 ,and we y s tb eapo i n to fC1,..., s e et ha tA mu Ck-1 .S i nc ei fno t,t he r ei sno g r i d i n toni n t e r i oro fs egmen tOA ,and po t he r eex i s t si,0 ≤i <k, s ucht ha tA l oca t e si nt heang l emadebyr ay sOCi uppo s et ha ti > andOCi+1 .Wemays F i .6 .2 g Ch i naNa t i ona lTe amS e l e c t i onTe s t 119 0,o t he rwi s et aket he g r aphs t r i cove rt hel i nex = y. ymme oca t e s S i nc et hea r eao ft hepa r a l l e l og r am ▱OA0AA1i s1,Ci l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ou t s i deof▱OA0AA1 ,Ci+1l oca t e sou t s i deof▱OA0AA1orCi+1 → +O→ = A1 . Inany way,wet akeBs ucht ha tO→ B = OC Ci+1 and i ucht ha tCi+1B'‖OA0 ,CiB'‖A0A ,t henAl oca t e son t akeB's ▱OCi+1B'Ci .Thu s,A l oca t e si ns i de o f ▱OCiBCi+1 ,wh i ch con t r ad i c t st hef ac tt ha tt hea r eaof▱OCiBCi+1i s1.The r e f or e o( A0 ,A1) forexpans i oni ss omeCi .Then t hei ns e r t edpo i n tAt weu s et hei nduc t i on hypo t he s e st o( A0 ,A )and ( A ,A1), r e s c t i ve l pe y. 2012 ( Na n c h a n i a n x i) g,J F i r s tDay 8: 00 12: 30,Ma r ch25,2012 1 Le heor t hoc en t e ro fan tH bet acu t e ang l ed△ABC wi t h ∠A > 60 °.Le tpo i n t s M and N beon e s c t i ve l s i de sAB and AC ,r pe y, s uch t ha t ∠HMB = 6 0 ° = ∠HNC.L e tOb et hec i r c umc en t e r o f△HMN .Le tpo i n t sD andA beont hes ames i deo fl i neBC , s ucht ha t△DBC i sr egu l a r( s ee F i .1 .1 g F i .1 .1).Pr ovet ha tpo i n t s H ,O and D a r eco l l i nea r. g ( s edbyZhangS i hu i) po Ma t hema t i c a lOl i adi nCh i na ymp 120 So l u t i on.Le tT bet heor t hoc en t e rof△HMN .Ex t endedl i ne s ofHM andCAi n t e r s ec ta tpo i n tP .Ex t endedl i ne sHN andBA i n t e r s e c ta tpo i n tQ . I ti sea s os eet ha tpo i n t sN ,M ,P andQ yt Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a r econcyc l i c. F i .1 .2 g eet ha t ∠PQH - ∠OHN = By ∠THM = ∠OHN ,wes ∠NMH - ∠THM = 9 0 °, t ha ti s, HO ⊥ PQ . ① Le tpo i n tR bes t r i ct opo i n tC ove rHP .ThenHC = ymme HR . By ∠HPC + ∠HCP = (∠BAC -60 °)+ ( 90 °- ∠BAC)= 30 °,wes eet ha t ∠CHR = 60 °.Hence△HCRi sr egu l a r. By ∠HPC = ∠HQB and ∠HCP = ∠HBQ ,wes eet ha t △PHC ∽ △QHB .Then △PHR ∽ △QHB ,s o △QHP ∽ △BHR . Deno t e by ∠ ( UV ,XY)t he ang l e be tween U→ V and X→ Y ( s i t i vean t i c l ockwi s e). po S i nc e ∠PHR = 150 °,wehave ∠ ( PQ ,RB ) = ∠ ( HP , HR ) = 150 °, △BCD and △RCH a r er egu l a r.So △BRC ≌ △DHC . Henc e ∠( RB , HD ) = ∠ ( CR , CH ) = - 60 °. Ch i naNa t i ona lTe amS e l e c t i onTe s t 121 Con s equen t l y, ∠( °-60 ° = 90 °, PQ,HD )= ∠ ( PQ,RB)+ ∠ ( RB,HD )= 150 t ha ti s ② Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. DH ⊥ PQ . By ① and ② ,po i n t sH ,O andD a r eco l l i nea r. 2 Pr he r eex i s tk ovet ha t,f orany g i veni n t ege rk ≥ 2,t d i s t i nc tpo s i t i vei n t ege r sa1 ,a2 ,...,ak s ucht ha tforany i n t ege r sb1 ,b2 ,...,bk wi t hai ≤bi ≤2 ai , i =1,2,..., k,andanynon -nega t i vei n t ege r sc1 ,c2 ,...,ck ,wehave ci < k∏i=1bi k ∏ b ,prov i ded k i=1 i ( s edbyChenYonggao) po ∏ k ci i=1 i b ∏ k i=1 i < b. So l u t i on.Wewi l lpr oveas t ronge rpropo s i t i on:foranyg i ven s i t i ve r ea lnumbe rkandanypo s i t i vei n t ege rn,t he r eex i s tn po i n t ege r sa1 ,a2 ,...,an , ai ,1 ≤i ≤ n -1, s a t i s f i ngai+1 > 2 y andf oranyr ea lnumbe r sb1 ,b2 ,...,bn ,ai ≤bi ≤2 ai , i =1, 2,...,n,andany non -nega t i vei n t ege r sc1 ,c2 ,...,cn ,we ci < havek ∏i=1bi n ∏ ci < b ,prov i ded ∏i=1bi n i=1 i n ∏ n i=1 i b. Le tk >1.Wepr I fn ovet hepr opo s i t i onbyi nduc t i ononn. = 1, c1 = 0, t hent akea1 >k, t heconc l u s i oni st r ue.Suppo s e t ha tt heconc l u s i oni st r ueforn ≥1.The r ea r ex1 <x2 < … < xn s a t i s f i ngxi+1 >2xi ,1 ≤i ≤n -1.Nowcons i de rt heca s eof y n +1.Takeapo s i t i vei n t ege rxn+1 > 2xn , s a t i s f i ng y æ2xn ö÷n xn+1 >k ç . è x1 ø 2xn ① Thenl e tai =t xi , i = 1,2,...,n +1,ti sas u f f i c i en t l y l a r n t ege r,s ucht ha t gei Ma t hema t i c a lOl i adi nCh i na ymp 122 n +1 > 2a2 …an+1 , an 1 ② n-1 k·2n-1an +1 < a1 …an . ③ and Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Int hef o l l owi ng,wewi l ls howt ha tt he s eai( 1 ≤i ≤n +1) mee tt her equ i r emen t.Le tbi ∈ [ ai ,2ai]( he 1 ≤i ≤n +1)bet r ea lnumbe r( no t i c et ha twehaveb1 <b2 < … <bn+1 ),andci( 1 ≤i ≤ n + 1)b enon -nega t i vei n t ege r s,s a t i s f i ng y ∏ n+1 i=1 i b. ∏ n+1 c i i=1 i b < t hen I f∑i=1ci ≤ n, n+1 n+1 ci n n-1 n-1 ≤k k∏bi bn bn+1 < a1 …anbn+1 +1 ≤ k·2 an+1 i=1 ≤ ∏b i=1 ∏b .(byusing ③ ) i=1 i I f∑i=1ci ≥ n +2, t hen n+1 n+1 n+1 ci i ≥b1+ ≥ a1+ b1 > 2a2 …an+1b1 ≥ n 2 n 1 I ti simpo s s i b l e. n n+1 ∏b .(byusing ② ) i=1 i I f∑i=1ci = n +1,wecons i de rt hr eeca s e s. n+1 t hen I fcn+1 ≥ 2, n+1 ∏b i=1 n+1 ci i ∏b i=1 ≥ bn+1 · æçb1 ö÷n-1 an+1 · æç a1 ö÷n-1 ≥ èbn ø bn 2 an è2an ø = xn+1 · æç x1 ö÷n-1 > 1.( byu s i ng ① ) 2xn è2xn ø i I ti simpo s s i b l e. Ch i naNa t i ona lTe amS e l e c t i onTe s t 123 I fcn+1 = 1, t hen n+1 ∏b i=1 n+1 ci i ∏b Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i=1 i ∏ = ∏ n ci i=1 i n i=1 i b b . bi ∈ [ No t et ha t xi ,2xi],1 ≤ i ≤ n.By t hei nduc t i on t hypo t he s e s,wes eet ha t n+1 ci < k∏bi i=1 I fcn+1 = 0, t hen n+1 ∏b . i=1 i n+1 ∏b i=1 n+1 i ∏b i=1 ci i ≥ bn+1 · æçbn ö÷n an+1 · æç an ö÷n ≥ èb1 ø bn an è2a1 ø 2 = xn+1 · æç xn ö÷n >k.( byu s i ng ① ) 2xn è2x1 ø Thu s,wehaveche ckedt ha ta1 ,a2 , ...,an+1 mee tt he r equ i r emen t. 3 Le tP ( x)=x2012 +a2011x2011 +a2010x2010 + … +a1x +a0be apo l a lo fdeg r ee2012o fr ea lcoe f f i c i en t swi t h1a s ynomi i t sl ead i ngcoe f f i c i en t.Fi ndt hemi n imumo fr ea lnumbe rc s ucht ha t|Imz|≤ c|Rez|,whe r eRez andImz a r e, r e s c t i ve l her ea landt heimag i na r r t so fanyr oo t pe y,t ypa o f a po l a l ob t a i ned by chang i ng s ome of t he ynomi ( ) ( coe f f i c i en t sofP x t ot he i roppo s i t enumbe r s. po s edby ZhuHuawe i) So l u t i on.Fi r s t,wepo i n tou tt ha tc π .Cons i de rt he 4022 ≥c o t Ma t hema t i c a lOl i adi nCh i na ymp 124 l a lP ( x)= x2012 -x.Chang i ngt hes i f f i c i en t s po ynomi gnofcoe o fP ( x),weob t a i nf ou rpo l a l sP ( x),- P ( x),Q ( x)= ynomi x2012 + x and - Q ( x).No x)and - P ( x)havet t et ha tP ( he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1006 1006 s s amer oo t s;oneo ft her oo t si sz1 = co π +is i n π. 2011 2011 Q( x)and - Q ( x)havet hes amer oo t s,anda r et heoppo s i t e numbe ro ft heroo t sofP ( x).Thu x)ha sQ ( sar oo tz2 = -z1 . Then, æ|Imz1 |,|Imz2 |÷ö π =c nç c ≥ mi . o t è|Rez1 | |Rez2 |ø 4022 π .Forany Nex t,wes howt ha tt hean swe ri sc = co t 4022 P( x)= x2012 +a2011x2011 +a2010x2010 + … +a1x +a0 , weob t a i napo l a l ynomi R( x)=b2012x2012 +b2011x2011 +b2010x2010 + … +b1x +b0 , x),whe r eb2012 = 1, bychang i ngs i omecoe f f i c i en to fP ( gnofs andforj = 1,2,...,2011, bj = mod4), |aj |,j ≡ 0,1( { -|aj |, mod4). j ≡ 2,3( fR ( x),wehave|Imz|≤ Wes howt ha t,f oreachroo tzo c|Rez|. Wepr ovet h i sr e s u l tbycon t r ad i c t i on.Suppo s et he r ei sa r oo tz0o fR ( x), t henz0 ≠0and s ucht ha t|Imz0|>c|Rez0|, e i t he rt heang l eo fz0andii sl e s st hanθ = π , ort heang l eof 4022 z0 and -ii sl e s st hanθ.Suppo s et ha tt heang l eo fz0andii sl e s s t hanθ;f ort heo t he rca s e,weneedon l i de rt hecon uga t e ycons j o fz0 .The r ea r etwoca s e s: Ch i naNa t i ona lTe amS e l e c t i onTe s t 125 I fz0i sont hef i r s tquadr an t( orimag i na r i s),s uppo s e yax z0 , i)=α <θ,whe z0 ,i)i t ha t ∠( r e ∠( st hel ea s tang l et ha t oian t i c l ockwi s e.For0 ≤j ≤2012, ro t a t e sz0t i fj ≡0,2( mod bjzj0 ,1)=j α < 2012 θ. 4), t hen ∠ ( α ≤ 2012 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. I fj ≡ 1,3( mod4),t hen ∠ ( bjzj0 ,i) = j α < 2011 θ and ∠( b1z0 ,i)=α.Thu s,t hepr i nc i la r tofbjzj 2π 0 ∈ [ pa gumen 1 2012 r t exang l eoft h i sang l e α,2π) ∪ 0, π -α .Theve 2 [ doma i ni s2012 α+ ] 1 1 π -α = π +2011 α < π.Andbjzj0 ,0 ≤j 2 2 ≤2 012,a r eno ta l lze ro,s ot he r es umcanno tbez e ro. I fz0i i,z0)=α sa tt hes e condquadr an t,s uppo s et ha t ∠( <θ, θ. I fj ≡ i fj ≡0,2( mod4), t hen ∠ ( 1,bjzj α <2012 0 )=j π i,bjzj0)=j α ≤ 2011 α < .Thu s,eve r 1,3( mod4), t hen ∠ ( y 2 π ,π +2011 α .S i nc e +2011 i nc i l ea r to fbjzj α 0 ∈ 0 pr p gumen 2 2 [ ] < π,a ndbjzj 012,a r eno ta l lze r o,s ot he i rs um 0 ,0 ≤ j ≤ 2 canno tbez e ro. π t . Summi ngup,t hel ea s tr ea lnumbe rc = co 4022 S e c ondDay 8: 00 12: 30,Ma r ch26,2012 4 Gi venani n t ege rn ≥ 4,Le tA ,B ⊆ { 1,2, ...,n}. Suppo s et ha tab +1i sape r f e c ts r enumbe rforanya ∈ qua A andb ∈ B .Provet ha t mi n{ og2n. |A |,|B |}≤l ( s edby Xi ongBi n) po Ma t hema t i c a lOl i adi nCh i na ymp 126 So l u t i on.Fi r s tweproveal emma. Lemma.Gi venani n t ege rn ≥ 4, l e tA ,B ⊆ { 1,2,...,n}. Suppo s et ha tab +1i sape r f e c ts r enumbe rforanya ∈ A and qua b ∈ B .Le ta,a ' ∈ A ,b,b ' ∈ B ,anda <a ',b <b ', t hena ' b ' Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. >5 ab. .5 Proo fo ft hel emma.Fi r s tno t i c et ha t( ab +1)( a ' b ' +1)> ( ab ' +1)( a ' b +1).So ( ab +1)( a ' b ' +1) > ( ab ' +1)( a ' b +1). S i nc etwos i de so fabovei nequa l i t r ea l li n t ege r s,wehave ya 2 ( ab +1)( a ' b ' +1)≥ ( ( ab ' +1)( a ' b +1)+1) . Byexpan s i on,weob t a i n ab +a ' b ' ≥ ab ' +a ' b +2 ( ab ' +1)( a ' b +1)+1 >a b ' +a ' b +2 ab '·a ' b. Bya <a ',b <b ',wehaveab ' +a ' b >2ab.Le ta ' b ' =λab, andcomb i n i ngt heabovei nequa l i t 1 +λ) ab > ( 2+ y,wehave( ab,s 2 λ) oλ > 3 +2 2 > 5 .5. Nowt u rnt ot heor i i n pr ob l em.Le tA = { a1 ,a2 ,..., g am },B = { b1 ,b2 ,...,bn },a1 <a2 < … <am ,b1 <b2 < … <bn .We ma uppo s et ha t2 ≤ m ≤ n.S i ncea1b1 + 1i sa ys r f ec ts r enumbe r,wehavea1b1 ≥ 3.Byt hel emma,a2b2 pe qua >5 s, .5 a1b1 > 4 ,ak+1bk+1 > 4akbk ,k = 2,...,m -1.Thu 2 n2 ≥ ambm ≥ 4m-2a2b2 > 4m . The r e for e,m ≤l og2n. 5 F i nda l li n t ege r sk ≥ 3,wi t ht hefo l l owi ngprope r t i e s: a t i s f i ng( m ,k)= ( n,k)= The r eex i s ti n t ege r sm andns y Ch i naNa t i ona lTe amS e l e c t i onTe s t 127 1andk| ( m -1)( n -1)wi t h1 < m <k,1 <n <kand m +n >k.( s edby YuHongb i ng) po So l u t i on.I fkha saf ac t oro fs r enumbe rg r ea t e rt han1,l e t qua Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t2 |k, t > 1, t hent ak i ngm =n =k - k +1,wes eet ha ts uch t kha st hepr ope r t i e s. I fkha snof ac t oro fs r enumbe r, i ft he r ea r etwopr ime s qua tk = ucht ha t( p1 ,p2s p1 -2)( p2 -2)≥ 4andp1p2 |k.Le i rwi s ed i f f e r en t,r ≥ 2. p1p2 …pr andp1 ,p2 ,...,pr be pa S i ncet he r ei sa tl ea s toneo f( p1 -1) p2p3 …pr +1and ( p1 - 2) scopr imewi t hp1 ( o t he rwi s ep1 d i v i de st he i r p2p3 …pr + 1i d i f f e r enc ep2p3 …pr ,wh i chi sa con t r ad i c t i on),t ak i ng t h i s m ,k) = 1. numbe ra st henumbe rm ,t hen,1 < m < k,( S imi l a r l akeanumbe rn o f( p2 -1) p1p3 …pr +1or y,wecant ( ucht ha t1 < n < k,( n,k) = 1.So p2 -2) p1p3 …pr + 1s m -1)( n -1),and p1p2 …pr | ( m +n ≥ ( p1 -2) p2p3 …pr +1 + ( p2 -2) p1p3 …pr +1 ( =k + ( p1 -2)( p2 -2)-4) p3 …pr +2 >k. Suchm ,n wi l ls a t i s f hecond i t i ons. yt I ft he r ea r enotwopr ime sp1 ,p2s ucht ha t( p1 -2)( p2 -2) ≥ 4,p1p2 |k, t heni ti sea s ove r i f uchi n t ege rk ≥ 3can yt ys on l sp,2p ( whe r epi sanoddpr ime).I ti sea s y ybe15,30ora t os eet ha t,i fk = p,2p,30, t hent he r ea r enom ,ns a t i s f i ng y t hecond i t i on s; i fk = 15,t hen m = 11,n = 13s a t i s f he yt cond i t i on s. Summi ngup,i n t ege rk ≥ 3s a t i s f i e st hecond i t i onsi fand on l fki sno tanoddpr ime,nordoub l eofanoddpr imeand yi nor30. Ma t hema t i c a lOl i adi nCh i na ymp 128 6 Suppo s et he r ea r ebee t l e son ache s s boa rdcons i s t i ng of 2012 × 2012 un i t s r e s. Each un i t s r e can qua qua ac commoda t ea tmo s tonebee t l e.Atamomen t,a l lbee t l e s f l andont heche s s boa rdaga i n.Forabee t l e,weca l l yandl Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t he ve c t orf r om i t sf l i ng un i tt oi t sl and i ng un i tt he y bee t l e s“ d i s l ac emen tve c t or”. Weca l lt hes um ofa l l p bee t l e s“ d i s l ac emen tve c t or s”t he “ t o t a ld i s l acemen t p p ve c t or s”. F i nd t he max imum l eng t ho f “ t o t a ld i s l acemen t p ve c t or”con s i de r i ngt henumbe ro fbee t l e sanda l lpo s s i b l e s i t i on so ff l i ngandl and i ng.( s edby QuZhenhua) po y po So l u t i on.Se tup a coord i na t e wi t h or i i na tt he c en t e r of g che s s boa rdO andt heg r i dl i nea st hecoord i na t el i ne.Deno t e t hes e to ft hec en t e r so fs r e sbyS,andt hes e twhe r et he qua bee t l e si n i t i a l l t andonbyM 1 ⊆S,andt hes e tt ha tt hebee t l e s ys l andonbyM 2 ⊆S.Le tf:M 1 → M 2bet heone t o onemapp i ng de f i nedbyabee t l e spo s i t i onva tt hebeg i nn i ngt ot hepo s i t i on u = f( v)off i s l acemen tvec t or i r s tl and i ng.Thu s,t het o t a ld p i sg i venby V = v)-v)= ∑ u - ∑ v. ∑ (f( v∈M 1 u∈M 2 ① v∈M 1 No t et ha tt her i t -hands i deof①i si ndependen to ff.We gh needon l of i ndt hemax imumo f|V|=|∑u∈M u - ∑v∈M v| yt 2 1 f ora l lM 1 ,M 2 ⊆ S,|M 1|=|M 2|.Wemays uppo s et ha tM 1 ∩ M 2 = ⌀, s i ncee l emen tofM 1 ∩ M 2 doe sno tchange|V |. Suppo s et ha t|V |a i ou s l t t a i nsi t smax imuma t( M 1 ,M 2),obv y V ≠ 0.Le tl i nel ⊥ V bea tpo i n tO . Lemma1.Lineldoesnotpas st hes e tof sanypo i n tofS.M 1i Sonones i deofl,andM 2i st hes e tofS ont heo t he rs i deofl. Ch i naNa t i ona lTe amS e l e c t i onTe s t Proo fo fLemma1.F i r s t,M 1 129 ∪ M 2 =S.O t he rwi s e,s i nc e|S| i seven,t he r ea r ea tl ea s ttwopo i n t sa andb wh i cha r eno ti n M 1 ∪ M 2 .Suppo s et ha tt heang l ebe tweena -bandV doe sno t exceed90 °,t hen|V + ( a -b)|>|V |.Soaddai n t oM 2 ,add Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t hen|V |wi l li nc r ea s e,wh i chi sacon t r ad i c t i on. bi n t oM 1 , Se cond,M 2 = - M 1 .Ot he rwi s e,t he r eex i s ta,b ∈S, s uch t ha ta,-a ∈ M 1 andb,-b ∈ M 2 .Suppo s et ha tt heang l e be tweena -bandV doe sno texc eed90 °.Pu tai n t oM 2 ,andb i n t oM 1 ,Vchange st oV +2( a -b),and|V +2( a -b) | >|V |, wh i chi sacon t r ad i c t i on. ldoe sno tpa s sanypo i n to fS.Ot he rwi s e,l e tlpa s s Th i rd, a,a ∈ M 1 ,-a ∈ M 2 , n t oM 2 ,-ai n t oM 1 ,V t henchangeai change st oV +4 a.No t i cet ha ta ⊥V , a|>|V|,wh i ch s o|V +4 i sacon t r ad i c t i on. Four t h,weshowt ha tM 2 i st hes e tofS onones i deofl ( t hes i det ha tV i st hes e tofS ont spo i n t i ngt o),and M 1 i he he rwi s e,t he r ei sa ∈ M 1 a tt hes i det ha tVi s o t he rs i deofl.Ot i n t i ngt o.Andt he r ei sab ∈ M 2 ont heo t he rs i de.Thent he po ang l ebe tweena -bandVi sl e s st han90 °.Changeai n t oM 2and bi n t oM 1 , t henVchange si n t oV +2( a -b), t hel eng t hofwh i ch i sg r ea t e r,wh i chi sacon t r ad i c t i on.Lemma1i snow proved. Lemma2.Le tSk = or|y|=k { (x,y)∈S |x|=k - 1 2 1 , k = 1,2,...,1006,lbeal i nepa s s i ngO anddoe sno t 2 } s spo i n t so fSk .Deno t ea l lpo i n t so fSk onones i deo flbyAk , pa heo t he rs i debyBk .Deno t eVk = ∑u∈A u a l lpo i n t so fSk ont k ∑v∈B v,thenthe maximum of|Vk |isobtained whenlis k sve r t i ca l( orhor i zon t a l). hor i zon t a l( orve r t i ca l),andVki Ma t hema t i c a lOl i adi nCh i na ymp 130 Proo fo fLemma2.Sk i sl oca t eda t t hebounda r fas r ewi t h2 kpo i n t s yo qua oneachs i de.Le tt hefourve r t i c e sof Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 1ö æ t he s r e be A çk - ,k - ÷ , qua 2 2ø è æ 1 1ö æ 1 B ç -k + ,k - ÷ ,C ç -k + ,2 2ø è 2 è k+ 1 1ö æ 1 ö÷ andD çk - ,-k + ÷ .By 2 2ø è 2ø F i .6 .1 g n t e r s ec t sAD a tpo i n tP s t r i c i t uppo s et ha tli ymme y,wemays wi t hnon -nega t i ves l ope.Le tP bel oca t edbe tweent het( 1 ≤t ) ≤kt t hofSk h( f romt opt obo t t om)o fSk onAD andt he( t +1) onAD ( t = 3).Thu s eeF i .6 .1,i nca s eo fk = 6, s, g → → → Vk = ( k -2)( k -1)j + ( 2 k -t)(- ( 2 k -1) i +tj)+ 2 2 → → t (( 2 k -1) i+( 2 k -t)j) → → = -2( 2 k -1)( k -t) i +2(- ( k -t) +3 k -3 k +1)j, → 2 2 → r e hor i zon t a l and ve r t i ca l un i t vec t or s, whe r e i and j a 2 2 , = u,0 ≤ u ≤ ( r e s c t i ve l t e( k -t) k -1) t hen pe y.Deno 1 2 2 2 2 k -1) u +u2 -2( 3 k2 -3 k +1) u+( 3 k2 -3 k +1) |Vk | = ( 4 =u - ( 3 2 k -2 k +1) u+( k -3 k +1). 2 2 2 2 1 Asa quadr a t i cf unc t i on ofu,u = k2 - k + i st he 2 ake si t smax imum s t r i cax i s.I ti sea s oknowt ha t|Vk|2t ymme yt a tu = 0.Sot = k,t shor i zon t a l,s oVk i sve r t i ca l. ha ti s,li Lemma2i spr oved. Turnt ot heor i i na lprob l em.Bys t r i c i t o g ymme y,weneedt on l s i de rt ha tt hes l opeofli snon -nega t i veandl e s st han1. ycon Le tM 1 ,M 2 bel oca t edontwos i de sofl.Deno t eM 2 ∩ Sk = Ch i naNa t i ona lTe amS e l e c t i onTe s t Ak ,M 1 ∩ Sk = Bk ,Vk = |V |= ∑ u - ∑v∈B v, t hen u∈Ak 1006 ∑ Vk ≤ k=1 131 k 1006 ∑ |V k=1 k |. ② So,|V|max ≤ ∑k=1|Vk|max. sa l lt he I fli shor i zon t a l,M 2i Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1006 i n t so fS ont heuppe rha l f -p l ane,M 1i sa l lpo i n t sofS ont he po ake si t s max imum,anda l lVk l owe rha l f l ane;each|Vk |t p ndeed i n tupwa rd.Andt heequa l i t f ② ho l d s.So|V |i po yo t ake st hemax imum|V |max = 2 ×10063 . 2013 ( J i a n i n,J i a n s u) gy g F i r s tDay 30,Ma r ch24,2013 8: 00 12: 1 Gi venanyn(> 1)copr imepo s i t i vei n t ege r sa1 ,a2 ,..., an ,deno t eA = a1 +a2 + … +an .Le tdi = ( A ,ai)( t he r ea t e s tcommond i v i s or), i = 1,2,...,n. g Le tDi bet heg r ea t e s tcommond i v i s orof { a1 ,a2 ,..., an } ai},i = 1,2, ...,n.Fi nd t he mi n imum of \{ ∏ n i=1 A -ai ( . po s edbyZhangS i hu i) diDi So l u t i on.Cons i de r D1 = ( a2 ,a3 ,...,an )andd2 = ( a2 ,A ) =( a2 ,a1 +a2 + … +an ). Le t( D1,d2)=d.Thend|a2,d|a3,...,d|an ,d|a1 + a2 + … +an .Thu s,d|a1 .Con s equen t l y, Ma t hema t i c a lOl i adi nCh i na ymp 132 d| ( a1 ,a2 ,...,an ). S i nc ea1 ,a2 ,...,an a r ecopr ime,wehaved = 1.No t e D1 ,d2)= 1.WehaveD1d2|a2 .So t ha tD1 |a2 ,d2|a2 and ( D1d2 ≤ a2 .S imi l a r l y,wehaveD2d3 ≤ a3 ,...,Dnd1 ≤ a1 . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Henc e, n ∏dD i=1 i i =( D1d2)·( D2d3)·…·( Dnd1) ≤ a2a3 …ana1 n ∏a . = i=1 Con s i de r i ng n ∏ (A -ai)= i=1 ≥ ① i n ∏ ( ∑a ) i=1 n j≠i j 1 n-1 ∏ ( (n -1)( ∏a ) ) i=1 n · =( n -1) andby ① and ② ,wes eet ha t n j≠i j ∏a i=1 i ② n A -ai n ≥( n -1) . i=1 diDi ∏ Ont heo t he rhand,i fa1 =a2 = … =an = 1, ∏ n i=1 A -ai n =( n -1) . diDi A -ai ( n Summi ngup,t hemi n imumo f∏i=1 i s n -1) . diDi n 2 Suppo s et ha tO andI a r et hec en t r e so ft hec i r cumc i r c l e t hr ad i u sRandr, andi nc i r c l eo f△ABC wi r e s t i ve l pec y,P Ch i naNa t i ona lTe amS e l e c t i onTe s t 133 ︵ tQP bet i st hemi dpo i n to fa r cBAC .Le hed i ame t e ro fO . n t e r s e c tBCa tpo i n tD ,andl e tt hec i r cumc i r c l eof Le tPIi n t e r s e c tt heex t endedl i neofPA a tpo i n tF .Le t △AIDi i n tE beon PD s ucht ha tDE = DQ .Provet ha t,i f po r ( 2 . po s edbyXi ong R Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ∠AEF = ∠APE , t hens i n2 ∠BAC = Bi n) F i .2 .1 g F i .2 .2 g So l u t i on.S i nc e ∠AEF = ∠APE , t hen△AEF ∽ △EPF .So i nc epo i n t sA ,I,D andF a AF ·PF = EF .S r econcyc l i c, 2 PA ·PF =PI·PD .Thu s, PF2 = AF ·PF +PA ·PF = EF +PI·PD . 2 ① S i nc ePQi st hed ime t e ro fc i r c l eOandpo i n tIi sonAQ ,we s eet ha tAI ⊥ AP .Cons equen t l y, Thu s,wehave ∠IDF = ∠IAP = 9 0 °. PF2 -EF2 = PD2 -ED2 . Comb i n i ng ① ,wehave PI·PD = PD2 -ED2 . Ma t hema t i c a lOl i adi nCh i na ymp 134 Thu s QD2 = ED2 = PD2 -PI·PD =ID ·PD . Con s equen t l y,wehave Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. △QID ∽ △PQD . ② S i ncePQi st hed i ame t e ro fc i r c l eO ,wes eet ha tBP ⊥BQ . Suppo s et ha tPQi st hepe rpend i cu l a rb i s ec t orofBCa tpo i n tM . No t et ha tIi st hei nc en t r eof△ABC .WehaveQI2 = QB2 = QM ·QP .Thu s, △QMI ∽ △QIP . ③ By ② and ③ ,wes eet ha t ∠IQD = ∠QPD = ∠QPI = ∠QIM .He nc e, MI ‖QD .Le tIK ⊥ BC be a t K .Then IK ‖PM ; t hu s, PM PD PQ = = . IK ID MQ Byt he Ci r c l e -Powe rTheor em andt heS i ne Theor em,we knowt ha t PQ ·IK = PM ·MQ = BM ·MC = t hu s,s i n2 ∠BAC = 3 æç 1 ö2 2 , BC ÷ = ( Rs i n∠BAC) è2 ø PQ ·IK 2R ·r 2 r = = . R R2 R2 Suppo s et he r ea r e101pe r s on ss i t t i nga roundar oundt ab l e i nana rb i t r a r r.Thekt hpe r s onpo s s e s s e skp i ece sof yorde ca r t s,k = 1, ...,101.Weca l li tat r an s i t i on i fone t r ans i t soneofh i sca r t st ooneofh i sad ac en tpe r s ons. j F i ndt hemi n imum po s i t i venumbe rk,s ucht ha twha t eve r t heorde ro ft hes ea t i ng,t he r ei swayofno mor et hank Ch i naNa t i ona lTe amS e l e c t i onTe s t 135 t r ans i t i onss ot ha teachpe r s onpo s s e s s e s51ca r t s.( s ed po by QuZhenhua) So l u t i on.Theansweri sk =4 2925. Le tt hec i r cumf e r enceoft het ab l ebe101,andt hed i s t ance Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. be tweentwo ad ac en t pe r s on s be 1.In t he fo l l owi ng, we j con s i de rt hel ea s tt r an s i t i ont ime s. rd sby [ iDeno t et he pe r s on whoi n i t i a l l s s e s s e sica ypo 51].So,i fp >0, t henwecant h i nkofpe r s on[ st hes our ce p]a st he rd s;andi fp < 0,pe r s on [ whos hou l ds endou tp ca p]i s i nkwhos hou l dr e c e i ve -pca rd s. Suppo s et ha ta tt heendo ft r ans i t i ons,eachpe r s onha s51 n i t i a l l l ong i ng t o s s e s s e sa ca rd u i ca rd s.I f pe r s on B po y be r s onA ,wecant r r i e sca rdu t oB bypa s s i ng h i nkt ha tA ca pe ︵ t hr ought he mi nora r cAB ,t hel eng t hoft herou t ei st hea r c ︵ l eng t h|AB |mean st het ime so ft r an s i t i on s. Le ts ea t i ng orde rbea sshowni nt hef i r e.Pe r s on [ i] gu t r an s i t sa l lh i sica rd st ope r s on [-i]( i = 1,2,...,50)wi t h rou t el eng t hi.Sot he r ea r ea l lt oge t he r12 + 22 + … + 502 = 42925t r an s i t i on s.Int hef o l l owi ng,wewi l lshowt ha tnol e s s t r an s i t i on scanmee tt her equ i r emen ti fpe r s on sa r es ea t edi nt h i s way. We u s e t he no t i on o f “ t en t i a l”.Le t po t en t i a la t po t he h i s t po s i t i on [ 50]be ghe 50; t he ne i s i t i on s ghbor po [49 ] and [48 ] each ha s t en t i a l49,...,t hel owe s t po s i t i on [- 49]and [- 50] po each ha s po t en t i a l 0. Then, t he t o t a l po t en t i a l a t t he F i .3 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 136 beg i nn i ngi s S = 101 ×50 + ( 100 +99)×49 + … + ( 2 +1)×0. Anda tt heendo ft r an s i t i on s,t het o t a lpo t en t i a li s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. T = 51 ×50 + ( 51 +51)×49 + … + ( 51 +51)×0. 925. Thed i f f e r encei sS -T = 42, S i nc ea f t e reacht r an s i t i on,t het o t a lpo t en t i a lchange sa t mo s t1,s oa tl ea s t42, 925t r an s i t i onsa r eneeded. Int hefo l l owi ng,weshowt ha twha t eve rbet heorde rof s ea t i ng,t he r ei sa lway s a way of no mor et han 42,925 t r ans i t i on ss ucht ha teach pe r s on po s s e s s e s51ca rd s.Toshow t h i s,weg i vetwol emma s. Lemma1.Le tc,a0 ,a1 ,...,an-1 bei n t ege r swi t ht he i rs um z e r o,andc ≥ 0,a0 ≤ a1 ≤ … ≤ an-1 . I fn + 1pe c],[ a0],[ a1],...,[ an-1], r s on sdeno t edby [ s s e s sN + c,N + a0 ,N + a1 , ...and N + an-1 ca rd s, po sapo s i t i vei n t ege r,s ucht ha tN +a0 > r e s c t i ve l r eN i pe y,whe 0.Le tt hepe r s on ss t andon0,1,...,n oft henumbe rax i s, s ucht ha t[ c]s t and sa tn.Thent he r ei sawayofno mor et han cn + ∑ i ai t r ans i t i ons,s ucht ha teach pe r s on po s s e s s e sN n-1 i=0 ca rd s. P r oo fo fLemma1.Suppo s et ha tan-1 ≥ … ≥as >0 ≥as-1 ≥ … ≥a0 , t heni nduc t i ononM =an-1 + … +as . I fM =0, t hen[ c] s s e scca rd st o[ ai], ai]ob rd s( 0 ≤i ≤ s ucht ha t[ t a i ns-aica pa s -1).Suppo s et ha tpe r s on [ ai]s t and sa txi( 0 ≤i ≤ n -1), sape rmu t a t i onof0,1,...,n - 1. t henx0 ,x1 ,...,xn-1 i Thu s,a ca rd t ha t pa s s e sf r om [ c]t o [ ai]need s n - xi t r ans i t i ons.Sot het o t a lt r an s i t i onsneededa r e Ch i naNa t i ona lTe amS e l e c t i onTe s t 137 s-1 s-1 s-1 i=0 i=0 i=0 ∑ (n -xi)(-ai)=cn + ∑ xiai ≤cn + ∑iai =c n+ n-1 ∑ia . i=0 i Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Nows uppo s et ha tt heconc l u s i oni st r uefori n t ege r ssma l l e r i de rt heca s eo fi n t ege rM .Suppo s et ha t t hanM .Cons an-1 = … = an-s > an-s-1 ,al > al-1 = … = a0 . an-1],...,[ an-s ]andape Thent he r ei sape r s oni n[ r s oni n [ ucht ha tt he i rd i s t ancei snomor et hann -s a0],...,[ al-1]s -l +1.We ma uppo s et ha tt hed i s t ancebe tween [ an-s ]and ys [ al-1] i snomor et hann -s -l +1, e tpe r s on [ an-s ]pa s s t henl aca rdut o[ al-1]bynomor et hann -s -l+1t r an s i t i on s.Nex t, byi nduc t i ononcand an-1 ≥ … ≥ an-s+1 > an-s -1 ≥ an-s-1 ≥ … ≥ al ≥ al-1 +1 > al-2 ≥ … ≥ a0 , wes eet ha tbyt ak i ngnomor et han L =cn + ( n -1) an-1 + … + ( n -s +1) an-s+1 + ( n -s)( an-s -1)+ ( n -s -1) an-s-1 + … + l al + ( l -1)( al-1 +1)+ ( l -2) al-2 + … +0·a0 t r an s i t i ons,each pe r s on po s s e s s e s N ca rd s.Add i t i on oft he t r an s i t i on so fca rdu,weknowt ha tt he r ea r enomor et han n-1 lL + ( n -s -l +1)=cn + ∑i ai i=0 t r ans i t i ons.Theproo fo fLemma1i scomp l e t ed. Lemma2.Foranypermut a t i ono f[-50],[-49],...,[ 49], [ c],deno 50]onac i r c l e,t he r ea lway sex i s t sa pe r s on [ t et he c]andt l i nepa s s i ngt hr ough [ heo r i i nbyl,s u cht ha tt hes umo f g Ma t hema t i c a lOl i adi nCh i na ymp 138 e a chs i deo fl ( i nc l udec )i nt heb r a cke t sha st hes ames i fc. gno Proo fo fLemma2.Le tt hepe rmu t a t i onont hec i r c l ebe[ a1], [ a2],...,[ a101]c l ockwi s e.Thent he r ei sad i r ec t edd i ame t e rl o ft hec i r c l es ucht ha t[ a1],[ a2],...,[ a50]a r eonones i deof Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. land [ a51],[ a52],...,[ a101]a r eont heo t he rs i de. l s e,i f∑i=1ai ≠0, I f∑i=1ai = 0, t hent akec =a51;e t hen 50 50 t hes umso fe a chs i deo flhav ed i f f e r en ts i fwer o t a t el180 ° gn.I ha tt hes um o feachs i dechange ss i c l ockwi s e,wes eet gn.So t he r eex i s t sa[ i ch mee t i ngt her equ i r emen tofLemma2. c]wh Take[ c] i nLemma2.Suppo s et ha tc ≥0( e l s echangeeach [ ai]by [-ai]andchangea l lt hed i r ec t i onso ft hea r c.Le tc = s ucht ha tt hes umofc1andt henumbe r son c1 +c2 ,c1 ,c2 ≥ 0, ones i deo f( no ti nc l udec) li sze r o.Deno t e50numbe r sont h i s s i debya0 ≤a1 ≤ … ≤a49 ,anddeno t e50numbe r sont heo t he r s i debyb0 ≤b1 ≤ … ≤b49 .Thent hes umo fc2andb0 ,b1 ,..., b49i sa l s oze r o.Us i ngLemma1t oc1 ,a0 ,a1 ,...,a49 andc2 , b0 ,b1 , ...,b49 ,r e s t i ve l t a i nt ha tt het r an s i t i on pec y,weob t ime sa r enomor et han 49 49 j=0 j=0 L = 50 c1 + ∑jaj +50 c2 + ∑j bj . b0 ,..., b49i sape rmu t a t i onof-50, S i ncec,a0 ,...,a49 , -4 9,...,50,byt heorde ri nequa l i t y 49 L = 50 c + ∑j( aj +bj ) j=0 ≤5 0 + 2 49 ∑j(2j -50 +2j -49) j=0 =4 2925. Summi ngup,t hel ea s tpo s i t i vei n t ege rk = 42925. Ch i naNa t i ona lTe amS e l e c t i onTe s t 139 S e c ondDay 8: 00 12: 30,Ma r ch25,2013 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 4 Le tp beapr ime,aandk bepo s i t i vei n t ege r s,s a t i s f i ng y a a ha tt he r eex i s t spo s i t i vei n t ege rn,n p <k <2p .Provet a 2a k <p s modp ).( s ed by Yu ucht ha tCn ≡ n ≡ k( po Hongb i ng) So l u t i on.Wearetoproveanext endedprob l em. Le tp beapr ime, aandkbepo s i t i vei n t ege r s,s a t i s f i ngpa y <k < 2 ha tf orany non -nega t i vei n t ege rb,t he r e p .Provet a ex i s t spo s i t i vei n t ege rn,n < pa+bs ucht ha tn ≡k( modpa )and k ≡k( Cn modpb ). I fb =0,pb =1, t aken =k -pa .Wepr ovebyi nduc t i on. ti s, Suppo s et ha tt heconc l u s i oni st r uef ori n t ege rb ≥ 0.Tha k ≡ t he r eex i s t sapo s i t i vei n t ege rn <pa+bn ≡k( modpa ),andCn k( modpb ). Le t1 ≤t ≤ p -1.Cons i de r k a+b = Cn +t p k-1 n +tpa+b -i . k -i i=0 ∏ Fori n t ege rm ,l e tP ( m )= pvp ,r( m )= ( m) m , whe r e P( m) vp ( m) i st henumbe rofpi nt hes t anda rdf ac t or i za t i ono fm . k -i)≤ a and eet ha tvp ( S i ncek -i < 2pa ≤ pa+1 ,wes n -i ≡k -i( modpa ).Hence, P( k -i)|n +tpa+b -i. Con s equen t l y, C k a+b n+t p n -i a+b-vp ( k-i) +t p P( k -i) = ∏ . r( k -i) i=0 k-1 Ma t hema t i c a lOl i adi nCh i na ymp 140 I fk -i ≠pa , t henvp ( k -i)≤a -1anda +b -vp ( k -i) ≥b +1. I fk -i = pa , t henvp ( k -i)= a.Hence, k a+b ≡ Cn +t p k-1 æ k-1 ö n -i n -i b ç ∏ ÷ ·t + p ∏ ( )( ) ç 0≤i≤k-1P( k -i) r( k -i)÷ i=0 P k -ir k -i è i≠k-pa ø r( n -i)ö÷ · b ( tp modpb+1). ( r ç 0≤i≤k-1 k -i)÷ è i≠k-pa ø Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. æ k ≡ Cn + ç k-1 ∏ t henpa | ( Th i si sbecau s et ha ti fk -i ≠ pa , n -i)- ( k -i). So vp ( n -i)=vp ( k -i). S i nc e ∏ k-1 r( n -i) i s copr ime t o p, we s ee t ha t r( k -i) 0≤i≤k-1 i≠k-pa k a+b ( Cn 0 ≤t ≤p -1)goe st hr ought hefo l l owi ngr ema i nde r sof +t p modu l a rpb+1 : k b +j Cn p ,j = 0,1,...,p -1. k ≡k( S i nceCn modpb ), t he r eex i s t sj( 0 ≤j ≤ p -1), s uch b k +j t ha tCn modpb+1).Tha ti s,t he r eex i s t st( 0 ≤t ≤ p ≡ k( k a+b ≡k( s u cht ha tCn modpb+1).Le tN =n +tpa+b .Then p -1), +t p N < pa+b+1 ,N ≡ n ≡k( modpa )andCkN ≡k( modpb+1). Theex t endedpr ob l emi spr ovedbyi nduc t i on. 5 Le tn ≥ 2anda1 ,a2 ,...,an ,b1 ,b2 ,...,bn benon - nega t i vei n t ege r s.Pr ovet ha t n n æç n ö÷n-1 æç 1 ö æ1 ö2 ai2 ÷ + ç ∑bi ÷ ≥ ∑ èn -1ø è n i=1 ø è n i=1 ø ( s edbyLeng Gang s ong) po n ∏ (a i=1 2 i n +bi ) . 2 1 ① Ch i naNa t i ona lTe amS e l e c t i onTe s t So l u t i on.Denot eλ = 141 æç n ö÷n-1 , n ≥ 2.Obv i ou s l λ > 1. y, èn -1ø 2 2 +bk f 1, ...,n},f Forg i veni ∈ { i xp = ak ork = 1, 2,...,nandf i xaj andbj ( hel e f t -hands i deof j ≠i).Thent Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ① = λ( 1 2 2 2 bi + ∑j≠ibj ) i saquad r a t i c p -bi + ∑j≠iaj )+ 2 ( n n 0, p ],wi t hl e ad i ngc oe f f i c i en tf un c t i ono fbi , bi ∈ [ λ 1 + < n n2 0.Thu s,i t smi n imumi st akena tt heendpo i n t s,t ha ti s, bi = 0o r ai = 0. So,wecans uppo s et ha taibi = 0, i = 1,2,...,n. Ca s e1.Eachai =0, t henbyt hemeanva l uei nequa l i t y,we have n æç 1 ö2 bi ÷ ≥ ∑ è n i=1 ø n ∏b i=1 2 n i . Ca s e2.Eachbi =0, t henbyt hemeanva l uei nequa l i t y,we have n n æ1 1 2ö λ ç ∑ ai ÷ ≥ ∑ ai2 ≥ è n i=1 ø n i=1 n ∏a i=1 2 n i . Ca s e3.Wemays uppo s et ha tb1 = … =bk = 0,ak+1 = … = an = 0,1 ≤k < n. Le ta1a2 …ak =ak ,bk+1 …bn =bn-k ,a,b ≥0.Thenbyt he meanva l uei nequa l i t y,wehave 2 2 a2 a2 ,bk+1 + … +bn ≥ ( n -k) b. 1 +a2 + … +ak ≥ k I ts u f f i c e st opr ovet ha t 2 2 k 2( n-k) λk 2 ( n -k) a + b2 ≥ an ·b n . 2 n n Byt hemeanva l uei nequa l i t eet ha t y,wes Thel e f t -hands i deo f ② Ma t hema t i c a lOl i adi nCh i na ymp 142 ② = λ 2 … λ 2 n -k 2 … n -k 2 + a + + a + b + b n n n2 n2 kt e rms n-kt e rms æn -k ö÷ è n ø ≥λ a · ç 2 k n k n n-k n 2( n-k) n ·b . So,i ts u f f i c e st os howt ha tλn Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. k æçn -k ö÷ è n ø n-k n ≥ 1,t ha ti s,t o æ n ö÷n-k k ≤λ . s how ç èn -k ø Inf ac t, n · n ·…· n · · ·…· 1 1 1 n -k n -k n -k nk-nt e rms n-kt e rms ≤ = nk -n)÷önk-k æn + ( è nk -k ø ç ( ) æç n ö÷ n-1 k k =λ . èn -1ø 6 I na p l ane wi t hc a r t e s i anc oo r d i na t e s,l e tP andQ b etwo r e i on so fc onv e xpo l i n c l ud i ngbounda r n t e r i o r) g ygon ( yandi who s ev e r t i c e sa r ea l li n t e e rpo i n t s( i. e.,t he i rcoord i na t e s g a r ea l li n t ege r s)andT = P ∩ Q .Provet ha ti fT i sno t emp t sno tcon t a i ni n t ege rpo i n t,t henTi sanon yanddoe degene r a t econvexquadr i l a t e r a l.( s edby QuZhenhua) po So l u t i on.S i ncet henon emp t n t e r s ec t i onT oftwoconvex yi c l o s ed po l si sa c l o s ed convex po l r a t ed ygon ygon or degene ea r et hr eepo s s i b l eca s e s. l he r po ygon,t ( 1)T i sa po i n t.ThenT mu s tbet heve r t exofP orQ , con t r ad i c t i ngt hef ac tt ha tT con t a i n snoi n t ege rpo i n t. ( 2)Ti sas egmen t.ThenT mu s tbet hei n t e r s ec t i onofan edgeo fP andanedgeofQ ,wh i chcon t a i nst heve r t exofP or Q ,wh i chi sacon t r ad i c t i on. Ch i naNa t i ona lTe amS e l e c t i onTe s t 143 ( 3)Ti sac l o s edconvexpo l ygon. saquadr i l a t e r a l. So,i tr ema i n st obeshownt ha tTi F i r s t,weno t et ha ti fT ha stwoad acen tedge sont heedge s j ofP ( orQ ), t hent hecommonve r t exo ft h i stwoedge smu s tbe Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. orQ ),wh i chi sacon t r ad i c t i on.Thu s,t he t heve r t exo fP ( bounda r fT i sf ormeda l t e rna t e l r to fanedgeofP yo ybyapa andt henapa r to fanedgeofQ ,andeachve r t exofT i st he s,t henumbe rofedge s i n t e r s e c tpo i n to fedge so fP andQ .Thu o fTi seven. Wes eet ha ti fanedgeeo fPi n t e r s ec t sanedgefo fQ , t hen e mu s ti n t e r s e c tano t he redgeo fQ ,o t he rwi s eT wi l lcon t a i nan i n t ege rpo i n t. In t he f o l l owi ng, we show by con t r ad i c t i on t ha tt he numbe ro fedge so fT canbe6ormor e. I fT ha sedge snol e s st han6,t hens uppo s eP con t a i nsk i n t ege rpo i n t sexcep tt heve r t i c e so fP . t henPi Ca s e1.I fk =0, sane l emen ti n t ege rt r i ang l e,ora r a l l e l og r am wi t ha r ea1.So,P canbel oca t edbe tweentwo pa n ti nt heopen r a l l e ll i ne sl1 ,l2 .Andt he r ei snoi n t ege rpo i pa doma i nΩbe tweenl1 andl2 .Atl ea s tt hr eeedge sofP a r et he edge sofT ,becau s eT ha sa tl ea s ts i xedge s. ( a)In ca s eo f P be i ng △ABC ( SeeF i .6 .1),DE ,FG and HI a r e g edge sofQ .D ,G andE mayco i nc i de wi t hF ,H andI,r e s t i ve l ne s pec y.Li FG ,HI,l1 andl2 form a convex i l a t e r a l.S i nc el i neDE doe sno t quadr i n t e r s e c ts egmen tBC ,wes eet ha tt he F i .6 .1 g i n t e r s e c tpo i n to fl i neDE andFG orofl i neDE andHIi si nΩ. Thu s,Q ha si n t ege rve r t exi nΩ,acon t r ad i c t i on. 144 Ma t hema t i c a lOl i adi nCh i na ymp ( b)Inca s eo fP be i ngapa r a l l e l og r am ▱ABCD .Le tAD , AB andBC bet hr eeedge so fP ( s eeFi .6 .2).Thei n t e r s e c t i on g oca t e si nΩ.Le tAB ,BC andCD be i n to fl i neEF andHGl po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t hr eeedge so fP ( s eeF i .6 .3),andt he r ei snoedgeo fQ on g AD .S imi l a rt ot heca s eo f( a),wecans eet hei n t e r s e c t i on fl i neEF andIJ l oca t e si n Ω, i n tofl i neEF and HG oro po wh i chi sacon t r ad i c t i on. F i .6 .2 g F i .6 .3 g Ca s e 2.k ≥ 1.Cons i de r i n t ege rpo i n tX onP o t he rt han t he ve r t i c e s.S i nce X ∉ T , fT t he r eex i s t sanedge MN o s ucht ha tT andX a r es epa r a t ed byl i neMN ( deno t ebyl )( s ee F i .6 .4).Thu s,MN i sa pa r t g F i .6 .4 g sont heedgeAB ofP ,N i sont heedge o fbounda r yofQ .M i CD ofP .A ,C andX a r eont hes ames i deofl (A andC may co i nc i de,bu tB andD dono tbyt hehypo t he s i st ha tT ha sa t fT onAB , l ea s ts i xedge s).Thu s,t he r ei sano t he rve r t exU o andt he r ei sano t he rve r t exV o fT onCD . i n t sX andve r t i c e so fP be l ow Deno t et heconvexhu l lofpo lbyP'.ThenP'andP co i nc i debe l owBD ,andt hepa r tofP' upBDi s△BXD .Compa r i ngT' = P' ∩ Q wi t hT ,wes eet ha t , T' ⊂ T andt hebounda r fT'i st hebounda r fT wi t hMN , yo yo Ch i naNa t i ona lTe amS e l e c t i onTe s t 145 MU andNV r ep l ac edbyM'N',M'U'andN'V',r e s t i ve l pec y. hes ame numbe rofedge s,andt he Tha ti s,T' and T havet numbe rofi n t ege rpo i n t so fP' o t he rt hanve r t i ce si sl e s st han t ha to fP .By af i n i t e procedu r el i ket h i s,wecan ob t a i na Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. convexpo l t hnoi n t e r i ori n t ege rpo i n t.By Ca s e1,i ti s ygonwi impo s s i b l e. 2014 ( Na n i n i a n s u) j g,J g F i r s tDay 8: 00 12: 30,Ma r ch23,2014 1 Le t O be t he c i r cumcen t e ro f △ABC and H A be t he t oBC .Theex t ens i ono fAO i n t e r s ec t s e c t i onofA on pro j tA'.The pro ec t i onsofA' t hec i r cumc i r c l eo f △BOC a j r e s t i ve l tO A bet on t oAB andAC a r eD andE , he pec y.Le c i r cumc en t e ro f △DH AE .De f i ne H B ,OB ,H C andOC s imi l a r l y. Pr ovet ha t H AO A , H BOB and H COC a r e concu r r en t. (po s ed byZhangS i hu i) So l u t i on.Le t T be t he s t r ymme y i n t o f A ove r BC , F be t he po , o e c t i ono fA' on t oBC and M be pr j t hepr o e c t i ono fT on t oAC . j S i nc eAC =CT ,weha v e ∠TCM = 2∠TAM .S i nce ∠TAM = π 2 F i .1 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 146 ∠ACB = ∠OAB ,weh ave ∠TCM = 2∠OAB = ∠A 'OB = ∠A'CF , and Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ∠TCH A = ∠A 'CF + ∠A'CT = ∠TCM + ∠A'CT = ∠A'CE . Be cau s eo ft he f ac tt ha t ∠CH AT , ∠CMT , ∠CEA', ∠CFA 'a r er i tang l e s,wehave, gh c o s∠TCH A CH A CH A ·CT cos∠A'CE = = = CM CT CM c o s∠TCM cos∠A'CF = CE ·CA' CE , = CA' CF CF i. e.,CH A ·CF = CM ·CE ,s r eont he o H A ,F ,M andE a s amec i r c l eω1 . hepro ec t i onofT on t oAB , S imi l a r l e tN bet t henH A , j y,l F ,N and D a r eont hes amec i r c l eω2 .S i nce A'FH AT and A'EMT a r ebo t hr i r apezo i d,t hepe rpend i cu l a rb i s ec t orof ghtt ta tt he mi dpo i n t K oft he t hes egmen t s H AF and EM mee s egmen tA'T ,i. e.,K i st hec en t e rofc i r c l eω1 andKF i st he r ad i u so fc i r c l eω1 .S imi l a r l r ea l s ot hecen t e rand y,K andKF a t her ad i u sofc i r c l eω2 ,r e s t i ve l s,ω1 andω2 a r et he pec y.Thu s ame,D ,N ,F ,H A ,E andM a r eont hes amec i r c l e.SoO Ai s t hemi dpo i n tK o fA'T ,O AH A ‖AA'. S i nce ∠H CAO + ∠AH CH B = π ∠ π, - ACB + ∠ACB = 2 2 we have AA' ⊥ H BH C ,t hu s O AH A ⊥ H BH C ,t he r e for e, O AH A ,OBH B andOCH C a l lpa s st hrought heor t hocen t e ro f △H AH BH C . 2 l oreve r Le tA1A2 …A101 bear egu l a r101 gon,andco y ve r t exr ed or b l ue.Le t N be t he numbe r of ob t u s e Ch i naNa t i ona lTe amS e l e c t i onTe s t 147 t r i ang l e ss a t i s f i ng t hef o l l owi ng cond i t i ons:Thet hr ee y ve r t i c e so ft het r i ang l emu s tbeve r t i ce soft he101 gon, bo t ht heve r t i c e swi t hacu t eang l e shavet hes ameco l or, andt heve r t exwi t hob t u s eang l eha sd i f f e r en tco l or. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 1)F i ndt hel a r s tpo s s i b l eva l ueo fN . ge ( 2)F i ndt henumbe rof way st oco l ort heve r t i ce ss uch t ha t max imum N i sach i eved. (Two co l or i ng sa r e heco l or sa r ed i f f e r en ton d i f f e r en ti ffors omeAi t t hetwoco l or i ngs cheme s.)( s edby QuZhenhua) po So l u t i on.Def i n i ngxi =0o sr edor r1depend sonwhe t he rAii b l ue.Forob t u s et r i ang l eAi-aAiAi+b ( ve r t exAii st heve r t exof t he s et hr eeve r t i ce ss a t i s f t heob t u s eang l e,i. e., a +b ≤ 50), y t hecond i t i onso ft hepr ob l emi fandon l f yi ( xi -xi-a )( xi -xi+b )= 1, o t he rwi s eequa l0,he r et hes ub s c r i tmodu l e s101.Thu s, p N = 101 ∑ ∑ (x i=1 ( a,b) i ① ( -xi-a ) xi -xi+b ). He r e ∑ (a,b)s t and sfort hes umma t i onofa l lpo s i t i v ei n t e e rpa i r s g ( a,b) s a t i s f i nga +b ≤50.The r ea r e49+48+ … +1 =1225s uch y s i t i vei n t ege rpa i r s.Expand i ng ① ,wehave po N = = 101 ∑ ∑ (x i=1 ( a,b) 101 i ∑ ∑ (x i=1 ( a,b) 2 i ( -xi-a ) xi -xi+b ) -xixi-a -xixi+b +xi-axi+b ) 101 101 50 2 k -1 -2( 50 -k)) xixi+k ∑ xi + ∑ ∑ ( =1 225 i=1 =1 225 n+ i=1 k=1 101 50 ∑ ∑ (3k -101)xx i=1 k=1 i i+k . ② He r eni st henumbe ro ft heb l ueve r t i c e s.Foranytwove r t exe s Ma t hema t i c a lOl i adi nCh i na ymp 148 Ai andAj ,1 ≤i ≤j ≤ 101, l e t d( Ai ,Aj )= d( Aj ,Ai)= mi n{ j -i,101 -j +i}. A1 ,A2 ,...,A101}bet hes e to fa l lb l ueve r t i ce s. Le tB ⊆ { Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Then ② canbewr i t t ena s 2 +3 N = 1225n -101Cn ∑ d( P ,Q ), { P ,Q}⊆B ③ whe r e{ P,Q}pa s st h r ougha l lt hetwo e l emen ts ub s e t so fB. Wi t hou tl o s so fg ene r a l i t s s umet ha tni se v en.Ot he rwi s e, y,wea chang et hec o l o ro fa l lv e r t i c e s,t hev a l u eo fN doe sno tchang e. Wr i t en = 2 t,0 ≤t ≤ 50,f r omonepo i n t,r enumb e ra l lt heb l u e v e r t i c e sbyP1,P2,...,P2t c l o ckwi s e.Then ∑ d( P ,Q )= { P ,Q}⊆B t t t-1 1 d( Pi ,Pi+t)+ ∑ ∑ ( d( Pi ,Pi+j )+ ∑ 2 i=1 j=1 i=1 d( Pi+j ,Pi+t)+d( Pi+t ,Pi-j )+d( Pi-j ,Pi)) 101 ( tt -1). 2 ≤5 t+ 0 ④ He r et hes ub s c r i to fPi modu l e s2 t,and weu s et hei nequa l i t p y d( Pi ,Pi+t)≤ 50and d( Pi ,Pi+j )+d( Pi+j ,Pi+t)+d( Pi+t ,Pi-j )+d( Pi-j ,Pi) ≤1 01. Comb i n i ng ③ and ④ ,wehave 101 æ ö 2 0 t + t( t -1)÷ +3ç5 N ≤ 1225n -101Cn 2 è ø 101 2 5099 t + t. 2 2 =- ⑤ ⑥ Ther i t -hands i deof ⑥ a t t a i n si t smax imum va l uewhen gh Ch i naNa t i ona lTe amS e l e c t i onTe s t 149 t = 25, t hu sN ≤ 32175. Ne x t, c on s i d e rt h e n e c e s s a r u f f i c i en tc ond i t i on o f y and s N =3 21 7 5.F i r s t,t = 2 5 me an st ha tt h e r ea r e5 0b l u ev e r t i c e s. S e c ond l o r1 ≤i ≤t,weha v ed( Pi,Pi+t)= 5 0.Whend( Pi, y,f Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 0,e u a l i t f⑤ ho l d s.Th e r e f o r e,t henumb e ro fwa st o Pi+t)= 5 q yo y c hoo s e5 0b l u ev e r t i c e ss u c ht ha tN a c h i e v e sma x imumi sequ a lt ot he numbe ro f way st o choo s e 25 d i a l sf r om t hel ong e s t 101 gona e r t i c e s. d i a l ss u cht ha tanytwod i a l sdono thav ec ommonv gona gona Edg i ngAiandAi+50 , i =1,2,...,101,wege tag r aphG . No t et ha t50and101a r er e l a t i ve l ime,s o1 +50 n( 0 ≤n ≤ ypr 100)f orm acomp l e t er e s i dues s t em modu l e101,i. e.G i sa y c i r c l ewi t h101edge s.The r e for e,t henumbe ro fway s( wr i t t en i eve smax imumi sequa l a sS )t oco l or50ve r t i c e sb l ueandN ach t ot henumbe ro fway st ochoo s e25edge so fGs ucht ha tanytwo o ft hemdono thavecommonve r t i c e s.Now,f i xoneedgeeof 4 G .S i nc et henumbe ro fway st ochoo s eedge shav i ngei sC2 no t 75 , 5 4 25 sC2 s oS = C2 S imi l a r l henumbe ro fway s hav i ngei 76 , 75 + C 76 . y,t t oha v e50r edv e r t i c e si sa l s oS,t hu st henumbe ro fway st oc o l o r t hev e r t i c e ss u cht ha tt hel a r e s tv a l u eo fNi sa ch i e v edi s2S. g s32175,t he Ins umma r hel a r s tpo s s i b l eva l ueofN i y,t ge 4 25 numbe ro fway st oco l ori s2S = 2( C2 . 75 + C 76 ) 3 Show t ha tt he r ea r e no 2 t up l e s( x,y)of po s i t i ve i n t ege r ss a t i s f i ngt heequa t i on y ( x +1)( x +2)…( x +2014)= ( y +1)( y +2)…( y +4028). ( s edbyLiWe i po gu) Proo f.Forn k = 2 ·m ( ki sanon -nega t i vei n t ege r,m i sodd), l e tv( n)= 2k . We pr ovet h i s by con t r ad i c t i on.As s ume ( x,y)i s one Ma t hema t i c a lOl i adi nCh i na ymp 2 Pa r tI Sho r t -An swe r Qu e s t i on s( e s t i on s1 8,e i tma r k s Qu gh e a ch) 1 x)= Ther angeo ff( x -5 - 24 -3x i s . So l u t i on.I ti sea s os eet ha tf( x) i si nc r ea s i ngoni t sdoma i n yt [ 5,8].The r e f or e,i t sr angei s [-3, 3]. Ma t hema t i c a lOl i adi nCh i na ymp 150 2 s i nxi acos2x -3) s -3.Thent he Themi n imumo fy = ( ai s . a n eo fr e a ln u m b e r g o s i t i v ei n t e e rs o l u t i ono ft heequa t i on.Le t pr g = So l u t i on.Le ts i nxv( hee r e s s i oni he o p { ( )nc }. hangedt = x +j xt+.T i) m a x v xst 1≤j≤2014 t)= (-a t2 +a -3) t,or g( t hen I f1 ≤j ≤ 2014,j ≠i, ( ) t3 + ( a -3) t. g t = -a v( x +j)=v( x +i + ( j -i))=v( j -i), Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t3 + ( a -3) t ≥ -3,wege t F r om -a s o )≥ 0, -( a t( t2 -1)-3( t -1v x +j) =v (( 2014 j)!·( 2013!). j -1)!)≤v( ∏ i a 1≤j≤2014,j≠( t -1)(t( t +1)-3)≥ 0. ) , ( )=ave -a ) ( ≤ weh t++ 3u 0 o rf4028!,t S i n c et i sam l t i l eo hu s s i n c e y j1 p ∏-j1=1≤(x0+,j ∏j=1t 2014 4028 a( t2 +t)≥ -3. ① 4028! 1007 >2 . x +i ≥v( x +i)≥v ! 2013 expr e s s i on ① a lway sho l d s;when0 <t ≤ Whent =0,-1, 4028 4028 The r e f or e,x > 21006 .So ( y +4028) > y2+j)= 1 ( - j=1≤t + 1,wehave0 <t2 +t ≤ 2;andwhen -1 <t < 0,∏ 4 2014 )> 21006·2014 .Wehavey +4028 > 2503 ,y > 2502 . ∏j=1(x +j, 3 ≤a ≤ 1 t < 0.The r e f or e 2. n 2 1( 1 Lemma.Le t0 ≤ xi < 1 ≤i ≤n). I fx = ∑i=1xi ,y = n 2 2 { , a x1≤u x t h e n i≤ nb i} 3 2m Th en m e ro f i n t e r a lpo i n t s( i. e.,t hepo i n t swho s ex g n 1 )w -coord andy i na t e sa r eb o t hi n t ege r s i t h i nt hea r ea ( no t n 1 -xi) ≥ 1 -x -y. 1 -x ≥ ∏ ( )enc i nc l ud i ngt hebounda r l o s edbyt her i ancho f iy =1 ghtbr -a = =1 xmm ndl i nex 0 0 i s . ho e rbf o l a y.B Pr o fo l e yp t h eAM GMi n e u a l i t hei n e a l i t y1a q y,t qu yon 2 2 , , So l u t i ol n. s mm e t r n l n e e dt oc on s i d rt h e a r to f t h e e f tB i s a s t o ov ew .eo The i ne u a l i t ont h e l e f th o l d s s i n c e ye y y y p y p q y n 1x bovet hex -a i s.Suppo s el i ney = k i n t e r c ep t st he t hea r eaa i∏ (1 -xi) n ≥ n n 1 ≥ n ( n 2 =1 ∑i=1 1 -xi ∑i=1 1 +xi +2xi ) n n n +nx +2∑i=1xi2 = ≥ 1 1 +x +y ≥ 1 -x -y. Thel emmai spr oved. Ch i naNa t i ona lTe amS e l e c t i onTe s t 151 4029 Backt ot hepr ob l em, l e tw =x +2015 >21007 ,z =y + 2 502 >2 ; t hent heequa t i oni sequ i va l en tt o 1 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1ö æ 2ö æ 2014ö÷ ö÷ 2014 ææ w · ç ç1 - ÷ ç1 - ÷ … ç1 wøè wø è w øø èè 1 9 ö æ 1 ö÷ æç 40272 ö÷ ö÷ 2014 1 - 2 ÷ … ç1 . 2 4 4 z øè z ø è 4 z2 ø ø 2 =z · ç ç1 - ææ èè Byt hel emma, 1 2015ö÷ 1 ö÷ æç 2ö æ 2014ö÷ ö÷ 2014 æ ææ 1 - ÷ … ç1 > w · ç ç1 w ç1 2w ø wøè wø è w øø è èè 2015 2·20142 ö÷ 2w ø w2 > w ç1 - æ è 2015÷ö 1 . 2w ø 8 > w ç1 - æ è The r e f or e,t he de c ima lo fw · 1 ææ 2 ö÷ … 1 ö÷ æç ç ç1 1 wøè wø èè æ æ 3 1ö 2014ö÷ ö÷ 2014 ç1 be l ong st oç , ÷ . w øø è è 8 2ø Ont heo t he rhand,byt hel emma 12 +32 + … +40272 ö÷ æ z2 ç1 è ø 4 z2 ·2014 ææ èè 12 +32 + … +40272 2·40274 ö÷ , 2 ( ø 4 z2 ·2014 4 z2) 2 >z ç1 - æ è t hu s 1 1 ö÷ æç 9 ö æ 40272 ö÷ ö÷ 2014 1 - 2 ÷ … ç1 2 z øè 4 z ø è 4 z2 ø ø 4 2 >z · ç ç1 - 1 1 ö÷ æç 9 ö æ 40272 ö÷ ö÷ 2014 ææ 4·20142 -1 2 1 - 2 ÷ … ç1 >z · ç ç1 2 èè 4 z øè 4 z ø è 4 z2 ø ø 12 z2 - 4·20142 -1 40274 12 z2 8 2 >z 2 4·20142 -1 1 . 8 12 >z - Ma t hema t i c a lOl i adi nCh i na ymp 152 4·20142 -1 S i nc ez2 i sani n t ege r,s ot hede c ima lpa r tof 12 1 ææ æ7 ö 40272 ö÷ ö÷ 2014 1 öæ 9 ö æ z · ç ç1- 2 ÷ ç1- 2 ÷ … ç1be l ong st o ç ,1÷ , 2 4 4 4 z øè z ø è z øø èè è8 ø 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. wh i chi sacon t r ad i c t i on. The r e f or e,t he r ea r eno2 t up l e s( x,y)o fpo s i t i v ei n t e e r s g s a t i s f i ng y 2014 ∏ (x +i)= j=1 4028 ∏ (y +i). j=1 S e c ondDay 8: 00 12: 30,Ma r ch24,2014 4 Gi ven an odd i n t ege rk > 3.Prove t ha tt he r e ex i s t i nf i n i t e l s i t i vei n t ege r sn,s ucht ha tt he r ea r e y many po n2 +1, ha ta l ld i v i de twopo s i t i vei n t ege r sd1 ,d2 t and 2 d1 +d2 =n +k.( s edby YuHongb i ng) po So l u t i on.Cons i de rt heDi ophan t i neequa t i on 2 2 (( +1) +1, k -2) xy = ( x +y -k) ① wep r ov et ha t① ha si n f i n i t e l s i t i v eodds o l u t i on s( x,y). y manypo Obv i ou s l 1,1)i sapo s i t i veodds o l u t i on,l e t( x1 ,y1)= y,( ( 1,1).As s umet ha t( xi ,yi)i sapo s i t i veodds o l u t i onof ① , andxi ≤ yi , l e txi+1 = yi ,yi+1 = ( k -1)( k -3) k -xi . yi +2 S i nc e ① canbewr i t t ena s 2 x2 - (( k -1)( k -3) k) x +( y +2 y -k) +1 = 0, by Vi e t a st heor em,( xi+1 ,yi+1) i sa l s oai n t ege rs o l u t i onof①. , , S i nc exi yi andk a r ea l lpo s i t i veoddi n t ege r s andk ≥ 5,s o xi+1i sapo s i t i veoddi n t ege r,and Ch i naNa t i ona lTe amS e l e c t i onTe s t 153 k -1)( k -3) k -xi ≡ -xi ≡ 1( mod2), yi+1 = ( yi +2 k -xi > yi > 0. yi+1 ≥ 8yi +2 i sapo s i t i veodds o l u t i onof① ,andxi + Thu s,( xi+1 ,yi+1) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. x1 ,y1)andt hecons t r uc t i onabove,wege t yi <xi+1 +yi+1 .By( as e r i e so fpo s i t i veodds o l u t i onso f① :( xi ,yi), i =1,2,..., s ucht ha tx1 +y1 < x2 +y2 < … . tn =xi + Foranyi n t ege rig r ea t e rt hank,xi +yi >k.Le sapo s i t i veoddi n t ege r,and yi -k,d1 =xi ,d2 =yi .Thenni 2 +1i d1 +d2 =n +k.S i nc e( k -2) seven,wecans howt ha td1 , n2 +1, d2a ss uchn r ebo t hd i v i s or so f andd1 +d2 =n +k.Thu 2 s a t i s f i e sa l lt hecond i t i on s,t he r e f or et he r eex i s ti nf i n i t e l y many s i t i veoddi n t ege r sns a t i s f i ngt hecond i t i on s. po y 5 Le tnbeag i veni n t ege rwh i chi sg r ea t e rt han1.Fi ndt he n) r ea t e s tcons t an tλ( s ucht ha tforanynon ze rocomp l ex g z1 ,z2 ,...,zn ,wehave n ∑ |z | k=1 k 2 ≥λ( n)mi n{ |zk+1 -zk | }, 1≤k≤n 2 whe r ezn+1 =z1 .( s edbyLeng Gang s ong) po So l u t i on.Le t ìïn ,2|n, ïï4 ( ) λ0 n = í n ,o t he rwi s e, ï 2 π ï4c î os 2 n wepr oveλ0( n) i st heg r ea t e s tcons t an t. I ft he r eex i s t sk( 1 ≤k ≤n) s ucht ha t|zk+1 -zk |= 0, t he i nequa l i t l d sobv i ou s l t hou tl o s so fgene r a l i t yho y.Sowi y,wecan a s s umet ha t Ma t hema t i c a lOl i adi nCh i na ymp 154 2 mi n{ |zk+1 -zk | }= 1. ① 1≤k≤n Unde rt h i sa s s ump t i on, i ts u f f i ce st o show t ha tt he mi n imumva l ueo f∑k=1 |zk |2i sλ0( n). n Whenni seven.S i nce Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. n 2 ∑ |zk | = k=1 ≥ n n 1 1 2 2 2 ( |zk | +|zk+1 | )≥ ∑ |zk+1 -zk | ∑ 2k 4 k=1 =1 n n 2 mi n{ |zk+1 -zk | }= , 4 1≤k≤n 4 andequa l i t l d swhen ( z1 ,z2 ,...,zn )= yho æ1 ç ,- 1 ,..., 2 è2 n 1 , 1 ö÷ , n t hu st hemi n imumv a l u eo f∑k=1|zk|2i s =λ0( n). 2 4 2ø sodd.Le t Nex t,cons i de rt hecond i t i onwhenni zk+1 ∈[ θk = a r 0,2π),k = 1,2,...,n. g zk Fora l lk =1,2,...,n, i fθk ≤ π 3π orθk ≥ , t henby ① , 2 2 2 2 2 sθk |zk | +|zk+1 | =|zk -zk+1 | +2|zk ||zk+1 |co ≥|zk -zk+1 | ≥ 1. 2 I fθk ∈ æç π ,3πö÷ , t henbyc o sθk < 0and ② , è2 2 ø 1 ≤|zk -zk+1 |2 =|zk | +|zk+1 | -2|zk | |zk+1 |cosθk 2 2 2 2 ≤( 1 + (-2c osθk )) |zk | +|zk+1 | )( θk . 2 2 2 =( i n2 |zk | +|zk+1 | )·2s The r e f or e, ② Ch i naNa t i ona lTe amS e l e c t i onTe s t 2 2 |zk | +|zk+1 | ≥ 155 1 . θk 2s i n2 2 ③ Nowcon s i de rt hef o l l owi ngtwocond i t i ons. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. æ π 3πö ( 1 ≤k ≤ n),θk ∈ ç , ÷ ,by ③ 1)I ff ora l lk( è2 2 ø n 2 ∑ |zk | = k=1 S i nc e ∏k=1 n n n 1 1 2 2 ( |zk | +|zk+1 | )≥ ∑ ∑ 2k 4 k=1 =1 1 .④ θk s i n2 2 zk+1 zn+1 = = 1,weh ave zk z1 n n zk+1 ö÷ æ +2mπ = 2mπ, r θk = a gç ∏ ∑ è k=1 zk ø k=1 ⑤ ( π mπ n -1) π ≤s =c i n o s . 0 <s i n 2 n 2 n n ⑥ whe r emi sapo s i t i vei n t ege r,andm <n.No t et ha tni sodd,s o x)= Le tf( 1 , x ∈ s i n2x π ,3π .I ti sea s oshowt ha t yt [4 4] x) i saconvexf unc t i on.By ④ andJ ens en sInequa l i t f( y,and comb i n i ng ⑤ and ⑥ ,wehave n 2 ∑ |zk | ≥ k=1 = l e t n 1 ∑ 4k =1 1 n· 1 ≥ n 4 θk ö÷ 1 2θk 2æ s i n s i n ç ∑k=1 èn 2 2ø n· 1 n· 1 ≥ =λ0( n). 4 4 π 2 mπ s i n cos2 n 2 n æ π 3πö ( 2)I ft he r eex i s t sj( 1 ≤j ≤n), s ucht ha tθj ∉ ç , ÷ , è2 2 ø { I = j θj ∉ æç π ,3πö÷ , j = 1,2,...,n . è2 2 ø } Ma t hema t i c a lOl i adi nCh i na ymp 156 By ② ,f orj ∈I,wehave|zj|2 +|zj+1|2 ≥1;andby ③ , forj ∉I,wehave Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 2 2 |zj | +|zj+1 | ≥ The r e f or e, n ∑ |z | k k=1 2 = ≥ = 1 1 ≥ . θ 2 j 2 2s i n 2 1 ( ∑I (|zj |2 +|zj+1|2)+ ∑I (|zj |2 +|zj+1|2)) 2 j ∈ j∉ 1 1 n -|I|) |I|+ ( 2 4 n +1 1( n +|I|)≥ . 4 4 ⑦ No t et ha t n +1 n · 1 π n ≥ ≥ ⇔c o s2 4 n n +1 4 2 2 π c o s n 2 ⇔s i n2 n π π 1 = 1 -c ≤1= . o s2 n n n +1 n +1 2 2 Theequa l i t l d swhenn = 3;whenn ≥ 5, yho s i n2 æ π ö÷2 π π2 · 1 1 , < ç < < nø è2 2 n 2 n n +1 n +1 n +1 n · ≥ t hei nequa l i t l s oho l d s.Sof oroddi n t ege rn ≥3, ya 4 4 1 c o s2 π n 2 .Comb i n i ng ⑦ ,wehave n ∑ |z | k=1 k 2 ≥ n· 1 =λ0( n). 4 2 π c o s 2 n Ont heo t he rhand,whenzk = 1 π 2co s 2 n i( n-1) kπ n ·e ,k = 1, Ch i naNa t i ona lTe amS e l e c t i onTe s t 157 2,...,n,wehave|zk -zk+1 |= 1,k = 1,2,...,n,and ∑ n Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. k=1 2 n). i eve si t smi n imumva l ueλ0( |zk | ach n) Summi ngup,t heg r ea t e s tλ( i s ìïn ,2|n, ïï4 λ0( n)= í n ,o t he rwi s e. ï 2 π ï4c o s î 2 n 6 Forpo s i t i vei n t ege rk >1, l e tf( k)bet henumbe rofway s o ff ac t or i ngk i n t o produc to f po s i t i vei n t ege r sg r ea t e r t han1. (The orde r off ac t or sa r e no tcoun t e r ed,f or examp l ef( 12) = 4,a s12canbef ac t or edi nt he s ef our way s:12,2 ×6,3 ×4,2 ×2 ×3.) Pr ovet ha t,i fni sapo s i t i vei n t ege rg r ea t e rt han1,pi sa n imef ac t orofn, n)≤ .( s edby YaoYi un) t henf( po j pr p So l u t i on.Weus eP ( n)t os t andf ort heb i s tpr imed i v i s or gge o fn,andde f i neP ( 1)=f( 1)=1.Wef i r s tpr ovetwol emma s. Lemma1.Forpos i t i vei n t ege rnandpr imep|n,wehavef( n) ≤ ∑ d n p d). f( Proo fo fLemma1.Forconveni ence,af ac t or i ngs a t i s f i ng y t hecond i t i oni sca l l edaf ac t or i ngf orshor t. Foranyf ac t or i ngo fn,wr i t en =n1n2 …nk , s i ncep|n, s o t he r eex i s t si ∈ { 1,...,k},s ucht ha tp |ni ( i ft he r ei smor e t hanones uchi,choo s e any one oft hem ),wi t hou tl o s sof h i sf ac t or i ngt oaf ac t or i ng r a l i t s s umet ha ti = 1.Mapt gene y,a n, o fd = d = n2n3 …nk . n1 Fortwod i f f e r en tf ac t or i ng so fn,n = n1n2 …nk andn = Ma t hema t i c a lOl i adi nCh i na ymp 158 n ' n ' ' whe r ep d i v i de sn1 andn '1).I fn1 = n '1 ,t hend = 1 2 …n l( n2 …nk andd =n ' ' r etwod i f f e r en tf ac t or i ng sofd (di sa ka 2 …n n n n ≠ =d d i v i s orof );i ' t hend = ', fn1 ≠n s ot he s etwo 1, n1 n ' 1 p Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. f ac t or i ng smapt oaf ac t or i ngofd andd ' ,r e s t i ve l pec y (d and n d 'a r ed i v i s or so f ).Thu s,f( n)≤ p oved. pr ∑ d d).Lemma1i s f( n p d Lemma2.Forpos i t i vei n t ege rn, n)= ∑d|n ( ), l e tg( t hen Pd n)≤ n. g( Proo fo fLemma 2.Induc t i on ont he numbe rofd i f f e r en t imed i v i s or sofn.I fn = 1, 1)= 1.I fn = pa ,pi sa t heng( pr , ime t hen pr a p -1 a ≤ 1 +p -1 = n. n)= 1 +1 +p + … +pa-1 = 1 + g( p -1 As s umet ha twhent henumbe rofd i f f e r en tpr imed i v i s or s o fni sk,wehaveg( n)≤ n.Cons i de rt hes i t ua t i onwhenn ha s k +1d i f f e r en tpr imed i v i s or s.Le tt hepr imef ac t or i za t i onofn ak k+1 1 ben = pa whe r ep1 < … < pk < pk+1 ,andwr i t en 1 …pk pk+1 , a ak 1 = mpk++1 . So a k+1 k+1 pk+1 -1 dpik+1 ( ) ( ) ( ) ( ) , = + = + σm gn gm gm ∑ ∑ pk+1 -1 d|m i=1 pk+1 a m )s whe r eσ( t and sfort hes um o fpo s i t i ved i v i s or so fm .By a s s ump t i on,g( m )≤ m ,s i nc e a a k+1 k+1 k a +1 pk+1 -1 æç pii -1ö÷pk+1 -1 ( ) = ∏ σm è i=1 pi -1 ø pk+1 -1 pk+1 -1 ≤ a +1 æç pii -1ö÷ ( ak+1 pk+1 -1) ∏ èi =1 pi+1 -1 ø k Ch i naNa t i ona lTe amS e l e c t i onTe s t 159 a +1 pii -1ö ak+1 ÷( -1) p ∏ èi ø k+1 pi =1 æ ≤ ç ≤ k k ( ∏ p ) (p ai i=1 ak+1 k+1 i -1) Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. =n -m , n)≤ n.Lemma2i sproved. s og( Backt ot he pr ob l em,i ts u f f i c e st o prove,f or po s i t i ve i n t ege rn,f( n)≤ n ho l d s. P( n) i fn =1, Byi nduc t i ononn, t heequa l i t l d s.As s umet ha t yho n f orn = 1,2,...,k,wehavef( n)≤ ( ).Then,i fn =k + Pn 1,byLemma s1and2andt hea s s ump t i on, k +1)≤ f( d ∑ d)≤ f( k+1 P( k+1) d ∑ d P( d) k+1 P( k+1) æ k +1 ö÷ k +1 ≤ . k +1)ø P ( èP ( k +1) =g ç Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naGi r l s Ma thema t i ca l Ol i ad ymp 2010 Sh i i azhuang j F i r s tDay8: 00~12: 00 Augus t10,2010 1 Le tn b eani n t e e rg r e a t e rt hantwo,andl e tA1,A2,..., g A2n bepa i rwi s ed i s t i nc tnonemp t ub s e t so f{ 1,2,...,n}. ys De t e rmi net hemax imumv a l u eo f∑i=1 n 2 |Ai ∩ Ai+1| . |Ai |·|Ai+1| ( He r e,wes e tA2n+1 =A1 .Foras e tX , l e t|X|deno t et he numbe ro fe l emen t si nX .) So l u t i onTheansweri sn. Ch i naGi r l s Ma t hema t i c a lOl i ad ymp Wecons i de reachs ummandsi = 161 |Ai ∩ Ai+1 | . |Ai |·|Ai+1 | semp t e t,t hensi = 0. I fAi ∩ Ai+1i ys I fAi ∩ Ai+1i snonemp t s eAi ≠ Ai+1 ,a tl ea s toneof y,becau Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ai andAi+1 ha smor et hanonee l emen t,t ha ti s,max{ |Ai |, cau s eAi ∩ Ai+1i sas ub s e to feacho fAi and |Ai+1 |}≥ 2.Be Ai+1 ,|Ai ∩ Ai+1 |≤ mi n{ |Ai |,|Ai+1 |}and si = ≤ |Ai ∩ Ai+1 | |Ai |·|Ai+1 | mi n{ |Ai |,|Ai+1 |} 1 ≤ . max{ n{ |Ai |,|Ai+1 |}·mi |Ai |,|Ai+1 |} 2 I tf o l l owst ha t 2n |Ai ∩ Ai+1 | ≤ ∑ Ai |·|Ai+1 | | i=1 2n 1 ∑2 i=1 = n. Th i suppe rboundcanbeach i evedwi t hs e t s A1 = { 1},A2 = { 1,2},A3 = { 2},A4 = { 2,3},..., A2n-2 = { n -1,n},A2n-1 = { n},A2n = { n,1}. 2 In t r i ang l e ABC ,AB = AC . Po i n tD i st he mi dpo i n tofs i de BC .Po i n tE l i e s ou t s i de t he ucht ha tCE ⊥ AB t r i ang l eABCs andBE = BD .Le t M bet he mi dpo i n to fs egmen tBE .Po i n t ︵ Fl i e sont he mi nora r cAD o f t hec i r cumc i r c l eo ft r i ang l eABD s ucht ha tMF ⊥ BE .Pr ovet ha t ED ⊥ FD . F i .2 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 162 So l u t i on1.Cons t r uc tpo i n tF1 s ucht ha tEF1 = BF1 a ndr ay DF1i spe rpend i cu l a rt ol i neED . I ts u f f i ce st oshowt ha tF = F1 orABDF1i scyc l i c;t ha ti s, ① Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ∠BAD = ∠BF1D . S e t ∠BAD = ∠CAD = x.Be c a u s eEC ⊥ ABandAD ⊥ BC, ∠ECB = 9 0 °- ∠ABD = ∠BAD = x. sa mi d l i neo f t r i ang l eBCE .In pa r t i cu l a r, No t et ha t MD i MD ‖EC and ② ∠MDB = ∠ECD = x. In i s o s c e l e st r i ang l e EF1M , wemays e t ∠EF1M = ∠BF1M = s eEM ⊥ MF1 andMD ⊥ y.Becau , DF1 ∠EMF1 = ∠EDF1 = 9 0 °, imp l i ng t ha t EMDF1 i s cyc l i c. y Cons equen t l y,wehave ∠EDM = ∠EF1M = y. ③ F i .2 .2 g Comb i n i ng ② and ③ ,weob t a i n ∠BDE = ∠EDM + ∠MDB = x +y. Be cau s e BE = BD ,we conc l ude t ha tt r i ang l e BED i s s e i s o s c e l e swi t h ∠MED = ∠BED = ∠BDE = x + y.Becau EMDF1i scyc l i c,wehave ∠MF1D = ∠MED = x +y.I ti s t henc l ea rt ha t ∠BF1D = ∠MF1D - ∠MF1B = x = ∠BAD , wh i chi s①. Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 163 So l u t i on 2. (Bas ed on work by She r r y Gong and Inna Zakha r ev i ch)Wema i n t a i nt heno t a t i onsu s edi nSo l u t i on1.Le t ωandO deno t et hec i r cumc i r c l eandt hec i r cumcen t e roft r i ang l e e tT bes tween ABD ,andl e condi n t e r s e c t i on ( o t he rt hanB )be Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. l i neBE andω.Ex t ends egmen tDE t hroughE t o mee tω a tS. ucht ha tDF2 ⊥ DE .Wewi l ls howt ha tF2 = i e sonωs Po i n tF2l F orF2B = F2E .Le tM 2 deno t et hefoo to ft hepe rpend i cu l a r f r omF2t ol i neBE .I ts u f f i c e st os howt ha tM 2i st hemi dpo i n t t ha ti s,EM 2 = M 2B . o fs egmen tBE , Be c a u s e ∠SDF2 = ∠EDF2 = 90 °,O i st he mi dpo i n tofSF2 . sc c l i c,∠BTD = Be c a u s eBTAD i y ∠BAD = ∠BCE = x.No t ea l s o t ha t BD = BE and ∠EBC = ∠DBT . We t r i ang l e BDT conc l ude t ha t and BEC a r e cong r uen tt oeacho t he r,imp l i ng y t ha tBE = ET .Hence,OE ⊥ F i .2 .3 g BT .Le tN bet hefoo to ft hepe rpend i cu l a rf romSt ol i neBE . No t e t ha t s egmen t s NE and EM 2 a r e t he r e s t i ve pec rpend i cu l a rpro ec t i onsofs egmen t sSOandOF2on t ol i neBE . pe j Be cau s eSO = OF2 ,NE = EM 2 .Be cau s eTE = EB , i ts u f f i c e s t os howt ha tTN = NE ,wh i chi sev i den ts i nce ∠STE = ∠SEB = ∠SDB = ∠EDB = ∠BED = ∠SET ands ot r i ang l eSETi si s o s ce l e swi t hSE = ST . 3 Pr ovet ha tf oreve r i venpo s i t i vei n t ege rn, t he r eex i s ta yg imep andani n t ege rm s ucht ha t pr ( a)p ≡ 5( mod6); Ma t hema t i c a lOl i adi nCh i na ymp 164 ( b)p n; ( c)n ≡ m 3( modp). So l u t i on 1. There are inf i n i t e l ime sp t ha ta r e y many pr cong r uen tt o 5 modu l o 6. (Th i si sas c i a l ca s e of t he pe Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Di r i ch l e t st heor emonpr ime si nana r i t hme t i cprog r e s s i ont ha t canbeea s i l i r ec t l r ei sa tl ea s tones uchpr ime yprovend y.The ( name l s et he r e we r e on l i n i t e l ime s y,5).Suppo yf y many pr cong r uen tt o5 modu l o6,andl e tt he i rpr oduc tbeP .Then6 P -1,wh r uen tt o5 modu l o6, i chi sl a r rt han P andcong ge mu s thave ano t he r pr ime d i v i s orcong r uen tt o 5 modu l o 6, wh i chi sacon t r ad i c t i on.Thu s,t he or i i na la s s ump t i on wa s g wr ong,and t he r ea r ei nf i n i t e l ime st ha ta r e y many odd pr s cong r uen tt o5 modu l o6.)In pa r t i cu l a r,ones uch pr imep i l a r rt hann.Thu s, a t i s f i e sbo t hcond i t i ons( a)and ( b).We ps ge canwr i t ep = 6 k +5fors e tm = omepo s i t i vei n t ege rk.Wes n4k+3 .Thent heFe rma t sLi t t l eTheor em,wehave m 3 ≡ n12k+9 ≡ n6k+4 ·n6k+4 ·n ≡ np-1 ·np-1 ·n ≡ n( modp). So l u t i on2.Se tm =n -1.Th enm 3 -n =n -3 n +2n -1, 3 2 wh i chi sr e l a t i ve l imet on.Anypr imed i v i s orp ofm 3 -ni s ypr r e l a t i ve l imet on, t ha ti s,ps a t i s f i e st hecond i t i on s( b)and ypr ( c).I tr ema i n st of i ndap t ha ti scong r uen tt o5 modu l o6. No t et ha t m3 -m = m ( m2 -1)= ( m -1) m( m +1), mod wh i chi sd i v i s i b l eby6.Henc e,m 3 -n ≡ m -n ≡ -1 ≡ 5( 6).Thu s,t he r ei sapr imed i v i s orp o fm 3 -nt ha ti scong r uen t t o5modu l o6,andt h i si st hepr imewes ought. Comment:Theor i i na lve r s i ono ft heprob l emi sa sfo l l ows. g Pr ovet ha tf oreve r i venpo s i t i vei n t ege rn,t he r eex i s ta yg Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 165 imep andani n t ege rm s ucht ha t pr ( a)p ≡ 3( mod4); ( b)p n; ( c)n ≡ m2( modp). Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. I twa si n s i r edbyas o l u t i ont ot heprob l em3ofIMO2010. p 4 whe r en ≥ 2)ber Le tx1 ,x2 ,...,xn ( ea lnumbe r swi t h 2 2 x2 1 +x2 + … +xn = 1. Pr ovet ha t n 2 2 k æ ö2 xk xk æçn -1ö÷2 1n · ≤ . ç ÷ 2 ∑ ∑ k=1 è ∑i=1ixi ø k èn +1ø k=1 k n De t e rmi newhent heequa l i t l d s. yho Comment: Expand i ng t he l e f t -hand s i de of t he de s i r ed i nequa l i t i ve s yg 2 k æ ö2 xk 1n · ç ÷ 2 ∑ k=1 è ∑i=1ixi ø k n n = n 2 xk -∑ ∑ k=1 k k=1 n = 2 xk ∑ k=1 k n = 2 xk ∑ k=1 k n = 2 xk ∑k k=1 - 2 2xk ∑ n ixi2 i=1 n + n 2 kxk ∑( ∑ n k=1 2 ixi2) i=1 n 1 1 1 2 kxk 2 + n 2∑ 2 2 2∑ x k ( ) k=1 k=1 i x i x i i ∑i=1 ∑i=1 n 2 1 + n 2 2 i x ∑i=1 i ∑i=1ixi n 1 . 2 i x i ∑i=1 n Wewan tt os howt ha t n 2 xk ∑ k=1 k n 2 xk æn -1÷ö2 1 ≤ ç = n ∑ 2 + ∑i=1ixi èn 2ø k=1 k n n 2 2 xk xk n 4 2∑ ∑ ) ( n +1 k=1 k k=1 k Ma t hema t i c a lOl i adi nCh i na ymp 166 or n xk n 4 ≤ 2∑ ( n +1) k=1 k 2 t ha ti s, 2 2 ( xk æç ö÷ n +1) 2 ≤ kxk . ( ) ∑ ∑ è k=1 k ø k=1 n 4 n Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 , 2 i x i ∑i=1 n n ① Wepr e s en ttwopr oo f so ft heabovei nequa l i t y. So l u t i on1.Werewr i t e① a s n n 2 xk æ ö÷ 4 nç ∑ ( ∑kxk2 ) ≤ (n +1)2. è k=1 k ø k =1 Byt heAM GMi nequa l i t y,wehave n n n n 2 2 xk nxk æ ö÷ æ ö÷ 2 = 4ç ∑ nç ∑ kxk 4 ( ) ( ∑kxk2 ) ∑ è k=1 k ø k=1 è k=1 k ø k =1 ≤ n n 2 2 nxk æç 2ö + ∑kxk ÷ ∑ è k=1 k ø k=1 n = I ts u f f i ce st os howt ha t æç æç n ö 2 ö÷2 +k ÷xk . ∑ èk ø ø =1 èk n +k ≤ n +1 k or 0 ≤ nk +k -k2 -n = ( n -k)( k -1), wh i chi sev i den t. Now,wecon s i de rt heequa l i t s e.No t et ha tt hel a s t y ca i nequ a l i t ss t r i c tf o r1 <k <n.Hen c e,wemu s thav ex2 = … = yi xn-1 = 0.Fort heAM GMi nequa l i t oho l d,wemu s thave yt Ma t hema t i c a lOl i adi nCh i na ymp 2 Pa r tI Sho r t -An swe r Qu e s t i on s( e s t i on s1 8,e i tma rk s Qu gh e a ch) 1 Ch i naGi r l s Ma t hema t i c a lOl i ad ymp x -5 - 24 -3x i s x)= Ther angeo ff( 167 . n ne So l u t i on.I ti sea os et ha tf( x) i si nc r ea s i ngoni t sdoma i n yt 2s nxk 2 2 2 2 = ∑kxk o rnx2 xn 1 +xn = x1 +n ∑ [ k 5,8].Th e r e f o r e,i t s r angei s [-3, 3]. k k=1 =1 1 2 2 2 wi t hx2 u s thavex andx2 = … = 1 +xn = 1.Wem 1 = xn = 2 Themi s i nx2 acos2x -3) i s -3.Thent he n imumo fy = ( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. xn-1 = 0. s r a ngeo fr ea lnumbe rai . ( = So l u t i on. Le ts i n t .T h) ee e s s i oni o u n2. By Lx e l l eY e Wx ew r i t e ④ st a shenchangedt pr ynn n t)= (-a t2 +a -3) t,or n 2 g( xk 2 2 ) ( n +1 -4n ∑ kxk ≥ 0. ∑ ka =1 =1 + ( t)= t3k ak3) t. g( N o t et ha tt el e f t -h a nds deo hea nequa l i t st he yi ) ( + ≥i a t3h a 3 t 3,ft we tbovei F r o mge d i s c r imi nan to ft hequa dr a t i c( i nt): -a t( t2 -1)-3( t -1)≥ 0, 2 2 2 2 ( ) ( ) … ( + + -0. t n +1) t+ 2t f t =n ( )2 (x )x )≥ ( -11 -a -n3 tx t ++ 1n 2 xn 2 x2 2 …-+ + + x0 . ) , ( 1, w eh a v e a t S i nc et -1 ≤ 2 nt +1 -3 ≤ 0 or 2 ( )≥ -3. +( ah t ① t)ha sar ea lr oo t.Be cau s et h e I ts u f f i c e st os howt a tft 2 2 2 ( ) ( , , ;x +… + = -1 < ≤ x n x① 2 x i s i c hi s l ead i n o e f f i c i e n to f t 0 ex e s i on a l w a sh o l d sn wn h) e, nw 0h t Wh e n 1 + 2 f gc pr y t) ≤ 0fors s i t i ve,i tr ema i n st o bes hownt ha tf( 1ome2t. po , <t < 0,≤t + 1 wehave0 <t2 +t ≤ 2;andwhen -1 4 2 wehave Be cau s ex2 1 + … +xn = 1, 3 t < 0.The r e f or e,-2 ≤ a ≤ 1 2. 2 2 + … +nxn ) t)= n( x2 x22 t2 - ( n +1)( x2 1 +2 1 +x2 + … + f( 3 2 x2 xn 2 2 ) +… + xn t + x2 1 + . 2 n Thenumbe ro fi n t eg r a lpo i n t s( i e.,t hepo i n t swho s ex n ord )wi ks 2a 2 2n -co an dy i n t e sa r ebo t hi t ex e r t h i nt hea r ea ( no t = ∑ nkxkt - (n +1)xkt +gk k=1 r l o s edbyt her i tbr ancho f i nc l ud i ngt hebounda y)enc gh n 2 2 1 i 2x -y = 1a nex = 100i s . hy rbo l a pe -ndl t = x nk t -1) . k( ∑ k k=1 So l u t i on.Bysymme t r l ocon s i de rt hepa r tof y,weon yneedt 1 p el 1 = x -a x i s .S up o s n eb k n t e r c ep t st he t hea r eaa b o v et he y , = ≤i 0 e ca u s ei f o r t I ti se a s t o s e e t h a t f y n n 2 eachs ummand 2 ( 1 xk k -1)( k -n) 2 ( ≤ 0, xk nk t -1)t - = k nk 168 Ma t hema t i c a lOl i adi nCh i na ymp comp l e t i ng ou r proo f.For t he equa l i t s e of t he g i ven y ca i nequa l i t s thaveequa l i t s efort heabovei nequa l i t y,wemu yca y f oreve r or2 ≤ k ≤ n - 1.I ti st henno t ha ti s,xk = 0f yk;t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. d i f f i cu l tt oob t a i nt ha t 2 x2 1 = x2 = 1 . 2 S e c ondDay 8: 00 12: 00,Augus t11,2010 5 Le tf( x)andg( x)bes t r i c t l nc r ea s i ngl i nea rf unc t i ons yi f rom Rt o Rs ucht ha tf( x)i sani n t ege ri fandon l f yi x)i sani n t ege r.Pr ovet ha tforanyr ea lnumbe rx, g( x)-g( x) i sani n t ege r. f( So l u t i on.Wecanwr i t ef( x)=ax +bandg( x)=cx +dfor s omer ea lnumbe r sa,b,c,d wi t ha,c > 0. Bys t r s s umet ha ta ≥c. ymme y,wemaya Wec l a imt ha ta =c.As s umeont hecon t r a r ha ta >c. yt Be cau s ea >c >0, t her ange so ffandga r ebo t hR.The r ei sa ucht ha tf( x0s x0)=ax0 +bi sani n t ege r.Henceg( x0)=cx0 +di sa l s oani n t ege r.Bu tt hen, 1ö æ f çx0 + ÷ = ax0 +b +1 aø è and 1ö æ c g çx0 + ÷ =cx0 +b + . aø è a Bu tt h i si simpo s s i b l ebe cau s ewecanno thavetwoi n t ege r s c ha thavepo s i t i ved i f f e r enc e wh x0)andg( x1)t i chi sl e s s g( a Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 169 t han1. The r e f or e,wecanwr i t ef( x)=ax +bandg( x)=ax +d f ors omer ea lnumbe r sa,b,d wi t ha > 0. s tbeani n t ege r,t ha ti s, Thenb -d = f( x0)-g( x0)mu Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. x)-g( x)=b -d f( i sani n t ege r. 6 Inacu t et r i ang l eABC , AB > AC .Le t M be t he mi dpo i n t of s i de BC .Theex t e r i orang l e b i s e c t or o f ∠BAC mee t sr ay BC a t P. Po i n t sK and F l i e on F i .6 .1 g t hel i nePA s ucht ha tMF ⊥ BC and MK ⊥ PA .Prove t ha tBC2 = 4PF ·AK . So l u t i on. Le t ω and O deno t et he c i r cumc i r c l e and t he c i r cumcen t e roft r i ang l eABC ,r e s c t i ve l e tN bet he pe y,andl ︵ mi dpo i n to fa r cBC ( no tcon t a i n i ngA ).No t et ha tl i neMN i s t hepe rpend i cu l a rb i s e c t orofs egmen tBC . Inpa r t i cu l a r,bo t hF andO l i eonl i ne MN .No t ea l s ot ha tr ayAN i st hei n t e r i or l i ngt ha tNA ⊥ FP or ∠NAF = 90 °.I t b i s e c t oro f ∠BACimp y fo l l owst ha tNF i sad i ame t e rofω.I ti sc l ea rt ha tMK ‖AN , f r om wh i chi tfo l l owst ha t AK FK = . MN FM Henc e, PK ·AK = PK ·FK ·MN . FM Ma t hema t i c a lOl i adi nCh i na ymp 170 No t et ha t MK i st hea l t i t udet ot he hypo t enu s ei nr i ght r es imi l a r t r i ang l eFMP , imp l i ngt ha tt r i ang l eFMK andFPM a y t oeacho t he randPF ·FK = FM 2 .Comb i n i ngt hel a s ttwo Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. equa l i t i e sy i e l d s PF ·AK = PF ·FK ·MN = FM ·MN . FM F i .6 .2 g By t he powe r o f a i n t t heor em po (or c r o s s chord t heor em),wehaveFM ·MN = BM ·MC = BC2 .Comb i n i ng 4 t hel a s ttwoequa l i t i e sg i ve st hede s i r edr e s u l t. 7 Le tn b e an i n t e e rg r e a t e rt han o r equ a lt o 3.Fo ra g rmu t a t i onp = ( x1,x2,...,xn )o f( 1,2,...,n),wes ay pe t ha txjl i e si nb e twe enxi andxki fi <j <k.( Fo re xamp l e, i nt hepe rmu t a t i on ( 1,3,2,4),3l i e si nb e twe en1and4, )S and4do e sno tl i ei nbe twe en1and2. e tS = { p1,p2,..., on s i s t so f( d i s t i n c t)pe rmu t a t i on spi o f( 1,2,...,n). pm }c Suppo s et ha t among e v e r h r e ed i s t i n c t numbe r si n{ 1, yt 2,...,n},oneo ft he s enumb e r sdoe sno tl i ei nbe twe ent he o t he rtwonumbe r si ne v e r rmu t a t i on spi ∈ S.De t e rmi ne ype t hema x imumv a l u eo fm . So l u t i on.Theansweri s2n-1 . Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 171 Wef i r s ts howt ha tm ≤2n-1 .Wei nduc tonn.Theba s eca s e n =3i st r i v i a l.( Indeed,s ay3doe sno tl i ei nbe tween1and2, t henwecanhaveS = {( 1,2,3),( 3,1,2),( 2,1,3),( 3,2, whe r ek ≥ 1)}.)As s umet ha tt hes t a t emen ti st r ueforn = k ( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 3).Nowcon s i de rn =k +1andas e tSk+1s a t i s f i e st hecond i t i ons oft hepr ob l em.No t et ha ti ft hee l emen tk +1i sde l e t edf rom eachpe rmu t a t i onpii nSk+1 , t her e s u l t i ngpe rmu t a t i on sqiforma s e tSkt ha ts a t i s f i e st hecond i t i on so ft heprob l em ( forn =k). I t s u f f i c e st oshow t hefo l l owi ng c l a im:t he r ea r ea t mo s ttwo d i s t i nc tpe rmu t a t i onsp andqi nSk+1 t ha tcan mapt ot hes ame rmu t a t i onr i by de l e t i ng t he e l emen tk + 1i nt he n Sk ( pe rmu t a t i onsp andq). pe Indeed,a s s umet ha tf or x1 ,x2 ,...,xk+1),p2 = ( p1 = ( y1 ,y2 ,...,yk+1), z1 ,z2 ,...,zk+1) p3 = ( i nSk+1 ,q1 =q2 = q3 = q.Bys t r s s umet ha t ymme y,wemaya 1,2,...,k).As s umet ha txa =yb =zc =k +1.Aga i nby q =( s t r s s umet ha t1 ≤ a <b <c ≤ k +1.( No t e ymme y,wemaya t ha tbe cau s eq1 = q2 = q3 = q,a,b,c a r ed i s t i nc t.) We con s i de rt hr eenumbe r sa,b,k +1.Wehavep1 = (...,k +1, a,...,b,...)( i npa r t i cu l a r,al i e si nbe tweenk +1andb), i e si n i npa r t i cu l a r,k +1l p2 = (...,a,...,k +1,b,...)( be tweenaandb ),andp3 = (...,a,...,b,...,k +1,...) ( i npa r t i cu l a r, bl i e si nbe tweenaandk +1).Henceeachoneof t henumbe r sa,b,k +1l i ei nbe tweent heo t he rtwonumbe r si n s omepe rmu t a t i on si nSk ,v i o l a t i ngt hecond i t i onsofSk .Thu s ou ra s s ump t i onwa swr onganda tmo s ttwoe l emen t si nSk+1can bemappedt oae l emen ti nSk ,e s t ab l i sh i ngourc l a im. I tr ema i n st obeshownt ha tm = 2n-1 i sach i evab l e.We Ma t hema t i c a lOl i adi nCh i na ymp 172 con s t r uc tpe rmu t a t i on p i nduc t i ve l 1)p l ac e1; ( 2)a f t e r y:( la r ep l ac ed,wep l ac el +1e numbe r s1,2,..., i t he rt ot hel e f t ort her i ft hea l lt henumbe r sp l aceds of a r.Becau s et he r e ghto a r etwopo s s i b l ep l ac e sf oreacho ft henumbe r s2,3,...,n,we Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. cancon s t r uc t2n-1s uchpe rmu t a t i ons.Foranyt hr eenumbe r s1 ≤ a <b <c ≤n, cdoe sno tl i ei nbe tweenaandb.Henc e,t h i ss e t rmu t a t i onss a t i s f i e st hecond i t i onsoft he prob l em, of2n-1 pe comp l e t i ngou rpr oo f. Comment:Theor i i na lve r s i ono ft heprob l emi sa sfo l l ows. g The r ea r enbooksa r r angedi nar owonashe l f.Al i br a r i an come spe r i od i ca l l ea r r ange st hebooksi naneworde r.I t yandr t u rn sou tt ha t,among anyt hr ee books,t he r ei sonet ha ti s neve rp l ac edanywhe r ebe tweent heo t he rtwo.Provet ha tt he t o t a lnumbe rofd i f f e r en torde r st ha toc cu ri sa tmo s t2n-1 . Toenhanc et hel eve lofd i f f i cu l t het e s tpape r,t he y oft l emde c i de st oa s kcon t e s t an t st of i ndt h i smax imumva l ue. prob 8 De t e rmi net hel ea s todd numbe ra > 5s a t i s f i ng t he y f o l l owi ng cond i t i ons: The r ea r e po s i t i vei n t ege r s m1, 2 2 m 2 ,n1 ,n2 s ucht ha ta = m2 a2 = m2 and 1 +n1 , 2 +n2 , m1 -n1 = m 2 -n2 . So l u t i on.Theansweri s261. No t et ha t 261 = 152 +62 ,2612 = 1892 +1802 ,15 -6 = 189 -180. Weknowt ha tt he r ei snonumbe ri nbe tween5and261t ha t s a t i s f i e st hecond i t i onoft hepr ob l em.As s umeont hecon t r a r y t ha tai ss uchanumbe r.Wemays e td = m 1 -n1 > 0.Becau s e ai sodd,m 1 andn1 haved i f f e r en tpa r i t od i sodd. y,ands Be cau s em 1 < 261,d ≤ 15; t ha ti s,t hepo s s i b l eva l ue so fda r e Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 173 1,3,5,7,9,11,13,15.Wewi l le l imi na t eeve r hem. yoneoft 2 2 +n2 o Wecanwr i t em 2 = n2 +danda2 = ( n2 +d) r 2 2 2 a2 -d2 = ( n2 +d) . ① Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. I fd = 1, t hen ① be come saPe l l sequa t i onx2 -2y2 = -1 wi t h( x,y)= ( 2 n2 +1,a).Th i sPe l l sequa t i onha smi n ima l 2k-1 1+ 2) s o l u t i on ( x,y)= ( 1,1)andx +y 2 = ( f orpo s i t i ve l ue so ft hes o l u t i onsoft h i sPe l l sequa t i ons i n t ege r sk.They va a r e5,29,169,985,....Thu s,t heon l s s i b l eva l ue sfora ypo a r e29and169.I ti sea s oche ckt ha tne i t he r29nor169can yt 2 2 +n1 .He bewr i t t eni nt hef ormof( n1 +1) nced ≠ 1. I fd i sa mu l t i l eo f3,t hen m 2 ≡ n2( mod3)anda2 ≡ p 2m 2 mod3).Be cau s e2i sno taquadr a t i cr e s i duemodu l o3,we 2( 2 conc l udet ha t0 ≡ m 2 ≡ n2 ≡ a( mod3).Hence,m 2 1 + n1 ≡ 0( mod3),f r om wh i chi tf o l l owst ha tm 1 ≡ n1 ≡ 0( mod3). 2 2 2 sa mu l t i l eo f9andm 2 sa Thu s, a = m2 1 + n1 i 2 + n2 = a i p mu l t i l eo f81.Wecanwr i t em 2 =3m', n2 =3n ',anda =9a ' p f ori n t ege r sm 2 ,n3 ,a '.Wehave m'2 + n '2 = 9a ' i n, 2 .A ga becau s e -1i sano taquadr a t i cr e s i duemodu l o3,wemu s thave m' ≡ n ' ≡ 0( mod3). I tf o l l owst ha td = 3( m' -n ') i sd i v i s i b l e by9.Thu s,d = 9. Becau s ea <261 =152 +62 ,n1 <6.The r e for e, n1 =3and 2 , tt hen ① become s92·577 = ( 2 a =122 +32 =153.Bu n2 +9) wh i chi simpo s s i b l e.Hence,d ≠ 3,or9,or15. I fd = 11ord = 13,be cau s e2i sno taquadr a t i cr e s i due modu l od,f r oma2 ≡ 2m 2 modd),weconc l udet ha t0 ≡ m 2 ≡ 2( modd)and0 ≡ m 1 n2 ≡a( modd). I tf o l l owst ha t2m 2 1 ≡a ≡0( ≡ n1 ≡ a( modd).Inpa r t i cu l a r, n1 ≥ d,m 1 ≥ 2dand 2 a = m2 d2 > 261. 1 +n1 ≥ 5 Ma t hema t i c a lOl i adi nCh i na ymp 174 Henc e,d ≠ 11or13. I fd = 5,wea l s o no t et ha t2i sno ta quadr a t i cr e s i de modu l o5.Byt hes amer ea s on i ngbe f or e,wehave Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. m 1 ≡ n1 ≡ m 2 ≡ n2 ≡ 0( mod5). 2 I fn1 ≥ 10, t henm 1 ≥ 15anda = m 2 02 +152 > 1 +n1 ≥ 1 261.Thu s, n1 =5,m 1 =10,anda =125.Bu tt hen ① become s 2 , 52 ·1249 = ( 2 n2 +5) wh i chi simpo s s i b l e.Henc e,d ≠ 5. I fd =7, t henbe cau s ea <261 =152 +62 ,wehaven1 ≤8. Thepo s s i b l eva l ue sofa a r et hen65,85,109,137,169,205, 2 =2 245.Bu tt hen ① be come s( 2 n2 +7) a2 -49.I ti sea s o yt che ckt ha tt he r ei snos o l u t i oni nt h i sca s e.Henc e,d ≠ 7. Comb i n i ngt heabove,weconc l udet ha t261i st heanswe r o ft h i sque s t i on. 2011 ( S h e n z h e n,Gu a n d o n g g) The10th Ch i naG i r l s Ma t hema t i ca lO l i ad washe l ddu r i ngJu l ymp y 28 Augus t3,2011 a tt he No.Th r ee Sen i o rH i lo f gh Schoo Shen z heni n Shen z hen,Guangdong P r ov i nce,Ch i na.Ar ound 39 t eamsf r om Ma i n l andCh i na,p l us9t eamsf r om Russ i a,t heUn i t ed ,t S t a t es,S i ngapo r e,Japan,e t c. o t a l l 8g i r ls t uden t sa t t ended y18 t hecompe t i t i on.Thecompe t i t i oncon s i s t so ftwor ounds — each l as t sf ou rhou r sandcon t a i n sf ou rp r ob l ems.Thet eamf r omShangha i H i lwont hef i r s tp l acei nt eamt o t a lsco r e.20pa r t i c i t s ghSchoo pan wongo l dmeda l s,4 0wons i l ve rmeda l s,and8 0b r on zemeda l s. Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 175 F i r s tDay 8: 00 12: 00,Augus t1,2011 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 1 1 = 1 F i nda l lpo s i t i vei n t ege r sns ucht ha tequa t i on + x y n ha sexac t l s i t i vei n t ege rs o l u t i ons( x,y)wi t hx ≤ y2011po s edby Xi ongBi n) y.( po So l u t i on.Fromtheg i venequa t i on,wehavexy -nx -ny = 2 0⇒ ( x -n)( s i de sx = y = 2 n,forany y -n)= n .Then,be x -n equa lt oa pr ope rd i v i s oro fn2 ,we wi l lge ta po s i t i v e i n t e e rs o l u t i on( x,y) s a t i s f i ngt her equ i r edc ond i t i on.The r e f o r e, g y n2s hou l dha v ee xa c t l r ope rd i v i s o r st ha ta r el e s st hann. y2010p αk 1 he r e p1 , ...,pk a Suppo s en = pα r e pr ime 1 …pk , w numbe r sd i f f e r en tf r omeacho t he r.Thent henumbe rofpr ope r d i v i s or so fn2l e s st hanni s ( α1 +1)…( 2 αk +1)-1 2 . 2 So ( 2 α1 +1)…( 2 αk +1)= 4021.S i nce4021i spr ime,we tk = 1,2 α1 +1 = 4021,andα1 = 2010. ge The r e for e, r epi n = p2010 ,whe sanypr imenumbe r. 2 Ass howni nF i .2 .1, t hed i a l s g gona AC,BD o f quad r i l a t e r a lABCD i n t e r s ec t a t po i n t E, t he mi dpe rpend i cu l a r s of AB , CD ( wi t hM ,N b e i ngt he i rmi dpo i n t s, r e s c t i ve l n t e r s e c ta t po i n t pe y)i F ,andl i ne EF i n t e r s e c t s wi t h BC , AD a t po i n t s P, Q, F i .2 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 176 r e s c t i ve l s eMF ·CD = NF ·AB ,DQ ·BP = pe y.Suppo AQ ·CP .ProvePQ ⊥ BC .( s edbyZheng Huan) po So l u t i on. As shown in Fi .2 .2, g conne c tpo i n t sA F ,B F ,C F , Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. and D F .Byt heg i vencond i t i on, r ebo t hi s o s ce l e s △AFB and△CFD a t r i ang l e swi t hFM andFN be i ngt he a l t i t ude st oeacht r i ang l e sba s e. S i nce MF · CD = NF · AB , △AFB ∽ △DFC .Then ∠AFB = F i .2 .2 g ∠CFD a nd ∠FAB = ∠FDC .Mor eove r, ∠BFD = ∠CFA . Fr omFB = FA ,FD = FC ,wehave△BFD ≌ △AFC ,wh i ch r e for e, mean st ha t ∠FAC = ∠FBD and ∠FCA = ∠FDB .The i n t sA ,B ,F ,E andpo i n t sC ,D ,E ,F a r eeachconcyc l i c. po Fromt heabover e s u l t,wege t ∠FEB = ∠FAB = ∠FDC = ∠FEC , wh i chimp l i e st ha tl i neEPi st heang l eb i s ec t oro f ∠BEC .Then EB BP ED QD = = .Int hes ameway,wehave . wehave EC CP EA AQ I fDQ ·BP = AQ ·CP , t henEB ·ED = EC·EA ,wh i ch sacyc l i cquadr i l a t e r a lwi t hF a st hec en t e rofi t s mean sABCDi c i r cumc i r c l e.Att h i st ime,s i nc e ∠EBC = 1∠ 1∠ DFC = AFB = ∠ECB , 2 2 wet henge tPQ ⊥ BC . 3 Suppo a t i s f cd =1. s epo s i t i ver ea lnumbe r sa,b,c,ds yab Prove 1 1 1 1 9 25 ( + + + + ≥ . po s edbyZhu a b c d a +b +c +d 4 Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 177 Huawe i) So l u t i on1.Fi r s t,wewi l lprovet ha t,wheneve rt he r ea r etwo numbe r samonga,b,c,dt ha ta r eequa l,t hei nequa l i t l d s. yho Wemaya s s umet ha ta = b andl e ts = a +b +c + d.Then Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. wehave 1 1 1 1 9 + + + + a b c d a +b +c +d = 2 c +d 9 2 9 2 + + = +a ( s -2a)+ a cd s a s = 9ö æ 2 2 -2 a3 + ças + ÷ . sø è a Wede f i net heexpr e s s i onabovea sf( s).Thenwes eet ha t eache st hemi n imumfors = s)r f( Whena ≥ 3 . a 2, howeve r,wehave 2 s = a +b +c +d ≥ 2a + 2 3 ≥ . a a 2 Att h i st ime,f( eache st he mi n imumfors = 2 s)r a+ . a Wet henhave 9ö 2 2ö 9 æ 2 æ 2 -2 -2 a3 + ças + ÷ = a3 +a2 ç2a + ÷ + sø a aø s è è a = 2 9 9 +2 =s + a+ a s s = 7 9 9 7 9 ·9 ≥ ×4 +2 s+ s+ s 16 16 16 16 s s = When0 < a < 2 ö 7 9 25æç ∵s = 2 ≥ 4÷ . a+ + = a ø 4 2 4è 2, wehave 2 Ma t hema t i c a lOl i adi nCh i na ymp 178 9ö 2 æ 2 2 2 -2 -2 +5 a3 + ças + ÷ ≥ a3 +6a = a+( a -2a3) sø a è a a > 2 2· 25 +5 a ≥2 a = 2 10 > . 5 a a 4 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Se cond,wecon s i de rt heca s et ha ta,b,c,d a r ed i f f e r en t ad f · f r omeacho t he r.Wemaya s s umet ha ta >b >c > d.I c s i ngt her e s u l tabove,wehave b·c·c =ab cd = 1,byu 1 1 1 1 9 25 + + + + ≥ . ad b c c ad 4 +b +c +c c c The r e f or e,weon l opr ovet ha t yneedt 1 1 1 1 9 + + + + a b c d a +b +c +d ≥ 1 1 1 1 9 + + + + . ad b c c ad +b +c +c c c ① Wehave ① ⇔ c 1 1 9 1 9 + + ≥ + + a d a +b +c +d ad c ad +b +2 c c ac +cd -c2 -ad 9 · ≥ ⇔ acd æçad ö÷ ( ) +b +2 c a +b +c +d èc ø ad æç ö -c÷ a +d c è ø ⇔ ( a -c)( c -d) 9 · ≥ acd æçad ö÷ ( ) +b +2 c a +b +c +d èc ø ( a -c)( c -d) c ⇔ 1 9 ≥ ad æad ö ( +b +2 c÷ a +b +c +d)ç èc ø Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 179 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. æad ö +b +2 c÷ ≥ 9ad ⇔( a +b +c +d)ç èc ø ⇐ ad +b +2 c≥ c ⇐ ad +3 c≥ c ad æ ö +b +2 c÷ ad ç ∵a +b +c +d > 9 c è ø ad . 9 Thepr oo fi scomp l e t e. So l u t i on2.Weprovei tbyu s i ngt head u s tmen tme t hod.We j f i ne maya s s umet ha ta ≤b ≤c ≤ d,andde 1 1 1 1 9 + + + + a,b,c,d)= . f( a b c d a +b +c +d F i r s t l l lpr ove y,wewi a,b,c,d)≥ f( ac ,b, ac ,d). f( Asama t t e ro ff ac t,expr e s s i on ②i sequ i va l en tt o ② 1 1 9 1 1 9 + + ≥ + + a c a +b +c +d ac ac 2 ac +b +d ⇔ 2 2 (a - c) 9(a - c) ≥ ac ( a +b +c +d)( 2 ac +b +d) 2 (∵ (a - c) ≥ 0) ③ ⇐( a +b +c +d)( 2 a c +b +d)≥ 9ac 2 ö æ d = ç ∵b +d ≥ 2 b ÷ è ac ø 2 öæ 2 ö æ c+ ÷ ç2 a ÷ ≥9 ⇐ ça +c + ac. è ac ø è ac ø Fr om1 = ab cd ≥a·a·c·c⇒ac ≤ 1⇒ 2 ≥2 a c and ac a +c ≥ 2 ac ,wehavet hef o l l owi ngcond i t i on: 2 öæ 2 ö æ c+ c+ ÷ ç2 a ÷ Thel e f t -hands i deo f③ ≥ ç2 a è ac ø è ac ø Ma t hema t i c a lOl i adi nCh i na ymp 180 ≥4 a her i -hands i deof③. c·4 ac =16ac >9ac =t ght The r e f or e,② ho l d s,wh i ch meansf( a,b,c,d)(wi t h a ≤b ≤c ≤ d )r eache si t smi n imumi fandon l fa =c ( i. e.a yi = b = c) .So we may a s s ume t ha t ( a, b,c, d) = æ 1 ,1 ,1 , 3 ÷ö ( t t ≥1).Thenweon l opr ovet ha t,f or yneedt èt t t ø Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ç a l lt ≥ 1, æ 1 1 1 3 ö÷ 25 t ≥ . fç , , , èt t t ø 4 ④ Wehave æ 1 1 1 3 ö÷ 25 t ≥ fç , , , èt t t ø 4 1 9 25 ≥ ⇔3 t+ 3 + 3 4 t 3 t + t ⇔12 t8 -25 t7 +76 t4 -75 t3 +12 ≥ 0 ⑤ 2 ( t -1) 12 t6 -t5 -14 t4 -27 t3 +36 t2 +24 t +12)≥ 0 ⇔( ⇐12 t6 -t5 -14 t4 -27 t3 +36 t2 +24 t +12 ≥ 0 ⇔( t -1)( 12 t5 +11 t4 -3 t3 -30 t2 +6 t +30)+42 ≥ 0. t5 +6 t ≥ 2 12 t5 ·6 t = 12 2t3 > 3 t3 ,and S i nc et ≥ 1,12 t4 +30 ≥ 2 11 t4 ·30 = 2 330t2 > 30 t2 , 11 t hen ( t - 1)( 12 t5 + 11 t4 - 3 t3 - 30 t2 + 6 t + 30)+ 42 > 0. The r e for e,⑤ ho l d s,wh i chj u s t i f i e s④. Thepr oofi scomp l e t e. 4 n( n ≥ 3) t ab l e t enn i s p l aye r s have a round rob i n — t ournamen t eachp l aye rwi l lp l aya l lt heo t he r sexac t l y onc e,andt he r ei sno dr aw game.Suppo s e,a f t e rt he t ou rnamen t,a l lt he p l aye r scanbea r r angedi nac i r c l e Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 181 s ucht ha t,f oranyt hr eep l aye r sA ,B ,C ,i fA ,B a r e l ea s e ad acen t,t hena tl ea s toneoft hem de f ea t edC .P j f i nda l lpo s s i b l eva l ue sofn.( s edbyFuYunhao) po So l u t i on.Wewi l lprovet ha tncanbeanyoddnumbe r sno tl e s s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t han3.Suppo s en =2 k +1,anoddnumbe rg r ea t e rt han3,and l aye r sa r er epr e s en t edbyA1 ,A2 ,...,A2k+1 .Le np tu sa r r ange k +1)de t hecompe t i t i onr e s u l ta sfo l l ows:Ai( 1 ≤i ≤ 2 f ea t ed Ai+2 ,Ai+4 ,...,Ai+2k ( wes t i l a t et ha tA2k+1+j = Aj ,j = 1, pu k + 1 )bu tl o s tt ot he o t he rp l aye r s.Then t he s e 2,...,2 l aye r scanbea r r angedi nac i r c l ei norde rA1 ,A2 ,...,A2k+1 , p A1 .Now,g i venanyt hr ee p l aye r sA ,B ,C wi t hA ,B be i ng ad ac en ti nt hec i r c l e,wemaya s s umet ha tA =At ,B =At+1 ,C j = At+r ( k +1,2 ≤r ≤2 k).Thene san 1 ≤t ≤2 i t he rrorr -1i evennumbe rno tl e s st han2 i chimp l i e st ha ta tl ea s toneof k,wh t hep l aye r sA ,B de f ea t edC . sanevennumbe rno tl e s s Ont heo t he rhand,s uppo s eni t han4,and t hen p l aye r scan be a r r angedi n ac i r c l e A1 , A2 ,...,An ,A1 t ha t mee t st her equ i r edcond i t i on.We may a s s umet ha tA1 de f ea t edA2 .Ac cord i ngt ot her equ i r emen t,a t l ea s toneo fA2 ,A3 de f ea t edA1 ,andt henA3 de f ea t edA1 ;bu t a tl ea s toneofA1 ,A2 de f ea t edA3 ,s oA2 de f ea t edA3 ,ands o f or t h.Wet henge tt ha t,f orany1 ≤i ≤n,Aide f ea t edAi+1and l o s tt o Ai-1 ( s t i l a t et ha tAn+1 = A1 ,A0 = An ).Wenow pu n -2 f or eAi-1 i n t o d i v i det hep l aye r sa f t e rAi andbe i r s— pa 2 eachcon s i s t so ftwoad ac en tp l aye r s.Thent he r ei sa tl ea s tone j l aye ri neach pa i rwhode f ea t edAi ,andt ha tmeans,be s i de s p n -2 Ai-1 , t he r ea r ea tl ea s t l aye r swhode f ea t edAi .ThenAi p 2 n n2 l o s ta tl ea s t game s.Sonp l aye r sl o s tt o t a l l tl ea s t game s. ya 2 2 182 Ma t hema t i c a lOl i adi nCh i na ymp 2 = Bu tt henumbe roft o t a lgame si sCn n( n -1) n2 < .Th i si sa 2 2 con t r ad i c t i on.The r e f or e,t he po s s i b l eva l ue so fn a r ea l lt he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. oddnumbe r sno tl e s st han3. S e c ondDay 00,Augus t2,2011 8: 00 12: 5 Gi venr ea lnumbe rα,p l ea s ef i nd t he mi n imum r ea lnumbe rλ = λ( α),s ucht ha tforanycomp l ex numbe r sz1 ,z2andr ea lnumbe rx ∈[ 0,1], i f|z1 |≤α|z1 -z2| t hen|z1 -xz2 |≤λ|z1 -z2|. ( s edbyLiShenghong) po So l u t i on.Asshowninthef i r e,i n gu F i .5 .1 g t hecomp l ex p l ane,po i n t s A ,B and C deno t et he comp l ex numbe r sz1 ,z2 andxz2 ,r e s t i ve l sobv i ou s l he pec y.C i y ont → → s egmen tOB .The Vec t or s BA and CA a r er epr e s en t ed by comp l exnumbe r sz1 -z2andz1 -xz2 , r e s t i ve l pec y.As|z1|≤ α|z1 -z2 |,wehave|O→ A |≤α|B→ A |.Then → → → |z1 -xz2 |max =|AC |max = max {|OA |,|BA |} = ma x {|z1 |,|z1 -z2 |} = ma x {α|z1 -z2 |,|z1 -z2 |} . The r e for e, λ( α)= max{ α,1}. 6 Ar et he r eanypo s i t i vei n t ege r sm ,ns ucht ha tm 20 +11ni s as r enumbe r? Proveyourconc l u s i on.( s edby Yuan qua po Hanhu i) Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 183 So l u t i on.As s umi ngt he r ea r epo s i t i vei n t ege r sm ,ns ucht ha t m 20 +11n =k2 wi t hk ∈ Z,wet henhave 11n =k2 - m 20 = ( k - m 10)( k + m 10), Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. wh i ch meanst ha tt he r ea r ei n t ege r sα, ucht ha t β ≥ 0s k - m 10 = 11α , { k + m 10 = 11β . ① ② I ti sobv i ou st ha tα <β.Sub t r ac t i ng ① f rom ② ,wehave 2m 10 = 11α ( 11β-α -1). Le tm = 11γm 1 ,whe r eγ,m 1 ∈ N* ,11 m 1 .Then 0 11β-α -1), 1110γ ·2m 1 1α ( 1 =1 0 wh i chimp l i e s10 γ =αand2m 1 1β-α -1. 1 =1 0 ≡ 1( ByFe rma t sLi t t l eTheor em,wehavem 1 mod11), 1 0 mod11).Bu t11β-α -1 ≡ 10(mod11),wh t hen2m 1 i chi s 1 ≡ 2( ac on t r ad i c t i on.Sot he p r ov ed c onc l u s i oni st ha tt he r ea r e no s i t i v ei n t e e r sm ,ns u cht ha tm20 +11ni sas a r enumbe r. po g qu 7 Suppo s en sma l lba l l shavebeen p l acedi n t on numbe r ed boxe sB1 ,B2 ,...,Bn .Eacht imewecans e l ec taboxBk anddot hef o l l owi ngope r a t i on s: ( 1)I fk = 1andt he r ei sa tl ea s toneba l li nB1 ,moveone ba l lf r omB1i n t oB2 . ( 2)I fk =nandt he r ei sa tl ea s toneba l li nBn ,moveone ba l lf r omBni n t oBn-1 . ( 3)I f2 ≤k ≤n -1andt he r ea r ea tl ea s ttwoba l l si nBk , moveoneba l lf r om Bk i n t o Bk+1 and oneba l li n t o Bk-1 ,r e s c t i ve l pe y. Pr ove t he fo l l owi ng: no ma t t e r how t he ba l l sa r e 184 Ma t hema t i c a lOl i adi nCh i na ymp d i s t r i bu t ed among t he boxe s or i i na l l ti sa lway s g y, i r ea l i zab l et ol e teachboxcon t a i nexac t l l lbyf i n i t e yoneba ope r a t i ons.( s edby Wang Xi nmao) po So l u t i on.Foranytwovec t or sx = ( x1 ,x2 ,...,xn )andy = Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( i ft he r eex i s t s1 ≤k ≤ ns ucht ha t y1 ,y2 ,...,yn ), x1 = y1 ,...,xk-1 = yk-1 ,xk > yk , x1 ,x2 ,...,xn )r wet hendeno t ei ta sx ≻ y.Le tx = ( epr e s en t t hed i s t r i bu t i ono ft heba l l samongt heboxe s.Thenxi sanon - t i onde f i nedi nt heque s t i on, nega t i vei n t ege rvec t or.Theope r a i fexe cu t ab l e,canbeexpr e s s eda sx +αk ,whe r eα1 = (-1,1, 0,...,0),αk = ( 0,...,0,1,-2,1,0,...,0)( 2 ≤k ≤ k-2 n -1),αn = ( 0,...,0,1,-1).Thenfork ≥ 2,wea lway s havex+αk ≻x.Sof oranyi n i t i a ld i s t r i bu t i onoft heba l l s,a f t e r af i n i t enumbe ro fope r a t i on soneve r ha tcon t a i ns k ≥ 2)t yBk ( a tl ea s ttwoba l l s,wecana r r i vea taba l ld i s t r i bu t i ony = ( y1 , s a t i s f i ngyk ≤1f ora l lk ≥2. I fa tt h i st imey2 = y2 ,...,yn ) y … =yn =1, t heprob l emi ss o l ved;o t he rwi s e,wehavey1 ≥2. st hesma l l e s tnumbe rs ucht ha tyi =0,wecant hen As s umi ngii doas e r i e sofope r a t i on sonB1 ,B2 ,...,Bi-1 : ( y1 ,1,...,1,0,yi+1 ,...,yn ) B1,B2,...,Bi-1 ( y1 ,1,...,1,0,1,yi+1 ,...,yn ) → B1,B2,...,Bi-2 ( y1 ,1,...,1,0,1,1,yi+1 ,...,yn )→ … → → B1 ( y1 ,0,1,...,1,yi+1 ,...,yn ) → ( y1 -1,1,...,1,yi+1 ,...,yn ) t o ge t( t i ng t he y1 - 1,1, ...,1,yi+1 , ...,yn ).Repea ope r a t i onsabove,wecanf i na l l r r i vea tt heba l ld i s t r i bu t i on ya ve c t ort ha tmee t st her equ i r emen t. Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 185 8 Ass howni nF i .8 .1,☉O i st hee s c r i bedc i r c l et ouch i ng g f△ABC a tpo i n tM ,andpo r eont he s i deBC o i n t sD ,E a s egmen t s AB , AC ,r e s t i ve l a t i s f i ng DE ‖BC ; pec y,s y st hei n s c r i bedc i r c l eo f△ADEt ☉O1i angen tt os i deDE a t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i n tN ;O1B ,DO i n t e r s e c ta tpo i n tF ,andO1C ,EO po i n t e r s ec ta tpo i n tG .P l ea s epr ovet ha tMN d i v i de ss egmen t FG equa l l s edbyBi anHongp i ng) y.( po F i .8 .1 g So l u t i on1.I fAB = AC ,t hent heg r aphi ss t r i cabou t ymme t heb i s ec t orof ∠BACandt heconc l u s i oni sobv i ou s.Sowemay a s s umet ha tAB > AC .Asshowni n Fi .8 .2,l e tL bet he g mi dpo i n tofBC , l i neO1Li n t e r s ec t i ng wi t hFG a tR ,andO1N beex t endedt oi n t e r s ec twi t hBC a tK .Dr awl i neAT t ha ti s rpend i cu l a rt o BC a tT andi n t e r s ec t s wi t hl i ne DE a tS. pe Conne c t i ng AO ,obv s on t he s egmen t AO .By i ou s l y O1 i Mene l au s t heor em,wehave F i .8 .2 g Ma t hema t i c a lOl i adi nCh i na ymp 186 O1F ·BD · AO O1G ·CE · AO = 1, = 1. FB DA OO1 GC EA OO1 ① O1F O1G , BD CE = = .So wh i ch S i nc eDE ‖BC ,wehave DA EA FB GC Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. FR BL = = 1.Th mean st ha tFG ‖BC .Then st he e r e f or e,Ri GR CL mi dpo i n to fFG .Wenowon l oprovet ha tM ,R ,N a r e yneedt co l l i nea r. For t h i s purpo s e, by t he i nve r s e of Mene l au s t heor em,weon l oprovet ha t yneedt O1R ·LM · KN = 1. RL MK NO1 ② O1R O1F OO1 · AD ( = = S i nc eFR ‖BL ,wehave t he RL FB AO DB s e cond equa l i t sj u s t i f i ed by ① ).So we on l o yi y need t ovet ha t pr OO1 ·AD ·LM · KN = 1. AO DB MK NO1 ③ S i nc eO1K ⊥ DE ,OM ⊥ BC ,AT ⊥ BC ,DE ‖BC ,t hen l i ne sO1K ,OM ,AT a r epa r a l l e l.Byt het heor em ofd i v i d i ng OO1 = t hes egmen t si n t opr opor t i ona lbypa r a l l e ll i ne s,wehav e AO MK .Sub s t i t u t i ngi ti n t o ③ ,wet henon l oprove yneedt MT AD ·LM · KN = 1. DB MT NO1 ④ S i nc e DE ‖BC , KN ⊥ DE ,ST ⊥ BC ,quadr i l a t e r a l KNST i sar e c t ang l eandt henKN = ST .Fu r t he rmor e,f rom AD AS = DS ‖BT wehave .Sub s t i t u t i ngt he s er e s u l t si n t o ④, DB ST wenowon l opr ove yneedt Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 187 NO1 LM = . MT AS ⑤ Le tBC = a,AC =b,AB =c.Wehave Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. BM = a +b -c ( a t hepr ope r t fane s c r i bedc i r c l e),BL = , yo 2 2 a2 +c2 -b2 a2 +c2 -b2 = BT =ccos∠ABC =c· . 2 2 ac a Then c -b LM BL -BM 2 a = = 2 = . MT BT -BM c -b2 +a( c -b) a +b +c a 2 Ont heo t he rhand, 2S△ADE NO1 AD + DE + AE DE a = = = . AS 2S△ADE AD + DE + AE a +b +c DE The r e f or e,⑤ ho l d s.Thepr oo fi scomp l e t e. So l u t i on2.Le tt he r ad i io f ☉O and ☉O1 ber andr1 , r e s c t i ve l i ou s l r eco l l i nea r. pe y.Obv y,O1 ,O andA a DE ∥ BC ⇒ AB AC = BD CE OF S△BOO1 = FD S△DBO1 OG S△OO1C = GE S△EOO1 üï ï ï A ö÷· 1 æç ABs i n OO1 ï 2ø 2è ï = ï OF 1 · OG r1 BD = ý⇒ 2 ï FD GE ï A ö÷· 1 æçACs i n OO1 ï 2ø 2è ï = 1 · ï r1 CE ï þ 2 ⇒FG ‖DE ‖BC . 188 Ma t hema t i c a lOl i adi nCh i na ymp Conne c tON andex t endi tt oi n t e r s e c t wi t h BC a tK .I f ∠ABC = ∠ACB , t henbys t r hepropo s i t i onho l d s.So ymme y,t nt hefo l l owi ng. wemaya s s umet ha t ∠ABC < ∠ACBi e s c t i ve l s ee Weconnec tOM ,O1M ,OB ,MD ,DO1 ,r pe y( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. F i .8 .3).S i nceO1N ‖OM ,wehave g F i .8 .3 g S△ONM = S△MOO1 = 1 · C -B , r OO1 ·s i n 2 2 ① C 1 i n BO ·OO1 ·s 2 OG OF S△BOO1 2 = = = 1· B GE DF S△BDO1 BD ·DO1 ·s i n 2 2 C r·OO1 ·s i n 2 = B r1 ·BD ·cos 2 ② B B ö÷ , æç r = BO ·cos ,r1 = DO1 ·s i n 2 2ø è S△DMN -S△MEN = = 1 NK ·DN - NE 2 B Cö æ 1 -r1c t o t ÷. ③ i nB · çr1co BD ·s 2 2ø è 2 Fr om ② and ③ ,wehave Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 189 OG S△DMN -S△MEN GE Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. C i n r·OO1 ·s B C æ ö 2 1 · -c o t ÷· o t = r1 BD ·s i nB · çc 2 2ø è B 2 r1 ·BD ·cos 2 =r·OO1 ·s i n =r·OO1 ·s i n B Cö B· C æo -c t o t ÷ s i n · çc 2 2ø 2 2 è C -B . 2 Comb i n i ngi twi t h ① ,wege t OG ( S△DMN -S△MEN )= 2S△MON . GE OG OG æçDF + EG ö÷ , S i nce t hen ∶2 = ∶ GE OE èOD OE ø æDF EG ö÷· OF · OG · + S△MND S△MEN = ç S△MON , èOD OE ø OD OE OF ·S△MND - DF ·S△MON OG ·S△MEN +EG ·S△MON = .④ OD OE Ont heo t he rhand, S△NMG = S△MEN ·OG +EG ·S△MON . OE OF ·S△MND - DF ·S△MON The r e f or e,S△NMG = . OD Int hes ameway, S△NMF = OF ·S△MND - DF ·S△MON . OD By ④ ,wehaveS△NMG = S△NMF .The r e f or e,MN d i v i de s s egmen tFG equa l l y. Thepr oofi scomp l e t e. Ma t hema t i c a lOl i adi nCh i na ymp 190 2012 ( Gu a n z h o u,G u a n d o n g g g) F i r s tDay Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 8: 00 12: 00,Augus t,10,2012 1 Le ta1 ,a2 , ...,an ben non -nega t i ve r ea l numbe r s. Provet ha t a1 a1a2…an-1 1 + +… + ≤ 1. ( ( ( 1+a1) 1+a1) 1+a1 ( 1+a2) 1+a2)…( 1+an ) ( s edby AiYi nghua) po So l u t i on.Le ta0 = 1.Wep rovet hefo l l owi ngi den t i t y: n k n aj-1 aj =1- ∏ ∑∏ k=1 j=1 1 +aj j=1 1 +aj ① byi nduc t i ononn. I ti sev i den tt ha t①i st r uef orn =1.Suppo s et ha t①i st r ue f orn -1,n ≥ 2, t henf orn, n k aj-1 ∑ ∏ 1 +a k=1 j=1 j = n-1 k n aj-1 ∑ ∏ 1 +a j k=1 j=1 =1=1- + n-1 aj-1 ∏ 1 +a j=1 j n aj aj-1 ö÷ æç -∏ ∏ + +aj ø èj 1 a 1 j =1 j=1 n aj ∏ 1 +a . j=1 j 2 Ass howni nt hef i r e,c i r c l e sΓ1 andΓ2 a r et angen t gu ex t e rna l l tpo i n tT .Po i n t sA andE a r eonc i r c l eΓ1 , ya l i ne sAB andDE a r et angen tt oc i r c l eΓ2 a tpo i n t sB and D ,r ta tpo i n tP .Prove e s c t i ve l ne sAE andBD mee pe y.Li t ha t Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 191 AB ED ; ( = 1) AT ET ( 2)∠ATP + ∠ETP = 180 °. ( s edby Xi ongBi n) po So l u t i on.Le tt heex t en s i onofAT mee tc i r c l eΓ2 a tpo i n tH , Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. andt heex t en s i ono fET mee t tpo i n tG .Theni c i r c l eΓ2a ti s ea s os ee t ha t AE ‖GH , yt t hu s △ATE △HTG , ∽ AT ET = con s equen t l .And y TH TG AH EG = f u r t he rwehave . TH TG Byt heCi r c l ePowe rTheor em,wehave F i .2 .1 g AB2 AT ·AH ET ·EG ED2 , = = 2 = · TH TH TG ·TG TH TG2 Henc e t he r e f or e, TH TG = AB ; ED AT HT AB = = . ET GT ED By t he S i ne Law, we have F i .2 .2 g AP s i n ∠ABP s i n ∠EDP EP = = = . AB s i n ∠APB s i n ∠EPD ED Henc e, AP AB AT , = = EP ED ET t ha ti s,PT i st he b i s e c t oro ft heex t e r i orang l e of ∠ATE . The r e f or e, Ma t hema t i c a lOl i adi nCh i na ymp 192 ∠ATP + ∠ETP = 1 80 °. 3 F i nd a l lt he pa i r so fi n t ege r s( a,b)s a t i s f i ng t he y fo l l owi ngcond i t i on:t he r eex i s t sani n t ege rd ≥ 2s uch Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t ha tan +bn +1i sd i v i s i b l ebydforanypo s i t i vei n t ege rn. ( s edbyChenYonggao) po So l u t i on.I fa +bi sodd,t henan +bn +1i seven,t hu sd =2. I fa +bi seven,t hena +b +1i sodd. sodd,s odi n n 2 2 2 =a +b +1 +2( S i nced|a +b +1,( a +b +1) a +b + ab -1),andd|a2 +b2 +1, t hen,d|2( ab -1), s owe 1)+2( haved|ab -1. Wes eet ha ta3 +b3 + 1 ≡ ( a +b)( a2 +b2 -ab)+ 1 ≡ (-1)(-1-1)+1 ≡3( modd), andd|a3 +b3 +1,henced|3, s od = 3. S i nce 2 2 2 ( = a +b -2 a -b) ab ≡ -1 -2 ≡ 0( mod3), wehavea ≡b( mod3).Thu s, 0 ≡a +b +1 ≡2 a +1( mod3),we mod3). s eet ha ta ≡ 1( The r e f or e,a ≡ b ≡ 1( mod3).Cons equen t l y,for any s i t i vei n t ege rn,wehave po an +bn +1 ≡ 1 +1 +1 ≡ 0( mod3). Summi ngup,t her equ i r edi n t ege rpa i r sa r eoft heforms: ( 2 r ekandla 2 k,2 l +1),( k +1,2 l),( 3 k +1,3 l +1),whe r e i n t ege r s. 4 The r ei sas t one ( o fgog ame)a te a chv e r t e xo fag i v enr e l a r gu 13 hec o l o ro fe a chs t onei sb l a cko rwh i t e.Pr ov e gon,andt t ha twema x chang et hepo s i t i ono ftwos t one ss u cht ha tt he ye Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 193 c o l o r i ngo ft he s es t one sa r es t r i c wi t hr e s c tt os ome ymme pe s t r i ca x i so ft he13 s edbyFuYunhao) ymme gon.( po So l u t i on.Takeanyver s s i ng t exA andt hes t r i cax i slpa ymme i t.The r ea r es i xpa i r so fve r t i ce ss t r i ct ol. I ft heco l orof ymme Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s t one sa teachpa i ro fve r t i c e si st hes ame,t hent heco l or i ngi s s t r i ct ol. ymme I ft he r ei son l i ro fs t r i cve r t i c e st ha tha s y one pa ymme d i f f e r en tco l ors t one s,t henexchanget hes t onea toneve r t exof t hepa i ri nd i f f e r en tco l ort oA wi t ht hes t oneonA . I ft he r ea r eexac t l i r sofve r t i ce si nd i f f e r en tco l or s, ytwopa t henexchanget hewh i t es t onei nve r t exo fapa i rwi t ht heb l ack s t onei nve r t exo ft heo t he rpa i r. Nows uppo s et ha tf oranyve r t exA andt hes t r i cax i s ymme s s i ngA , t he r ea r ea tl ea s tt hr eepa i r sofs t r i cve r t i ce si n pa ymme d i f f e r en tco l or s.Wes ha l ls howt h i si simpo s s i b l e. I ft he r ea r exb l acks t one sandy wh i t es t one s,t henx +y = 13 ;wi t hou tl o s sofgene r a l i t e tx beoddandy beeven. y,l I ft hes t oneonAi si nb l ack,t hent her ema i n i ngs t one sa r e eveni nb l ackand wh i t e,r e s c t i ve l he r ea r eeven pe y.Hence,t i r so fve r t i ce si nd i f f e r en tco l or s,t ha ti s,t he r ea r ea tl ea s t pa fou rpa i r sofve r t i c e si nd i f f e r en tco l or s. S imi l a r l ft hes t oneonA i si n wh i t e,t hent he r ea r ea t y,i l ea s tt hr eepa i r so fs t r i cve r t i ce si nd i f f e r en tco l or s. ymme S i nc eeachpa i ro fve r t i c e si ss t r i ct ooneax i s,wes ee ymme t ha tt henumbe ro fpa i r si nd i f f e r en tco l or si sa tl ea s t4x +3y. Ont heo t he rhand,t henumbe rofpa i r si nd i f f e r en tco l or s t hu s i sexac t l yxy, xy ≥ 4x +3y = x +39, t ha ti s,x( ti tcon t r ad i c t st o y -1)≥ 39,bu Ma t hema t i c a lOl i adi nCh i na ymp 194 2 æx + ( y -1)ö÷ =3 x( 6. y -1)≤ ç 2 è ø S e c ondDay Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 00,Augus t,11,2012 8: 00 12: 5 As s hown i n F i .5 .1, t he g i n s c r i bedc i r c l e☉I o f△ABC i st angen tt ot hes i de sAB and AC a t po i n t s D and E , st he r e s c t i ve l pe y. And O i c i r cumc en t r eo f△BCI.Pr ove t ha t ∠ODB = ∠OEC .( s ed po byZhuHuawe i) So l u t i on.S st hec i r c umc en t r e i n c eOi o f△BCI,wes eet ha t F i .5 .1 g ∠BOI = 2∠BCI = ∠BCA . Int hes amemanne r,∠COI = ∠CBA .Hence, ∠BOC = ∠BOI + ∠COI = ∠BCA + ∠CBA = π - ∠BAC . Thu s,fou rpo i n t sA ,B ,O andC a r econcyc l i c.ByOB = OC ,weknowt I tcana l s obes eenf rom ha t ∠BAO = ∠CAO .( t hewe l l -knownconc l u s i ont ha tpo i n t ︵ Oi st he mi dpo i n tof a r c BC ( no t i r cumc i r c l eo f con t a i npo i n tA )onc △ABC .) Comb i ned wi t ht hef ac t t ha tAD = AE ,AO = AO ,wehave △OAD ≅ △OAE .Hence, ∠ODA = ∠OEA ,t he r e f or e, ∠ODB ∠OEC . = F i .5 .2 g Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 6 195 The r ea r enc i t i e s( n > 3)andtwoa i r l i necompan i e si na coun t r tweenanytwoc i t i e s,t he r ei sexac t l y.Be yone2 wayf l i tconne c t i ngt hem wh i chi sope r a t edby oneof gh Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t hetwo compan i e s. A f ema l e ma t hema t i c i an p l ans a t r ave lr ou t e,s ot ha ti ts t a r t sandend sa tt hes amec i t y, s s e st hr ougha tl ea s ttwoo t he rc i t i e s,andeachc i t pa yon t her ou t ei sv i s i t edonce.Shef i nd sou tt ha twhe r eve rshe s t a r t sand wha t eve rr ou t es he choo s e s,s he mu s tt ake f l i t so fbo t hcompan i e s.F i ndt hemax imumva l ueofn. gh ( s edbyLi ang Yi ngde) po So l u t i on.Cons i de reachc i t save r t exandeacha i r l i nea san ya edgei naco l orcor r e s i ngt ot hea i r l i necompanyi tbe l ong s. pond Thent hea i r l i ner ou t echa r toft h i scoun t r ega rdeda sa ycanber two co l orcomp l e t eg r aph wi t h n ve r t i ce s.By t he prob l em s t a t e s,any c i r c l econ t a i n sedge so fbo t hco l or s.Tha ti s,t he s ubg r apho feachco l orha snoc i r c l e.I ti swe l l -knownt ha tf or t hes imp l eg r aphwi t hou tc i r c l e,t henumbe ro fedge si sl e s st han t henumbe ro fve r t i c e s.Thu s,t henumbe ro fedge soft hes ame co l ori snomor et hann -1, t ha ti s,t het o t a lnumbe ro fedge si s nomor et han2( n - 1).Ont heo t he rhand,t henumbe ro fa /2.Hence, comp l e t eg r aphwi t hnve r t i c e si sn( n -1) n( n -1)≤ 2( n -1), t ha ti s, n ≤ 4. I fn = 4,deno tt hr ee t ef ou rc i t i e sbyA ,B ,C andD .Le rou t e sAB ,BCandCD beope r a t edbyt hef i r s tcompanyandt he o t he rt hr eerou t e sAC ,AD andBD byt hes e condcompany.We s eet ha tt herou t echa r to feachcompanyha snoc i r c l e.So,t he max imumnumbe ro fni s4. 7 Le ta1 ≤ a2 ≤ … beas equenc eofpo s i t i vei n t ege r ss uch t ha tr/ ar =k +1fors omepo s i t i vei n t ege r skandr.Prove Ma t hema t i c a lOl i adi nCh i na ymp 196 t ha tt he r eex i s t sapo s i t i vei n t ege rss ucht ha ts/ as = k. ( s edbyJ ac ekFabr i,USA) po ykowsk So l u t i on.Le t)=t -kat .Theng( r)=r -kar tg( =ar > 0. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. No t et ha tg( 1)= 1 -ka1 ≤ 0.Sot hes e t{ t|t = 1,2,...,r, sno temp t ts bet t)≤ 0}i he max ima le l emen toft he g( y.Le s e t;t hens <r.Henc e,g( s +1)> 0.Ont heo t he rhand, s +1)=s +1 -kas+1 ≤s +1 -kas = g( s)+1 ≤ 1. ① g( Thu s,0 < g( s +1)≤ 1.Cons s +1)= 1.And equen t l y,g( s +1)≤g( s)+1 ≤1.Wehaveg( s)=0, by ① ,1 =g( t ha ti s, s/as =k. 8 F i ndt henumbe ro fi n t ege r ski nt hes e t{ 0,1,2,..., æ2012ö 2012}s uch t ha tt he comb i na t i on numbe r ç ÷ = è k ø 2012! i sa mu l t i l eo f2012. ( s ed by Wang p po !( k 2012 -k)! Bi n) So l u t i on.Fac t or i z i ng2012 = 4 ×503,wes eet ha tp ime.I fki sno tamu l t i l eo fp, t hen pr p =5 03i sa æ4p -1ö æ2012ö æ4p ö ( 4p)! 4p = ×ç ç ÷= ç ÷= ! ( ÷. ) ! k èk -1 ø è k ø è k ø k × 4p -k æ2012ö 2012! Hence ç i samu l t i l eo fp. ÷ = !( p è k ø k 2012 -k)! I fki samu l t i l eo fp, t he r ea r eon l i veca s e s: k = 0,p, p yf 2p,3p,4p.And wes eet ha ti na l lca s e s,t hecomb i na t i on numbe r s æ4p ö æ4p ö ç ÷= ç ÷ = 1, è0ø è4p ø æ4p ö æ4p ö 4( 3p +1)( 3p +2)…( 3p + ( p -1)) ≡ 4( modp), ç ÷= ç ÷= ( ) ! p -1 èp ø è3p ø Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 197 and 2p +2)…( 2p + ( 6[( 2p +1)( p -1))] æ4p ö [( 2p + ( 2p + ( 2p + ( 2p -1))] p +1))( p +2))…( ç ÷= ( 2p -1)] p -1)![( p +1)( p +2)…( è2p ø Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ≡ 6( modp) a r eno tmu l t i l eofp. p Now wedeno t et heb i na r r a lofnon -nega t i vei n t ege r ynume nby n=( arar-1 …a0) 2 = r j ∑ aj2 ,ands(n)= j=0 whe r eaj = 0or1f orj = 0,1,...,r. r ∑a , j=0 j Thent hepowe rm o ff ac t or2mi nf ac t or i za t i onofn ! canbe expr e s s edby n n n + +… + m +… 2 4 2 =( arar-1 …a2a1) arar-1 …a3a2) 2 +( 2 + … +ar = ar × ( 2 -1)+ar-1 × ( 2- -1)+ … +a1 × r r 1 ( 21 -1)+a0 × ( 20 -1) = n -s( n). æ2012ö 2012! = I ft hecomb i na t i onnumbe rç ÷= ! ( è k ø k × 2012 -k)! ( k + m )! i sodd ( m =2012 -k), t heni tmeanst ha tt hepowe r s k! × m ! of2i nf ac t or so ft henume r a t orandt hedenomi na t oroft he f r ac t i onabovea r et hes ame.Then k + m -s( k + m )=k -s( k)+ m -s( m ), or s( k + m )=s( k)+s( m ). Ma t hema t i c a lOl i adi nCh i na ymp 198 Th i smean st ha tt heb i na r i t i ono fk +m =2012ha sno yadd ca r r i ng. y S i nc e2012 = ( 11111011100) i tcons i s t sofe i t1 sand 2, gh t hr ee0 s.I ft he r ei snoca r r i ngi nt headd i t i ono fk + m = y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 11111011100) t henont heb i tof0i n( 11111011100) k,m 2, 2, a r e0; ont heb i to f1,onei s0,t heo t he ri s1,s ot he r ea r etwo cho i c e s:1 =1 +0and1 =0 +1.Thu s,t he r ea r e28 =256ca s e s t ha tt heb i na r i t i ono ftwonon -nega t i vei n t ege r sk + m = yadd 2012 ha s no ca r r i ng.Tha ti s,t he r ea r e 256 comb i na t i on y numbe r st ha ta r eodd,andt her ema i n i ng2013 - 256 = 1757 comb i na t i onnumbe r sa r eeven. I ft hecomb i na t i onnumbe r æ2012ö ( k + m )! 2012! = ç ÷= ! ( ) ! k! × m ! è k ø k × 2012 -k i sevenbu tno tamu l t i l eo f4,t heni tmeanst ha tt hepowe rof p 2i nt henume r a t ori sg r ea t e rt hant ha ti nt hedenomi na t orby1. Thu s, k + m -s( k + m )=k -s( k)+ m -s( m )-1, or s( k + m )=s( k)+s( m )-1. Tha ti s,t he r ei son l r r i ngi nt heb i na r i t i onof yoneca y yadd k +m =2012.Theca r r i nghappen sa ttwob i t sa s01+01 =10. y By k + m = 2012 = ( 11111011100) 2, wes eet ha tt heca r r i ngcanon l tt hef i f t h( f romt he y yhappena h i s tb i tt ot hel owe s tb i t)ands i x t hb i t sora tt hen i n t hand ghe 7 i s t2 = 128ca s e s.Tha ti s,t he r ea r e256 t en t hb i t s.Sot he r eex Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 199 comb i na t i on s who s e va l ue sa r e even numbe r s bu t no tt he mu l t i l e so f4. p Thu s,t he r ea r e2013 - 256 - 256 = 1501comb i na t i ons who s eva l ue sa r e mu l t i l e so f4.Now,wegobackt ocons i de r p Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t heca s e swhe r eki sno tt hemu l t i l eofp = 503. p æ4p ö æ4p ö Wes eet ha tç sno ta mu l t i l eof4.For ÷= ç ÷ = 1i p è0ø è4p ø æ4p ö æ4p ö ( 4p)! he powe rof2i ss( 3p)ç ÷= ç ÷ = !( )! ,t p)+s( 3p p èp ø è3p ø æ4p ö ( 4p)! s( 4p)=s( 3p)=7,andf or ç t hepowe rof ÷ = ( )!( )! , 2 p 2p è2p ø 2i ss( 2p)+s( 2p)-s( 4p)=s( s,t he r ea r et hr ee p)= 8.Thu comb i na t i onnumbe r swh i cha r emu l t i l e sof4bu tno tmu l t i l e s p p o fp = 503, t ha ti s,t henumbe rofks ucht ha tt hecomb i na t i ons æ2012ö r emu l t i l e sof2012i s1501 -3 = 1498. ç ÷a p è k ø 2013 ( Z h e n h a i,Z h e i a n j g) F i r s tDay ( 8: 00 12: 00Augus t12,2013) 1 hec l o s ed doma i n ont he p l anede l imi t ed by Le tA bet t hr eel i ne sx = 1,y = 0,andy =t( 2x -t),whe r e0 < t < 1.Provet ha tt hes u r f aceo fanyt r i ang l ei n s i det he doma i nA wi t hP ( t, t2)andQ ( 1,0)a stwoo fi t sve r t i c e s canno texc eed 1 . 4 200 Ma t hema t i c a lOl i adi nCh i na ymp So l u t i on. I t i s ea s o y t ob s e r vet ha tt hedoma i ni sa c l o s ed t r i ang l e. I t s t hr ee Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. æt ö ve r t i c e sa r e B ç ,0÷ ,Q ( 1, è2 ø 0)andC( 1, t( 2-t)).P i cka i n tX i n s i de t he △BQC , po t hen t he a r ea o f △PQX i s equa lt oha l fo ft heproduc to f F i .1 .1 g PQ wi t ht hed i s t anc ef r om X t oPQ .Sot hea r ea o fPQX t ake si t smax imum va l ue whent hed i s t anc ef rom X t o PQ i s max imi zed,i. e.,whenX co i nc i de swi t hB orC . Thea r eao f△PQBi s 1 æç1 - t ö÷ 2 1 1 t = ( 2 -t) t2 ≤ ( 2 -t) t 2ø 2è 4 4 ≤ 1 æç2 -t +tö÷ 1; = 2 ø 4è 4 2 t hea r eao f△PQCi s 1( 1 ( 1 -t)( 2 t -t2)= 2 t 1 -t)( 2 -t) 2 4 ≤ t +1 -t +2 -tö÷ 1 æç2 1 = . 3 ø 4è 4 3 Henc e,i nt hedoma i nA ,anyt r i ang l ewi t hP ,Q a stwoof i t sve r t i c e scanno thaveana r eat ha texc eed s 2 1 . 4 Ass howni nF i .2 .1,Inat r apezo i dABCD ,AB ‖CD , g ☉O1i st angen tt ot hes egmen t sDA ,AB ,BC ,☉O2 i s tP bet he t angen tt ot hes egmen t sBC ,CD ,DA .Le Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 201 t angen tpo i n to f ☉O1 wi t h AB ,and Q bet het angen t t h CD .Pr i n to f ☉O2 wi ovet ha tAC ,BD ,PQ a r e po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. concu r r en t. F i .2 .1 g F i .2 .2 g So l u t i on.Le hei n t e r s e c t i onofl i ne sAC ,BD ,andj o i n tR bet O1A ,O1B ,O1P ,O2C ,O2D ,O2Q ,PR ,QR .Asshown i nF i .2 .2. g AsBA andBC a r et angen tl i ne sof☉O1 , ∠PBO1 = ∠CBO1 = 1∠ ABC . 2 1∠ S imi l a r l BCD . y,wehave ∠QCO2 = 2 Fr om AB ‖CD ,weknowt °. ha t ∠ABC + ∠BCD = 180 The r e f or e,∠PBO1 + ∠QCO2 = 90 °,anda sRt△O1BP and O1P CQ = Rt△CO2Q a r es imi l a r,wehave . BP O2Q O2Q AP = S imi l a r l .By mu l t i l i ngt he s etwo y,wehave p y O1P DQ AP CQ , AP = = i den t i t i e s,weob t a i n wh i chi nt u r nimp l i e s BP DQ AP +BP CQ AP CQ , = i. e., . CQ + DQ AB CD r e Aga i nbyAB ‖CD ,weknowt ha t△ABR and △CDR a Ma t hema t i c a lOl i adi nCh i na ymp 202 AR CR AP CQ , AR = = = weha v e s imi l a r,s o .Compa r i ngwi t h AB CD AB CD AP CR .Meanwh i l e, △PAR i ss imi l a rt o △QCR a s ∠PAR = CQ ∠QCR .Thu s ∠PRA = ∠QRC .So P ,R ,Q a r eco l l i nea r. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. The r e for e,AC ,BD ,PQ a r econcu r r en t. 3 Inag r oupo fm g s,anytwooft heme i t he r i r l sandn boy know eacho t he r,ordono tknow eacho t he r.Forany twoboy sandtwog i r l s,a tl ea s toneboyandoneg i r ldo no tknoweacho t he r.Pr ovet ha tt henumbe rofboy i r l g n( n -1) i r st ha tknoweacho t he ri sa tmo s tm + . pa 2 So l u t i on.Fromthehypothes i s,f oranytwoboy s,t he r ei sa t henumbe rof mo s toneg i r lt ha tknowsbo t ho ft hem.Le txibet i r l st ha tknowexac t l s,1 ≤i ≤n.So ∑i=1xi = m .By g yiboy n coun t i ng t he numbe r o f t he above two boy s one g i r l comb i na t i on s,wehave i( i -1) n( n -1) xi ≤ . 2 2 ∑ i≥2 Thenumbe ro fboy i r lpa i r st ha tknoweacho t he ri st hen g i( i -1) ixi = m + ∑ ( i -1) xi ≤ m + ∑ xi ∑ 2 i=1 i=2 i=2 n n n n( n -1) . 2 ≤m + 4 F i ndt henumbe ro fpo l a l sf( x) = ax3 +bx t ha t ynomi s a t i s f hefo l l owi ngcond i t i onsbe l ow: yt ( 1,2,...,2013}; 1)a,b ∈ { ( 2) t he d i f f e r enc eo fany two numbe r samong f( 1), Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 2),...,f( 2013) i sno tamu l t i l eof2013. f( p So l u t i on.2013i sf ac t or i zeda s2013 =3×11×61.Le tp1 203 =3, t ebyait her e s i dueofa modu l opi , p2 = 11,p3 = 61.Wedeno Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. bybi t her e s i dueo fb modu l opi ( i = 1,2,3),a,b ∈ { 1, 2,...,2013}.Byt heCh i ne s eRemi nde rTheor em,wehavea b i e c t i ono f( a,b)wi a1 ,a2 ,a3 ,b1 ,b2 ,b3). t h( j Now,l e tfi( l la x) = aix3 + bix,i = 1,2,3.Weca l a l “good modu l o n ”i ft he r e s i due s of f( 0), po ynomi n -1)modu l ona r ea l ld i s t i nc t. 1),...,f( f( I ff( x)= ax3 +bxi sno tgood modu l o2013,t hent he r e ex i s t sx1 ≢ x2 ( mod 2013)s uch t ha tf( x1) ≡ f( x2)( mod modpi).Le her e s i due s 2013).Suppo s ex1 ≢ x2( tu1 andu2bet ofx1 andx2 modu l opi ,r e s c t i v e l modpi)and pe y.Thenu1 ≢ u2( modpi), u1)≡ fi( u2)( x) s ofi( i sno tgoodmodu l opi . fi( I ff( x)=ax3 +bxi sgoodmodu l o2013,t henforeve r yi, x)i sgood modu l opi .Ther ea s oni sa sfo l l ows.Forany fi ( d i s t i nc tpa i rr1 ,r2 ∈ { 0,1,...,pi -1}, t he r eex i s tx1 ,x2 ∈ { 1,2,...,2013} s ucht ha tx1 ≡r1 ,x2 ≡r2( modpi)andx1 ≡ 2013ö÷ 2013ö÷ , æ æ x2 ç mod .Nowf ( x1)≡ f( x2)ç mod bu tf( x1) ≢ è è pi ø pi ø x2)( mod2013),s of( r1)≢ f ( r2)( modpi). f( Henc e, we need t o de t e rmi ne t he numbe r of good l a l sfi( x)modu l opi . po ynomi Forp1 = 3,byFe rma t st heor em,agoodpo l a l ynomi x)≡ a1x +b1x ≡ ( a1 +b1) x( mod3) f1( i sequ i va l en tt os ayt ha ta1 +b1i sno td i v i s i b l eby3.The r ea r ei n t o t a ls i xs uchf1( x). x) Fori =2,3, i ffi( i sgoodmodu l opi , t henf oranyuand v ≢ 0( modpi),fi( u +v)≢ fi( u -v)( modpi), i. e., Ma t hema t i c a lOl i adi nCh i na ymp 204 u +v)-fi( u -v)= 2 v[ ai( 3u2 +v2)+bi] fi( i sno td i v i s i b l e bypi .I fai ≠ 0,t her e s i due s modu l opi of pi -1 e l emen t si nt hes e t sA = 3 and aiu2 |u = 0,1,..., 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. { } -1 pi B = (-bi -aiv2)|v = 1,2,..., 2 { } do notcoincide, and|A|+|B|=pi .SoA ∪Bf ormsacomp l e t er e s i dues s t em y i rs um mu s t be a mu l t i l eo f pi ,i. e., modu l o pi . The p pi-1 2 pi-1 2 2 2 ∑u=03aiu + ∑v=1 (-bi -aiv )≡ 0(modpi). æpi -1ö÷ 1·pi -1·pi +1· = s Now,12 +22 + … + ç pii 2 2 6 è 2 ø 2 pi -1 ·bii amu l t i l eo fpi , sa l s oamu l t i l eo fpi .Hence, s op p 2 bii sd i v i s i b l ebypi , i. e.,exac t l s0. yoneofai ,bii x) = bix i sobv i ou s l I fai = 0,bi ≠ 0,t henfi ( ygood. The r ea r epi -1s uchgoodpo l a l s. ynomi henfi( x) = aix3 .Forp2 = 11 ,by I fai ≠ 0,bi = 0,t 7 21 =x ≡ x( x3) mod11),s oforx1 ≠ Fe rma t st heor em,( 3 x2( mod11)andx3 mod11),f2( x)=a2x3i sgood.The r e 1 ≠x2 ( a r ei nt o t a l10s uchpo l a l s. ynomi s43 =64 ≡125 =53( mod61),f3( Forp3 =61,a x)=a3x3 canno tbegood. The r e for e,t het o t a lnumbe rt ha twea r el ook i ngf ori s6 × ( 10 +10)×60 = 7200. S e c ondDay 8: 00 12: 00,Augus t13,2013 5 Gi venpo s i t i ver ea lnumbe r sa1 ,a2 ,...,an .Provet ha t Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 205 t he r eex i s tpo s i t i ver ea lnumbe r sx1 ,x2 ,...,xns ucht ha t ∑ x = 1,andt ha tf oranypo s i t i ver ea lnumbe r sy1 , n i=1 i a t i s f i ng ∑i=1yi = 1,oneha s y2 ,...,yn s y n Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. n n aixi 1 ≥ ai . ∑ ∑ 2i i=1 xi +yi =1 So l u t i on.Le txi = n ∑ n i=1 i a aixi = i +yi ∑x i=1 n ai .Then ∑xi = 1.Mor eove r, i=1 n n xi2 . i +yi ∑a ∑x i=1 i i=1 Foranypo s i t i venumbe r sy1 ,y2 ,...,yn wi t h ∑i=1yi = n 1.Byt heCauchy Schwa r zInequa l i t y, n xi2 = i=1 xi +yi 2∑ Henc e, n n xi2 ≥ i=1 xi +yi i=1 aixi = ∑ x i +yi i=1 6 n ∑ (xi +yi)∑ n ( ∑x ) n i=1 i 2 = 1. n xi2 1 ai . i ≥ ∑ + 2i x i y i=1 =1 ∑ai ∑ i=1 n Le tS beas ub s e tofm e l emen t sof{ 0,1,2,...,98},m ≥ 3, s ucht ha tf oranyx,y ∈S, t he r eex i s t sz ∈S wi t hx +y ≡ 2 z( mod99).F i nda l lpo s s i b l eva l ue sofm . So l u t i on.Le tS = { s1 , s2 ,..., sm }.AsS '={ 0, s2 -s1 ,..., sm -s1}s a t i s f i e sa l s ot hehypo t he s i s,we maya s s ume wi t hou t l o s so fgene r a l i t ha t0 ∈ S.Foranyx,y ∈ S,50( x +y)≡ yt z( ak i ngy = 0,wehavet ha tforanyx ∈ S, mod99)∈ S.Byt 50x ∈ S.As50and99a r ecopr ime,t he r eex i s t sa po s i t i ve mod99).Hence, i n t ege rks ucht ha t50k ≡ 1 ( x +y ≡ 50k ( x +y)( mod99)∈ S. Ma t hema t i c a lOl i adi nCh i na ymp 206 Le td = gcd( 99,s1 ,s2 , ...,sm ).Theabovea r t gumen t henimp l i e st ha td ∈ S,t he r e for eS = { 0,d,2d,...}.For any po s i t i ve f ac t or d < 99 of 99,t h i sS s a t i s f i e sa l lt he r equ i r emen t s.Henc e,a l lt hepo s s i b l eva l ue sofm a r e3,9,11, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 33,99. 7 As s hown i n F i .7 .1, ☉O1 and ☉O2 a r et angen t g ex t e rna l l t po i n t T . The quadr i l a t e r a l ABCD i s ya i ne sDA andCB a r et angen tt o i n s c r i bedi n ☉O1 .Thel tpo i n t sE andF ,r ☉O2 a e s c t i ve l heb i s ec tof pe y.BN ,t ∠ABF , i n t e r s ec t st hes egmen tEF a tpo i n tN .Li neFT i ch doe sno tcon t a i n B )a t i n t e r s e c t st he a r c AT (wh i n tM .Pr ovet ha tM i st heexocen t e rof△BCN . po F i .7 .1 g F i .7 .2 g So l u t i on.Le tP bet hei n t e r s e c t i onofl i neAM wi t hEF .Jo i n AT ,BM ,BP ,BT ,CM ,CT ,ET ,TP .Asshowni nFi .7 .2. g AsBFi st angen tt o☉O2 a tF ,wehave ∠BFT = ∠FET . As☉O1i st angen tt o ☉O2 a tT ,wehave ∠MBT = ∠FET . Henc e,∠MBT = ∠BFM .So△MBT and△MFB a r es imi l a r, t he r e f or eMB2 = MT ·MF .Thes amea r tg i ve sMC2 = gumen MT ·MF . Nowaga i nf romt ha t☉O1i st angen tt o☉O2a tT ,wehave ∠MAT = ∠FET .S o A ,E ,P and T a r econcyc l i c,wh i ch Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 207 imp l i e st ha t ∠APT = ∠AET .AsAEi st angen tt o☉O2 a tE , wehave ∠AET = ∠EFT .Thu s, ∠MPT = ∠PFM ,and ss imi l a rt o△MF .The r e f or e,MP2 = MT ·MF . △MPTi Fr omt heabovea r t,wehave MC = MB = MP , gumen Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. wh i ch meanst ha tM i st heexc en t e rof △BCP .So ∠FBP = 1∠ CMP .Meanwh i l e, ∠CMP = ∠CDA = ∠ABF ,and we 2 have ∠FBN = 1∠ e.,P ABF .Hence,∠FBN = ∠FBP ,i. 2 andN co i nc i de. 8 Le tn ≥ 4beanevennumbe r.Att heve r t i ce sofar egu l a r n i t ei n an a rb i t r a r i s t i nc tr ea l gon we wr y way n d numbe r s.S t a r t i ngf r om oneedge,wenamea l lt heedge s sca l l ed i nac l ockwi s ewaybye1 ,e2 ,...,en .Anedgei “po s i t i ve”,i ft he d i f f e r enc e of t he numbe r sa ti t s endpo i n tandi t ss t a r tpo i n ti spo s i t i ve.As e toftwoedge s { ei , ej } i +j),andamongt i sca l l ed “ c r o s s i ng”,i f2|( he fou r,t henumbe r s wr i t t ena tt he i rve r t i ce s,t hel a r s t ge andt het h i rdl a r s tone sbe l ongt ot hes ameedge.Prove ge t ha tt henumbe rofc r o s s i ng sandt henumbe rofpo s i t i ve edge shaved i f f e r en tpa r i t y. So l u t i on1.Wi t hou tl o s so fgene r a l i t s s umet ha t y,we maya t henumbe r swr i t t enont heve r t i c e sa r e1,2,...,n.Le tA be t henumbe ro fc r o s s i ng s,andB bet henumbe ro fpo s i t i veedge s. We wi l l pr ovet ha tt he pa r i t fS r ema i nst hes amei f we yo exchangenumbe r siandi +1.Wed i s t i ngu i shtwoca s e s. CaseI.The number si andi + 1a r e wr i t t en on ad ac en t j ve r t i ce s,i. e.,t heendpo i n t so fedgeek .Once weexchangei andi +1,t henumbe ro fpo s i t i veedge si smod i f i edby1,t hu s Ma t hema t i c a lOl i adi nCh i na ymp 208 t hepa r i t fB i schanged.Ont heo t he rhand,t heon l yo ytwo edges ub s e tt ha twi l lbe comeanewc ro s s i ng ( orchangef r om a ek-1 ,ek+1}( c ro s s i ngt oanon c r o s s i ng)i s{ t hes ub i nd i ce sa r et o beunde r s t ood modu l on ),a l lt heo t he rtwo edges ub s e t swi l l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. no tbea f f e c t ed.Sot hepa r i t fA wi l lchange. yo CaseI I.Thenumber siandi +1a r ewr i t t enonnon ad ac en t j ve r t i c e s. As s ume t ha tt hey a r e wr i t t en on t he ( common) endpo i n t so fej ,ej+1 and ofek ,ek+1 ,r e s t i ve l e we pec y.Onc exchangeiandi+1, eve r s i t i veedgewi l lr ema i npo s i t i ve,s o ypo a r enon -po s i t i veone s.The r e f or e,B i sunchanged.Now,f or numbe rofc ro s s i ng s, i fatwo edges ub s e tdoe sno ti nvo l vea tt he s amet imeiandi +1, t henwhe t he ri ti sac ro s s i ngorno ti sno t a f f e c t edbyt heope r a t i on.Sot heon l edges ub s e t st obe ytwo con s i de r eda r et hetwot ha thavebo t hiandi+1wr i t t enont he i r ve r t i c e sandt hes umo ft heedgenumbe ri seven.Theywi l lbo t h rt heexchangei ft heya r eno tbe for e,and be comec r o s s i nga f t e v i c e ve r s a.Henc e,t hepa r i t fA r ema i nst hes ame. yo Nowobv i ou s l r t t e rncanbeob t a i nedf romaf i n i t e y,eve ypa numbe ro fs uchexchangei fwes t a r tf rom wr i t i ng1,2,...,n con s e cu t i ve l nac l ockwi s e way,andi nt hei n i t i a ls i t ua t i on, yi B = n -1,A = 0.SoA andB haved i f f e r en tpa r i t y. So l u t i on2.St a r t i ngf rom oneve r t ex,wedeno t et henumbe r s wr i t t enont heve r t i c e sbyx1 ,x2 ,...,xn i nac l ockwi s eway. Wemaya s s umet ha tt henumbe r swr i t t enont heendpo i n t sof r exi ,xi+1 ,i = 1,2, ...,n,whe r exn+1 = x1 . edgeei a Appa r en t l spo s i t i vei fandon l fxi+1 -xi >0.Now,l e tA yeii yi , bet he numbe r of po s i t i ve edge s and B bet he numbe r of c r o s s i ng s.Wr i t e n β= ∏ (x i=1 i+1 -xi ) . Ch i naGi r l s Ma t hema t i c a lOl i ad ymp 209 A Asni seven,t hes i fβi sj u s t(-1) . gno Ont heo t he rhand,{ ei ,ej }i sac ro s s i ngi fandon l f2| yi i +jand Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( xj -xi)( xj -xi+1)( xj+1 -xi)( xj+1 -xi+1) < 0. hel e f t -hands i deo fabovei nequa l i t Wedeno t ebyf( ei ,ej )t y, obv i ou s l ei , ej )=f( ej , ei).Th i squan t i t snega t i vei fand yf( yi on l ft hetwo edges e ti sac r o s s i ng.Now,de f i ne yi α= = ∏ ( xj -xi)( xj -xi+1)( xj+1 -xi)( xj+1 -xi+1) 1≤i<j≤n 2| i+j e ,e ), ∏ f( 1≤i<j≤n 2| i+j i j B t hent hes i fαi s(-1) .Le tu sca l cu l a t et hes i gno gnofαβ.For 1 ≤i < j ≤n,con s i de rt het ime sofappea r anc e,andt hes i gn o fxj - xi i e s c t i ve l i s t i ngu i sh s eve r a l nα andβ,r pe y. We d ca s e s. CaseI. r soncei nβwi t hapo s i t i ves i j -i =1,xj -xiappea gn, andonc ei nα. I fi >1, i tappea r si nf( ei-1 , ei+1)wi t hapo s i t i ve s i fi =1,x2 -x1appea r si nf( e2 , en )wi t hanega t i ves i gn.I gn. Theproduc toft he s enumbe r sha sanega t i ves i gn. CaseI I.2 ≤j -i < n -1,thenxj -xi d oe sno tappea ri nβ, c e.Among ( i - 1,j - 1)and ( i - 1,j), andappea ri nαtwi t he r ei sexac t l i rt ha ti so ft hes amepa r i t i, yonepa y,among ( i,j), t he r ei sa l s oexac t l uchpa i r.Thes i j -1)and ( yones gn o fxj -xii neachappea r anc ei sa lway spo s i t i ve.Fori = 1, i t s appea r ancei nf( ej-1 ,en )orf( ej ,en )come swi t hanega t i ve i,j)( i= s i e,t he r ei sanega t i ves i i r( gn.Henc gnforeachpa 1,j = 3,4,...,n -1).Th i spa r toft heproduc tha st hes i gn n-3 (-1) = -1. 210 Ma t hema t i c a lOl i adi nCh i na ymp CaseI I I. i = 1,j = n, t henxn -x1 appea t ha r soncei nβ wi nega t i ves i ei nα ( en-1 ,e1 ))wi t ha po s i t i ve i nf( gn,andonc Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s i i spa r to ft hepr oduc tha sanega t i ves i gn.Th gn. Inbr i e f,t hes i fα snega t i ve,i. e.,A +Bi sodd. gno βi Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naWe s t e rn Ma thema t i ca l Ol i ad ymp 2010 ( Ta i i) yuanShanx 28 29Oc t ob e r,2010 1 Suppo s et ha tm andka r enon -nega t i vei n t ege r s,andp = m 22 +1i sapr imenumbe r.Provet ha t ( a)22 m+1 k p ≡ 1( modp k+1 ); ( b)2m+1pki st hesma l l e s tpo s i t i vei n t ege rns a t i s f i ngt he y cong r uenc eequa t i on2n ≡ 1( modpk+1). m+1 k So l u t i on.Wewanttoprovethat22 p = p +tk +1f ors ome k 1 i n t ege rtk no td i v i s i b l ebyp.We proceedbyi nduc t i ononk. Ma t hema t i c a lOl i adi nCh i na ymp 212 2 Whenk = 0, i tfo l l owsf r om22 = p -1t ha t22 = ( p -1) = m m t0 = p -2. i nt h i sca s e, p( p -2)+1, Fori nduc t i ves t ep,s uppo s et ha t22 m+1 k k ≥ 0andp tk , t hen =( 22 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. m+1 k+1 22 m+1 k p p = p æp ö p = p +tk +1wh e r e k 1 k+1 p ) p =( p tk +1) ∑ çès ÷ø (p s tk ) k+1 s=0 p æp ö k+1 s p( p -1)(k+1 ) 2 p tk + ∑ ç ÷ ( p tk ). 2 s=3 ès ø = 1 +p·p +tk + k 1 Ask ≥ 0, t henforanys ≥ 2,wehave ( k +1) s ≥ 2( k +1)= 2 k +2 ≥k +2, s o22 m+1 k+1 p =p +tk+1 +1,wh e r etk+1 ∈ Z+ andp k 2 tk+1 .I tf o l l owsf r om ma t hema t i ca li nduc t i ont ha t( a)ho l d s. Nex t,wepr ove ( b).Wr i t e2m+1pk =n ℓ +r ,whe r eℓ,r ∈ Zand0 ≤r <n.Theni tfo l l owsf rom ( a)t ha t1 ≡22 m+1 k p ≡2 n ℓ+r ℓ ·2r ≡ 2r ( ≡( modpk+1).As0 ≤r <n, i tfo l l owsf romt he 2n ) de f i n i t i ono fnt ha tr =0, i. e., n|2m+1pk .Byt heFundamen t a l Theor em o f Ar i t hme t i c,n = 2tps .I ft ≤ m ,t hen22 m ( 2 t k 2p ) m-t 2 ≡ 1( modp).Ont heo t he rhand,2 m k 2 p k p = ( 2 ) m 2 k p = ≡ (-1) ≡ -1( modp),wh i chi sacon t r ad i c t i on,andhencet = k p m +1,i. e.,n = 2m+1ps . 2 Ass hown i nF i .2 .1,AB i sa g d i ame t e rofac i r c l e wi t h cen t e r O .Le tC andD betwod i f f e r en t i n t sont hec i r c l eont hes ame po , hel i ne st angen t s i deo fAB andt t ot hec i r c l ea t po i n t sC and D mee ta tE .Segmen t sAD andBC mee ta tF .Li ne sEF andAB mee t F i .2 .1 g Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 213 a tM .Provet ha tE ,C ,M andD a r econcyc l i c. So l u t i on1.Asshownin Fi t .2 .2,j o i nOC ,OD andOE .I g f o l l owsf rom ∠OCE + ∠EDO = 90 s °+90 ° = 180 °t ha tECODi Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. cyc l i c.I fO = M , t henECMDi scyc l i c.Wemaya s s umet ha tO ≠ Mi nt hefo l l owi ng.Le tG bet hei n t e r s ec t i onofBC andAD , tH2 bet andH1bet hei n t e r s ec t i onofGF andAB .Le hefoo tof oAB .I tf o l l owsf romAC ⊥ BG andBD ⊥ rpend i cu l a ro fEt pe AG t ha tF i st heor t hoc en t e rof △BAG ,s oGH2 ⊥ AB .As ∠CH 2D = ∠CBF + ∠DAF = 1 80 °-2∠BGA ,and ∠COD = 180 °180 ° - ∠BOC - ∠AOD = 180 °- ( 180 ° -2∠GBA )- ( 2∠GAB )=2(∠GBA + ∠GAB)-1 8 0 °-2∠BGA , t hen ∠CH2D = ∠COD , r econcyc l i c.I tfo l l owsf rom s oC ,O ,H2 andD a ∠ECO = ∠EDO = ∠EH 1O =9 0 °t ha tE ,C ,O ,H1andD a r e concyc l i c.Henc e,H1 = H2 ,i. hey mee ta t e.EF ⊥ AB andt M .Cons equen t l r econcyc l i c. y,E ,C ,M andD a So l u t i on2.AsshowninFi .2 .2,j o i n g EO ,CO , DO ,CA ,CD .I tf o l l ows f r om ∠COE = ∠CAFt ha tRt△COE ∽ CE CO = Rt△COF ,ands o .As ∠ECF CF CA =9 0 °- ∠BCO = ∠OCA , s o△ECF ∽ △OCA ,andhenc e ∠CFE = ∠CAO . As ∠BFM = ∠CFE ,s o ∠FBM + ∠BFM = ∠FBM + ∠CAO = 9 0 °. Hence,EM ⊥ AB ,i tf o l l owst ha tO , F i .2 .2 g M ,D ,E ,C a r econcyc l i c. 3 De t e rmi nea l lpo s s i b l eva l ue sofpo s i t i vei n t ege rn,s uch i f f e r en t3 e l emen ts ub s e t sA1 ,A2 ,..., t ha tt he r ea r end An oft hes e t{ 1,2,...,n},wi t h|Ai ∩ Aj |≠ 1fora l l Ma t hema t i c a lOl i adi nCh i na ymp 214 i ≠j. So l u t i on.The s e to f po s i t i vei n t ege r ss a t i s f i ng t he g i ven y cond i t i oncon s i s t sofa l lpo s i t i ve mu l t i l e sof4.Wef i r s tp r ov e p t ha tn =4 k( k ∈ Z+ ) s a t i s f i ngt hec ond i t i on.De f i neA1,A2,..., y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. A4k a sfo l l ows:A4i-j = { 4 i -3,4 i -2,4 i -1,4 i} i -j}, \{ 4 f or a l l1 ≤i ≤kand0 ≤j ≤ 3. hens uchns ub s e t s Se cond,wewan tt oprovet ha ti f4 n ,t dono tex i s t.Suppo s et ot he con t r a r ha tA1 , ...,An be yt d i f f e r en t3 e l emen ts ub s e t sf u l f i l l i ngt hecond i t i ons t a t edi nt he ob l em.Le tA1 = { a,b,c},cons i de ra l lt heg i ven3 e l emen t pr s ub s e t swi t hnon emp t n t e r s e c t i on,we maya s s umet ha tt hey yi a r eA2 ,...,Am a f t e rr e l abe l i ng.Le tU = A1 ∪ A2 ∪ … ∪ Am . Wed i v i dei n t ot hefo l l owi ngd i f f e r en tca s e s: ● ● ● t henm = 1 <|U |. I f|U |= 3, æ4ö I f|U |= 4, t henm ≤ ç ÷ = 4 =|U |. è3ø t henweprovet ha tm <|U|a sf o l l ows. I f|U|≥5, Suppo s et ha tm ≥|U|, t henf o rany2 ≤i, e|A1 ∩ j ≤ m ,wehav Ai |=2, eAi ∩ Aj ≠ ⌀. |A1 ∩ Aj|=2.As|A1|=3,wehav Andi tfo l l owsf r om|Ai ∩ Aj |≠ 1t ha t|Ai ∩ Aj |= 2.I t l mean st ha tanytwod i s t i nc ts ub s e t so fA1 ,...,Am haveon y twocommone l emen t s. Con s i de rt hef ou ri n t e r s e c t i on sA1 ∩ A2 ,A1 ∩ A3 ,A1 ∩ A4 , A1 ∩ A5 ,i tf o l l owsf romt hep i -ho l epr i nc i l et ha tt he r e geon p a r etwoi n t e r s e c t i on st ha ta r eequa l;byr e l abe l i ng aga i n,we maya s s umet ha tt heya r eA2 = { a,b,d},A3 = { a,b,e}. Thenf orany4 ≤ i ≤ m ,wehavea,b ∈ Ai ;o t he rwi s e,Ai c on t a i n sa tl e a s toneo fao rb, andt heo t he rt h r e ee l emen t sc,dand Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 215 i chi simpo s s i b l e.He n c e|U|=m +2, e, i nt h i sc a s e, |Ai|≥4,wh wh i c hc on t r ad i c t st o|U|= m. Fr omt hea r umen tabov e,onec and i v i det hes ub s e t sA1,..., g Ani n t os eve r a lg r oups,andanytwos ub s e t si nt hes ameg r oup Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. havenon emp t n t e r s e c t i on s.Mor eove r,t henumbe rofs ub s e t s yi i nt hes ame g roupi sno t mor et hant he numbe rofe l emen t s appea r edi nt h i sg roup.Ast henumbe rofs ub s e t si sn,wh i chi s equa lt henumbe ro fe l emen t si n{ 1,2,...,n},s oeachg roup ha sexac t l ou rs ub s e t s.I tf o l l owst ha t4|n ,wh i chcon t r ad i c t s yf t heor i i na la s s ump t i on4 n. g 4 Le ta1,a2,...,an , b1, b2,..., bn benon -ne a t i v enumb e r s g s a t i s f i ngt hef o l l owi ngc ond i t i on ss imu l t aneou s l y y: ( 1)∑i=1( ai +bi)= 1 ; n ( ai -bi)= 0 ; 2)∑i=1i( n ( 3)∑i=1i2( ai +bi)= 10. n Pr ovet ha tmax{ ak ,bk }≤ So l u t i on.For any 1 10 f ora l l1 ≤k ≤ n. 10 +k2 ≤ k ≤ n,i tf o l l owsf rom t he g i ven cond i t i onsandCauchy sInequa l i t ha t yt 2 ( ≤ kak ) ≤ n ( ∑ia ) i i=1 n 2 n = n ( ∑ib ) i=1 ( ∑ib ) ( ∑b ) 2 i=1 = 1 0- ( i i=1 i 2 i n n i=1 i=1 2 ∑iai ) (1 - ∑ai ) ≤( 10 -ka2ak )( 1 -ak ) 2 =1 0- ( 10 +k2) ak +k2ak . Ma t hema t i c a lOl i adi nCh i na ymp 216 I tf o l l owsf r omt ha tak ≤ 10 10 , and .S imi l a r l bk ≤ y, 10 +k2 10 +k2 henc et her e s u l tf o l l ows. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 5 Le tkbeani n t ege randk > 1.De f i neas equence { an }a s f o l l ows: a0 = 0,a1 = 1andan+1 =kan +an-1forn = 1, 2,.... De t e rmi ne,wi t h proo f,a l l po s s i b l ek for wh i cht he r e ex i s tnon -nega t i vei n t ege r sℓ,m ( ℓ ≠ m )and po s i t i ve i n t ege r sp,qs ucht ha taℓ +kap = am +kaq . So l u t i on.Theansweri sk = 2. I fk = 2 ,t hena0 = 0,a1 = 1,a2 = 2, s oa0 +2 a2 =a2 + 0,2)and ( 2,1). 2 a1 = 4.Hence,( ℓ,m )= ( p,q)= ( Fork ≥ 3 ,i tf o l l owsf romt her ecur r en tr e l a t i ont ha tt he s equence{ an } i ss t r i c t l nc r ea s i ng,andk|an+1 -an-1fora l ln ≥ yi 1.Inpa r t i cu l a r,f orn ≥ 0,wehave a2n ≡ a0 ≡ 0( modk),a2n+1 ≡ a1 ≡ 1( modk). (* ) Suppo s et he r eex i s tℓ,m ∈ N,andp, ucht ha tℓ ≠ m and q ∈ Z+s aℓ +kap =am +kaq .Wemaya s s umet ha tℓ < m ,andd i v i dei n t o t hef o l l owi ngca s e s. ( a)p <ℓ < m :Themaℓ +kap ≤aℓ +kaℓ-1 <kaℓ +aℓ-1 = aℓ+1 ≤ am a s s ump t i on. < am + k aq , wh i ch con t r ad i c t st he or i i na l g ( b) ℓ =p < m : t henwehaveam +kaq = I fℓ =p = m -1, aℓ +kap = ( k +1) am-1 ,andmodu l ok wehaveam ≡am-1 modk, wh i chcon t r ad i c t s(* ). I fℓ = p < m -1,t henaℓ +kap < am-2 +kam-2 = am < am +kaq ,wh i chcon t r ad i c t st heor i i na la s s ump t i on. g ( c) ℓ < p < m :Int h i sca s e,f romq wehaveaq > 0,t hen Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 217 aℓ +kap ≤kap +ap-1 =ap+1 ≤am <am +kaq ,wh i chcon t r ad i c t s t heor i i na la s s ump t i on. g ( ℓ < m ≤ p:Thenap > d) aℓ +kap -am =aq . I tf o l l ows k Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. f r om kaq +am =kap +aℓ ≥kap ① ap k -1 am ≥ ap = aq ≥ ap ap . k k k ② t ha t No t et ha t ap =kap-1 +ap-2 ≥kap-ℓ , ③ hence ap > aq ≥ k -1 ap ≥ ( k -1) ap-1 ≥ ap-1 . k ④ Thenbyi nc r ea s i ngpr ope r t an }and ④ ,wehaveaq = yof{ ap-1 ,s oi nequa l i t i e s① ③ t u rni n t oequa l i t i e s.I tfo l l owsf rom ②t ha tm = p,andf r om ③t ha tp = 2.Hence, aq =ap-1 =a1 = 1.Andi tf o l l owsf rom ① t ha tℓ = 0.Fromaℓ +kap =am + kaq ,wehavek2 =k +k = 2 k,s ok = 2,wh i chi simpo s s i b l e. Sok = 2i st heon l o l u t i on. ys 6 As s hown i n F i .6 .1, g △ABC i sa r i t ang l ed gh t r i ang l e,∠C =90 °.Dr aw ac i r c l ec en t e r eda tB wi t h r ad i u s BC.Le tD b ea i n tont hes i deAC,and po DE be t ang en tt ot he F i .6 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 218 c i r c l ea tE.Thel i net h r oughC pe rpend i cu l a rt oAB mee t s tpo i n tF .Li neAF mee t sDE a tpo i n tG .Thel i ne l i neBE a t hroughA pa t sDE a tH .Pr ovet ha tGE = r a l l e lt oBG mee GH . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on. Le t K be t he i n t e r s e c t i ono fABandDE,and M bet hei n t e r s e c t i ono fABand CF. J o i nFK ,AE andME.I n i tf o l l owsf r omCM ⊥ △ABC, AB t ha tBM ·BA = BC2 = BE2,s o △BEM ∽ △BAE, and△BEM = ∠BAE . F i .6 .2 g As ∠FMK = ∠FEK = 90 °,s o MFEK i scyc l i c,and ∠BEM = ∠FKM .I t fo l l ows f rom ∠BAE = ∠BEM = ∠FKM , s oFK ‖AE ,andhenc e KA EF , KA ·BF = = 1. i. e., KB BF KB FE ① Ast hel i neEGAi n t e r s e c t s△EBK ,wehave EG ·KA ·BF =1 GK AB FE ② HK AK , = AsBG ‖AH ,wehave s o KG KB HG AB = . GK BK ③ EG = 1, Fr om ① ③ ,wehave ands oEG = HG . HG 7 The r ea r en( n ≥ 3)p l aye r si nat ab l et enn i st ournamen t, i nwh i chanytwop l aye r shaveama t ch.P l aye rAi sca l l ed Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 219 no tou t r f o rmedbyp l ay e rB,i fa tl e a s toneo fp l a e rAs pe y o s e r. l o s e r si sno taBsl De t e rmi ne,wi t hproo f,a l lpo s s i b l eva l ue sofn, s ucht ha t t hef o l l owi ngca s ecou l d happen:a f t e rf i n i sh i ng a l lt he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ma t che s,eve r l aye ri sno tou t r f ormedbyanyo t he r yp pe l aye r. p So l u t i on.Theansweri sn = 3o rn ≥ 5. ( 1)Forn = 3 ,s uppo s eA ,B andC a r et hr eep l aye r s,and t her e s u l to ft hr eema t che sa r ea sf o l l ows:A wi n sB ,B wi nsC , andC wi n sA .The s er e s u l t sobv i ou s l a t i s f hecond i t i on. ys yt ( uppo s et ha tt hecond i t i onho l d s,i. e.,i n 2)I fn = 4 ,s v i ew o ft her e s u l t so fa l l ma t che s,eve r l aye ri s no tou t yp r f ormedbyanyo t he rp l aye r.I ti sobv i ou st ha tnoneoft he s e pe f ourp l aye r swi n si nh i st hr eegame s,o t he rwi s e,t heo t he rt hr ee l aye r s wi l lbe no tou t r f ormed by t h i sp l aye r.S imi l a r l p pe y, noneo ft he s et hr eep l aye r sl o s e si nh i st hr eegame s.I tf o l l ows t ha teachp l aye rwi n soneortwoma t che s. Fort hep l aye rA ,a nsB andD ,bu s s umet ha tA wi tl o s e st o C, t henbo t hB andD wi nC ;o t he rwi s e,t hey wou l dno tou t - r formedA .Fort hel o s e ri nt hema t chB v s.D ,heon l ns pe y wi , C ands ot hel o s e ri simpo s s i b l et obeno tou t -pe r formedbyt he wi nne r.Con s equen t l orn = 4, t heg i vencond i t i oncou l dno t y,f happen. ( 3) For n = 6,one can cons t r uc tt he t ou rnamen tr e s u l t s by mean so ft hef o l l owi ngd i r ec t ed r aph,i n wh i ch each b l ack do t g r epr e s en t sa p l aye r,and · → r epr e s en t sama t ch wi t ht her e s u l t t ha tp l aye r·wi n sp l aye r. F i .7 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 220 ( 4)I ft he r eex i s tt ou rnamen tr e s u l t ss ucht ha teacho fn l aye r sAi( 1 ≤i ≤n) i sno tou t -pe r formedbyanyo t he rp l aye r, p wewi l lprovet ha tt hes ameho l d sforn +2p l aye r sa sf o l l ows: Suppo s et ha tM andN a r et headd i t i ona lp l aye r st ot heor i i na ln g Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. l aye r s.Con s t r uc tt hegamer e s u l t so fM andN a sfo l l ows: p Ai → M ,M → N ,N → Ai f o ra l li = 1,2,...,n,andt heg amer e s u l t s amongAisa r es t i l lt heo r i i na lone s.Now we g wan tt oche ckt ha tt he s en +2p l ay e r ss a t i s f he yt i v enc ond i t i on.Fo rany p l ay e rG ∈ { A1, g A2,...,An },t heni ts u f f i c e st oc on s i de rt he l a e r sG,M ,N ,andi tr edu c e st ot hec a s en p y F i .7 .2 g = 3. Onec anche ckt ha te a cho ft he s et h r e ep l ay e r sG,M ,Ni sno t ou t r f o rmed by any on eo ft h eo t h e rtwo p l a e r s.He n c e,t he pe y t o u r n ame n tr e s u l to ft h e s en +2p l a e r ss a t i s f i e st her e i r e dc ond i t i on. y qu Inpa r t i cu l a r,i tf o l l owsf r om ( 1)andt hei nduc t i ons t epof ( 4)t ha tt her equ i r edcond i t i onho l d sforanyoddn wi t hn ≥3. Mor eove r,i tf o l l owsf r om ( 3)andt hei nduc t i ons t ep of ( 4) t ha tt her equ i r edcond i t i onho l d sf oranyevenn wi t hn ≥6,and i tcomp l e t e st hepr oo f. 8 De t e rmi nea l lpo s s i b l eva l ue so fi n t ege rkforwh i cht he r e b +1 a +1 + =k. ex i s tpo s i t i vei n t ege r saandbs ucht ha t a b So l u t i on.Theansweri st ha tk = 3o r4. Fi xa po s s i b l e va l uek,among a l lt he pa i r s( A ,B )of s i t i vei n t ege r ss a t i s f i ng po y B +1 A +1 + =k, A B Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 221 choo s eany( a,b) s ucht ha tbi st hesma l l e s t.Thent hequadr a t i c equa t i on x2 + ( 1 -kb) x +b2 +b = 0 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ha sani n t eg r a lr oo tx = a.Le tx = a ' bet hes e condroo t,i t f o l l owsf roma +a ' =kb -1t ha ta ' ∈ Z ,andf rom a ·a ' =b( b +1) t ha ta ' > 0.Hence,wehave b +1 a ' +1 + =k. a ' b Andi tf o l l owsf r omt hea s s ump t i ononbt ha t a ≥b,a ' ≥b. Sooneo faanda 'i sequa lt ob.Wi t hou tl o s so fgene r a l i t y,we maya s s umea =b, s ok = 2 + andk = 3or4,r e s c t i ve l pe y. 2, ands ob|2i. e., b = 1or2, b I fa = b = 1 ,t henk = 4;i fa = b = 2,t henk = 3. Cons equen t l k = 3or4i st heon l o l u t i on. y, ys 2011 ( Y u s h a n, J i a n x i) g F i r s tDay 8: 00 12: 00,Oc t obe r29,2011 1 Gi vent ha t0 < x,y < 1,de t e rmi ne,wi t h proof,t he xy( 1 -x -y) s edbyLi u max imumva l ueo f( .( po x +y)( 1 -x)( 1 -y) Ma t hema t i c a lOl i adi nCh i na ymp 222 Sh i x i ong) So l u t i on.Whenx =y = 1 1, t heva l ueofexpr e s s i oni s . 8 3 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. xy( 1 -x -y) 1 ≤ f orany Wewi l lpr ovet ha t( x +y)( 1 -x)( 1 -y) 8 0 < x,y < 1a sf o l l ows. 1 -x -y) xy( 1 ≤0 < I fx +y ≥ 1, t hen ( . 1 -x)( 1 -y) 8 x +y)( I fx +y < 1, t henl e t1 -x -y = z > 0, i tfo l l owsf rom AM GMInequa l i t ha t yt xy( 1 -x -y) xyz = ( x +y)( x +y)( z +x) 1 -x)( 1 -y) ( y +z)( ≤ xyz 1 = . 8 · · 2 xy 2 yz 2 zx 1 Inconc l u s i on,t hemax imumva l ueoft heexpr e s s i oni s . 8 1,2, ...,2011}beas ub s e ts a t i s f i ngt he Le tM ⊆ { y 2 fo l l owi ngcond i t i on:Foranyt hr eee l emen t si nM ,t he r e ex i s ttwo o ft hem a andb,s ucht ha ta |b orb |a . De t e rmi ne,wi t h pr oo f,t he max imum va l ueof| M |, whe r e|M |deno t e st henumbe ro fe l emen t sofM .( s ed po byFengZh i gang) So l u t i on.OnecancheckthatM ={ 1,2,22 ,23 ,...,210 ,3, 3 ×2,3 ×22 ,...,3 ×29}s a t i s f i e st hecond i t i on,and|M |= 21. Suppo s et ha t|M |≥ 22,andl e ta1 < a2 < … < ak bet he e l emen t so fM ,whe r e|M |=k ≥22.Wef i r s tprovet ha tan+2 ≥ 2 an fora l ln;o t he rwi s e,wehavean < an+1 < an+2 < 2 an for Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 223 s omen <k +2, t henanytwooft he s et hr eei n t ege r san ,an+1 , an+2 dono thaveany mu l t i l er e l a t i onsh i i chcon t r ad i c t s p p,wh t hea s s ump t i on. a2 ≥ 4,a6 I tf o l l owsf r omt hei nequa l i t ha ta4 ≥ 2 yabovet ≥2 a4 ≥8,… a22 ≥2a20 ≥2 >2011,wh i chi sacon t r ad i c t i on! Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 11 Henc e,t hemax imumva l ueo f|M |i s21. 3 i veni n t ege r. Le tn ≥ 2beag ( 1)Pr ovet ha tonecana r r angea l lt hes ub s e t soft hes e t { 1,2, ...,n }a sas equenc eofs ub s e t sA1 ,A2 , ..., A2n ,s ucht ha t|Ai+1|=|Ai|+1or|Ai|-1,whe r ei = 1,2,...,2n andA2n+1 = A1 . ( 2)De t e rmi ne,wi t hproo f,a l lpo s s i b l eva l ue soft hes um ∑ n 2 i=1 i (-1) r eS ( S( Ai),whe Ai)= ∑x∈A xandS(⌀ ) i = 0, f oranys ub s e ts equenceA1 ,A2 ,...,A2n s a t i s f i ng y t hecond i t i oni n( 1).( s edbyLi ang Yi ngde) po So l u t i on.( 1)Weproveby ma t hema t i ca li nduc t i ont ha tt he r e ex i s t sas equenc eA1 ,A2 ,...,A2ns ucht ha tA1 = { 1},A2n = ⌀ ands a t i s f i e st hecond i t i oni n( 1). Whenn = 2,t hes equenc e{ 1},{ 1,2},{ 2},⌀ of { 1,2} work s. As s umet ha twhenn = k,t he r eex i s t ss uchas equenc eB1 , B2 ,...,B2k ofs ub s e t sof{ 1,2,...,k}.Asf orn =k +1,one cancon s t r uc tas equenceo fs ub s e t so f{ 1,2,...,k + 1},a s fo l l ows: A1 = B1 = { 1}, Ai = Bi-1 ∪ { k +1}, i = 2,3,...,2k +1, Aj = Bj-2k ,j = 2k +2,2k +3,...,2k+1 . Onecanea s i l ha tt hes equencef u l f i l l st her equ i r ed ycheckt Ma t hema t i c a lOl i adi nCh i na ymp 224 cond i t i on ss t a t edabove.Byi nduc t i onwehaveproved ( 1)forn ≥ 2. ( 2) We wi l ls howt ha tt hes um i s0,i ndependen toft he a r r angemen t.Wi t hou tl o s so fgene r a l i t s s umet ha t y,we maya Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1},o t he rwi s es h i f tt hei ndexcyc l i ca l l tfo l l owsf rom A1 = { y.I ha tt he i rpa r i t i e sa r ed i f f e r en t, |Ai+1 | =|Ai|+1or|Ai|-1t andhenc et hepa r i t i e soft hei ndexl abe lofanys ub s e tandi t s ca rd i na l i t r et hes ame. ya i Ai)= ∑A ∈PS ( A )- ∑A ∈Q I tfo l l owst ha t∑i=1(-1) S( n 2 A ),whe i s t so fa l ls ub s e t so f{ 1,2,...,n}wi S ( r eP cons t h i s t sofa l ls ub s e t sof{ 1,2, evennumbe r so fe l emen t s,andQcons ...,n}wi t hoddnumbe r sofe l emen t s. l lk -e l emen ts ub s e t s, x Foranyx ∈ { 1,2,...,n},amonga k-1 appea r si nexac t l ft hem,hencei tcon t r i bu t e st ot hes um yCn-1 o ∑ s S( A )- ∑A ∈Q S ( A )a A ∈P - Cn-1 + Cn-1 - Cn-1 + … + ( -1)Cn1 -1)- = 0. -1 = - ( 0 1 2 n n 1 n 1 i The r e f or e,∑i=1(-1) S( Ai)= 0. n 2 4 Ass howni nF i .4 .1,AB andCD a r e g twochord si nt hec i r c l e☉O mee t i ng a tpo i n tE ,and AB ≠ CD . ☉I i s t angen tt o☉Oi n t e rna l l tpo i n tF , ya andi st angen tt ot hechord sAB and CD a tpo i n t sGandH , r e s t i ve l l pec y. i sal i nepa s s i ngt hroughO ,mee t i ng AB,CD a tpo i n t sP,Q,r e s c t i v e l pe y, s ucht ha tEP = EQ .Li neEF mee t s t hel i nela tpo i n tM .Pr ovet ha tt he F i .4 .1 g Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 225 l i net hroughM and pa r a l l e lt ot hel i neAB i st angen tt o s edbyLiQi u sheng) t hec i r c l e☉O .( po So l u t i on.AsshowninFi .4 .2,dr awal i nepa r a l l e lt oAB and g i ch mee t st he common t angen tt oc i r c l e ☉O a t po i n tL ,wh Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t angen tl i net ot he s etwo c i r c l e sa ta po i n tS.Le tR bet he o i ns e t sLFandGF. i n t e r s e c t i ono fl i ne sFSandBA ,andj gmen F i .4 .2 g F i r s t,wepr ovet ha tt hepo i n t sL ,G andF a r eco l l i nea r. As bo t hSL andSF a r et angen tt o☉O ,SL = SF ;a sbo t hRG and RF a r et angen tt o☉I,RG = RF . AsSL ‖RG , we have ∠LSF = ∠GRF , s o ∠LFS = 180 °- ∠LSF 180 °- ∠GRF = 2 2 = ∠GFR , andhenc eL ,G andF a r eco l l i nea r. S imi l a r l ss hown i n F i .4 .3, y,a g dr awal i nepa r a l l e lt oCD andt angen t t o☉O a tpo i n tJ,t r e henF ,H andJ a F i .4 .3 g 226 Ma t hema t i c a lOl i adi nCh i na ymp co l l i nea r.Le tt angen tl i ne st ot hec i r c l e tt he po i n t sL andJ mee tEF a t ☉O a i n t sM 1 andM 2 ,r e s c t i ve l he po pe y.Int , f o l l owi ng wepr ovet ha tt hepo i n t sM 1 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. and M 2 co i nc i de.I tfo l l owsf r om t he homo t he t en t e r eda tF mapp i ng ☉O yc hen t o☉It ha tLJ ‖GH ,t = M 1E LG = GF EF M 2E , JH = andhenc eM 1 and M 2 HF EF co i nc i de.Deno t et h i spo i n tbyK . F i na l l tt o pr ove t ha t y, we wan i n t sM andK a l s oco i nc i de.I ts u f f i ce s po F i .4 .4 g i e sont hel i nel. t os howt ha tK l Ass howni nF i .4 .4,j o i nKO ,a sKL andKJ a r et angen t g t o☉O , s o ∠LKO = ∠JKO . No t et ha tKO b i s e c t s ∠LKJ, i tf o l l owsf r om KL ‖AB and KJ ‖CD t ha tt hel i neKO mee t sl i ne sAB andCD wi t ht hes ame sj u s tt hel i nel,i. ang l e so fi n t e r s e c t i on,andhenc eKOi e.,K l i e sont hel i nel. Henc e,K i sj u s tt hei n t e r s e c t i onpo i n tM ofl i neEF andl. S e c ondDay 8: 00 12: 00,Oc t obe r30,2011 5 De t e rmi ne,wi t hpr oo f,whe t he rt he r ei sanyoddi n t ege r n ≥ 3andn d i s t i nc tpr imenumbe r sp1 ,p2 ,...,pn s uch t ha ta l lpi + pi+1( i = 1,2,...,n,andpn+1 = p1)a r e r f ec ts r e s? ( s edbyTaoP i ng s heng) pe qua po Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 227 So l u t i on.Theansweri snega t i ve.Suppo s et ha tt he r eex i s todd i n t ege rn ≥ 3andn d i s t i nc tpr imenumbe r sp1 ,p2 , ...,pn s a t i s f i ngt heg i vencond i t i on. y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. I fa l lp1 ,p2 ,...,pn a r eodd,t heni tf o l l owsf rom t he i vencond i t i ont ha ta l lt hes umspi +pi+1a r emu l t i l e so f4,s o g p t hepr imenumbe r sp1 ,p2 ,...,pn modu l o4appea rt obe1and sodd. 3a l t e rna t i ve l tcon t r ad i c t st hef ac tt ha tni y,andi I fone o f p1 ,p2 , ...,pn i s2,t hen wi t hou tl o s s of r a l i t s s umet ha tp1 = 2.Asbo t hp1 + p2 and y,we maya gene r epr e f e c ts r e sandbo t ha r eodd,i tf o l l owst ha tp2 pn +p1 a qua andpn a r econg r uen tt o3modu l o4.S imi l a rt ot hed i s cu s s i oni n l o t hef i r s tca s e,weknowt ha tt hepr ime sp2 ,p3 ,...,pn modu sodd,wh i chi sa 4appea rt obe1and3a l t e rna t i ve l on -1i y,s con t r ad i c t i on. Henc e,t he r ea r e no odd i n t ege rn ≥ 3 and n pr ime s s a t i s f i ngt heg i vencond i t i ons. y 6 Le ta,b,c > 0,pr ovet ha t 2 2 2 ( ( ( a -b) b -c) c -a) + + ( c +a)( c +b) ( a +b)( a +c) ( b +c)( b +a) ≥ 2 ( a -b) . 2 a +b +c2 2 ( s edbyLiShenghong) po 1 1( 2 2 + +( So l u t i on1.I tf o l l owsf r om ( a -2 b) a -2 c) b- c) ≥ 0t ha t 2 2 2 3( a2 +b2 +c2)≥ 2a2 +2ab +2 b c +2ac = 2( a +b)( a +c), s owe have ( a + b)( a + c) ≤ wehave 3(2 a + b2 + c2).S imi l a r l y, 2 Ma t hema t i c a lOl i adi nCh i na ymp 228 3 2 ( b +a)( b +c)≤ ( a +b2 +c2), 2 3 2 and ( c +a)( c +b)≤ ( a +b2 +c2). 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Henc e, 2 2 2 ( ( ( a -b) b -c) c -a) + + ( ) ( ) ( ) ( ) ( ) ( c +a c +b a +b a +c b +c b +a) ≥ 2 2 2 +( +( 2·( a -b) b -c) c -a) 2 2 2 3 a +b +c 2· ≥ 3 = 1( 2 2 ( + a -b) b -c +c -a) 2 a2 +b2 +c2 2 ( a -b) . 2 a +b +c2 2 So l u t i on2.I tf o l l owsf r om CauchyInequa l i t ha t yt [ 2 2 2 ( ( ( a -b) b -c) c -a) · + + ( c +a)( c +b) ( a +b)( a +c) ( b +c)( b +a) ] [( c +a)( c +b)+ ( a +b)( a +c)+ ( b +c)( b +a)] 2 ≥( |a -b|+|b -c|+|c -a|) 2 2 ≥( a -b) . |a -b|+|b -c +c -a|) = 4( And ( c +a)( c +b)+ ( a +b)( a +c)+ ( b +c)( b +a) =( a2 +b2 +c2)+3( ab +b c +ca)≤ 4( a2 +b2 +c2), hence 2 2 2 ( ( ( a -b) b -c) c -a) + + ( c +a)( c +b) ( a +b)( a +c) ( b +c)( b +a) ≥ ≥ 2 4( a -b) ( c +a)( c +b)+ ( a +b)( a +c)+ ( b +c)( b +a) 2 2 ( 4( a -b) a -b) = 2 . 2 2 2 2 4( a +b +c ) a +b +c2 Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 7 229 Ass howni nF i .7 .1,AB > g AC ,andt f hei nc i r c l e ☉I o △ABCi st angen tt oBC ,CA and AB a t po i n t sD ,E and Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. e s t i ve l tM bet F ,r he pec y.Le mi dpo i n t o f s i d e BC, and AH ⊥ BC a tt he po i n t H. f ∠BAC The b i s e c t o r AI o i n t e r s e c t st hel i ne sDE ,DF a t F i .7 .1 g i n t sK ,L ,r e s c t i ve l po pe y. r econcyc l i c.( s edbyBi an Provet ha tM ,L ,H andK a po Hongp i ng) So l u t i on.Jo i nCL ,BI,DI,BK ,ML andKH .Ex t endCLt o mee tAB a tpo i n tN . Asbo t hCD andCE a r et het angen tt o☉I,s oCD = CE . As ∠BIK = ∠BAI + ∠ABI = = 1 (∠ BAC + ∠ABC) 2 1( 180 °- ∠ACB )= ∠EDC = ∠BD , 2 s oB ,K ,D andIa r ecyc l i c. As ∠BKI = ∠BDI = 90 °,i. e.,BK ⊥ AK ;s imi l a r l y, CL ⊥ AL .AsALi st heb i s e c t oro f ∠BAC , s oLi st hemi dpo i n t o fCN .AsM i st hemi dpo i n to fBC , s oML ‖AB . S i nce ∠BKA = ∠BHA =90 °, i tf o l l owst ha tpo i n t sB ,K , H andA a r ecyc l i c,s o ∠MHK = ∠BAK = ∠MLK ,henc eM , L ,H andK a r ecyc l i c. 8 De t e rmi ne,wi t hpr oo f,a l lpa i r s( a,b)ofi n t ege r s,s uch t ha tf oranypo s i t i vei n t ege rn,oneha sn | ( an +bn+1). Ma t hema t i c a lOl i adi nCh i na ymp 230 ( s edbyChenYonggao) po So l u t i on.Theso l u t i onpa i r scon s i s tof ( 0,0)and (-1,-1). I foneo faandbi s0,i ti sobv i ou st ha tt heo t he ri sa l s o0. Nowwea s s umet ha tab ≠0, s e l ec tal a r imeps ucht ha t gepr Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 2 i tf o l l owsf r om Fe rma t sLi t t l eTheor emt ha t p >|a +b |, ap +bp+1 ≡a +b2( modp). Asp| ( ap +bp+1)andp >|a +b2|,wehavea +b2 = 0. Thenwes e l ec tano t he rpr imeqs ucht ha tq >|b +1|and ( tn = 2 q,b)= 1.Le q.Thenwehave 2 2 1 4 2 1 2 1 q +b q+ =b q +b q+ =b q+ ( an +bn+1 = (-b2) b2q-1 +1). I tf o l l owsf r omn| ( ha t an +bn+1)and ( q,b)= 1t b2q-1 +1). q| ( 2 , ·b +1 ≡b +1( Asb2q-1 +1 ≡ ( bq-1) mo dq) andq >|b +1|, i tf o l l owst ha tb +1 = 0, i. e., oa =-b2 = -1. b =-1,ands Inconc l u s i on,t he r ea r eon l o l u t i onpa i r s( 0,0)and ytwos (-1,-1). 2012 ( Ho h h o t, I n n e rMo n o l i a) g F i r s tDay 8: 00 12: 00,Sep t embe r28,2012 1 F i ndt hel ea s t po s i t i vei n t ege r m ,s uch t ha tforeve r y imenumbe rp > 3, pr 2 105|9p -29p + m . ( s edby Yang Hu) po Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 231 So l u t i on.As105 =3×5×7, t heor i i na lprob l emi sequ i va l en t g 2 t ot ha to ff i nd i ngt hel ea s tpo s i t i vei n t ege rm s ucht ha t9p i v i deds imu l t aneou s l 29p + m canbed yby3,5,7. S i nc ep2 andp havet hes ameodd evencha r ac t e r,wehave p p -( -1) +m ≡ m ( 9p -29p + m ≡ (-1) mod5). Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 2 2 Thenwege tm ≡ 0 ( mod5). Asp > 3i sanoddnumbe r,wehave p + m ≡ m +1( 9p -29p + m ≡ - (-1) mod3). 2 The r e f or e,m ≡ 2 ( mod3). Wehavea l s op2 ≡1( mod3) f orp >3,orp2 =3 k +1.Then 9p -29p + m ≡ 23k+1 -1 + m ≡ 8k ·2 -1 + m 2 ≡ m +1( mod7). The r e for e,m ≡ 6 ( mod7). Ins umma r y,wehave mod5), ìïm ≡ 0( ï mod3), ím ≡ 2( ïï îm ≡ 6( mod7). Theni ti sea s of i ndt ha tt hel ea s tpo s i t i vei n t ege rmi s20. yt 2 Pr ovet ha t,among anyn ve r t i c e so far egu l a r2 n -1 n ≥ 3),t he r ea r et hr eeone s,wh i ch a r et he l po ygon ( ve r t i ce sofani s o s ce l e st r i ang l e.( s edbyZouJ i n) po So l u t i on.S i nc ei ti sea s ove r i f i r ec t l hea s s e r t i onfort he yt yd yt a pen t agon)andn = 4 ( ahep t agon),we may ca s e sn = 3 ( a s s umet ha tn > 4i nt hef o l l owi ngd i s cu s s i on. Byr educ t i ont oab s u rd i t s s umewecans e l e c tn ve r t i c e s y,a Ma t hema t i c a lOl i adi nCh i na ymp 232 i nar egu l a r2 n -1po l ucht ha tnot hr eeof ygonA1A2A3 …A2n-1s t hem con s t i t u t e sani s o s c e l e st r i ang l e. We ma rkt he s e po i n t s wi t h co l or r ed and t he r ema i nde r n - 1 one s wi t hb l ue, r e s c t i ve l pe y. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ed,andd i v i det heo t he r2 i n t s Wemayl e tA1 ber n - 2po i n t on - 1pa i r s( s eet hef i r e):( A2 ,A2n-1),( A3 ,A2n-2), gu ...,( An ,An+1). Thetwopo i n t si neach i rcanno tbebo t hr eda s pa t hey p l u sA1 cons t i t u t e an i s o s c e l e st r i ang l e,andmu s t beoner edandoneb l uef or F i .2 .1 g edpo i n t s. t he r ea r eexac t l ynr As s umi ngA2i sr ed,t henA2n-1i s tbeb l ue sb l ue,andA3 mu a s△A1A2A3 i sani s o s c e l e st r i ang l e.The r e for e A2n-2 i sr ed, r e bo t h b l ue, a s f r om wh i ch we i nf e rt ha t A5 , A2n-4 a △A2n-2A2A5 , △A2n-4A2n-2A1 a r etwoi s o s ce l e st r i ang l e s.Bu t r et hetwopo i n t si napa i r,t hey mu s tbeoner edand A5 ,A2n-4a oneb l ue.Th i si sacon t r ad i c t i on! Thea s s e r t i onforn >4i st hen a l s ot r ue.Thepr oofi scomp l e t e. 3 Le tE be a g i ven s e t wi t hn e l emen t s.Suppo s e A1 , A2 ,...,Ak a r ek d i s t i nc tnon emp t ub s e t sofE ,wi t h ys t hep r ope r t ha t,f o rany1 ≤i < j ≤k,e i t he rAi ∩ Aj = yt ⌀ oronei nc l ude st heo t he r( i. e.,Ai ⊂ Aj orAj ⊂ Ai ). Fi ndt hemax imumva l ueo fk.( s edbyLengGang s ong) po So l u t i on.Wec l a imt ha tt hemax imumva l ueofki s2 n -1.To h i s,wea tf i r s tg i veanexamp l et ha ts a t i s f i e sk =2 n -1. provet Wemayl e tE = { 1,2,...,n},and Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp Ai = { { i} 233 f or1 ≤i ≤ n, { i -n +1} forn +1 ≤i ≤ 2n -1. 1,2,..., I ti sea s os eet ha tt heypo s s e s st heg i venprope r t yt y. Now wewi l lpr ovet ha tk ≤ 2 n -1byi nduc t i on. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i ti sobv i ou s l r ue. Whenn = 1, yt As s umet ha ti ti st r uef orn ≤ m -1.Thenwhenn = m ,we con s i de rt hes e tt ha tcon t a i n st hemo s te l emen t samongA1 ,A2 , ...,Ak exc l ud i ngE .We maya s s umet ha ti ti sA1 con t a i n i ng t(≤ m -1)e l emen t s.Thenwecand i v i deA1 ,A2 ,...,Aki n t o t hr eeca t egor i e s: ( e tE ; 1)t hes ( 2)s e t st ha ta r ei nc l udedi nA1 ;and ( semp t 3)s e t swho s ei n t e r s e c t i onwi t hA1i y. ( Ca t egor i e s( 1)and ( 2)maybeemp t y.) Thent henumb e ro fs e t si nc a t e r 1)i sno tg r e a t e rt han1. go y( Byi nduc t i on,t henumbe ro fs e t si nca t egor 2)i sno t y( r ea t e rt han2 t -1. g Thenumbe rofe l emen t scon t a i nedi nt heun i onoft hes e t s i nca t egor 3)i sno tg r ea t e rt hanm -t.Thenbyi nduc t i on,t he y( numb e ro fs e t si nt ha tc a t e r sno tg r e a t e rt han2( m -t)-1. go yi Int o t a l, k ≤1+ ( 2 t -1)+ [ 2( m -t)-1]= 2m -1. The r e f or e,t hepr opo s i t i oni st r uef orn = m . Byi nduc t i on,weconc l udet ha tforanys e tE ofne l emen t s, wea lway shavek ≤ 2 n -1.Th i scomp l e t e st heproo f. 4 Le tP beanyi nne rpo i n to fanacu t e△ABC ;E ,F bet he t ol i ne sAC ,AB ,r e c t i onpo i n t sofP on e s t i ve l pro j pec y; and t he ex t ended l i ne s o f BP , CP i n t e r s ec tt he Ma t hema t i c a lOl i adi nCh i na ymp 234 c i r cumc i r c l eo f△ABC a tpo i n t sB1 ,C1 ( B1 ≠ B ,C1 ≠ C),r t et her ad i ioft he e s t i ve l tR andr deno pec y.Le c i r cumc i r c l eandi nc i r c l eo f △ABC ,r e s c t i ve l pe y.Prove Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. EF r, , ≥ t ha t and whent heequa l i t l d s,de t e rmi ne yho B1C1 R s edbyLiQi u sheng) comp l e t e l hepo s i t i on so fP .( po yt So l u t i on.Ass eeni nF i .4 .1, g we make PD ⊥ BC wi t h t endl i ne i n t e r s e c t i ngpo i n tD ,ex AP t o i n t e r s e c t wi t h t he c i r cumc i r c l eo f △ABC a tpo i n t A1andconnec tDE ,DF ,A1B1 , A1C1 . S i n c e P , D , B, F a r e c onc c l i c, we have ∠PDF = y F i .4 .1 g ∠PBF ; s i nc eP ,D ,C ,E ,wehave ∠PDE = ∠PCE .Then ∠FDE = ∠PDF + ∠PDE = ∠PBF + ∠PCE = ∠AA1B1 + ∠AA1C1 = ∠C1A1B1 . In t he s ame way, we have ∠DEF The r e f or e,△DEF ~ △A1B1C1 . = ∠A1B1C1 . Ther ad i u so ft hec i r c umc i r c l eo f △A1B1C1 i sa l s o R ,and l e tt he r ad i u so ft hec i r cumc i r c l eo f△DEF EF R' = beR'.Wet henhave . B1C1 R Nowl e tt hei n c en t e ro f△DEF b e O '.Conne c t AO ', BO ', CO ' ( s e e F i .4 .2).Weha v e g F i .4 .2 g Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp S△ABC = 235 ( AB +BC +CA )·r 2 = S△O'AB +S△O'BC +S△O'AC ≤ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. = AB ·O'F BC ·O'D CA ·O'E + + 2 2 2 ( AB +BC +CA )·R' . 2 l i t l d si fandon l f The r e f or e,R' ≥r.Theequa yho yi O'D ⊥ BC ,O'E ⊥ CA ,O'F ⊥ AB , wh i chimp l i e sP = O'andPi st hei ncen t e rof△ABC . EF r ≥ The r e for e, andt heequa l i t l d si fandon l yho yP B1C1 R i st hei nc en t e rof△ABC . Thepr oo fi scomp l e t e. S e c ondDay 8: 00 12: 00,Sep t embe r29,2012 5 Le tH andO bet heor t hoc en t e randc i r c umc en t e ro fa c u t e t r i ang l e△ABC,r e s c t i v e l r enon c o l l i ne a r). pe y (A ,H ,O a Suppo s eD i st he pr o e c t i ono fA on t ol i neBC ,andt he j rpend i cu l a rb i s ec t oroft hes egmen tAO mee t sl i neBCa t pe E .Provet ha tt hemi dpo i n tN o fOH i sont hec i r cumc i r c l e o f △ADE . ( po s ed by FengZh i gang) So l u t i on.Ass eeni nFi .5 .1, g ol e ti ti n t e r s ec t weex t endHD t wi t ht hec i r cumc i r c l eo f△ABC a t po i n t H',l i nk FN , DN , BH ,BH',OH',and deno t e F i .5 .1 g 236 Ma t hema t i c a lOl i adi nCh i na ymp t hemi dpo i n tofAO a sF . st heor t hoc en t e r,wehave AsH i ∠CBH' = ∠CAH' = ∠CBH . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. st hemi dpo i n to fHH'.Ont heo t he rhand, The r e f or e,Di Ni st he mi dpo i n to f HO ,s st he med i anof △HOH'. oDN i The r e f or e,DN = 1 OH'. 2 t hen S i nc eOH' = OA andFi st hemi dpo i n to fOA , DN = 1 1 OH' = OA = AF . 2 2 I ti sea s os ee FN ‖AH .Then AFND i sani s o s ce l e s yt t r apezo i dandA ,F ,N ,D a r econcyc l i c. Fu r t he rmor e,f r om ∠ADE =90 °= ∠AFE weknowA ,F , D ,E a r ea l s oconcyc l i c.ThenA ,F ,N ,D ,E a r econcyc l i c, wh i chimp l i e st ha tt hec i r cumc i r c l eo f△ADE c ro s s e sN a tt he oofi scomp l e t e. mi dpo i n tofOH .Thepr Remark.Bythefac t st ha tt her ad i u soft hen i ne -po i n t c i r c l eo fat r i ang l ei sha l fo ft ha to ft hec i r cumc i r c l eandN i s t hec en t e ro ft hen i ne -po i n tc i r c l eof △ABC ,wecan ge ta l s o sani s o s c e l e st r ape zo i d. t ha tAFNDi 6 Sequenc e{ an }i sde f i nedbya0 = 2 an 1, , an+1 = an + 2 2012 n = 0,1,2,....Fi ndi n t ege rk, s ucht ha tak <1 <ak+1 . ( s edbyBi anHongp i ng) po So l u t i on.Fromtheg i vencond i t i on,wehave 1 = a0 < a1 < … < a2012 . 2 Weno t et ha t Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 237 1 2012 1 1 , = = an+1 an ( an +2012) an an +2012 or Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 1 1 = . an an+1 an +2012 Byt het e l e s cop i ngs um,wehave 1 1 = a0 an Then 1 = 2a2012 The r e f or e, n-1 ∑a i=0 i 2011 1 . +2 012 1 < ∑ 012 i=0 ai +2 2011 1 ∑ 2012 = 1. i=0 a0 < a1 < … < a2012 < 1. Thenwehave 1 = 2a2013 2012 1 > ∑ 012 i=0 ai +2 wh i ch mean sa2013 > 1. 2012 1 ∑ 1 +2012 = 1, i=0 Con s equen t l i ndt ha tk = 2012. y,wef 7 Gi venann ×ng r i d,weca l ltwoc e l l si ni tad ac en ti ft hey j have a common s i de. Att he beg i nn i ng,each ce l li s a s s i r+1.Anope r a t i onont heg r i di sde f i ned gnednumbe a sf o l l ows:onechoo s e sac e l l,andt henchange st hes i s gn o fe v e r ri ni t sad a c en tc e l l s( bu tno tchang et hes i ynumbe j gn o ft henumb e ri ni t s e l f).F i nda l lt hei n t e e r sn ≥2, s u cht ha t g a f t e raf i n i t eo fope r a t i on s,a l lt henumbe r si nt hec e l l so ft he r i da r echang edt o -1.( s edbyShenHuyu e) g po So l u t i on.Wewi l lpr ovet ha tn mee t st her equ i r edcond i t i oni f andon l fi ti sanevennumbe r. yi Wedeno t et hec e l li nt hei -rowandj -co l umnoft heg r i da s 238 Ma t hema t i c a lOl i adi nCh i na ymp Aij ( i,j ∈ { 1,2,...,n}). k,k ∈ N* ,wema a t i s f i ngi +j ≡ Whenn =2 rkeachAijs y 0( mod2)wi t hco l orr ed ( e s en t edbyshadeda r ea s),andt ha t pr Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s a t i s f i ngj -i ≡ 3( mod4)andj -i ≢j +i( mod4)wi t hb l ue y ( )( ) e s en t edbyob l i i nea r ea s s eeFi .7 .1 . pr quel g F i .7 .1 g F i .7 .2 g I nt h i swa ans e et ha te v e r e l lad a c en tt oab l u eonei s y,wec yc j r ed,andt he r ei se x a c t l l u ec e l la r ounde a chr edone. yoneb Now wedot heope r a t i ononeachb l uece l l.Thenumbe ri n eachr edc e l li st henchangedf r om +1t o -1,wh i l et henumbe r s i nt her ema i n i ngc e l l sa r eunchanged. S i nc eni sanevennumbe r,wecanro t a t et heg r i da roundi t s cen t e rO an t i c l ockwi s e by 90 °.Then a l lt her edc e l l soft he r o t a t edg r i dcove rexac t l l lt hec e l l st ha ta r eno tr edi nt he ya or i i na lg r i d( s eeFi .7 .2). g g Wedot heope r a t i on saga i nf ort heor i i na lg r i dona l lt he g c e l l st ha ta r ecove r edbyb l uec e l l so ft hero t a t edg r i d.Thenwe s eet ha ta l lt henumbe r swi t hva l ue +1i nt her ema i n i ngce l l sof t heor i i na lg r i da r echangedt o -1,wh i l et henumbe r si nt he g o t he rc e l l sa r eunchanged. The r e f o r e,whenni sane v ennumbe r,a l lt henumb e r si nt he c e l l so ft heg r i dc anb echang edt o -1byaf i n i t eo fope r a t i on s. Whenni sodd,wedeno t et henumbe ri nce l lAiia sM i( i= Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 239 1,2,..., n),anddeno t et henumbe r ofope r a t i on soneacho ft he i rad a c en t j c e l l sa sx1,x2,...,xn-1 ,y1,y2,..., Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. e s c t i ve l s eeF i .7 .3). yn-1 ,r pe y( g Af t e rf i n i t eope r a t i on s,i ti sea s y t os eet ha t: M 1i schangedf r om +1t o -1i f andon l sodd; yx1 +y1i F i .7 .3 g M 2i schangedf rom +1t o -1i fandon l yx1 +y1 +x2 +y2 i sodd; M 3i schangedf rom +1t o -1i fandon l yx2 +y2 +x3 +y3 i sodd; ︙ M n-1i schangedf r om +1t o -1i fandon l yxn-2 +yn-2 + sodd,and xn-1 +yn-1i M ni schangedf r om +1t o -1i fandon l sodd. yxn-1 +yn-1i S i nceni sodd,t hes umo fnoddnumbe r si ss t i l lodd.Then, (x1 +y1 ) + (x1 +y1 +x2 +y2 ) + (x2 +y2 +x3 +y3 ) + … + (xn-2 +yn-2 +xn-1 +yn-1 ) + (xn-1 +yn-1 ) = 2( x1 +x2 + … +xn-1 +y1 +y2 + … +yn-1) i sodd.I ti simpo s s i b l e! The r e f or e,a l lt henumbe r si nt hec e l l sofag i venn ×ng r i d canbechangedf r om +1t o -1a f t e raf i n i t eo fope r a t i onsi f andon l fni sanevennumbe r. yi 8 F i nd a l lt he pr ime numbe r s p,for wh i ch t he r ea r e n+1 + i nf i n i t e l s i t i vei n t ege r sn,s ucht ha tp |n y manypo n ( n +1) .(po s edbyChenYonggao) So l u t i on.nn+1 n + ( n + 1) i sa lway sanodd numbe rforany Ma t hema t i c a lOl i adi nCh i na ymp 240 s i t i vei n t ege rn, s ot her equ i r edp wi l lno tbe2.Now wea r e po i ngt o pr ovet ha t,f oranypr imenumbe rp ≥ 3,t he r ea r e go i nf i n i t e many po s i t i vei n t ege r st ha t mee tt he cond i t i on. We e s en ttwoproo f si nt hef o l l owi ng. pr Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Proo f1.Foranypr l e tn =pk -2beanodd imenumbe rp ≥3, numbe r.Then n k 1 pk-1 pk-2 ≡( -2) +( -1) ≡ 2p - -1 nn+1 + ( n +1) k ·2k-1 -1 ≡ 2k-1 -1( ≡( 2p-1) modp). Nex t,l e tk -1 = ( t. p -1) n ≡ 0( modp). Thennn+1 + ( n +1) The r e for e,whenn = p( t + p - 2 (whe r eti sany p - 1) n s i t i vei n t ege r),wehavep|nn+1 + ( n +1) . po Proo f2.Gi venanypr imenumbe rp ≥3,weh ave( 2,p)=1. ByFe rma t sLi t t l eTheor em,weknow2p-1 ≡ 1( modp).Le tn t , , , , = p -2 wh e r et = 1 2 3 ....Wehave n p -1 p -2 ≡( -2) +( -1) nn+1 + ( n +1) t t 1 p ≡ 2p - -1 ≡ ( 2p-1) t-1 t t-2 … + +p+1 +p ≡ 1 -1 ≡ 0( modp). -1 Thepr oo fi scomp l e t e. 2013 ( L a n z h o u,Ga n s u) F i r s tDay 00,Augus t17,2013 8: 00 12: 1 Dot he r eex i s ti n t ege r sa,b andc s ucht ha ta2b c + 2, Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 241 ab2c +2,ab c2 + 2 a r e pe r f ec ts r e s. ( s ed by Li qua po Qi u s heng ) So l u t i on. No.Suppos et he con t r a r ha tt he r ea r es uch yt i n t ege r sa,bandc. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. I foneo ft hemi seven,s aya, c +2 ≡ 2 ( t hena2b mod4), sa pe r f ec t wh i chcon t r ad i c t st he a s s ump t i ont ha ta2b c + 2i s r e.Wemayt hena s s umet ha ta,bandca r eodd,s ot hey qua a r ee i t he r1 or3 (mod4).I tfo l l owsf rom t he P i -ho l e geon Pr i nc i l et ha ttwooft hem a r econg r uen tmodu l o4.Re l abe li f p nec e s s a r s s umet ha ta ≡b( mod4), s oab c2 +2 ≡c2 + y,wemaya 2 ≡ 1 +2 ≡ 3 ( mod 4),wh i ch v i o l a t e st he pe r f ec ts r e qua a s s ump t i on. 2 Le tnbeani n t ege r, n ≥2,andx1 ,x2 ,...,xn ∈ [ 0,1]. Provet ha t ∑ kxkxl ≤ 1≤k<l≤n n n -1 kxk . ∑ 3 k =1 ( s edbyLeng Gang s ong ) po So l u t i on.Asx1 ,x2 ,...,xn 3 ∑ kx x 1≤k<l≤n k l = ∑ ∈[ 0,1],xixj ≤xi , s owehave 3 kxkxl ≤ 1≤k<l≤n ∑ ( kxk +2 kxl ). 1≤k<l≤n nt hel a s ts umi s For1 ≤k ≤ n, t hecoe f f i c i en to fxki 2[ 1 +2 + … + ( k -1)]+k( n -k)=k( n -1), s owehave 3 ∑ kxkxl ≤ 1≤k<l≤n ∑ ( kxk +2 kxl )= 1≤k<l≤n n ∑kx =( n -1) k=1 k , andhenc et hede s i r edi nequa l i t l d s. yho n ∑k(n -1)x k=1 k Ma t hema t i c a lOl i adi nCh i na ymp 242 3 In △ABC ,po i n tB2 i st her e f l ec t i on oft hecen t e rof B-exc i r c l ewi t hr e s tt ot he mi dpo i n tofs i deAC ,and pec i r c l ewi t h i n tC2i st her e f l e c t i ono ft hecen t e rofC-exc po r e s c tt ot he mi dpo i n to fs i de AB .The A-exc i r c l e pe Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t ouche ss i de BC a t po i n tD .Pr ovet ha tAD ⊥ B2C2 . ( s edbyBi anHongp i ng) po So l u t i on.Le tA1bet hec en t e rof A-exc i r c l e. By t he prope r t i e s abou tc en t e r so f ex c i r c l e s,t he f o l l owi ngs e t soft hr ee po i n t sa r e B1 ,A ,C1},{ A1 ,C , co l l i nea r:{ B1}and{ C1 ,B ,A1},andA1A ⊥ B1C1 . F i .3 .1 g Ont hep l ane,choo s epo i n tP → → hen i t s uch t ha t C2P = B2C ,t →t → .AsBC → =C → f o l l owsf r omB2→ C = AB a tC2→ P = AB nd 1 h 1 2 1A a t hepo i n t sC1 ,B ,A1 a r eco l l i nea r,t he po i n t sB ,C2 ,P a r e → .I P =C B tf o l l owsf rom co l l i nea r,s oB→ 1 ∠AC1B = 1 80 °= 1 °- ∠BAC ö÷ æç180 °- ∠ABC ö÷ æç180 2 2 è ø è ø ∠BAC + ∠ABC 2 = 180 °- ∠ACB = ∠BCA1 2 tA1D andA1A bet hea l t i t ude sof t ha t△A1BC ∽ △A1B1C1 .Le t he △A1BC and △A1B1C1 ,r e s c t i ve l t hr e s c tt ot he pe y,wi pe B1C1 A1A = oppo s i t es i de s,s o . BC A1D BP A1A →, = P =C1B henA1A ⊥ B1C1 , s o . I fBP ⊥ I fB→ 1 t BC A1D A1A ,t henBC ⊥ A1D ,s o wehave △BPC ∽ △A1AD ,and → = C→ henceCP ⊥ AD .Aga i nbyC2→ P = B2→ C ,oneha sB2C P, 2 Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 243 andhenc eAD ⊥ B2C2 . The r ea r en( n ≥ 2)co i nsi nar ow.I foneoft heco i nsi s 4 head,s e l e c tanoddnumbe ro fcons e cu t i veco i n s( oreven Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1co i n)wi t ht heonei nheadont hel e f tmo s t,andt hen f l i l lt hes e l e c t ed co i n sups i de down s imu l t aneou s l pa y. sa l l owedi fa l lnco i nsa r et a i l s. Th i si samove.Nomovei Suppo s en c o i n sa r ehe ad sa tt hei n i t i a ls t ag e,de t e rmi nei f 2n+1 t he r ei sawayt oc a r r t mov e s.( s edbyGuB i n) you po 3 So l u t i on.Theansweri spo s s i b l e. Fo ranyc on f i u r a t i ono ft hec o i n s,wede f i neac o r r e s i ng g pond 01 s i ft hei equenc ec1c2 ... cn ofl eng t hna sfo l l ows: ci =1, -t h co i nf r omt hel e f ti shead,o t he rwi s e, ci = 0.I ti sea s os ee yt i n sa sone t o onecor r e s o t ha tt hes t a t u so ft hen co pondencet s uch01 s equenc e s,s oi nt hef o l l owi ng,we wi l lcons i de rt h i s s equenc emode li n s t ead. In i t i a l l hes equenc ei s11…11,deno t edby1n (wi t hn y,t con s e cu t i ved i i t sof1).S imi l a r l t edby0n ( wi t h g y,00…00,deno nd i i t so f0).Forany01 s equenc ewi t ha tl ea s tad i i t“ 1”, g g con s i de rt hef o l l owi ng move:l oca t et hef i r s td i i t“ 1”f rom g r i tt ol e f ti nt hes equenc e,t hent aket he01 s ub s equence gh f r oml e f tt or i ts t a r t i ngt h i s“ 1”o fmax ima loddl eng t h,and gh changet he01 pa r i t nt h i ss ub s equencej u s tl i kef l i i ngt he yi pp co i n si namove.Deno t ebyant het o t a lnumbe rofmove si nt he ways t a t edabove.Wec l a im: an = t os eet ha ta1 =1 = an = 2n+1 .Whenn =1, i ti sea s y 3 22 , oc eedbyi nduc t i on.As s umef ormu l a pr 3 2n+1 2k+1 i. e., ho l d sf orn =k, ak = . 3 3 Ma t hema t i c a lOl i adi nCh i na ymp 244 Now,wed i s cu s st heca s en =k +1, i. e.,wewan tt of i nd t het o t a lnumbe ro fmove sa sde s c r i bedabovei ft hes equencei s 1k+1 , i nsi nar ow. i. e,t he r ea r ek +1co I fki sodd,t henbyi nduc t i onhypo t he s i s,t hes equence1k+1 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( i i t so f0)a f t e r i i t so f1)change st o10k ( wi t hkd wi t hk +1d g g ak move sa ss t a t edabove.Af t e ranadd i t i ona lmove,i tchange s t o01k-10 (wi i i t sof1),t hena f t e rapp l i ngak - 1 t hk - 1d g y move s,t hes equencechange st o0k+1 ( wi t hk +1d i i t so f0).In g f ac t,r e ca l lt ha ti ns equenc eo fak move sf rom1k t o0k by,t he f i r s tmovei sf r om1k t o1k-10.The r e f or e,wehave ak+1 = 2ak = 2 2k+1 2k+1 -1 2k+2 -2 2k+2 = 2· = = . 3 3 3 3 I fki seven,byt hei nduc t i ona s s ump t i oni tt ake sak move s f r om1k+1t henapp l i t i ona lf rom10k t o01k ,and o10k ,t yanadd f i na l l hei nduc t i on a s s ump t i on aga i n,i tt ake sak move s y by t f r om01k t o0k+1 .Thenwehave ak+1 = 2ak +1 = 2 = 2k+1 2k+1 -2 +1 = 2· +1 3 3 2k+2 -1 2k+2 = . 3 3 Byi nduc t i ont ha tan = 2n+1 , hencet he r eex i s t sa wayt o 3 maket her equ i r ednumbe ro fmove s. S e c ondDay 8: 00 12: 00,Augus t18,2013 5 1,2,3,...,n} Anon emp t e tA ⊆ { i sca l l edagoods e t ys Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 245 nx.Deno o r e eni f|A |≤ mi t ebyan t henumbe rof fde g x∈A or e t so fdeg r een.Pr ovet ha tan+2 = an+1 +an +1f goods anypo s i t i vei n t ege rn.( s edbyLiWe i po gu ) So l u t i on1.Le e tofdeg r een,and|A |= k, tA beagoods Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. oA ⊆ { he t henmi nx∈Ax ≥ k,s k,k + 1, ...,n}.Hencet k numbe ro fgoods e t sofdeg r een wi t hk e l emen t si sCn I t -k+1 . f o l l owst ha tan = ∑ [ n2+1 ] k=1 1 2 3 k Cn n +C n-1 + C n-2 + … . -k+1 = C I fni t hen seven, n = 2m , 2 m+1 a2m+2 = C1 2m+2 + C 2m+1 + … + Cm+2 0 1 m +1 =( C1 C2 Cm 2m+1 + C 2m+1 )+ ( 2m + C 2m )+ … + ( m+1 + Cm+1 ) m+1 2 2 =( C1 C1 2m+1 + C 2m-1 + … 2m + … + Cm+1 )+ ( 2m + C + Cm+1)+ C2m+1 0 m = a2m+1 +a2m +1. I fni sodd, n = 2m -1, t hen 2 m m+1 a2m+1 = C1 2m+1 + C 2m + … + Cm+2 + Cm+1 0 1 m m-1 m =( C1 C2 Cm 2m + C 2m )+ ( 2m-1 + C 2m-1)+ … + ( m+1)+ C m +1 + C 2 m 2 =( C1 C1 2m + C 2m-1 + … + Cm+1 )+ ( 2m-1 + C 2m-2 + … + Cm2m +1 + Cm )+ C m 1 m 0 = a2m +a2m-1 +1. Ins umma r heequa l i t l d sfora l l y,t yan+2 = an+1 +an +1ho s i t i vei n t ege r sn. po Remark.Le tFn bet hen -t ht e rmo fFi bonac c is equence.From [ n2 ] k t hecomb i na t or i a li den t i t r i ve y ∑k=0 Cn-k = Fn ,onecan de t ha tan = Fn+1 -1. So l u t i on2.I fn = 1, t he r ei son l e to fdeg r ee1, yonegoods name l 1}, s oa1 = 1. y{ hent he r ea r eon l e t sofdeg r ee2: I fn = 2,t ytwogoods Ma t hema t i c a lOl i adi nCh i na ymp 246 { 1},{ 2},s oa2 = 2. r et henumbe r soft henon emp t Re ca l lt ha tan ,an+1a ygood r e s t i ve l s ub s e t sA o 1,2,...,n +1}, f{ 1,2,...,n}and{ pec y, s a t i s f i ng|A |≤ mi nx∈Ax. y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Con s i de rt heca s en +2:Foranynon emp t e tA of ygoods deg r een +2,Ai sas ub s e to f{ 1,2,...,n +2}, t henwehave t hef o l l owi ngt hr eeca s e s: ( sno tcon t a i nt hee l emen tn +2; a)A doe ( b)A con t a i n sn +2,andha sa tl ea s t2e l emen t s; ( c)A = { n +2}. Int hefo l l owi ng,wefocu sont henumbe rofgoods e t sof n( a),i tfo l l owsf rom t s( a)and ( b).Foranygoods e tA i ype nx∈Ax andmaxx∈Ax ≤ n +1t ha tA i sagoods e tof |A | ≤ mi deg r een +1.Conve r s e l e tofdeg r een +1i sa l s oa y,anygoods t he r e f or e,t he r ea r eexac t l e tofdeg r een +2, yan+1 good goods s e t so ft a). ype ( a1 ,a2 ,...,ak ,n +2}ofdeg r ee Foranygoods e tA = { n +2oft b),whe r ea1 < a2 < … < ak < n +2.Asa1 = ype ( mi nx∈Ax ≥|A |≥ 2,onecancons i de rt henon emp t e tA' = ys { a1 -1,a2 -1,...,ak -1},whe r e1 ≤a1 -1 <a2 -1 < … < ak -1 <n +1, andA's a t i s f i e s|A'|=|A |-1 ≤a1 -1 = mi nx∈A'x,henc sagoods e tofdeg r een.Conve r s e l eA'i y,any e tofdeg r eencanber epr e s en t edi nt heformA' = { a1 goods 1,a2 -1,...,ak -1},whe r eA = { a1 ,a2 ,...,ak ,n +2} i s agoods e to fdeg r een +2o ft b),s ot hecor r e s s ype ( pondencei one t o onebe tweenA andA',andt he r ea r eexac t l e t s yan goods ft b). ofdeg r een +2o ype ( Ac cord i ngt ot hed i s cu s s i onont henumbe rofgoods e t sof t s( a),( b)and ( c),wehavean+2 = an+1 +an +1. ype Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 6 247 Ass howni nF i .6 .1,PA ,PB a r e g t angen tt ot hec i r c l e wi t hc en t e rO a tA andB ,po i n tC ( d i f f e r en tf rom A ,B ) i sonmi nora r cAB .Thel i ne Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. lt hr oughpo i n tC andpe rpend i cu l a r t st he ang l eb i s ec t or s t o PC mee ∠AOC a nd ∠BOC a tpo i n t sD and E ,r e s c t i ve l ovet ha tCD = pe y.Pr CE .( s edby HeYi i e) po j So l u t i on1.AsshowninFi .6 .2,Li ne g F i .6 .1 g PC mee t st hec i r c l e ☉O a tano t he rpo i n t F .Jo i nBC ,BE ,BF ,OF ,r e s t i ve l pec y. r e on ☉O , OE b i s ec t s S i nc e B, C a ∠BOC ,a nd henc e OE pe rpend i cu l a r l y s oCE = BE .ByFO = BO and b i s e c t sBC , PC ⊥ DE ,wehave ∠ECB = 9 0 ° - ∠FCB = 9 0 ° - 1∠ BOF = ∠OBF ,and s o△CEB ∽ 2 △BOF ,henc e CE CB = . BO BF F i .6 .2 g ① AsPBi st ang en tt o☉O, s o ∠PBC = ∠PFB,and△PCB ∽ △PBF ,andhenc e CB PC = . BF PB ② PC By ① and ② ,we have CE = BO · .Bys t r ymme y, PB PC CD =AO· . I tf o l l owsf r omAO =BO,PA =PBt ha tCD =CE. PA Ma t hema t i c a lOl i adi nCh i na ymp 248 So l u t i on2.Asshownin Fi .6 .3.Jo i n g AB ,AC ,BC .OD mee t sAC a tpo i n tM . As A , C a r e on ☉O ,s o OD b i s ec t s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ∠AOC , ∠DMC = 9 0 °andMC = 1 AC . 2 AsPC ⊥ DE , s oc os∠ACD =s i n∠ACP , and CD = MC AC = .③ c os∠ACD 2s i n ∠ACP F i .6 .3 g Le tR bet o her ad i u sof☉O .AsPA i st angentt o ☉O ,s ∠ABC = ∠CAP ,andt heni tfo l l owsf rom ③ andt heS i ne Lawt ha t CD = i n ∠ABC CP 2Rs s i n ∠CAP = R· = R· . 2s i n ∠ACP s i n ∠ACP AP CP , Bys t r sCE = R · andi tfo l l owsf rom ymme y,oneha BP AP = BPt ha tCD = CE . 7 Labe lt hes i de so far egu l a rn -goni nc l ockwi s ed i r ec t i oni n t e rmi nea l li n t ege r sn ( n ≥ 4) orde rwi t h1,2,...,n.De s a t i s f i ngt hef o l l owi ngtwocond i t i ons: y ( -gona 1)n -3non i n t e r s e c t i ngd i a l si nt hen r es e l e c t ed, gona wh i chs ubd i v i d et hen -goni n t on - 2non ov e r l app i ng t r i ang l e s,and ( 2)eacho ft hecho s enn -3d i agona l si sl abe l edwi t han i n t ege r,s ucht ha tt hes um ofl abe l ed numbe r son t hr ees i de so feacht r i ang l e si n( 1)i sequa lt ot he o t he r.( s edbyZouJ i n) po So l u t i on. The requ i r ed i n t ege r sa r et ho s es a t i s f i ng bo t h y cond i t i on s: n ≥4andn ≢2( mod4).Suppo s et ha tns a t i s f i e st he Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 249 cond i t i ons ( 1)and ( 2),wef i r s tpr ovet ha tn ≢ 2( mod4)a s hes umo ft hel abe l si nt hr ees i de sofany fo l l ows.Deno t ebySt t r i ang l e,andbym t hes um oft hel abe l si nt hen -3d i agona l s cho s en.Add i ng up t hes ums o fl abe l si nt hr ees i de s of a l l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t r i ang l e si nt hes ubd i v i s i oni n( a),eve r i agona lappea r stwi ce yd i nt ho s et r i ang l e s,i tf o l l owst ha t( n -2) S =( 1+2+ … +n)+ 2m .Suppo s en ≡2( mod4), t hen( n -2) Si sevenbu t( 1+2+ … +n)+2m i sodd,wh i chi sacon t r ad i c t i on,s on ≢ 2( mod4). Le tu sr ema rkt ha tt hecond i t i onont her egu l a rn -goni nt he ob l emcanber e l axedt oconvexn -gon, s owecans imp l i f he pr yt wr i t i ng upo ft hes o l u t i on. F i .7 .1 g Int hef o l l owi ng,wepr ovet ha ta l li n t ege r sn wi t hn ≥4and n ≢ 2( mod4)s a t i s f i t i ons ( 1)and ( 2). ycond TheF i .7 .1showsl abe l i ngf ort heca s en =4,5and7,and g onecanve r i f i r ec t l ha tbo t hcond i t i onsho l d. yd yt In t he f o l l owi ng, we show t ha ti fn s a t i s f i e s bo t h cond i t i on s,s odoe sn +4. Asshowni nFi .7 .2,l abe loned i agona lwh i chs ubd i v i de s g n t oaconvexn t heg i ven ( n +4) -goni -gonandaconvexhexagon ( non r egu l a ranymor e).Asns a t i s f i e scond i t i ons ( 1)and ( 2), ubd i v i det heconvexn -goni n - 2)t onecans n t o( r i ang l e ss uch t ha tt hes umo ft het hr eel abe l si neacht r i ang l ei sequa lt ot he 250 Ma t hema t i c a lOl i adi nCh i na ymp s amenumbe rS.Onecana l s os ubd i v i det hehexagonwi t ht hr ee n -1,n +1andS -2n -5.Onecan d i agona l swi t hl abe l s:S -2 checkt ha tt hes um oft het hr eel abe l si neach oft he s ef our t r i ang l e si sa l s oS,henc en +4a l s os a t i s f i e scond i t i on s( 1)and Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 2).Soi tf o l l owst ha tanyi n t ege rns ucht ha tn ≥ 4andn ≢ 2( mod4)s a t i s f i e scond i t i on s( 1)and ( 2). F i .7 .2 g Ins umma r ho s ei n t ege r sns a t i s f i ngcond i t i on s( 1)and y,t y ( 2)a r eexac t l i venbyn ≥ 4andn ≢ 2( mod4). yg 8 ucht ha t( 2n -n2)|( an -na ) F i nda l lpo s i t i vei n t ege r sas s edbyYang Mi ng l i ang) f ora l lpo s i t i vei n t ege r sn ≥5.( po So l u t i on.Theonl swe r sf oraa r e2and4. yan Fi r s t,weprovet ha tai seven.I tfo l l owsf rom t heg i ven cond i t i onbychoo s i nganeveni n t ege rn ≥ 6. Nex t,weprovet ha ta ha snoodd pr imef ac t or.Suppo s e t hecon t r a r e tpbeanoddpr imef ac t orofa. I fp =3, l e tn = y,l sno td i v i s i b l e 8, t hen2n -n2 = 192ha saf ac t or3,bu tan -nai by3,con t r ad i c t i ngt o( 2n -n2)| ( an -na ),hencepi sno t3. Ch i na We s t e rn Ma t hema t i c a lOl i ad ymp 251 I fp = 5, l e tn = 16, t hen2n -n2 = 64110ha saf ac t or5, sno td i v i s i b l eby5,con t r ad i c t i ng( 2n -n2)|( an bu tan -nai na ),hencepi sno t5. I fp ≥ 7,l e tn = p - 1,i tf o l l owsf rom Fe rma t sLi t t l e Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Theor emt ha t2p-1 ≡ 1( modp). 2 As( modp), s op|( 2n -n2).Mor eove r,s i nce p -1) ≡1( a a ai sevenandp|a, s ona ≡ ( modp)and p -1) ≡ (-1) ≡1( n sno td i v i de ( an - na ),con t r ad i c t i ng p|a ,andhencep doe ( an -na ). 2n -n2)| ( Fi na l l ovet ha tai s2or4.Fort h i s, l e ta =2t whe r e y,wepr ti sapo s i t i vei n t ege r,t heni tf o l l owsf rom ( 2n -n2) 2tn -n2 ) |( t and ( 2n -n2)| ( ha t( 2n -n2)| ( n2 -n2t). 2tn -n2t)t t I fwechoo s ent obes u f f i c i en t l a r tfo l l owsf romt he yl ge,i t t n2 t ha tn2 -n2t =0,hence2t =2 t. t =1andt = n =0 n→ ∞ 2 f ac tl im 2a r eobv i ou ss o l u t i ons. I ft ≥3, t henbyt heBi nomi a lTheor em,wehavet =2t-1 = t-1 ( >1 + ( 1 +1) t -1)=t,wh i chi simpo s s i b l e.Atl a s t,one canea s i l ha ta =2anda =4s a t i s f hecond i t i oni nt he yt ycheckt ob l em,s ot hes o l u t i onsf oraa r e2and4. pr Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naSou thea s t e rn Ma thema t i ca lOl i ad ymp 2010 ( L u k a n h a n h u a,T a i wa n) g,C g F i r s tDay 8:00 12:00,Augues t17,2010 1 Le ta,b,c ∈ { 0,1,2,...,9}.Thequadr a t i cequa t i on ax2 +bx +c =0ha sar a t i ona lr oo t.Provet ha tt het hr ee d i i tnumbe rab ci sno tapr imenumbe r. g So l u t i on.Weprovebycont r ad i c t i on.I fab c = pi sa pr ime numbe r,t her a t i ona lr oo to fqu ad r a t i ce a t i onf( x)=ax2 +bx + qu c = 0i sx1 ,x2 = -b ± b2 -4ac 2 .Obv i ou s l aci sa y,b - 4 2 a Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 253 r f e c ts r enumbe r,andx1 ,x2 a r ea l lnega t i ve,and pe qua Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Thu s, So, x)= a( x -x1)( x -x2). f( 10)= a( 10 -x1)( 10 -x2). p = f( ap = ( 20 a -2ax1)( 20 a -2ax2). 4 I ti sea s os eet ha t( 20 a -2ax1)and ( 20 a -2ax2)a r ea l l yt s i t i vei n t ege r s.Cons equen t l 20 a -2ax1)orp|( 20 apo y,p|( ax2).I fp | ( 20 a - 2ax1),t henp ≤ 20 a - 2ax1 ,s o,80 2 8x1 -10x2 +x1x2 ≤ 0,wh i ch con t r ad i c t st o x1 ,x2 < 0. 20 a -2ax1) i sno tt r ue. S imi l a r l y,p| ( 2 Forany s e tA = { a1 ,a2 , ...,am },deno A) = t eP( a1a2 ... am .Le tA1 ,A2 ,...,andAn bea l l99 e l emen t 9 s ub s e t so f{ 1,2, ...,2010},n = C9 rove t ha t 2010 .P 2010 ∑ n P( Ai). i=1 So l u t i on1.Foreach99 e l emen t ss ub s e t,Ai = { a1 ,a2 ,..., 1,2,...,2010}un i l r e s st o99 e a99}of{ l emen t s que ycor pond b1, b2,..., b99}o s ub s e tBi = { f{ 1,2,...,2010}bybk =2011- ak ,k = 1,2,...,99. S i nc e∑k=1( ak +bk )=99×2011i sodd,wes eet ha tAi ,Bi 99 a r ed i f f e r en ts ub s e t so f{ 1,2,...,2010}.WhenAi t akea l l99 e l emen t ss ub s e to f{ 1,2,...,2010}, s oa r eBi .Mor eove r P( Ai)+P ( Bi) = a1a2 …a99 + ( 2011 -a1)( 2011 -a2)…( 2011 -a99) (-a2)…(-a99)( ≡ a1a2 …a99 + ( -a1) mod2011) ≡ 0( mod2011). Ma t hema t i c a lOl i adi nCh i na ymp 254 Thu s, n Ai)= 2∑P ( i=1 n n i=1 i=1 ∑P (Ai)+ ∑P (Bi)≡ 0(mod2011), ∑P (A ). henc e2011 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. n i=1 i So l u t i on2.Le tf( n)= ( n -1)( n -2)…( n -2010)-n2010 - 2010!,whe r en ∈ Z. S i nc e2011i spr ime,by Fe rma t sLi t t l eTheor em,n2010 ≡ 1( mod 2011). By Wi l s on s Theor em, we have 2010! ≡ -1( mod2011).Thu s, ( i)I f2011 n, t hen n)≡ ( n -1)( n -2)…( n -2010)≡ 0( mod2011). f( ( i i)I f2011|n, t hen n)≡ ( 2011 -1)( 2011 -2)…( 2011 -2010)-20112010 -2010! f( ≡2 010! -2010!( mod2011) ≡ 0( mod2011). Sof( n)≡ 0( mod2011)ha s2011s o l u t i onsi nt hes en s eof mod2011. S i ncef( i sapo l a loforde r2009,andf ora l ln ∈ n) ynomi Z,2011|f( n),wes eet ha teachcoe f f i c i en toff( n)canbe d i v i dedby2011. Tu rn t o t he or i i na l pr ob l em, g coe f f i c i en t of t e rm 2011 3 ∑ n P( Ai). ∑ n P( Ai) i s t he i=1 wi t h orde r 1911 of f( n), t hu s i=1 Ass howni nF i .3 .1.Le tt hei n s c r i bedc i r c l eIof△ABC g t ouchBCandABa tD andF, n t e r s e c tt he r e s c t i v e l tIi pe y.Le Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 255 s e t sAD andCFa tH and gmen K ,r e s c t i v e l ha t pe y.Prove t FD × HK = 3. FH × DK Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on 1. Suppos et ha tt he l eng t hso fs egmen t sa r eAF = x, BF = y, CD = z,t hen by S t ewa r t sTheor em,wehave AD2 = = F i .3 .1 g BD · CD · AC2 + AB2 -BD ·DC BC BC 2 2 +z( x +z) x +y) y( -yz y +z 2 4xyz . y +z =x + AF2 x2 = . ByTangen t Se can tTheor em,wehaveAH = AD AD Thu s, HD = AD - AH = S imi l a r l y,wehave KF = 4xyz AD2 -x2 = . AD AD ( y +z) 4xyz . CF ( x +y) S i nc e△CDK ∽ △CFD ,wes eet ha t DK = DF ×CD DF = z. CF CF Byt hef ac to f△AFH ∽ △ADF ,wehave FH = DF × AF DF = x. AD AD Ma t hema t i c a lOl i adi nCh i na ymp 256 Byt heCo s i neLaw, DF2 = BD2 +BF2 -2BD ·BFcosB æ è =2 y ç1 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. = 2 2 2 ( -( x +y) x +z) ö÷ y +z) + ( 2( x +y)( ø y +z) 4xy2z . ( x +y)( y +z) Thu s 4xyz 4xyz · KF × HD CF ( x +y) AD ( y +z) = DF ·DF FH × DK x z AD CF = 16xy2z = 4. x +y)( DF ( y +z) 2 App l i ngPt o l emy sTheor emt ocyc l i cquadr i l a t e r a lDKHF , wehave KF ·HD = DF ·HK +FH ·DK . KF × HD FD × HK = 4,weob = 3. Comb i n i ngwi t h t a i n FH × DK FH × DK So l u t i on 2. Fi r s t we pr ove a l emma. Lemma.Asshownin Fi .3 .2. g I ft hec i r c l et ouche sAB andACa t e s c t i ve l B and C ,r sa pe y,Q i n t e r s ec t s i n tont hec i r c l e.AQ i po t he c i r c l e a t po i n t P, t hen F i .3 .2 g wehave PQ ·BC = 2BP ·QC = 2BQ ·PC . Proo fo ft hel emma.ByPto l emy sTheor em,wes eet ha t PQ ·BC = BP ·QC +BQ ·PC . Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 257 S i n c eABt ang e n t st ot h ec i r c l e,wes e et ha t ∠ABP = ∠AQB. Fu r t he rby ∠BAP = ∠QAB ,wehave△ABP ∽ △AQB . Con s equen t l y, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. BP AP AB = = . BQ AB AQ Thu s, æçBP ö÷2 AP AB AP = × = . èBQ ø AB AQ AQ S imi l a r l y, æçCP ö÷2 AP = . èCQ ø AQ æCP ö÷2 æBP ö÷2 , = ç Thu s,ç t ha ti sBP ·QC = BQ ·PC .So èCQ ø èBQ ø PQ ·BC = 2BP ·QC = 2BQ ·PC . Thel emmai spr oved. Tu rnt ot heor i i na lpr ob l em. g Le tc i r c l eIi n t e r s e c tACa tpo i n tO anddr aws egmen t sHO ,OK ,OD andFO .Asshowni nF i .3 .3. g By Pt o l emy s Theor em, we have Thu s, F i .3 .3 g KF ·HD = DF ·HK +FH ·DK . FD × HK KF × HD = 3⇔ = 4. FH × DK FH × DK S i n c eCQandCDa r et ang en tt oc i r c l eI,byLemma,wes e et ha t KF ·DO = 2DK ·FO , HD ·FO = 2FH ·DO . Ma t hema t i c a lOl i adi nCh i na ymp 258 Mu l t i l t i on sbyeachs i de,wehave p yabovetwoequa KF ·HD ·DO ·FO = 4DK ·FH ·DO ·FO , Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. KF × HD = 4. t ha ti s, FH × DK 4 Le taandbbepo s i t i vei n t ege r ss ucht ha t1 ≤a <b ≤100. I ft he r eex i s t sapo s i t i vei n t ege rks ucht ha tab|( ak +bk ), a,b)i sgo t e rmi net he t henwes ayt ha tt hepa i r( od.De numbe rofgoodpa i r s. So l u t i on.Le t( a,b)=d,a =sd,b =td,( s, t)=1, t >1, t hens td2|dk ( sk +tk ).Sok ≥2ands t|dk-2( sk +tk ).S i nc e( s t, sk +tk )= 1,wehaves t|dk-2 .The r e for e,anypr imef ac t orof s tcanbed i v i dedbyd. I ft he r ei sapr imef ac t orp o fsortnol e s st han11,t henp d i v i de sd.So p2 |a orp2 |b,bu i chi sa tp2 > 100,wh con t r ad i c t i on. Sot hepr imef ac t oro fs t maybe2,3,5or7. I ft he r ea r ea tl ea s tt hr eepr imef ac t or sofs tamong2,3,5, e s st han5.Andd > 7,t hent he r ei sapr imef ac t oro fsortnol 2 × 3 × 5 = 30,s ot ha ta orb ≥ 5d > 100,wh i chi sa con t r ad i c t i on.The pr imef ac t ors e tofs t canno tbe { 3,7}, o t he rwi s e, aorb ≥ 7 ×3 ×7 > 100,wh i chi sacon t r ad i c t i on. S imi l a r l hepr imef ac t ors e tofs tcanno tbe { 5,7}. y,t The r e f or e,t hepr imef ac t ors e to fs tcanon l 2},{ 3}, ybe { { 5},{ 7},{ 2,3},{ 2,5},{ 2,7}or { 3,5}. ( i)I ft hep r imef a c t o rs e to fs ti s{ 3,5},t hend c anon l ybe 15.Then, s =3, t =5. Sot h e r ei son eg oodp a i r( a, b)= ( 4 5,7 5). ( i i)I ft hepr imef ac t ors e tofs ti s{ 2,7},t hend canon l y be14.Thens =2, t =7ors =4, t =7.Sot he r ea r etwogood Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 259 i r s( a,b)= ( 28,98)and ( 56,98). pa ( ti s{ 2,5},t hend canon l i i i)I ft hepr imef ac t orofs ybe 10or20. Ford = 10, t hens = 2, t = 5;s = 1, t = 10;s = 4, t= Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 5;s = 5, t = 8. t hens = 2, t = 5;s = 4, t = 5. Ford = 20, The r ea r es i xgoodpa i r s. ( i v)I ft hepr imef ac t oro fs ti s{ 2,3},t hend canon l ybe 6,12,18,24or30. Ford =6, s =1, t =6; s =1, t =12; s =2, t =3; s= 2, t = 9;s = 3, t = 4;s = 3, t = 8;s = 3, t = 16;s = 4, t = 9; s = 8, t = 9;s = 9, t = 16. Ford =12, s =1, t =6; s =2, t =3; s =3, t =4; s= 3, t = 8. Ford = 18,s = 2, t = 3;s = 3, t = 4. t = 3;s = 3,t = 4.d = 30,s = 2, Ford = 24,s = 2, t = 3. The r ea r e19goodpa i r s. ( v)I ft hepr imef ac t ors e tofs ti s{ 7},t hens =1, t =7,d canon l ybe7or14. So,t he r ea r etwogoodpa i r s. ( v i)I ft hepr imef ac t ors e to fs ti s{ 5},t hens = 1, t = 5, dcanon l he r ea r efourgoodpa i r s. ybe5,10,15or20.So,t ( v i i)I ft hepr imef ac t ors e to fs ti s{ 3}t hen wehavet he fo l l owi ng: whens = 1, t = 3,dcanon l ybe3,6,...,or33; whens = 1, t = 9,dcanon l ybe3,6or9; whens = 1, t = 27,dcanon l ybe3. The r ea r e15goodpa i r s. ( v i i i)I ft hepr imef ac t ors e to fs ti s{ 2},t henwehavet he Ma t hema t i c a lOl i adi nCh i na ymp 260 f o l l owi ng: t = 2,dcanon l whens = 1, ybe2,4,...,or50; whens = 1, t = 4, l t hendcanon ybe2,4,...,or24; whens = 1, t = 8, t hendcanon l ybe2,4,...,or12; Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t hendcanon whens = 1, t = 16, l ybe2,4or6; t = 32, t hendcanon l whens = 1, ybe2. The r ea r e47goodpa i r s. The r e for e,t he r ea r ea l lt oge t he r1 +2 +6 +19 +2 +4 + 15 +47 =96goodpa i r s. S e c ondDay 8:00 12:00,Augues t18,2010 5 Ass howni nF i .5 .1.Le tC g be t he r i l e o f ght ang △ABC . M 1 and M 2 a r e twoa rb i t r a r i n t si ns i de y po △ABC , and M i s t he mi dpo i n t o f M 1M 2. The e x t en s i on so fBM 1,BM and n t e r s e c tAC a BM 2i tN1,N F i .5 .1 g M 1N1 M 2N2 MN + ≥2 andN2r e s c t i v e l r ov et ha t . pe y.P BM 1 BM 2 BM So l u t i on.AsshowninFi .5 .2. g he Le t H1 , H2 and H be t o e c t i onpo i n t sofM 1 ,M 2 and pr j M onl i neBC , r e s c t i ve l pe y.Then M 1N1 H1C , = BM 1 BH1 F i .5 .2 g Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 261 M 2N2 H2C , = BM 2 BH2 H1C + H2C MN HC = = . BM BH BH1 +BH2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Suppo s et ha tBC = 1,BH1 = xandBH2 = y.Wehave M 1N1 H1C 1 -x , = = BM 1 BH1 x M 2N2 H2C 1 -y , = = BM 2 BH2 y MN HC 1 -x +1 -y = = . x +y BM BH Thu s,t hei nequa l i t r et oprovei sequ i va l en tt o y wea 1 -x +1 -y 1 -x 1 -y , + ≥2 x x +y y wh i chi sequ i va l en tt o 1 1 4 , 2 + ≥ ≥0 x -y) t ha ti s,( x x +y y wh i chi sobv i ou s l r ue. yt 6 Le tN* bet hes e to fpo s i t i vei n t ege r s.De f i nea1 =2,and f orn = 1,2,..., nλ an+1 = mi { 1 1 … 1 1 + + + + < 1, λ ∈ N* . a1 a2 an λ } 2 -an +1f Pr ovet ha tan+1 = an orn = 1,2,.... So l u t i on.Bya1 = 2,a2 = m i nλ { 1 1 + < 1, λ ∈ N* , a1 λ } 1 1 1 1 1, <1, = con s i de r + t hen <1λ >2,hencea2 = a1 λ λ 2 2 3.Sot heconc l u s i oni st r uef orn = 1. Suppo s et ha tt heconc l u s i oni st r uefora l li n t ege rn ≤ k - 1 1 … 1 1 + + + + 1( k ≥ 2).I fn =k, t henak+1 = mi nλ a1 a2 ak λ { Ma t hema t i c a lOl i adi nCh i na ymp 262 < 1, λ ∈N * } .Considering Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 1 … 1 1 + + + + < 1, a1 a2 ak λ t ha ti s,0 < 1 … 1 ö÷ , æ1 1 + + + <1- ç wehave ak ø èa1 a2 λ λ> 1 . 1 1 … 1 1a1 a2 ak Int hef o l l owi ng,wes howt ha t 1 = ak ( ak -1). 1 1 … 1 1a1 a2 ak Byt hei nduc t i on hypo t he s e s,f or2 ≤ n ≤ k,an = an-1 ( an-1 -1)+1,wehave 1 1 1 1 , = = an -1 an-1( an-1 -1) an-1 -1 an-1 1 1 1 = t he r e f or e .Byt ak i ngt hes um,wehave an-1 an-1 -1 an -1 k 1 ∑a i=2 i-1 1 , t ha ti s ak -1 =1k 1 ∑a i=1 i 1 1 1 + =1. ak -1 ak ak ( ak -1) =1- Con s equen t l y, The r e for e, 1 1 … 1 1 1a1 a2 ak = ak ( ak - 1) . 1 1 … 1 1 + + + + < 1, λ ∈ N* ak+1 = mi n λ| a1 a2 ak λ { = ak ( ak -1)+1. } Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 263 Byi nduc t i ononn,f ora l lpo s i t i vei n t ege rn,wehave,an+1 = 2 -an +1. an The r ea r e2 nr ea lnumbe r sa1 ,a2 ,...,an ,r1 ,r2 ,..., 7 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. andrn s a t i s f i nga1 ≤a2 ≤ … ≤an and0 ≤r1 ≤r2 ≤ … y ∑ ∑ ≤rn .P r ovet ha t aiaj mi n( ri ,rj )≥ 0. n n i=1 j=1 So l u t i on.Wr i t eama t r i xo fn ×ne l emen t sa sfo l l ows: æa1a1r1 ç ça2a1r1 ç A1 = ça3a1r1 ç ︙ ç èana1r1 a1a2r1 a1a3r1 a2a2r2 a2a3r2 a3a2r2 a3a3r3 ︙ ︙ ana2r2 ana3r3 … a1anr1 ö ÷ … a2anr2 ÷ … a3anr3 ÷ . ÷ ︙ ÷ ⋱ ÷ … ananrn ø S i nce n n ri ,rj )= ∑ ∑aiajmin( i=1 j=1 n n r1 ,rj )+ ∑a2ajmi n( r2 ,rj ) ∑a1ajmin( j=1 j=1 +… + n + i t skt ht e rmi s n r ,r )+ … ∑aa min( j=1 n j r ,r )= aar ∑aa min( k j k k j r ,r ), ∑aa min( j=1 n j=1 k j k 1 1 j n j +aka2r2 + … +akakrk +akak+1rk + … +akanrk wh i chi st hes um o fe l emen t so ft hekt hrow o fA1 ,k = 1, 2,...,n. The r e for e,∑i=1 ∑j=1aiajmi n( ri ,rj )i st hes um ofa l l e l emen t so fA1 . n n 264 Ma t hema t i c a lOl i adi nCh i na ymp Ont he o t he rhand,t hes umma t i on can a l s o be donea s f o l l ows:Taket hee l emen t so ff i r s tco l umnandt hef i r s trowof A1 , s umup;deno t et her e s t( n -1)× ( n -1)e l emen tby ma t r i x A2 , t hent aket hef i r s tco l umnandt hef i r s trowofA2 , s umup, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. deno t et her e s tby ma t r i xA3 ,...,s owege t n n ri ,rj )= ∑ ∑aiajmin( i=1 j=1 = n ∑r (a k=1 n n ∑r ( ( ∑a ) k=1 k i=1 +rn i i=k n ( ∑a ) i 2 n ( ∑a ) i=n n r ∑( k 2 +r2 i - … -rn-1 k=1 n ∑ai ) - ( ∑a ) ) 2 i=k+1 k=1 =r1 = n ∑rk ( (ak + n = +2 ak ( ak+1 +ak+2 + … +an )) 2 k k 2 n - ( ∑a ) ) i=k+1 n ( ∑a ) i i=2 -r1 n i i=2 2 2 n ( ∑a ) i=3 -r2 i 2 +… n ( ∑a ) i=3 i 2 2 n i=k i i 2 +r3 n ( ∑a ) -rk-1) 2 i ( ∑a ) ( ∑a ) i=n i=k+1 i 2 ≥0 ( whe r er0 = 0). 8 Gi vene i tpo i n t sA1 ,A2 ,...,A8onac i r c l e,de t e rmi ne gh t hesma l l e s t po s i t i vei n t ege rn s uch t ha tamong anyn t r i ang l e swi t hve r t i ce si nt he s ee i i n t s,t he r ea r etwo ghtpo wh i chhaveacommons i de. So l u t i on.Fi r s t,wecon s i de rt he max ima lnumbe ro ft r i ang l e s wi t hnocommons i de. Con s i de rt hemax ima lnumbe ro ft r i ang l e swi t hnocommon s i depa i rwi s e.The r ea r eC2 8chord sby conne c t i ng e i t 8 = 2 gh Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 265 i n t s.I feachchordon l l ong st o po ybe onet r i ang l e,t hent he s echord scan Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. on l ormr ≤ yf [238] =9triangleswith no common s i de pa i rwi s e. Bu ti f t he r ea r en i nes uch t r i ang l e s,t hen t he r ea r e27ve r t i c e s.So,t he r ei sone i n ti ne i t po i n t s, wh i chi st he po gh common ve r t ex o ff ou rt r i ang l e s. F i .8 .1 g hen e i tedge sa r econne c t edt o Suppo s et ha tpo i n ti sA8 ,t gh s evenpo i n t sA1 ,A2 , ...,A7 .So,t he r e mu s tex i s tanedge A8Ak ,wh i chi st hecommons i deo ftwot r i ang l e s,wh i chi sa con t r ad i c t i on.Sor ≤ 8. Ont heo t he rhand,whenr = 8,wecan makes uche i ght t r i ang l e s,s eef i e.Deno t et het r i ang l e sbyt hr eeve r t i ce sa s: gur ( 1,2,8),( 1,3,6),( 1,4,7),( 2,3,4),( 2,5,7),( 3,5, s9. 8),( 4,5,6)and ( 6,7,8).Sot hemi n ima lnumbe ro fni 2011 ( N i n bo,Zh e i a n g j g) F i r s tDay l 8:00 12:00,Ju y27,2011 1 I fmi nx∈R ax2 +b = 3, f i nd x2 +1 ( 1)t her angeo fb; Ma t hema t i c a lOl i adi nCh i na ymp 266 ( 2)t heva l ueofaf org i venb.( s edbyLuXi ng i ang) po j ax2 +b So l u t i on1.Denot ef( x)= . I ti sea s os eet ha ta > yt 2 x +1 0.Byf( 0)=b,wes eet ha tb ≥ 3. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( i)I fb -2 a ≥ 0, ax2 +b b -a 2 = a x +1 + ≥ 2 a( x)= b -a) = 3, f( 2 x +1 x2 +1 e u a l i t l d si fa x2 +1 = q yho b -a , b -2 a t ha ti s, i fx = ± . 2 a x +1 Theva l ueo faf org i venbi sa = whenb = 3,a = 3 . 2 b - b2 -9, e s i a l l pec y 2 ( i i)I fb -2 a <0, l e t x2 +1 =t( t ≥1).f( x)=g( t)= b -a o, a t+ mono t on i ca l l nc r ea s i ngwhent ≥ 1,s yi t mi nf( x)= g( 1)= a +b -a =b = 3,whena > x∈R 3 . 2 Summi ngup,wege t( 1)t her angeofbi s[ 3,+ ∞ ). 3 b - b2 -9 ( 2)I fb =3, t hena ≥ ; i fb >3, t hena = . 2 2 ax2 +b So l u t i on2.Le tf( x)= . I ti sea s os eet ha ta > 0. yt 2 S i nc emi nx∈R 2 x +1 ax +b = 3, andf( 0)=b,wes eet ha tb ≥ 3. x2 +1 a ö÷ æ 2 b -2 ax çx a è ø '( x)= . f 3/2 ( x2 +1) ( a ≤0, l e tf x)=0,wehavet i)I fb -2 '( hes o l u t i onx0 = t henf '( x)<0;andi t henf '( x)>0. 0,andi fx <0, fx >0, Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 267 b 0)=bi st hemi n ima lva l ue.Henceb = 3,anda ≥ . f( 2 ( a >0, '( x)=0,wehavet i i)I fb -2 l e tf hes o l u t i onsx0 = Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 0,x1,2 = ± b -2a . a sno tt he mi n ima lva l ue,wh i ch I ti sea s os eef( 0)= bi yt imp l i e sb > 3;andf( x1,2) i st hemi n ima lva l ue b -2a +b a· a ) ( = = 3⇒2 a( , x b -a) = 3 f 12 b -2a +1 a ⇒a2 -ab + 9 b - b2 -9, = 0⇒a = 4 2 t ha ti s, b > 3anda = b - b2 -9 . 2 Summi ngup,wege t ( 1)t her angeo fbi s[ 3,+ ∞ ). 3 b - b2 -9 ( 2)I fb =3, t hena ≥ ; i fb >3, t hena = . 2 2 2 Le ta,bandcb ec op r imepo s i t i v ei n t e e r ss ot ha ta2|( b3 + g c3),b2|( a3 +c3)andc2|( a3 +b3).Fi ndt heva l ue sofa, bandc.( s edby Yang Xi aomi ng) po So l u t i on.Bythecond i t i onoft heprob l em,wehavea2|( a3 + b3 +c3),b2 | ( a3 +b3 +c3)andc2| ( a3 +b3 +c3).S i ncea,b andca r ecopr ime,wes eet ha ta2b2c2 | ( a3 +b3 +c3). Wi t hou tl o s sofgene r a l i t uppo s et ha ta ≥b ≥c,s o y,s 3 a3 ≥ a3 +b3 +c3 ≥ a2b2c2 ⇒a ≥ b2c2 , 3 Ma t hema t i c a lOl i adi nCh i na ymp 268 and 2 b3 ≥b3 +c3 ≥ a2 ⇒2 b3 ≥ b4c4 18 ⇒b ≤ 4 . 9 c Wes eet ha ti fc ≥ 2⇒b ≤ 1,t her e s u l tcon t r ad i c t sb ≥c. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Thu s, c = 1. 1,1,1) i s I fc = 1andb = 1, t hena = 1, s o( a,b,c)= ( as o l u t i on. I fc = 1,b ≥ 2anda = b,t i chi sa henb2 |b3 + 1,wh con t r ad i c t i on! I fb ≥ 2anda >b >c = 1, t hen 2 b a2b2 | ( a3 +b3 +1)⇒2a3 ≥ a3 +b3 +1 ≥ a2b2 ⇒a ≥ , 2 andbyc = 1, 4 b a2 | ( b3 +1)⇒b3 +1 ≥ a2 ≥ ⇒4 b3 +4 ≥b4 . 4 Forb >5, t hei nequa l i t snos o l u t i on.Takeb =2,3,4, yha 5,wes eet ha tt hes o l u t i on sa r ec = 1,b = 2,a = 3. The r e f or e,a l ls o l u t i on sa r e( a,b,c)= ( 1,1,1),( 1,2, 3),( 1,3,2),( 2,1,3),( 2,3,1),( 3,2,1)and ( 3,1,2). 3 Le ts e tM = { 1,2,3,...,50}.Fi nda l lpo s i t i vei n t ege r n, s ucht ha tt he r ea r ea tl ea s ttwod i f f e r en te l emen t saand bi nanys ub s e twi t h35e l emen t sofM , s ucht ha ta +b =n ora -b = n.( s edbyLiShenghong) po So l u t i on.T akeA ={ 1,2,3,...,35}, t henf o ranya,b ∈ A , a -b ≤ 34,a +b ≤ 34 +35 = 69. Int hefo l l owi ng,wes howt ha t1 ≤ n ≤ 69.Le tA = { a1 , Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 269 a2 ,...,a35},wi t hou tl o s so fgene r a l i t uppo s et ha ta1 < a2 y,s < … < a35 . ( i)I f1 ≤ n ≤ 19,by 1 ≤ a1 < a2 < … < a35 ≤ 50, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 2 ≤ a1 +n < a2 +n < … < a35 +n ≤ 50 +19 = 69, andby Di r i ch l e t sDr awe rTheor em,t he r eex i s t1 ≤i,j ≤ 35 ( i ≠j)s ucht ha tai +n = aj , t ha ti s, ai -aj = n. ( i i)I f51 ≤ n ≤ 69,by 1 ≤ a1 < a2 < … < a35 ≤ 50, 1 ≤ n -a35 < n -a34 < … < n -a1 ≤ 68, andby Di r i ch l e t sDr awe rTheor em,t he r eex i s ta tl ea s t1 ≤i, i ≠j)s ai +aj = n. ucht ha tn -ai = aj , t ha ti s, j ≤ 35( ( i i i)I f20 ≤ n ≤ 24, s i nc e 50 - ( 2 n +1)+1 = 50 -2n ≤ 50 -40 = 10, ha t wes eet ha tt he r ea r ea tl ea s t25e l emen t si na1 ,a2 ,...,a35t be l ongt o[ 1,2 n]. The r ea r ea tmo s t24e l emen t si n{ 1,n +1},{ 2,n +2},..., { n,2 n} s u cht ha t{ ai ,aj }= { i,n +i}.Henc e, aj -ai =n. ( i v)I f25 ≤n ≤34, s i nc e{ 1,n +1},{ 2,n +2},...,{ n, 2 n}havea tmo s t34e l emen t s,by Di r i ch l e tDr awe rTheor em, t he r eex i s t1 ≤i,j ≤ 35( i ≠j)s ucht ha tai =i,aj = n +i, t ha ti saj -ai = n. ( v)I fn =35, t he r ea r e33e l emen t s{ 1,34},{ 2,33},..., { 17,18},{ 35},{ 36},...,{ 50}.Hence,t he r eex i s t1 ≤i, j≤ 35( i ≠j)s ucht ha tai +aj = 35. ( v i)I f36 ≤ n ≤ 50, i fn = 2 k + 1,{ 1,2 k},{ 2,2 k - 1},...,{ k,k + 1}, { k +1},...,{ 50}; 2 i f18 ≤k ≤20,50- ( 2 k +1)+1 =50-2 k ≤50-36 =14; Ma t hema t i c a lOl i adi nCh i na ymp 270 i f21 ≤k ≤24,50- ( 2 k +1)+1 =50-2 k ≤50-42 =8, i ≠j) k +1 = n. t he r eex i s t1 ≤i, s ucht ha tai +aj =2 j ≤35( I fn = 2 k,{ 1,2 k -1},{ 2,2 k -2},...,{ k -1,k +1}, { k},{ k},{ k +1},...,{ 50}; 2 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i f18 ≤k ≤19,50- ( 2 k +1)+3 ≤16 k -1 ≤19-1 =18; i f20 ≤k ≤ 23,50 - ( k +1)+3 ≤ 50 -2 k +2 ≤ 12 k -1 2 ≤2 3 -1 = 22; k +1)+3 ≤50-2 k +2 ≤4 k -1 ≤ i f24 ≤k ≤25,50- ( 2 25 -1 = 24, t he r ee x i s t1 ≤i,j ≤ 35( i ≠j) s u cht ha tai +aj = 2 k. 4 Suppo s et ha tal i nepa s s i ngt hec i r cumc en t r eO of△ABC i n t e r s ec t sABandACa tpo i n t sM andN , r e s t i ve l pec y,and E andF a r et he mi dpo i n t so fBN andCM ,r e s t i ve l pec y. Pr ovet ha t ∠EOF = ∠A .( s edbyTaoP i ng sheng) po So l u t i on.Wes howt ha tt h ea bo v ec on c l u s i oni st r u ef o ranyt r i ang l e. I f △ABC i sr i ang l ed.Theconc l u s i oni sobv i ou s.In ght , , , f ac ts eeFi .4 .1 whe r e ∠ABC = 90 °.So t hec i r cumc en t r eO g i st hemi dpo i n tofAC ,OA = OB andN = O .S i nceF i st he mi dpo i n tofCM ,wes i neOF ‖AM .Hence eet ha tt hemed i anl ∠EOF = ∠OBA = ∠OAB = ∠A . I f△ABCi sno tr i t ang l ed,s eeFi .4 .2andFi .4 .3. gh g g F i .4 .1 g F i .4 .2 g F i .4 .3 g Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 271 F i r s tweg i veal emma. Lemma.Le tA andB betwo po i n t sont hed i ame t e rKL of c i r c l e☉O wi t hr ad i u sR ,andOA = OB = a.Seef i e. gur Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Le t CD and EF be two chord s pa s s i ng A and B , r e s c t i ve l s eCE and DF i n t e r s ec tKL a t M and N , pe y.Suppo r e s c t i ve l pe y.ThenMA = NB . Proo fo ft hel emma.Asshownin F i .4 .4.Suppo s et ha tCD ∩ EF = P . g Th i nko ft ha tl i ne sCE andDFi n t e r s e c t l a u s The o r em,wehav e △PAB.By Mene AC ·PE ·BM = 1, CP EB MA BF ·PD ·AN = 1. FP DA NB F i .4 .4 g Then MA AC ·AD ·PE ·PF ·BM = . NB BE BF PC PD AN ① Byt heIn t e r s e c t i ngChordTheor em,wege t PC ·PD = PE ·PF . ② So AC ·AD = AK ·AL = R2 -a2 = BK ·BL = BE ·BF . ③ MA MB , = By ① ,③ ,wehave t ha ti s NB NA MA MA + AB AB = = = 1. NB NB + AB AB Thu s,MA = NB . Now,r e t urn t ot he or i i na l prob l em,s ee F i .4 .5 and g g Ma t hema t i c a lOl i adi nCh i na ymp 272 F i .4 .6.Ex t endMN t od i ame t e rKK1 ,t akepo i n tM 1 onKK1 g tCM 1 ∩ ☉O = A1 ,andl s ucht ha tOM 1 = OM .Le e tA1B i n t e r s e c tKK1 a orM 1N1 = tN1 .Byt hel emma,MN1 = M 1N ( s, MN ont her i tf i e).So,Oi st hemi dpo i n tofNN1 .Thu gh gur Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. OE and OF a r et he med i anl i ne o f △NBN1 and △MCM 1 , r e s c t i ve l s,wehave ∠EOF = ∠BA1C = ∠A . pe y.Thu F i .4 .5 g F i .4 .6 g S e c ondDay 8:00 12:00,Ju l y28,2011 5 Le tAA0 ,BB0 andCC0 beangu l a rb i s ec t or sof △ABC . Le tA0A1 ‖BB0 andA0A2 ‖CC0 ,whe r eA1 andA2l i eon AC andAB,r e s c t i v e l pe y, andl e tl i neA1A2i n t e r s e c t BC a tA3.The po i n t sB3 and C3 a r e ob t a i ned s imi l a r l y. P r ov e t ha t i n t s A3, B3, C3 a r e po co l l i nea r.( s edby Tao po P i ng sheng) So l u t i on. By the Mene l au s F i .5 .1 g Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 273 Inve r s eTheor em,weneedon l os howt ha t yt AB3 ·CA3 ·BC3 = 1. B3C A3B C3A ① Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. S i nc el i ne A1A2A3 i n t e r s e c t s △ABC , by Mene l au s CA3 ·BA2 ·AA1 = 1. So Theor em,wehave A3B A2A A1C CA3 A2A ·A1C = . A3B BA2 AA1 ② S imi l a r l y,wehave AB3 B2B ·B1A , = B3C CB2 BB1 BC3 C2C ·C1B = . C3A AC2 CC1 ByBA2 = ③ ④ BC0 · AA0 · BA0 andAA2 = AC0 ,wehave BC AI AA2 AA0 ·AC0 ·BC = . BA2 BA0 ·BC0 AI Mor eove r,byAA1 = ⑤ AA0· CA0· AB0andCA1 = CB0 ,we AI CB have A1C CA0 ·CB0 ·AI = . AA1 AA0 ·AB0 BC Thenby ② ,⑤ and ⑥ ,wehave æCA0 ÷ö2 CA3 CA0 ·AC0 ·CB0 = = ç . èA0B ø A3B BA0 BC0 AB0 S imi l a r l y,wehave æ CB0 ö÷2 , CB3 = ç èB0A ø B3A ⑥ Ma t hema t i c a lOl i adi nCh i na ymp 274 æAC0 ö÷2 AC3 = ç . èC0B ø C3B ⑦ f△ABC a S i nc et hr eeang l eb i s e c t or sAA0 ,BB0 ,CC0 o r e Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. concu r r en t,byCeva sTheor em,wehave AB0 ·CA0 ·BC0 = 1. B0C A0B C0A ⑧ Thu s,by ⑦ and ⑧ ,wehave æAB0 ·CA0 ·BC0 ö÷2 AB3 ·CA3 ·BC3 = ç = 1, èB0C A0B C0A ø B3C A3B C3A t ha ti s①. 6 Gi vennpo i n t sP1 ,P2 ,...,Pn onap l ane,l e tM beany hep l ane.Deno t eby|PiM |t he i n tons egmen tAB ont po d i s t ancebe twe enPi andM , i = 1,2,3,..., n.Pr ov et ha t ∑ n i=1 x{ ∑i=1|PiA |,∑i=1|PiB|} . | PiM |≤ ma n n ( s edbyJ i n Mengwe i) po → So l u t i on.Le tO bet heor i i n.Then wehaveOM g ( B, t∈( 0,1). 1 -t)O→ → = tOA + → → |PiM | =|OM -OPi | → → -( → =|tOA + ( 1 -t)O→ B -tOP 1 -t)OP i i| → → → ≤t|OA -OPi |+ ( 1 -t)|O→ B -OP i| =t|PiA |+ ( 1 -t)|PiB |. Henc e, n n n → → ∑ |P M | ≤t∑ |P A |+ (1 -t)∑ |P B | i=1 i i i=1 i i=1 n n { ∑ |P A |,∑ |P B |} . ≤ ma x i=1 i i=1 i Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 7 275 Suppo s et ha tt hes equenc e{ an }de f i nedbya1 = a2 = 1, an =7an-1 -an-2 ,n ≥ 3.Provet ha tan + an+1 + 2i sa s edby Tao r f ec ts r ef oranypo s i t i vei n t ege rn.( po pe qua ) P i ng sheng Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on.I ti swe l lknownt ha tt hes o l u t i onoft hes equence canbeob t a i nedbys o l v i ngtwogeome t r i cs equenc e.Wege tan = C1λ1 +C2λ2 ,wh e r e n λ1 = n æ3 + 5 ö2 æ3 - 5 ö2 7 + 45 7 - 45 ÷ , λ2 = ÷ , = ç = ç 2 2 è 2 ø è 2 ø λ1 andλ2a r es o l u t i on soft heequa t i onλ2 -7 λ +1 =0,a1 = 1 = C1λ1 + C2λ2 ,a2 = 1 = C1λ2 λ2 1 +C2 2. The r e for e, n-1 -1 an +an+1 +2 =λn λ1 +λ2 λ2 +λ2 1 C1 ( 1 )+λ2 C2 ( 2 )+2 =λ1- +λ2- +2 n 1 n 1 n-1 2 æ æ3 + 5 ö ÷ = çç èè 2 ø 2 ø n-1 é æ3 + 5 ö ÷ = êê ç ëè ö æ æ - 5 ön-1 ö2 ÷ + ç ç3 ÷ ÷ +2 ø èè 2 ø ø n-1 2 æ3 - 5 ö ÷ +ç è 2 ø ùú 2 ú = xn , û ön-1 æ æ3 - 5 ön-1 ÷ + ç whe r exn = ç3 + 5 ÷ i st hes o l u t i onoft he è 2 ø è 2 ø s equence { xn }ofpo s i t i vei n t ege r sx1 =2,x2 =3,xn =3xn-1 xn-2 ,n ≥ 3. 8 Con s i de r12f i r e sont hec l ockf acea s12po i n t s.Co l or gu l l ow,b l ueandg r een.Each t hemi nf ou rco l or s:r ed,ye co l or i su s ed f or t hr ee po i n t s. Conf i e n convex gur , i l a t e r a l swi t hve r t i ce si nt he s epo i n t ss ucht ha t quadr ( 1)t he r ea r enos amec o l o ro fv e r t i c e sf o re a chquad r i l a t e r a l. ( 2)amonganyt hr eeo ft he s e quadr i l a t e r a l s,t he r ei sa 276 Ma t hema t i c a lOl i adi nCh i na ymp co l oro fve r t i ce ss ucht ha tt heve r t i ce soft ha tco l or a r ed i f f e r en t. s edbyTaoP i ng sheng) F i ndt hel a r s tnumbe ro fn.( po ge So l u t i on.Weus eA ,B ,C,D t or epr e s en tt he s efourco l or s, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. r e s c t i ve l hepo i n t si nt hes ameco l orbyl owe rl e t t e r sa s pe y,andt a1,a2,a3;b1,b2,b3;c1,c2,c3 andd1,d2,d3r e s c t i ve l pe y. Nowcon s i de rco l orA .I f,i nn quadr i l a t e r a l s,t henumbe r r en1 ,n2 ,n3r ofpo i n t sa1 ,a2 ,a3i nco l orA a e s c t i ve l hen pe y,t fn ≥ 10, n1 +n2 +n3 = n.Suppo s et ha tn1 ≥n2 ≥n3 .I t hen n1 +n2 ≥ 7.Cons i de rt he s es even quadr i l a t e r a l s( i t sA co l or ft henumbe r sofpo i n t sb1 ,b2 ,b3i n ve r t exi se i t he ra1 ora2 ),i co l orB a r em 1 ,m 2 ,m 3 , r e s t i ve l henm 1 +m 2 +m 3 =7. pec y,t t henm 3 Bys t r i c i t uppo s et ha tm 1 ≥ m 2 ≥ m 3 , ymme y,wemays ≤ 2, t ha ti s,m 1 + m 2 ≥ 5. l orpo i n ti se i t he r Con s i de rt he s ef i vequadr i l a t e r a l s( i t sAco a1 ora2 ,andi t sBco l orpo i n ti se i t he rb1orb2 ),i ft henumbe r s r ek1 ,k2 ,k3 ,r o fpo i n t sc1 ,c2 ,c3i nco l orC a e s c t i ve l hen pe y,t k1 +k2 +k3 =5.Bys t r i c i t uppo s et ha tk1 ≥k2 ymme y,wemays ≥k3 , t henk3 ≤ 1, t ha ti s, k1 +k2 ≥ 4. Con s i de rt he s ef our quadr i l a t e r a l s,deno t ed a s T1 ,T2 , i t sco l orA po i n ti se i t he ra1 ora2 , i t sco l orB po i n ti s T3 ,T4 ( e i t he rb1 orb2 ,andi t sco l orC po i n ti se i t he rc1 orc2 ).S i nce he r e a r e two t he r ea r e on l hr ee po i n t si n co l or D ,t yt i l a t e r a l st ha thavet hes ameco l orD po i n t.Suppo s et ha t quadr t hes ameco l orD po i n to fT1 ,T2i sd1 . Then,i nt hr eequadr i l a t e r a l sT1 ,T2 ,T3 ,wha t eve rbet he l or o ft he ve r t ex, t he r e a r e r epea t ed po i n t s, wh i ch co ( ) , con t r ad i c t scond i t i on 2 .Henc e n ≤ 9. t r uc t i on of Wes how t he max ima lnumbe rn = 9by cons t he s en i nequadr i l a t e r a l s. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 277 Wedr aw t hr ee “ conc en t r i cannu l u s”wi t hf our po i n t son eachr ad i u sr epr e s en t i ngf ou rve r t i c e sandt heco l or.So n i ne r ad i ir epr e s en tn i nequadr i l a t e r a l s,wh i chs a t i s f i t i on ( 1). ycond Nex t,wes howt ha tt heya l s os a t i s f i t i on ( 2).Take ycond anyt hr eer ad i i( ort hr eequadr i l a t e r a l s). I ft he s et hr eer ad i icomef r om aconcen t r i cannu l u s,f or eachco l orexc ep tA , t he r ea r et hr eepo i n t s. I ft he s et hr eer ad i icomef r om t hr ee concen t r i c annu l i, t henf orco l orA , t he r ea r et hr eepo i n t s. I ft he s et hr eer ad i icomef r omtwoconcen t r i cannu l i,ca l l t he s et hr ee f i r e s F i .1, F i .2 and Fi .3. The r ad i u s gu g g g d i r e c t i on sa r e ca l l ed “ up r ad i u s”, “ l e f tr ad i u s”and “ r i ght r ad i u s”,anddeno t ed,r e s c t i ve l ft hr ee pe y,byS,Z andY .I r ad i ihavet hr eed i r e c t i on s,t hent he r ea r et hr eepo i n t so fco l or Bi nt hr ee quadr i l a t e r a l s.I ft het hr ee r ad i ihave on l y two d i r e c t i on s,t hent he r ea r ea l lc a s e sa ss howni nt het ab l e sbe l ow, whe r e1,2and3s t andf o rF i .1,F i .2andF i .3,r e s c t i v e l g g g pe y. He r e, t he co l or i n f i r e means t ha t t he t hr ee gu i l a t e r a l shaveco l orwi t hd i f f e r en tf i e s. quadr gur S 1,21,2 2 Z Y 1 2 1,21,2 C 1 2 1 1,21,2 S 1,21,2 1 Z Y 2 1 1,21,2 D 2 1 2 1,21,2 Ma t hema t i c a lOl i adi nCh i na ymp 278 S 1,31,3 1 Z Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Y 3 1 1,31,3 C Y 2 3 2,32,3 C 1 3 1,31,3 S 2,32,3 3 Z 3 2 3 2 2,32,3 S 1,31,3 3 Z Y 1 3 1,31,3 Y 3 2 3 1 1,31,3 D S 2,32,3 2 Z 1 2,32,3 3 2 3 2,32,3 D Thu s,t hemax ima lnumbe ro fni s9. 2012 ( Pu t i a n,F u i a n) j F i r s tDay 8:00 12:00,Ju l y27,2012 1 F i nd at r i l e( l,m ,n)( 1 < l < m < n)o f po s i t i ve p i n t ege r ss ucht ha t ∑k=1k, ∑k=l+1k, ∑k=m+1k f orm a l m n t r i cs equenc ei norde r.( s edbyTaoP i ng s heng) geome po t t( t +1) So l u t i on.Fort ∈ N* ,denot eSt = ∑k=1k = .Le t 2 l m n k=1 k=l+1 k=m+1 ∑k = Sl ,∑k = Sm -Sl ,∑ k = Sn -Sm , formageome t r i cs equenc ei norde r.Then 2 , Sl ( Sn -Sm )= ( Sm -Sl ) t ha ti sSl ( Sn +Sm -Sl )= S2m .Thu s,Sl |S2m , t ha ti s ① Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 279 2 2 l( l +1)|m2( m +1) . Le tm +1 =l( akel =3.Thenm =11andSl = l +1)andt S3 =6,Sm =S11 =66.Sub s t i t u t ei ti n t o ① ,wehaveSn =666, Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. n( n +1) =6 t ha ti s 66.Son = 36. 2 The r e f or e,( l,m ,n)= ( 3,11,36) i sas o l u t i ons a t i s f i ng y t hecond i t i on. Remark.Theso l u t i oni sno tun i l e,t he r ea r e que.Forexamp o t he rs o l u t i on sa s( 8,11,13),( 5,9,14),( 2,12,62),( 3,24, 171).(Wemays howt ha tt henumbe ro fs o l u t i onsi si nf i n i t e.) 2 hei nc i r c l eof△ABC .Thec i r c l e☉Ii n t e r s ec t s Le t☉Ibet s i de sAB ,BC andCA a tpo i n t sD ,E andF ,r e s t i ve l pec y. n t e r s e c t sl i ne sAI,BI andDI a tpo i n t sM ,N Li neEF i e s t i ve l ha tDM ·KE = DN ·KF . andK ,r pec y.Provet ( s edbyZhangPengcheng) po So l u t i on.I ti sea s os eet ha t po i n t sI,D ,E and B a r e yt concyc l i cand ∠AID = 9 0 °- ∠IAD , ∠MED = ∠FDA = 9 0 °- ∠IAD . So ∠AID = ∠MED ,t hu s po i n t sI,D ,E and M a r e concyc l i c. Henc e,f i vepo i n t sI,D ,B ,E , Ma r econcyc l i cand ∠IMB = ∠IEB =9 0 °, t ha ti sAM ⊥ BM . S imi l a r l i n t sI,D ,A ,N y,po andF a r econcyc l i candBN ⊥ AN . Le tl i ne sAN andBM i n t e r s ec ta t i n t G . We s ee po i n tI i st he po F i .2 .1 g Ma t hema t i c a lOl i adi nCh i na ymp 280 or t hoc en t e ro f△GAB andID ⊥ AB ,s opo i n t sG ,IandD a r e co l l i nea r. r econcyc l i c,wes eet ha t S i nc epo i n t sG ,N ,D and B a ∠ADN = ∠G . Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i s ec t s ∠MDN , t hu s S imi l a r l y,∠BDM = ∠G .SoDK b DM KM = . DN KN ① S i nc epo i n t sI,D ,EandM a r econcyc l i c,andpo i n t sI,D , N andF a r econcyc l i c,wes eet ha t KM ·KE = KI·KD = KF ·KN . The r e for e, KM KF = . KN KE ② DM KF , = By ① and② ,wes eet ha t t ha ti sDM ·KE = DN KE DN ·KF . 3 Forpo s i t i vecompo s i t e numbe rn,deno t e byf( n)and n)t hes um oft hesma l l e s tt hr eepo s i t i ved i v i s or so fn g( andt hel a r s ttwo po s i t i ved i v i s or so fn,r e s t i ve l ge pec y. ucht ha tg( n)equa n)t Fi nda l ln s l sf( os omepowe rof s i t i vei n t ege r s.( s edby HeYi i e) po po j So l u t i on.I fni sodd,t hena l lf ac t or sofna r eodd.Sof( n)i s oddandg( n)i seven.g( n)canno tbef( n)t os omepowe rof seven.Thesma l l e s ttwod i v i s or s s i t i vei n t ege r. The r e for eni po ofna r e1and2,andt hel a r s ttwod i v i s or so fna r enandn/2. ge Le td bet het h i rdsma l l e s td i v i s orofn.I ft he r eex i s t sk ∈ n)= fk ( n), N s ucht ha tg( t hen * Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 281 3 n k k k =( =( ≡d ( mod3). 1 +2 +d) 3 +d) 2 S i nc e3 3 n, wes eet ha t3|dk .So3|d,ands i ncedi st he 2 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t h i rdsma l l e s t,wes eet ha td = 3. 3 Thu s, n =6k ,wege tn =4×6k-1 .S i nce3|n,wes eet ha t 2 k ≥ 2. Summi ngup, n = 4 ×6l ( l ∈ N* ). 4 Le tr ea lnumbe r sa,b,candds a t i s f y x)= acosx +bcos2x +ccos3x +dcos4x ≤ 1 f( foranyr ea lnumbe rx.F i ndt heva l ue sofa,b,candd s ucht ha ta + b - c + d t ake st he max imum numbe r. ( s edbyLiShenghong) po So l u t i on.S i nc e 0)= a +b +c +d, f( π)= -a +b -c +d, f( t hen æπö a b d, -c fç ÷ = è3 ø 2 2 2 2 4 æπö a +b -c +d = f( 0)+ f( π)+ f ç ÷ ≤ 3 3 3 è3 ø æπö i fandon l ff( 0)= f( π)= f ç ÷ = 1, t ha ti s,i fa = 1,b + yi è3 ø hent heequa l i t l d s.Le tt = c osx, d = 1andc = - 1,t y ho -1 ≤t ≤ 1.Th en x)-1 = cosx +bcos2x -cos3x +dcos4x -1 f( ( =t + ( 1-d) t -1)- ( t -3 t)+d( t -8 t +1)-1 2 4 8 2 3 4 2 ]≤0,∀t ∈ [-1,1], [-4 = 2( 1-t2) d t2 +2 t+ ( d -1) Ma t hema t i c a lOl i adi nCh i na ymp 282 t ha ti s 4d t2 -2 t+ ( 1 -d)≥ 0,∀t ∈ (-1,1). T ak i ngt = 1/2 +ε,|ε|< 1/2, t henε[( ε]≥ 0, 2d -1)+4d Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 1 e et ha td = . I fd = , t hen |ε|< 1/2.Sowes 2 2 2 ≥ 0. t2 -2 t+ ( 1 -d)= 2 t2 -2 t +1/2 = 2( t -1/2) 4d So,t hemax ima lnumbe ro fa +b -c +di s3,and( a,b, c, 1 1ö æ d)= ç1, ,-1, ÷ . 2 2ø è S e c ondDay 8:00 12:00,Ju l y28,2012 5 Anon -nega t i venumbe rmi sca l l edas ixma t chnumb e r.I f m andt hes um ofi t sd i i t sa r ebo t h mu l t i l e so f6,f i nd g p t henumbe r oft hes i x ma t ch numbe r sl e s st han 2012. ( s edbyTaoP i ng s heng) po So l u t i on.Le tn = d1d2d3d4 = 1 000d1 +100d2 +10d3 +d4 , d1 ,d2 ,d3 ,d4 ∈ [ 0,1,2,...,9],andS( n) = d1 + d2 + d3 +d4 . Ma t cht henon -nega t i ve mu l t i l e sof6l e s st han2000i n t o p 167pa i r s( x,y),x +y = 1998,s ucht ha t ( 0,1998),( 6,1992),( 12,1986),...,( 996,1002). Fo re a chpa i r( x,y), l e tx =a1a2a3a4,y =b1b2b3b4, t hen a1 +b1)+100( a2 +b2)+10( a3 +b3)+ ( a4 +b4) 1000( = x +y = 1 998. S i nc ex,y a r eeven,a4 ,b4 ≤ 8.Soa4 +b4 ≤ 16 < 18. Thu s,a4 +b4 = 8.S i ncea3 +b3 ≤ 18 < 19,a3 +b3 = 9. Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 283 S imi l a r l t a i na2 +b2 = 9anda1 +b1 = 1.Thu s, y,wecanob S( x)+S( a1 +b1)+ ( a2 +b2)+ ( a3 +b3)+ ( a4 +b4) y)= ( = 1 +9 +9 +8 = 2 7. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. x)andS( Con s equen t l he r ei son l fS( ha ti s y)t y,t yoneo r ea l lmu l t i l e sof t hemu l t i l eo f6.( Th i si sbe cau s et ha tx,ya p p 3,s oa r eS( ti s,t he r ei son l x)andS( y).)Tha yoneofxandy wh i chi sas i xma t chnumbe r. The r e f or e,t he r ea r e167s i xma t chnumbe r sl e s st han2000, andt he r ei sj u s tones i xma t chnumbe rbe tween2000and2011. The r e f or e,t hean swe ri s167 +1 = 168. 6 F i ndt hemi n imum po s i t i vei n t ege rns ucht ha t 3 n -2011 n -2012 < 2012 2011 ( s edbyLi uGu ime i) po 3 n -2013 n -2011 . 2011 2013 ≤4 023,t hen So l u t i on.Wes eet ha ti f2012 ≤ n 3 3 n -2011 2012 n -2012 n -2013 n -2011 ≥ 0a < 0. nd 2011 2011 2013 Ot he rwi s e, 3 n -2011 ≤ 2012 n -2013 ≥ 2011 3 n -2012 ⇔n > 4023 2011 n -2011 ⇔n ≥ 4024. 2013 Thu s,i fn ≥ 4024, t hen n -2011 n -2012 <0≤ 2012 2011 s4024. Sot hemi n imumo fni 3 3 n -2013 n -2011 . 2011 2013 Ma t hema t i c a lOl i adi nCh i na ymp 284 7 In△ABC wi t hAB =1, l e tD bea po i n ton AC s ucht ha t ∠ABD = ∠C a ndl e tE bea i n tonAB s ucht ha tBE = po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. DE .Le tH beapo i n tonDE s ucht ha tAH ⊥ DEandM be t hemi dpo i n to fCD . I fAH = F i .7 .1 g 2 - 3,f i ndt hes i z eo f ∠AME .( s edby Xi ongBi n) po So l u t i on.Le t ∠ABD = ∠C =αa nd ∠DBC =β. I ti sea s o yt s eet ha t ∠BDE =α,∠AED = 2 α, ∠ADE = ∠ADB - ∠BDE = ( α +β)-α =β, AB = AE +EB = AE +EH + HD . Henc e, AB AE +EH HD 1 +cos2 α = + = +c o tβ AH AH AH s i n2 α =c o tα+ co tβ. ① Dr aw l i ne s EK ⊥ AC and EL ⊥ BD wi t h peda l s K and L , r e s t i ve l s t he pec y. Then, L i mi dpo i n to fBD .Comb i n i ng wi t h t heS i neTheor em,weob t a i n EL DEs i n ∠EDL s i nα = = EK DEs i n ∠EDK s i nβ = F i .7 .2 g BD LD = . CD MD Thu s, c o tα = LD MD MK DK = = =c o t ∠AME -c o tβ. ② EL EK EK EK Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 285 By ① ,② andknowncond i t i on s,wehave c o t ∠AME = AB 1 = = 2 + 3. AH 2- 3 Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. The r e f or e,∠AME = 15 °. 8 Le andPn = { 1,2,..., tm b eapo s i t i v ei n t e e r, n =2m -1, g i n t sonnumbe rax i s.A g r a s s hoppe r n}bet hes e to fn po ump sbe tweenad ac en tpo i n t sonPn .F i ndt he max ima l j j numbe rofm s ucht ha tf oranyx,y ∈ Pn ,t henumbe rof way st ha tag r a s s hoppe rj ump i ngf romxt oyby2012s t eps i seven ( s s i ngx ory ont hewayi spe rmi t t ed).( s ed pa po byZhangS i hu i) So l u t i on.I fm ≥1 1,t henn = 2 -1 > 2013.S i ncet he r ei s m on l r a s s hoppe rj umpsf r om po i n t1t opo i n t2013by yonewayag 2012s t ep s,wes eet ha tm ≤ 10. Int hef o l l owi ng,wes howt ha tt heanswe ri sm = 10.To s howt h i s,wewi l lpr oveas t r onge rpr opo s i t i onbyi nduc t i onon m :foranyk ≥ n = 2m -1andanyx,y ∈ Pn , t henumbe rof way st heg r a s s hoppe rj umpsf r om po i n txt oybyks t epsi seven. I fm = 1, t henumbe ro fway si s0,whe r e0i seven. 2 l+1 I fm =l, t henumbe ro fway si seven.Then,f ork ≥ n = -1, t he r ea r et hr eek i ndo fr ou t e sf rom po i n txt opo i n ty byks t ep s.Wes howt ha tt henumbe rofway si sevenf oreach k i ndo fr ou t e. ( 1)Therou t edoe sno tpa s spo i n t2l .Sopo i n t sxandybo t h a r eonones i deofpo i n t2l .Byt hei nduc t i onhypo t he s e s,t he r e a r eevenr ou t e s. ( 2)Therou t epa s s e spo i n t2lj u s tonce. tt hei -t hs t ep s. Suppo s et ha tt heg r a s s hoppe ri sa tpo i n t2l a 286 Ma t hema t i c a lOl i adi nCh i na ymp ( i ∈{ 0, 1,..., k}, i =0me an sx =2l , i =k me an sy =2l ).We s howt ha t,f oranyi, t henumbe ro frou t e si seven. Suppo s et ha tt her ou t ei sx,a1 ,...,ai-1 ,2l ,ai+1 ,..., Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ak-1 ,y.Di v i dedi ti n t otwos ub r ou t e s:f rom po i n tx t opo i n t ai-1 ofi -1s t ep sandf r om po i n tai+1t fk -i -1s t eps opo i n tyo ( f ori = 0ork,on t ep s). l ub r ou t eofk -1s yones I fi -1 < 2l -1andk -i -1 < 2l -1, t henk ≤ 2l+1 -2, wh i chcon t r ad i c t sk ≥ n = 2l+1 -1.So,wemu s thavei -1 ≥ 2l -1ork -i -1 ≥ 2l -1.Byt hei nduc t i onhypo t he s e s,t he r e a r e even way sf or a s ub r ou t e.So, by t he Mu l t i l i ca t i on p Pr i nc i l e,t henumbe ro fway si seven. p ( 3)Ther ou t epa s s e spo i n t2l nol e s st hantwot ime s. o2l , t henumbe ro fway si s Con s i de rt hes ub r ou t e sf rom2lt even,s i ncewecancon s i de rt herou t e ss t r i ct o2l .Soby ymme t he Mu l t i l i ca t i onPr i nc i l e,t henumbe ro fway si seven. p p Summi ngup,t hemax ima lmi s10. 2013 ( Y i n t a n, J i a n x i) g g F i r s tDay 8:00 12:00,Ju l y27,2013 1 Le ta, bber ea lnumbe r ss ucht ha tt heequa t i onx3 -ax2 + bx - a = 0ha son l ea lr oo t s.Fi ndt he mi n imum of yr 2 a3 -3ab +3a . b +1 So l u t i on.Le tx1 ,x2 andx3 bet her ea lroo t soft heequa t i on Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 287 x3 -ax2 +bx -a =0.By Vi e t a sFormu l a,wehave x1 +x2 +x3 = a,x1x2 +x2x3 +x1x3 =b,x1x2x3 = a. 2 ≥3( By( x1 +x2 +x3) x1x2 +x2x3 +x1x3),wehavea2 ≥ 3 b,andbya = x1 + x2 + x3 ≥ 3 x1x2x3 = 3 a ,wehave Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 3 3 a ≥3 3.Thu s, a3 -3ab +3a a( a2 -3 b)+a3 +3a 2 = b +1 b +1 ≥ a3 +3a a3 +3a ≥ 2 b +1 a +1 3 =3 a ≥ 9 3. I fa =33, b =9, t hent heequa l i t l d swheneachroo ti s yho equa lt o 3. Summi ngup,t hean swe ri s9 3. 2 Le t☉I bet hei nc i r c l eo f △ABC wi t h AB > AC . ☉I t angen tt o BC and AD a t D and E ,r e s t i ve l pec y.The t angen tl i neEP o f☉Ii n t e r s ec t st heex t endedl i neofBCa t P .Segmen tCFi spa r a l l e lt oPE andi n t e r s ec t sAD a tpo i n t F .Li neBFi n t e r s e c t s☉Ia tpo i n t sM andN s ucht ha tM i s ons egmen tBF .Segmen tPM i n t e r s ec t s ☉I a tt heo t he r i n tQ .Pr ovet ha t ∠ENP = ∠ENQ . po So l u t i on.Suppos et ha t☉I t ouche sAC andAB a tS and T, r e s c t i ve l s e pe y. Suppo t ha t ST i n t e r s ec t s AI a t i n tG ,wes eet ha tIT ⊥ po AT and TG ⊥ AI.So we Ma t hema t i c a lOl i adi nCh i na ymp 288 haveAG ·AI = AT2 = AD·AE , t hu spo i n t sI,G ,EandD a r e concyc l i c. S i nceIE ⊥ PE andID ⊥ PD ,wes eet ha tpo i n t sI,E ,P andD a r econcyc l i c.Hence,po i n t sI,G ,E ,P and D a r e Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. concyc l i c. The r e f or e, ∠IGP = ∠IEP = 90 °,t ha ti s,IG ⊥ PG . r eco l l i nea r. Henc e,po i n t sP ,S andT a Li ne PST i n t e r s e c t s △ABC . By Mene l au s Theor em, wehave AS ·CP ·BT = 1. SC PB TA S i nc eAS = AT ,CS = CD andBT = BD ,wehave PC ·BD = 1. PB CD ① Le tt heex t en s i ono fBNi n t e r s e c tPE a tpo i n tH .Thenl i ne BFH i n t e r s ec t s△PDE .By Mene l au s Theor em, PH ·EF ·DB = 1. HE FD BP EF PC , = S i nc eCF pa r a l l e lt oBE , wehave FD CD PH ·PC ·DB = 1. HE CD BP ② By ① and ② ,wehavePH = HE .Hence,PH2 = HE2 = PH HN , = HM ·HN .Thu s,wehave △PHN ∽ △MHP and HM PH ∠HPN = ∠HMP = ∠NEQ .Fu r t he r,s i nce ∠PEN ∠EQN , t he r e for e ∠ENP = ∠ENQ . 3 Le tt hes equenc e{ an }bede f i nedby = Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 289 2 n +( -1) an ( a1 = 1,a2 = 2,an+1 = n = 2,3,...). an-1 Pr ovet ha tt hes um o fs r e sofanytwoad acen tt e rms qua j o ft hes equencei sa l s oi nt hes equence. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on.Byan+1 = 2 n +( -1) an 2 ,wehavean+1an-1 = an + an-1 n ( (-1) n = 2,3,...,),s o 2 2 n 1 2 -1)- -an-2 an -an-2 anan-2 -an an -2 -1 + ( = = an-1 an-1an-2 an-1an-2 = 2 an an-3 an-1 -an-3 -1 -an-1 = an-1an-2 an-2 = … = a3 -a1 = 2, a2 an = 2an-1 +an-2( n ≥ 3),a1 = 1,a2 = 2. t ha ti s, The r e f or e, an = C1λn λn λ1 +λ2 = 2,λ1λ2 = - 1, 1 + C2 2, a1 = 1,a2 = 2n ∈ N+ . Then,s i nceλ1λ2 = -1andλ2 = 2 -λ1 ,wehave 1 = C1λ1 +C2λ2 { 2 = C1λ2 λ2 1 +C2 2 ⇒ ⇒ λ2 = C1λ1λ2 +C2λ2 2 { { 2 = C1λ2 λ2 1 +C2 2 2 -λ1 = -C1 +C2λ2 2 λ2 2 = C1λ2 1 +C2 2 1 +λ2 ⇒C1( 1 )=λ1 . Thu s,bys t r i ccond i t i on,wehaveC2( 1 +λ2 2 )=λ2 . ymme So,s i nce1 +λ1λ2 = 0,wehave 2 2 2 n 2 n +an+1 = C1( an 1 +λ2 λ2 1 +λ2 λ2 1) 1 +C2 ( 2) 2 + n ( 2C1C2( λ1λ2) 1 +λ1λ2) n+1 2n 1 +C2λ2 + = a2n+1 . C1λ2 1 4 i v i dedi n t otwo Suppo s et ha t12ac r oba t sl abe l ed1 12d 290 Ma t hema t i c a lOl i adi nCh i na ymp c i r c l e sA andB ,wi t hs i xpe r s on si neach.Le te a cha c r oba t t andont hes hou l de r so ftwoad a c en ta c r oba t so fA .We i nBs j c a l li tat owe ri ft hel abe lo fe a cha c r oba to fBi sequ a lt ot he s umo ft hel ab e l so ft hea c r oba t sunde rh i sf e e t.How many Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. d i f f e r en tt owe r sc ant he y make? ( Remark. Wet r ea ttwot owe r sa st hes amei fonecan be ob t a i nedbyr o t a t i onorr e f l ec t i onoft heo t he r.Forexamp l e, t hef o l l owi ngt owe r sa r et hes ame,whe r et hel abe l si n s i det he c i r c l er e f e rt ot hebo t t om ac r oba t,t hel abe l sou t s i det hec i r c l e r e f e r st ot heuppe rac r oba t.) So l u t i on.Denot et hes um o fl abe l sofA andB byx andy, r e s c t i ve l s,wehave pe y.Theny = 2x.Thu 3x = x +y = 1 +2 + … +12 = 78,x = 26. Obv i ou s l t eA = { 1,2,a, y,1,2 ∈ A and11,12 ∈B .Deno b,c,d},whe r ea <b <c < d.Thena +b +c +d = 23,and a ≥ 3,8 ≤ d ≤ 10 ( i fd ≤ 7, t hena +b +c +d ≤ 4 +5 +6 + 7 = 22,wh i chi sacon t r ad i c t i on.) ( 1)I fd = 8, t henA = { 1,2,a,b,c, 8},c ≤7,a +b +c =15.Thu s,( a,b, c)= ( 3,5,7)or ( 4,5,6),t ha ti s,A = { 1,2, 3,5,7,8}orA = { 1,2,4,5,6,8}. I fA = { 1,2,3,5,7,8}, t henB = { 4, 6,9,10,11,12}.S i nc eB con t a i n s11,4,6and12,t he r ei s on l owe rt ha t,i nA ,8and3,3and1,1and5,5and7 yonet Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 291 a r ead ac en t. j I fA = { 1,2,4,5,6,8}, t henB = { 3,7,9,10,11,12}. r e S imi l a r l eet ha t,i nA ,1and2,5and6,4and8a y,wes ad acen t,r e s c t i ve l r ea r etwoa r r angemen t s,t ha ti s, j pe y.The Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t he r ea r etwot owe r s. ( t henA = { 1,2,a,b,c,9},c ≤ 8,a +b + 2)I fd = 9, c = 14,whe r e( a,b,c)= ( 3,5,6)or ( 3,4,7),t ha ti s,A = { 1,2,3,5,6,9}orA = { 1,2,3,4,7,9}. 1,2,3,5,6,9}, t henB = { 4,7,8,10,11,12}. I fA = { nA mu s tbead ac en t Toob t a i n4,10and12i nB ,1,3,and9i j i rwi s e,i ti simpo s s i b l e! pa I fA = { 1,2,3,4,7,9}, t henB = { 5,6,8,10,11,12}. Toob t a i n6,8and12i nB ,2and4,1and7,9and3mu s tbe ad ac en ti nA ,r e s c t i ve l r ea r etwoa r r angemen t s,t ha t j pe y.The i s,t he r ea r etwot owe r s. ( 3)I fd =10, t henA = { 1,2,a,b, c, 10},whe r ec ≤9,a +b +c =13.Thu s,( a, b,c)= ( 3,4,6), t ha ti s,A = { 1,2,3,4, 6,10}andB = { 5,7,8,9,11,12}.To Ma t hema t i c a lOl i adi nCh i na ymp 292 ob t a i n8,9,11and12i nB , 6and2,6and3,10and1,10and 2mu s tbead acen t,r e s t i ve l r ei son l owe r. j pec y.The yonet Summi ngup,t he r ea r es i xd i f f e r en tt owe r sa l lt og e t he r. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. S e c ondDay 8:00 12:00,Ju l y28,2013 5 x x x ,whe Le tf( x)= ! + ! + … + r e[ x] i st he 1 2 2013! [ ] [ ] [ ] r ea t e s ti n t ege rno g r ea t e rt hanx.Ca l lani n t ege rn a g ft heequa t i onf( x)= nha sar ea ls o l u t i on ri goodnumbe x.Fi ndt henumbe rofgood numbe r si nt hes e t{ 1,3, 5,...,2013}. So l u t i on.Fi r s t,wepo i n tou ttwoobv i ou sf ac t s: ( a)I fmi sapo s i t i vei n t ege randxi sr ea l,t hen x] [mx ] = [ [m ]. ( b)Foranyi n t ege rlandpo s i t i veevennumbe rm ,wehave [2lm+1] = [2ml] . Le tm =k!( k = 1,2,...,2013) i n( a)ands ummi ngup, wehave x)= f( [ x x] ∑ [k! ] = ∑ [ k! ] = f([x]), 2013 2013 k=1 k=1 t ha ti s,f( x)=nha sar ea ls o l u t i oni fandon l ff( x)=nha s yi ani n t ege rs o l u t i on. So,weon l i de rxa sani n t ege r. S i nce ycons 2013 x ö÷ æ x +1 , x +1)-f( x)= [x +1] - [x ] + ∑ ç f( ! ! ø ≥1 k k è k=2 [ ] [ ] ① Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 293 wes eet ha tf( x)( x ∈ Z) i smono t onou s l nc r ea s i ng.Now,we yi f i ndi n t ege r saandb, s ucht ha t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a -1)< 0 ≤ f( a)< f( a +1)< … f( < f( b -1)< f( b)≤ 2013 < f( b +1). i nce No t et ha tf(-1)< 0 = f( 0),s oa = 0.S 1173)= f( 6 1173 ∑ [ k! k=1 ] = 1173 +586 +195 +48 +9 +1 =2 012 ≤ 2013, 1174)= f( 6 1174 ∑ [ k! k=1 ] = 1174 +587 +195 +48 +9 +1 =2 014 > 2013, wes eet ha tb = 1173. Sot hegoodnumbe r si n{ 1,3,5,...,2013}a r et heodd numbe r si n { 0),f( 1),...,f( 1173)}. f( l( l = 0,1,...,586) Le tx = 2 i n ①.By ( b),wehave Thu s, [2lk+! 1] = [k2l! ] (2 ≤k ≤ 2013). 2013 l +1 2 l ö÷ æ 2 = 1, 2 l +1)-f( 2 l)= 1 + ∑ ç f( k! k! ø k=2 è [ ] [ ] t ha ti s,t he r ei se x a c t l e ri nf( 2 l)andf( 2 l + 1). yoneoddnumb 1174 =5 The r e f or e,t he r ea r e 87oddnumbe r si n{ 0),f( 1), f( 2 ...,f( 1173)},t ha ti s,t he r ea r e587goodnumbe r si nt hes e t { 1,3,5,...,2013}. 6 Le tnb eani n t e e rg r e a t e rt han1.Deno t et hef i r s tnp r ime si n g 294 Ma t hema t i c a lOl i adi nCh i na ymp i n c r e a s i ngo r de rbyp1,p2 ,...,pn (i. e.,p1 = 2,p2 = p1 p2 … pn ).Le 3,... tA = p1 i nda l lpo s i t i v ei n t e e r sx p2 pn .F g A s u cht ha t i se v enandha se x a c t l i s t i nc tpo s i t i v ed i v i s o r s. yxd x Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. So l u t i on.By2x |A ,not et ha tA p2 … pn = 4·p2 pn .We may αn 2 s uppo s et ha tx = 2α1pα whe r e0 ≤α1 ≤ 1,0 ≤αi ≤ pi 2 …pn , ( i = 2,3,...,n).Then,wehave A 2α p2-α2 … pn-αn = 2 - 1p2 . pn x A s Henc e,t henumbe rofd i f f e r en td i v i s or sof i x ( 3 -α1)( p2 -α2 +1)…( pn -αn +1). Weknowt ha t α α α ( 3 -α1)( p2 -α2 +1)…( pn -αn +1)= x = 2 1p22 …pnn . ① Byi nduc t i on on n,wesha l l provet ha tt he a r r ay ( α1 , α2 ,..., αn )s a t i s f i ng ①i s( 1,1,...,1)( n ≥ 2). y ( hen ① b e c ome s( 3 -α1)( 4 -α2)= 2α13α2 , 1)I fn = 2,t whe r eα1 ∈ { i c hha sno 0,1}.I fα1 = 0, t h e n3( 4 -α2)= 3α2 wh I fα1 =1, t h e n2( 4-α2)=2·3α2 .Weha v eα2 = i n t e e rs o l u t i onα2. g 1.Thu s,( α1, α2)= ( 1,1) .Tha ti s, t hec on c l u s i oni st r u ef o rn =2. ( 2)Suppo s et ha tt heconc l u s i oni st r uef orn =k -1.( k≥ 3), t henwhenn =k,① become s ( 3 -α1)( p2 -α2 +1)…( pk-1 -αk-1 +1)( pk -αk +1) ② α α αk-1 αk = 2 1p22 …pk1 pk . I fαk ≥ 2,con s i de r i ng 0 < pk -αk +1 < pk , 0 < pi -αi +1 ≤ pi +1 < pk ( 1 ≤i ≤k -1), Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 295 wes eet ha tt hel e f t -hands i deo f② canno tbed i v i dedbypk ,bu t t her i -hand s i de o f ② i s a mu l t i l eo f pk , wh i ch i sa ght p con t r ad i c t i on. I fαk = 0, t hen ② become s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ( 3 -α1)( p2 -α2 +1)…( pk-1 -αk-1 +1)( pk +1) k-1 . = 2 1p22 …pk1 α α α ③ No t et ha tp2 ,p3 ,...,pk a r eoddpr ime,t hu s,ont heone hand,pk +1i seven.Sot hel e f t -hands i deo f③i seven.Ont he tt hen3 o t he rhand,t her i ts i deof③i sodd.Soα1 = 1.Bu gh α1 = 2,s ot hel e f t -hands i deof ③ i sa mu l t i l eof4,bu tt he p r i t -hands i deo f③i sno t,wh i chi sacon t r ad i c t i on. gh Byt heabovea r t,wemu s thaveαk = 1,andi n ②, gumen Thu s, α pk -αk +1 = pkk = pk . α α -1 α ( 3 -α1)( . p2 -α2 +1)…( pk-1 -αk-1 +1)= 2 1p22 …pk-k1 Byt hei nduc t i onhypo t he s e s, α1 =α2 = … =αk-1 = 1. Thu s, α1 =α2 = … =αk-1 =αk =1, t ha ti s,t heconc l u s i on i st r uef orn =k. By ( 1)and ( 2),weconc l udet ha t( α1 ,α2 ,...,αn )= ( 1, 1,...,1), s ot hei n t ege rr equ i r edi sx =2p2 …pn =p1p2 …pn . 7 Cu to f facorne ro f2×2un i ts r e sf rom a3×3un i t qua s r e s,t her ema i n i ngf i r ei sca l l edaho rn ( Fi .7 .1i sa qua gu g horn).Now,pu ts omehorn s wi t hou tove r l app i ng ona boa rdo f10 × 10 un i ts r e s( Fi .7 .2)s ucht ha tt he qua g bounda r i e soft he horn co i nc i de wi t ht he g r i do ft he boa rd.F i ndt he max imum o fks ucht ha twha t eve rt hek Ma t hema t i c a lOl i adi nCh i na ymp 296 hornspu tont heboa rd,onecana lway spu tano t he rhorn Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ont heboa rd.( s edby HeYi i e) po j F i .7 .1 g So l u t i on.Fi r s t,wehavekmax F i .7 .2 g < 8, t h i si sbe cau s eo ft ha ti fwepu te i t gh hornsa si nF i .7 .3,t hen no mor e g horncanbepu t. Ne x t,wes howt ha t,a f t e rpu t t i ng any s even horn s, t he r ei sa lway s roomf orano t he rhorn. Con s i de rf ou r4×4 s r e si n qua F i .7 .3 g f ourcorne r so f 10 × 10 s r e s. Then any horn can on l qua y , × i n t e r s e c toneo ft he s e4 4s r e s.So s evenhornsa r ep l aced qua on10×10s r e s.Byt he Di r i ch l e t Dr awe r Theor em,t he r e qua ex i s t sa4×4s r eSs ucht ha tt he r ei sa tmo s tonehornH t ha t qua i n t e r s e c t sS,andH c anb ec on t a i nedbya3×3s r e s,s oS ∩ Hi s qua c on t a i nedbya3×3s a r e sonac onne ro fS. qu We can pu t a horn on S a ss hown i n F i .7 .4. g Summi ngup,wehavekmax = 7. F i .7 .4 g Ch i naSou t he a s t e rn Ma t hema t i c a lOl i ad ymp 8 297 Le tn ≥ 3bei n t ege r.Suppo s et ha tα,β,γ ∈ ( 0,1)and s a t i s f ak ,bk ,ck ≥0 ( k =1,2,...,n) k +α) ak y∑k=1( n ∑ ≤α, n k=1 ( k +β) bk ≤βand ∑k=1( k +γ) ck ≤γ.Fi nd n Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ucht ha t∑k=1( k +λ) akbkck ≤λ. t hemi n imumofλs n α , γ , , , So l u t i on.Le b1 = β , c1 = ai bi ci = ta1 = 1 +α 1 +β 1 +γ 0( i = 2,3,...,n).Wes eet ha ta l lcond i t i on sa r es a t i s f i ed. So,wemu s thave α · β · γ ( ≤λ, 1 +λ) 1 +α 1 +β 1 +γ t ha ti s, α βγ λ≥( . 1 +α)( 1 +β)( 1 +γ)-α βγ α βγ = λ0 .We w i l lshow Deno t e( ) ( 1 +α 1 +β)( 1 +γ)-α βγ t ha t,foranyak ,bk ,ck ,k = 1,2,...,n,wehave n k +λ ) abc ∑( 0 k=1 ≤λ0 , k k k ① wh i chs a t i s f i e sa l lt hecond i t i on s. Byt hecond i t i on so ft heprob l em, 1 n k +β æçk +α k +γ ö÷ 3 ak · bk · ck ∑ γ ø β k=1 è α 1 n n 1 1 n k +β ö÷ 3 æç k +α ö÷ 3 · æç k +γ ö÷ 3 æ ak ck ≤ ç∑ ≤ 1, bk · ∑ ∑ è k=1 α ø è k=1 γ ø è k=1 β ø whe r eweu s et heHö l de r sInequa l i t fxi ,yi ,zi ≥ 0( i = 1, y:i 2,...,n), t hen n ( ∑xyz ) i=1 i i i 3 ≤ n n n ( ∑x ) ( ∑y ) ( ∑z ) . i=1 3 i i=1 3 i i=1 3 i ② Ma t hema t i c a lOl i adi nCh i na ymp 298 The r e f or e,t opr ove① ,i ts u f f i c e st os howt ha t,f ork =1, 2,...,n,wehave 1 æk +α k +β k +γ ö3 k +λ0 · · ·akbkck ÷ , akbkck ≤ ç λ0 γ è α ø β Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t ha ti s 2 k +α)( k +β)( k +γ)ö÷ 3 æ( k +λ0 (akbkck ) 3 ≤ ç . λ0 α γ è ø β 1 ③ Inf ac t, λ0 = α βγ ( ) 1 + α +β +γ + ( α β +βγ +γα) α βγ k2 + ( α +β +γ) k+( α β +βγ +γα) ≥ = Thu s, kα βγ . ( ) ( + + k α k β)( k +γ)-α βγ ( k +α)( k +β)( k +γ) k +λ0 ≤ . λ0 α βγ ④ Ands i nc e( k +α) ak ≤α,( k +β) bk ≤β,( k +γ) ck ≤γ,we have 2 (akbkck ) 3 ≤ 2 α æç ö÷ 3 βγ . k +α)( k +β)( k +γ)ø è( ⑤ By ④ and ⑤ ,wes eet ha tequa t i on ③ ho l d s,t hu s① ho l d s. α γ β S ummi ngup,weha v eλmin =λ0 = ( . ( ( 1+α) 1+β) 1+γ)-α γ β Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Int e rna t i ona lMa thema t i ca l Ol i ad ymp 2011 ( Ams t e r d am,H o l l a n d) F i r s tDay 9:00~13:30,Ju l y18,2011 1 Gi venany s e tA = { a1 ,a2 ,a3 ,a4}offour d i s t i nc t s i t i vei n t ege r s,wedeno t et hes uma1 +a2 +a3 +a4 = po sA .Le tnA deno t et henumbe rofpa r e s( i, t h1 ≤i < j)wi i chai +aj d i v i de ssA .Fi nd sa l ls e t sA off our j ≤ 4forwh d i s t i nc tpo s i t i vei n t ege r swh i chach i evet hel a r s tpo s s i b l e ge va l ueofnA .( s edby Mex i co) po Ma t hema t i c a lOl i adi nCh i na ymp 300 So l u t i on.( Ba s edont hes o l u t i onbyJ i nZhaorong)Le ts e tA = { s i t i vei n t ege r swi t ha1 < a2 < a3 < a1 ,a2 ,a3 ,a4}offourpo a4 .S i nce Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1 1 sA = ( a1 +a2 +a3 +a4)< a2 +a4 < a3 +a4 <sA , 2 2 a2 +a4 ,a3 +a4 dono r e for e, td i v i desA .The nA ≤ C2 4 -2 = 4. Ont heo t he rhand,i fA = { 1,5,7,11},nA = 4,t he l a r s tpo s s i b l eva l ueo fnA i s4. ge Nex t,wewi l lf i nda l ls e t sA o ffourpo s i t i vei n t ege r swi t h nA = 4. td i v i desA ,and Fi r s t,wes eet ha ta2 +a4 ,a3 +a4 dono 1 sA ≤ max{ a1 +a4 ,a2 +a3}<sA . 2 Thu s, andt hen 1 sA = max{ a1 +a4 ,a2 +a3}, 2 a1 +a4 = a2 +a3 . Bya1 +a3 |sA , a1 +a3),whe sapo s i t i ve l e tsA = k( r eki i n t ege r.Andbya1 +a3 < a2 +a3 weknowt ha tk > 2. As2( a2 +a3)=sA =k( a1 +a3), a2 = andf rom a2 = 1( ka1 + ( k -2) a3), 2 1( ka1 + ( k -2) a3)< a3 , 2 wehavek < 4.Thu k = 3.Cons s, equen t l y,wehave 2( a2 +a3)= 2( a1 +a4)= 3( a1 +a3)=sA , f r om wh i ch wecande r i vea2 = 1( 1( 3 a1 + a3),a4 = a1 + 2 2 In t e rna t i ona lMa t hema t i c a lOl i ad ymp 301 3 a3). a1 +a2),wehave Andbya1 +a2 |sA , l e tsA =l( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t ha ti s, 1 æ ö a1 +a3)÷ , 3 a1 +a3)=lça1 + ( 3( 2 è ø ( 5 6 -l)a3 = ( l -6) a1 . S i nc ea1anda3a r epo s i t i v ei n t e e r sanda1 <a3, r5. l =3,4o g I fl = 3,wehavea2 = a3 ,wh i chi sacon t r ad i c t i on. I fl = 4,wehavea3 = 7 a1 ,cons equen t l tfo l l owst ha t y,i a2 =5a1 ,a4 = 11a1 . equen t l I fl = 5,wehavea3 = 19 a1 ,cons y,wehavea2 = 11 a1 ,a4 = 29a1 . I ti sea s ove r i f ha t,whenl = 4,5,eachofa1 +a2 , yt yt a1 +a3 ,a1 +a4 anda2 +a3 d i v i de ssA . Tos um up,a l ls e t sA o ff ou rd i s t i nc t po s i t i vei n t ege r s wh i chach i evet hel a r s tpo s s i b l eva l ueofnA = 4a r eA = { a, ge 5 a,7a,11a},andA = { a,11a,19a,29a},whe r eai sany s i t i vei n t ege r. po 2 Le tS beaf i n i t es e to fa tl ea s ttwo po i n t si nt hep l ane. As s umet ha tnot hr eepo i n t sofSa r eco l l i nea r.Awi ndmi l l i sapr oc e s st ha ts t a r t swi t hal i nelgo i ngt hroughas i ng l e i n tP ∈ S.Thel i ner o t a t e sc l ockwi s eabou tt hepo i n tP po un t i lt hef i r s tt imet ha tt hel i ne mee t ss omeo t he rpo i n t be l ong i ngt oS.Th i spr oc e s scon t i nue si nde f i n i t e l y. Showt ha twecanchoo s eapo i n tPi nS andal i nelgo i ng t hr oughP s ucht ha tt her e s u l t i ng wi ndmi l lu s e seachpo i n t ofS a sap i vo ti nf i n i t e l ime s.( s edbyBr i t i sh) y manyt po So l u t i on.( Ba s edont hes o l u t i onby Zhou Ti anyou)Cons i de r eachl i neℓ ha sad i r e c t i on,wh i chva r i e scon t i nuou s l he ywhent Ma t hema t i c a lOl i adi nCh i na ymp 302 l i ne ℓ r o t a t e s.Fi r s t,cons i de rt heca s e when|S |= 2 k +1 i so l d. Int hepr oc e s so fawi ndmi l l,wes ayal i ne ℓi sonagood s i t i on,i ft he r ea r ek po i n t so fS oneachs i deo ft hel i ne ℓ po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. when ℓ j u s tl eave sonepo i n to fS. Wecons i de rtwogood po s i t i on sa r eequa li f ℓ pa s s e st he s ametwo po i n t s wi t ht hes amed i r e c t i on,o t he rwi s et hey a r e d i f f e r en tgood po s i t i on s.I ti sea s os eet ha tt henumbe ro f yt s i t i on si sf i n i t ei na l l wi ndmi l l s. Gi ven a po s i t i ve good po d i r e c t i ons ucha sx -ax i s,deno t ea l lgoodpo s i t i onsi nt heorde r ofc l ockwi s eang l e si n[ 0,2π)byℓ1 ,ℓ2 ,...,ℓm .Wepr oceed t opr ovei nt hr ees t ep s. Fi r s t l i ven anygood po s i t i on,wecans eet he r ei sa t y,g mo s tonegoodpo s i t i on.S i ncei ft he r ewe r etwogoodpo s i t i ons ℓi andℓj t ha thadt hes amed i r ec t i on,t hent hey we r epa r a l l e l andd i dno tco i nc i de,t henumbe ront her i i det ot hel i ne sℓi ghts andℓjs hou l dbekork -1, i tcou l dno tbepo s s i b l et obo t hl i ne s ℓi andℓj . Se cond l i n ti nS,t he r eex i s t ss omeℓi pa s s i ng y,foranypo t hr oughP a sf o l l ows. Takeanypo i n tP ∈ S,andl i ne ℓ pa s s i ngt hroughP bu t no tt hr ought heo t he rpo i n to fS. I ft henumbe rofpo i n t st ot he r i ts i deo fl i ne ℓi ss,t hent henumbe rofpo i n t st ot hel e f t gh s i deofl i ne ℓi s2 k -s.Thed i f f e r enceoftwonumbe r si s2 k- 2 s.Now,ro t a t e ℓ abou tP c l ockwi s e,wi t hs i nc r ea s i ng or k -2 s dec r ea s i ngby 1 when ℓ i s pa s s i ng a po i n t,s ot ha t2 change s2.When ℓ r o t a t e s180 °,2 k -2 sbecome si t soppo s i t e he r eex i s t samomen tt ha tan ℓ ofwh i ch numbe r.The r e f or e,t t henumbe r so f po i n t sontwos i d e sa r eequ a l.Deno t et hel a s t s s i ngpo i n tbyQ, t hent he r ei sagoodpo s i t i onℓi pa s s i ngPandQ. pa In t e rna t i ona lMa t hema t i c a lOl i ad ymp 303 Th i rd l ndmi l ls t a r t i ngf romagoodpo s i t i onmee t st he y,awi r equ i r emen to ft hepr ob l em. Wi t hou tl o s so fgene r a l i t t a r twi t hℓ1 .And we y,we mays s howt ha tt he nex t mee t i ng po i n ti sj u s tℓ2 ,s ot he wi ndmi l l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t i nue si nf i n i t e l ime s.Bys t eptwo, mee t sa l lℓ1 andcon y manyt weknowt ha teachpo i n tofSi su s eda sap i vo ti nf i n i t e l y many t ime s. s s i ngP andQ wi t hP a sap i vo tand Suppo s et ha tℓ =ℓ1 pa sj u s tl eav i ngQ .Suppo d i v i d i ngSequa l l s et ha tℓ me t ywhenℓ1i tandu s eR a sa p i vo t,t hen ℓ r ema i nd i v i de sS i n tR nex po equa l l eav i ngP wh i chi sva l i dforR onanys i deo fP . ywhen ℓl Th i ss howst ha tℓi ss t i l lagoodpo s i t i onwheni tme tR and deno t e st h i sgood po s i t i ona sℓ '.I tr ema i nst obes hownt ha t i ncei f t he r ei snogoodpo s i t i onbe tween ℓ andℓ ', s oℓ ' =ℓ2 .S ℓ2 wa ',t aked i r ec t edl i neℓ ″ sagoodpo s i t i onbe tween ℓ andℓ s s i ngP andpa r a l l e lt o ℓ2 ,t henℓ ″d i v i de sS equa l l ot ha t pa y,s t he r ea r ea tl ea s tk + 1po i n t sont hel e f torr i i deo f ℓ2 , ghts wh i chcon t r ad i c t st hef ac tt ha tℓ2i sagoodpo s i t i on. Now,con s i de rt he ca s e when | S |= 2 ki s even.The a r tabovei ss t i l lva l i d wi t hs u i t ab l er ev i s i on.Wes ay a gumen l i ne ℓi sonagood po s i t i oni ft he r ea r ek po i n t sofS ont he r i ts i deo ft hel i neℓ when ℓj u s tl eave sonepo i n to fS.Then gh t hef i r s tands e conds t ep scanbepr ovens imi l a r l het h i rd y.Fort s t ep,s t a r t i ngf r omagood po s i t i on,wecans hows imi l a r l ha t yt t henex tmee t i ngpo i n tby ℓ i ss t i l lagoodpo s i t i on.Theon l y s l i i f f e r enc et ot heoddca s ei st os howt ha tt he r ei snogood ghtd s i t i onbe tween ℓ1 and ℓ2 . po Taked i r e c t edl i neℓ ″ pa s s i ngP and pa r a l l e lt o ℓ2 ,t hen t he r ea r ekpo i n t sofSt ot her i fℓ ″. I fℓ2 we r eont her i ghto ght Ma t hema t i c a lOl i adi nCh i na ymp 304 s i det oℓ ″, t hent he r ewou l dbenomor et hank -2po i n t s,s oℓ ″ wou l dno tbeagoodpo s i t i on.I fℓ2 we r eont hel e f ts i det oℓ ″, ot her i ts i det oℓ t hent he r ewou l dbea tl ea s tk +1t oℓ ″,s ″ gh wou l dno tbea good po s i t i on.Thu s,we provedf ort heeven Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ca s e. 3 Le tf :R → Rbear ea l va l uedf unc t i onde f i nedont hes e t o fr ea lnumbe r st ha ts a t i s f i e s x +y)≤ yf( x)+f( x)) f( f( ① ovet ha tf( x)= 0fora l l fora l lr ea lnumbe r sx andy.Pr x ≤ 0.( s edbyBe l a r u s) po So l u t i on.( Ba s edont hes o l u t i onby Wu Mengx i) Le ty = f( x)-xi ni nequa l i t henwehave y ① ,t t hu s, x))≤ ( x)-x) x)+f( x)), f( f( f( f( f( ( x)-x) x)≥ 0. f( f( Con s equen t l ea lnumbe rx,wehave y,foranyr ② ( x))-f( x)) x))≥ 0. f( f( f( f( No t et ha tbyt ak i ngy = 0,① imp l i e sf( x) ≤ f( x)). f( The r e for e, x))≥ 0,orf( x))= f( x)< 0. f( f( f( ③ Wef i r s ts how t ha tf( x) ≤ 0forany r ea lnumbe rx by con t r ad i c t i on.I ft he r ei sar ea lnumbe rx0s ucht ha tf( x0)>0, t henf oranyr ea lnumbe ry,wehavef( x0 + y) ≤ yf( x0)+ x0)) f( f( ,wehavef( x0)).Thu s,foranyy < x0 +y) f( f( x0) f( In t e rna t i ona lMa t hema t i c a lOl i ad ymp 305 x0)) f( f( ,wehavef( z) ( x f 0) <0. So,f oranyr ea lnumbe rz <x0 - x0)) f( f( ,wes eet ha t x0) f( < 0.Th e r e for e,f orz < mi n 0,x0 - { } Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. z < 0andf( z)< 0. Thu s,by ② ,wehave ④ x0)) x0)) f( f( f( f( ≤ x0 z)≤z < mi . n 0,x0 f( ( ( ) f x0) f x0 { } Con s equen t l s, z))=f( x0 + ( z)-x0))<0.Thu f( f( y,f( by ② and ④ ,wehave z))= f( z)< 0. f( f( ⑤ z +y)≤ yf( z)+f( z))= ( z). f( f( y +1) f( ⑥ x0)≤ ( z). 1 +x0 -z) f( f( ⑦ Henc e,foranyr ea lnumbe ry,by ① and ⑤ ,wehave Le ty = x0 -zi n ⑥ ,t hen Tak i ngzt obes u f f i c i en t l t i ves ucht ha t1 +x0 -z > ynega x0) < 0, wh i ch i sa 0,t hen by ⑤ and ⑦ , we have f( con t r ad i c t i on. The r e f or e,foranyr ea lnumbe rx, x)≤ 0andf( x))≤ 0. f( f( ⑧ Le ty = - x i n ① ,t henf( 0) ≤ - xf( x)+ f( x)) ≤ f( -xf( x)by ⑧. Thu s,weon l os howt ha tf( 0)=0, t henforx <0, yneedt x)≥ 0,andby ⑧ ,weprovedf( x)= 0fora l lx ≤ 0. f( Inf ac t,i ff( t)=tha snonega t i ves o l u t i on,t henforany r ea lnumbe rx,wehave Ma t hema t i c a lOl i adi nCh i na ymp 306 x))≠ f( x)andf( x)))≠ f( x)). f( f( f( f( f( ⑨ Sof( equen t l x))≥0andf( x)))≥0by ③.Cons f( f( f( y, x))= 0andf( x)))= 0by ⑧.Tha ti s,f( 0)= 0. f( f( f( f( I ff( t) = t ha sa nega t i ves o l u t i on,wes how t ha tt i s Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t henby ① , un i i ncei ft he r ea r etwos o l u t i onst1 <t2 <0, que.S wehave t2 = f( t2)≤ ( t2 -t1) t1)+f( t1))= ( t2 -t1) t1 +t1 , f( f( t hu s( 1 - t1) ≤ 0, wh i ch i s a con t r ad i c t i on. t2 - t1)( Fu r t he rmor e,wes howt ha tt he r eex i s t sr ea lnumbe rxs ucht ha t x)≠t ( o t he rwi s ef ora l ly, t ≤y t +t,wh i chf a i l swheny < f( 0),hence ⑨ ho l df ort h i sx. S e c ondDay 9:00~13:30,Ju l y19,2011 4 Le tn > 0beani n t ege r.Wea r eg i venaba l anc eandn n-1 0 1 we i t so fwe i r et op l aceeach gh ght2 ,2 ,...,2 .Wea oft hen we i t sont heba l anc e,onea f t e rano t he r,i ns uch gh awayt ha tt her i sneve rheav i e rt hant hel e f tpan. ghtpani Ateachs t ep,wechoo s eoneoft hewe i sun t i la l loft he ght we i st ha thaveno tye tbeenp l acedont heba l anc eand ght l ac ei tone i t he rt hel e f tpanort her i t i la l lt he p ghtpan,un we i shavebeenp l ac ed. ght De t e rmi net henumbe rofway si nwh i cht h i scanbedone. ( s edbyI r an) po So l u t i on.( Ba s edont hes o l u t i onby YaoBowen)Thenumbe r o fway si s( 2 n -1)!! = 1 ×3 ×5 × … × ( 2 n -1). s Wepr ovet hepr ob l embyi nduc t i ononn.Theca s en = 1i c l ea r,pu tonewe i tont hel e f tpan.Hence,t he r ei son l gh yone In t e rna t i ona lMa t hema t i c a lOl i ad ymp 307 way. Suppo s et ha ti nt heca s en =k, t henumbe rofway si s( 2 k- 1)!! . t hou ta f f e c t i ngt her e s u l t, Thenfort heca s en = k +1,wi Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. mu l t i l t hwe i sby1/2,t hek + 1we i t sofwe i ta r e p ybo ght gh gh 1/2,1,2,...,2k-1 .S i nc eforanypo s i t i vei n t ege rr,wehave 2r > 2r-1 +2r-2 + … +1 + r-1 1 i ≥ ±2 . 2 i∑ =-1 The heav i e r pan i s whe r e t he heav i e s t we i s. ght i The r e for e,t heheav i e s twe i heba l ance mu s tbeont he ghtont l e f tpan.Int hefo l l owi ng,cons i de rt hepo s i t i onoft hewe i ght 1/2i nt heproc e s s i on. ( a)I fwet akewe i t1/2f i r s t,t heni tmu s tpu tont hel e f t gh her ema i n i ngk we i t shave( 2 k -1)!! way st odo. pan.Thent gh , , , ( ) / bI fwet akewe i ts t ept = 2 3 ... k +1,we ght1 2a c anpu ti te i t he ront hel e f tpano ront her i tpanb e c a u s ewe i t gh gh 1/2i st hel i t e s tone,s ot henumb e ro fway si s( 2 k -1)!! gh Tos umup,whenn =k +1, t he r ea r et o t a l l 2 k -1)!! + y( k ×( 2 k -1)!! = ( 1+2 k)( 2 k -1)!! = ( 2 k +1)!! way s,wh i ch comp l e t e st hei nduc t i on. 5 Le tf beaf unc t i onf romt hes e tofi n t ege r st ot hes e tof s i t i vei n t ege r s.Suppo s et ha t,f oranytwoi n t ege r sm and po n, t hed i f f e r enc ef( m )-f( n)i sd i v i s i b l ebyf( m -n). Pr ovet ha t,foranytwoi n t ege r swi t hf( m )< f( n),t he numbe ri sd i v i s i b l ebyf( m ).( s edbyI r an) po So l u t i on.( Ba s edont hes o l u t i onby Long Zi chao)Ac cord i ng t ot heprob l em,wehavef( k)|f( x +k)-f( x),∀x ∈ Zfor anyi n t ege rk ≠ 0.Cons equen t l k)|f( x +t k)- f( x), y,f( Ma t hema t i c a lOl i adi nCh i na ymp 308 ∀x ∈Z,t ∈ Z.Sof( k)|f( m )-f( n)foranym ,ns ucht ha t k|m -n.Thu s, n)|f( m +n)-f( m ), f( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. m - (-n))|f( m )-f(-n). f( ① ② i nce Wenowshowt ha tf oranyi n t ege rn,f( n)=f(-n).S o t he rwi s e,wi t hou tl o s so fgene r a l i t ft he r ei sani n t ege rn ≠ y,i 0, s ucht ha tf( n)> f(-n)> 0,t hen0 < f( n)-f(-n)< n),wh i chcon t r ad i c t sf( n)|f( n)-f(-n). f( The r e for e,by ② ,wehave m +n)|f( m )-f( n). f( ③ t henby ③ ,wehave0 <f( I f0 <f( m )<f( n), m +n)< n).Comb i n i ng ① and ③ ,wehavef( m +n)=f( m ).Then f( by ③ ,wege tt her e s u l tf( m )|f( n).Forf( m )= f( n),t he r e s u l ti sobv i ou s. 6 Le t△ABC beanacu t et r i ang l ewi t hc i r cumc i r c l eΓ.Le tl beat angen tl i net o Γ,andl e tla ,lb andlc bet hel i ne s ob t a i nedby r e f l ec t i ngl t ot hel i ne sBC ,CA and AB , r e s c t i ve l ha tt hec i r cumc i r c l e oft het r i ang l e pe y.Show t st angen tt ot hec i r c l e Γ. f ormedbyl i ne sla ,lb andlc i ( s edbyJ apan) po So l u t i on.( Ba s edont hes o l u t i onby Chen Li n)Le tP bet he o Γ.Deno t et hes t r i cpo i n t sofP t o t angen tpo i n to flt ymme BC ,CA ,andAB byPa ,Pb andPc ,r e s t i ve l s epo i n t s pec y.The a r eonc i r c l e sΓa ,Γb andΓcs t r i ct oΓt oBC ,CA andAB , ymme r e s c t i ve l e, ℓa , ℓbandℓca r et angen tt oΓa ,ΓbandΓca t pe y.Henc i n t sPa ,Pb andPc ,r e s c t i ve l po pe y. Deno t ela ∩lb = C ',lb ∩lc = A',lc ∩la = B'. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. In t e rna t i ona lMa t hema t i c a lOl i ad ymp 309 F i .6 .1 g De f i net heor i en t edang l eo fl i nem t ol i nen by Themagn i t udeo f f r om m t on. ( m ,n). ( m ,n)i st heang l ero t a t edan t i c l ockwi s e Weconc l udet ha t ( r eco l l i nea r. 1)Po i n t sPa ,Pb Pc a Inf ac t,t he mi dpo i n t sofPPa ,PPb ,PPc a r et he peda l i n t so fPt oBC,CAandAB, r e s t i ve l lpo i n t sa r e pec y.Thepeda po , co l l i nea rbyS ims on sTheor em.SoPa Pb andPca r eco l l i nea r. ( 2) Deno t et hec i r c umc i r c l e so f △A'PbPc , △B 'PcPa and △C 'PaPb by Γ1,Γ2 and Γ3,r e s c t i v e l i r c umc i r c l e so f pe y.Andc △A 'B 'C 'byΩ.Thenf ou rc i r c l e sΓ1,Γ2,Γ3 andΩa r ec opunc t a l. In f ac t,i ti sj u s tt he Mi l Theor em for comp l e t e que i l a t e r a lA'PcB'PaC 'Pb .Deno t et hecopunc t a lpo i n tbyQ . quadr ( ) , , r e on c i r c l e s Γ1 Γ2 and Γ3 , 3 Po i n t s A B and C a r e s t i ve l pec y. ︵ Inf ac t,c i r c l e sΓb andΓc i n t e r s ec ta tpo i n tA ,andAPc = ︵ ︵ t a t eanang l eo f ( AP = APb .Ro PcA ,PbA) abou tpo i n tA , t hen lc ,lb )= Γc → Γb ,Pc → Pb ,andt ang en tl i nelc →lb .So, ( ( PcA ,PbA)and ( lc , lb )= ( i chme an sf ou r PcA',PbA'),wh Ma t hema t i c a lOl i adi nCh i na ymp 310 i n t sa r ec on c c l i c.Tha ti s,po i n tA i sonc i r c l eΓ1.S imi l a r l po y y, r eonc i r c l e sΓ2 andΓ3,r e s c t i v e l i n t sB andCa pe y. po ( 4)Po i n tQi sonc i r c l eΓ. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Byde f i n i t i ono fpo i n tQ , ( AQ ,BQ )= ( AQ ,PcQ )+ = ( APb ,BPa ) = = = =2 = ( APb ,PbPc )+ ( APb ,AC)+ ( AC ,AP )+ ( AC ,BC)- ( AC ,BC). ( PcQ ,BQ ) ( PcPa ,BPa ) ( AC ,BC)+ ( AC ,BC)+ ( AP ,BP ) ( BC ,BPa ) ( BP ,BC) Henc e,fou rpo i n t sA ,B ,C andQ a r econcyc l i c,t ha ti s, i n tQi sonc i r c l eΓ. po ( 5)Ci r c l eΓi st angen tt oc i r c l eΩa tpo i n tQ . n t e r s e c tc i r c l eΩa tpo i n t sQandA″, Le tl i neQAi t henweha v e ( A″B',B'Q )= = = = = = = = ( A″B',B'A')+ ( A″Q ,QA')+ ( A'B',B'Q ) ( A'B',B'Q ) ( AQ ,A'Q )+ ( A'B',B'Q ) ( AB ,BPc )+ ( A'B',B'Q ) ( AB ,BPc )+ ( PcB ,BQ ) ( APc ,A'Pc )+ ( AB ,BPc )+ ( AB ,BQ ). ( A'B',B'Q ) ( PcB',B'Q ) ︵ ︵ r eequa li n Th i ss howst ha tt hedeg r ee so fA″Q andAQ a c i r c l eΩ.No t et ha tpo i n t sA ,Q andA″a r eco l l i nea r,andpo i n t Qi st hes amepo i n tonc i r c l e sΩandΓ.The r e for e,t het angen t l i ne soft hetwoc i r c l e sa tpo i n tQ co i nc i de.Tha ti s,t hec i r c l e s ΩandΓa r et angen ta tQ . In t e rna t i ona lMa t hema t i c a lOl i ad ymp 2012 311 ( Ma rD e lP l a t a,A r e n t i n e) g F i r s tDay Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 9:00~13:30,Ju l y10,2012 1 Gi vent r i ang l e ABC ,t he po i n tJ i st he cen t r eo ft he i sexc i r c l ei st angen tt ot he exc i r c l eoppo s i t eve r t exA .Th s i deBC a tM ,andt tK andL , ot hel i ne sAB andAC a ta tF ,andt r e s c t i ve l i ne sLM andBJ mee hel i ne s pe y.Thel KM andCJ mee ta tG .Le tSbet hepo i n tofi n t e r s ec t i onof e tT bet l i ne sAF andBC ,andl hepo i n toft hei n t e r s e c t i on o fl i ne sAG andBC . Provet ha tM i st hemi dpo i n tofST .( Theexc i r c l eof △ABC oppo s i t et heve r t exAi st hec i r c l et ha ti st angen tt o t hel i nes egmen tBC ,t ot her ayAB beyondB ,andt ot he r ayAC beyondC .) So l u t i on.Le t ∠CAB =α,∠ABC =β,∠BCA =γ. S i nc eAJi s t heb i s e c t o ro f ∠CAB, ∠JAK = ∠JAL =α/2.S i nc e ∠AKL = ∠ALJ =9 0 °,po i n t sK andLa r eont hec i r c l eω wi t hd i ame t e rAJ. F i .1 .1 g st heb i s e c t oro f ∠KBM ,wehave ∠MBJ =90 °S i nc eBJi Ma t hema t i c a lOl i adi nCh i na ymp 312 imi l a r l ° - γ/2. y, ∠CML = γ/2 and ∠MCJ = 90 β/2.S Con s equen t l y,∠LFJ = ∠ MBJ - ∠BMF = ∠MBJ - ∠CML =9 0 °-β/2 -γ/2 = α/2 = ∠JAL .Hence,po i n tF i sont he c i r c l eω.S imi l a r l i n tG i sa l s oont hec i r c l eω.S i nceAJi s y,po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t hed i ame t e ro fc i r c l eω,wehave ∠AFJ = ∠ AGJ = 90 °. r es t r i cabou tt heb i s ec t orof Segmen t sAB andBC a ymme ex t e rna lang l eof ∠ABC .AndbyAF ⊥ BF andKM ⊥ BF ,we s eet ha ts egmen t sSM andAK a r es t r i cabou tBF ,SM = ymme AK . S imi l a r l i nc eAK = AL ,wehaveSM = y,TM = AL .Ands TM ,name st hemi dpo i n tofST . l y,M i 2 Le n t ege r,andl e ta2 ,a3 ,...,an bepo tn ≥3beani s i t i ve ovet ha t( 1+ r ea lnumbe r ss ucht ha ta2a3 …an = 1.Pr 2 3 n n ( …( >n . 1 +a3) 1 +an ) a2) So l u t i on.BytheAM GMinequa l i t y,wehave 1 1 æ 1 ök k ( + +… + +ak ÷ = ç 1 +ak ) k -1 èk -1 k -1 ø æ 1 ö÷k-1 ak èk -1ø k ≥k · ç f ork = 2,3,...,n. Thu s, 2 3 n ( ( …( 1 +a2) 1 +a3) 1 +an ) ≥ 2a2 ·3 2 =n . n æ 1 ö÷2 · 4 æç 1 ö÷3 ·…· n æç 1 ö÷n-1 a3 4 a4 n an è2ø è3ø èn -1ø 3ç Theequa l i t l d swhenak = yho 1 , k =2,...,n,wh i ch k -1 i sno tt heca s es i ncea2a3 …an = 1. 2 3 n n ( …( >n . Henc e,( 1 +a2) 1 +a3) 1 +an ) In t e rna t i ona lMa t hema t i c a lOl i ad ymp 313 3 Thel i a r sgue s s i nggame i sa game p l ayed be tweentwo u l e so ft he gamedependontwo l aye r sA andB .Ther p i cha r eknownt obo t hp l aye r s. s i t i vei n t ege r skandn wh po , Att hes t a r toft hegame A choo s e si n t ege r sx andN Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. wi t h1 ≤ x ≤ N .P l aye rA keepsx s e c r e t,andt r u t hf u l l y t e l l sN t o p l aye r B . Now p l aye rB t r i e st o ob t a i n s t i ons a s i nforma t i on abou t x by a s k i ng p l aye r A que f o l l ows:eachque s t i oncons i s t sofBs c i f i ngana rb i t r a r pe y y fpo s i t i vei n t ege r s( s s i b l c i f i edi ns ome s e tS o po y ones pe ev i ou sque s t i on),anda s k i ngA whe t he rx be l ong st oS. pr s ka s manys uch que s t i onsa she wi she s. P l aye rB may a Af t e reachque s t i on,p l aye rA mu s timmed i a t e l ri t yanswe wi t hye sorno,bu ti sa l l owedt ol i ea smanyt ime sa sshe wan t s;t he on l e s t r i c t i on i st ha t,among any k + 1 yr con s ecu t i veanswe r s,a tl ea s tonean swe ri st r u t hf u l. Af t e rB ha sa s keda smanyqu e s t i on sa shewan t s,hemu s t s c i f e to fX o fa tmo s tnpo s i t i v ei n t e e r s.I fxbe l ong st o pe yas g X, t henB wi n s;o t he rwi s e,hel o s e s.Pr ov et ha t: 1 .I fn ≥ 2k , t henB cangua r an t eeawi n; 2 . Fora l ls u f f i c i en t l a r t he r eex i s t sani n t ege rn > yl gek, 1 .99k s ucht ha tB canno tgua r an t eeawi n. So l u t i on.Werephras et her u l eo ft hegamea sfo l l ows:Gi ven twopo s i t i vei n t ege r skandn,p l aye rAt e l l saf i n i t es e tT = { 1, 2,...,N }t op l aye rB andkeep sonee l emen txs e c r e t.P l aye r B choo s e saf i n i t es ub s e tS o fT anda sksp l aye rA whe t he rx be l ong st oS.P l aye rA an swe r si twi t hye sornoandi sa l l owed t ol i ea tmo s tcon s e cu t i vekt ime s.Af t e ra s k i ngf i n i t eque s t i ons, i fp l aye rB cans i f ub s e tofT con t a i n i ngx wi t ha tmo s tn pec yas e l emen t s,t henp l aye rB wi n s. ( 1)I f N > 2k ,we wi l lshow t ha tp l aye rB can a lway s Ma t hema t i c a lOl i adi nCh i na ymp 314 de t e rmi nes omee l emen ty ∈ T s ucht ha ty ≠ x.Thu s,wecan r e s t r i c tN ≤ 2k .Con s equen t l fn ≥ 2k , t henp l aye rB wi ns. y,i Deno t eT = { 1,2,...,N },p sksp l aye rA s ame l aye rB a Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s t i onkt ime s:I sx = 2k +1?I ft heanswe r sa r ea l lno,t hen que k , ≠ + = x 2 1 y.Oncean answe ri s ye s t hent he nex tf i r s t s t i ono fp l aye rB i s:I sx ∈ { t ∈ Z|1 ≤ t ≤ 2k-1}? I fA que k-1 { , an swe r sye st henwehavet of i nday ∈ t ∈ Z|2 +1 ≤t ≤ 2k }s ucht ha ty ≠ x.I fA an swe r sno,t henwehavet of i nday k-1 { } , ∈ t ∈ Z|1 ≤t ≤ 2 s ucht ha ty ≠ x. Int h i sway byeach educ e ha l ft hes i z e oft hes e t an swe rof p l aye r A ,wecan r con t a i n i ngy.When A an swe r sk que s t i ons,wecanconc l ude k fa = x,t i e s t ha tt he r ei saun i henA l quea,1 ≤ a ≤ 2 .I con s e cu t i vek +1t ime s.Soy = a ≠ x. ( 2) We pr ove t ha tf or any 1 < λ < 2,i fn = k+1 [( λ ]-1,t hen p l aye r B canno t gua r an t ee a wi n. 2 -λ) Spec i a l l akeλ,1 .99 <λ <2, f ors u f f i c i en t l a r n t ege rk, y,t yl gei wehave n = [( 2 -λ) λk+1]-1 > 1 .99k , wh i chi st her equ i r edconc l u s i on. P l aye rA choo s e sT = { 1,...,n +1},andchoo s e sanyx ∈ T .Deno t et hemax imumnumbe ro fcons ecu t i vel i e sbymi when x =i.Andde f i neϕ = ∑i=1λmi .Thes t r a t egyofAi st ochoo s e n+1 t ol i eorno ts ucht ha tϕt ake st hesma l l e rva l ue.Wewi l lshow t ha tϕ < λk+1 a tanyt imewi t ht hes t r a t egys t a t edabove.Thu s mi ≤kforeachi ∈ T .SoBcanno tde t e rmi newhe t he ri =xor tgua r an t eeawi n. no t.Thu s,p l aye rB canno 1 1 n( λϕ +n +1) ϕ = mi ϕ1 ,ϕ2)≤ 2 ( ϕ1 +ϕ2)= 2 ( < 1 (k+2 ( λ + 2 -λ) λk+1)=λk+1 . 2 In t e rna t i ona lMa t hema t i c a lOl i ad ymp 315 Att hebeg i nn i ng,mi =0, s oϕ =n +1 <λk+1 .Suppo s et ha t a f t e rs eve r a lan swe r s,ϕ < λk+1 ,andB a s ks:I sx ∈ S? The an swe r“ s”or “ no”cor r e s i ng t otwo numbe r sofϕ, ye pond r e s c t i ve l sa sfo l l ows:ϕ1 = pe y,i ∑ Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i∉S + ∑ ∑ λmi+1 .Byde f i n i t i on, i∈S 1 + ∑i∉Sλmi+1andϕ2 = i∈S 1 1 n( λϕ +n +1) ϕ = mi ϕ1 ,ϕ2)≤ 2 ( ϕ1 +ϕ2)= 2 ( < 1 (k+2 ( λ + 2-λ) λk+1)= λk+1 . 2 S e c ondDay 9:00~13:30,Ju l y11,2012 4 F b, i nda l lf unc t i on sf:Z→Zs ucht ha t,fora l li n t ege r sa, ct ha ts a t i s f t hef o l l owi ngequa l i t l d s: ya +b +c = 0, yho 2 a)+f2( b)+f2( c) f ( =2 a) b)+2f( b) c)+2f( c) a). f( f( f( f( ① ( He r eZdeno t e st hes e to fi n t ege r s.) So l u t i on.Le ta =b =c = 0y i e l d3f2( 0)= 6f2( 0).Hence, 0)= 0. f( ② a)= f(-a),∀a ∈ Z. f( ③ 2 = 0.He Le tb = -a,c = 0y i e l d( a)-f(-a)) nce,f f( i sodd,t ha ti s, 2 2 2 + f( =2 + Le tb = a,c = - 2 ay i e l d2f( a) 2 a) a) f( 4f( a) 2 a).Hence, f( 2 a)= 0orf( 2 a)= 4f( a),∀a ∈ Z. f( ④ I ff( r)=0fors omer ≥1, t henl e tb =r, c = -a -ry i e l d Ma t hema t i c a lOl i adi nCh i na ymp 316 2 ( = 0.Th a +r)- f( a)) a ti s,fi sape r i od i cf unc t i onof f( r i odr: pe a +r)= f( a),∀a ∈ Z. f( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a)= 0,∀a ∈ Z,i Es c i a l l ff( 1)= 0.Andi nt h i s pe y,f( a)= 0,∀a ∈ Z,s ca s e,f unc t i onf( a t i s f i e st hecond i t i on ①. 1)=k ≠0.Thenby ③ ,f( 2)=0,or Nows uppo s et ha tf( 2)= 4 2)= 0, t henfi k.I ff( sape r i od i cf unc t i ono fpe r i od f( 2: 2 n)= 0,f( 2 n +1)=k,∀n ∈ Z. f( t henby ④ ,f( 2)= 4 4)= 0,orf( 4)= 16 I ff( k ≠ 0, k. sape r i od i cf unc t i onofpe r i od4,and I ff( 4)= 0, t henfi e, 3)= f(-1)= f( 1)=k.Henc f( n)= 0,f( n +1)= f( n +3)=k, 4 4 4 f( 4 n +2)= 4 k,∀n ∈ Z. f( 2 I ff( 4)= 16 k ≠ 0, l e ta = 1,b = 2,c = -3y i e l df( 3) 10 kf( k2 = 0. 3)+9 Thu s, 3)∈ { k,9 k}. f( ⑤ 3)∈ { 9 k,25 k}. f( ⑥ Le ta =1, b =3, c =-4y i e l df2( 3)-34 kf( k2 =0. 3)+225 Thu s, Henc e,by ⑤ and ⑥ ,f( 3)= 9 k. Now wes howt ha tf( x) = kx2 ,∀x ∈ N,byi nduc t i on. Forx = 0,1,2,3,4,wehavef( x)=kx2 .Nows uppo s et ha t x)=kx2forx = 0,1,...,n( n ≥ 4),t henl e ta = n -1, f( b = 2,c = -n -1i n①y i e l d In t e rna t i ona lMa t hema t i c a lOl i ad ymp 317 2 2 ,k( }. n +1)∈ { k( n +1) n -1) f( Andl e ta = n -1,b = 2,c = -n -1i n①y i e l d 2 2 ,k( }. n +1)∈ { k( n +1) n -3) f( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 2 n +1)=k ( n +1) f orn ≥4.The r e f or e,f( x)= Thu s,f( l lx ∈ N.By ③ ,f( l lx ∈ Z. kx2fora x)=kx2fora Tos um up a l lt heaboveca s e s,we now checkt hef i na l r e s u l t: x)= 0,f2( x)=kx2 , f1( x)= f3( 0,x ≡ 0( mod2), { mod2), k,x ≡ 1( mod4), ìï0,x ≡ 0( ï mod2), x)= ík,x ≡ 1( f4( ïï î4 k,x ≡ 2( mod4). a t i s f fa,b,ca Obv i ou s l r eeven, y ①.Forf3 ,i y,f1 ,f2 s a)= f( b)= f( c)= 0s a t i s f ft he r ea r etwoodd t henf( y ① ;i andoneeveni na,bandc,t henbo t hs i de sof ① a r eequa lt o 2 k2 . Forf4 , s i nc ea +b +c = 0,( a),f( b),f( c))ha son l f( y f ourca s e s:( 0,k,k),( 4 k,k,k),( 0,0,0)and ( 0,4 k,4 k), obv i ou s l heya l ls a t i s f y,t y ①. 5 Le t△ABC beat r i ang l ewi t h ∠BCA = 90 °,andl e tD be tX beapo i n ti nt he t hefoo toft hea l t i t udef r om C .Le i n t e r i oroft hes egmen tCD .Le tK bet he po i n tont he ucht ha tBK = BC .S imi l a r l e tL bet he s egmen tAX s y,l i n tont hes egmen tBXs ucht ha tAL = AC .Le tM bet he po i n to fi n t e r s ec t i onofAL andBK . po Showt ha tMK = ML . Ma t hema t i c a lOl i adi nCh i na ymp 318 So l u t i on.Le tpo i n t sC 'andC bes t r i cove rt hel i neAB , ymme andω1 andω2 bec i r c l e swi t hc en t r e sA andB ,r ad i u sAL and BK , r e s c t i v e l i nc eAC ' = AC pe y.S Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. = AL a nd BC ' = BC = BK , i n t sCandC 'a r ebo t honc i r c l e s po ω1 andω2 .S i nce ∠BCA = 90 °, r et angen tt o l i ne sAC and BC a c i r c l e sω2 andω1 ,r e s c t i ve l t pe y,a t he r i n t C . Le t K1 be ano po i n t e r s e c t i onpo i n tofl i neAX and i f f e r en tt oK ,andL1be c i r c l eω2d ano t he ri n t e r s e c t i on po i n to fl i ne BX andc i r c l eω1 d i f f e r en tt oL . F i .5 .1 g Byt heCi r c l e -Powe rTheor em,wehave XK ·XK1 = XC ·XC' = XL ·XL1 , henc eK1 ,L ,K ,L1 a r efou rconcyc l i cpo i n t sa tc i r c l edeno t ed byω3 . Byapp l i ngCi r c l e -Powe rTheor emt oc i r c l eω2 ,wehave y AL2 = AC2 = AK ·AK1 . Th i simp l i e st hel i neALi st ang en tt oc i r c l eω3a tpo i n tL.By t hes imi l a ra r t,l i neBKi st ang en tt oc i r c l eω3 a tpo i n tK . gumen The r e f or e,MK andMLa r etwot angen tl i ne sf r om po i n tM t oc i r c l eω3 , t hu sMK = ML . 6 F i nd a l l po s i t i vei n t ege r sn f or wh i ch t he r e ex i s t non nega t i vei n t ege r sa1 ,a2 ,a3 ,...,an ,s ucht ha t 1 1 1 1 2 n + + … + a = a + a + … + a = 1. ① 2a1 2a2 3n 2n 31 32 In t e rna t i ona lMa t hema t i c a lOl i ad ymp 319 So l u t i on.Suppos et ha tns a t i s f i e scond i t i on ① ,t ha ti s,t he r e ex i s tnon -nega t i vei n t ege r sa1 ≤ a2 ≤ a3 ≤ … ≤ an ,s ucht ha t n -a ∑2 i = i=1 n ∑i·3 -ai i=1 = 1. Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Mu l t i l i ngbo t hs i de so ft hes e condequa l i t p y y byan ,and t hent ak i ng mod2,wehave n 1 ∑i = 2n(n +1)≡ 1(mod2), i=1 henc e, n ≡ 1,2( mod4).Int hefo l l owi ng,wesha l lshowt ha t na r ea l li n t ege r ss uch t h i si sa l s oas u f f i c i en tcond i t i on.Tha ti s, t ha tn ≡ 1,2( mod4). Weca l las e to ff i n i t er epea t ab l epo s i t i vei n t ege r sB = { b1 , b2 ,...,bn }a r e s ea s i b l e”,i s“ ft hee l emen t sofB cancor f pond t onon -nega t i vei n t ege r sa1 ,a2 ,...,an s ucht ha t n -a ∑2 i = i=1 n ∑b3 i=1 i -ai = 1. sf ea s i b l e,r ep l ac i ngany No t eanimpor t an tf ac tt ha ti fBi e l emen tbo fB bytwopo s i t i vei n t ege r suandv wi t hu +v =3 b, sa l s of ea s i b l e.In f ac t,i fb t hen t he r e s u l t i ng s e t B' i cor r e s st o non -nega t i vei n t ege ra,wet akebo t hu andv pond cor r e s i ng t o a + 1, keep i ng o t he r s cor r e s pond pondence unchanged.S i nc e 2-a-1 +2-a-1 = 2-a , u·3-a-1 +v·3-a-1 =b·3-a , B'i sf ea s i b l e.I fB'canbeob t a i nedbys uchf i n i t er ep l acemen t s t et hem byB f r omB ,wedeno B'.Pa r t i cu l a r l fb ∈ B , y,i t henwecanr ep l acebbyband2 b, t hu s,B B ∪{ 2 b}. Wesha l ls howt ha tf oreachpo s i t i vei n t ege rn ≡ 1,2( mod Ma t hema t i c a lOl i adi nCh i na ymp 320 4),Bn = { 1,2,...,n} i sf ea s i b l e.B1i sf ea s i b l e,s i nc ewecan t akea1 = 0.B2i sf ea s i b l e,s i nceB1 n ≡ 1( mod4), t hens i nc eBn B2 .I fBni sf ea s i b l ef or Bn+1 ,Bn+1i sa l s of ea s i b l e. sf ea s i b l ef orn ≡ 1( Thu s,i ts u f f i ce st os howt ha tBn i mod Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. sf ea s i b l e,s i nc e 4).B5i B2 { 1,3,3} B9i sf ea s i b l e,s i nce { 1,2,3,4,6,9} B5 sf ea s i b l e,s i nc e B13i B9 { 1,3,4,5} B5 . { 1,2,3,5,6,7,9} B9\{ 8} B9. B13 , { 1,2,3,4,5,6,7,9,11,13} Thel a s ts t epi sob t a i ned by append i ng2 bs eve r a lt ime s. sf ea s i b l e, Append i ng8,10and12s uc ce s s i ve l tt ha tB17i y,wege s i nce B6 B5 ∪ { 7,11} B12 ∪ { 15} B8 ∪ { 11} B17\{ 10,14,16} La s t l howt ha tB4k+2 y,wes andcomp l e t et hepr oo f. B7 ∪ { 9,11,15} B17 . n t ege rk ≥ 2, B4k+13foranyi Bys uc c e s s i ve l i ng4 k +4,4 k +6,...,4 k +12,we yappend no t et ha t ( /2 ≤ 4 4 k +12) k +2. Int her ema i n i ngs i xoddnumbe r s4 k +3,4 k +5,...,4 k+ 13,deno t et henumbe r st ha ta r e mu l t i l e sof3byu1 ,v1 ,t he p s umoftwoo fwh i chi sa mu l t i l eo f3byu2 ,v2 andu3 ,v3 . p Thens ub s t i t u t ebi = 1( ui +vi)byui ,vifori =1,2,3.No t e 3 t ha tbii seven,s oweappendbi/2andge tB4k+13 . In t e rna t i ona lMa t hema t i c a lOl i ad ymp 2013 321 ( S a n t aMa r t a,Co l omb i a) F i r s tDay Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 9:00~13:30,Ju l y23,2013 Pr ovet ha tf oranypa i ro fpo s i t i vei n t ege r skandn,t he r e 1 ex i s tkpo s i t i vei n t ege r sm 1 ,m 2 ,...,mk ( no tnece s s a r i l y d i f f e r en t)s ucht ha t 1+ 1 ö÷ æç 1 ö÷ … æç 1 ö÷ æ 2k -1 1+ 1+ = ç1 + . m1 ø è m2 ø è mk ø è n ( s edbyJ apan) po So l u t i on.Me thod1.Byi nduc t i ononk,t heca s eofk ① = 1i s t r i v i a l.Suppo s et ha tt hepropo s i t i oni st r uefork = j - 1,we s howt ha tt heca s eo fk =ji sa l s ot r ue. sodd,t ha ti s, n =2 t -1fors I fni omepo s i t i vei n t ege rt, no t et ha t 2j -1 2( t +2j-1 -1) 2 · t = 1+ 2 t -1 2 t 2 t -1 æ è 2j-1 -1ö÷ æç 1 ö÷ , 1+ 2 t -1ø t øè = ç1 + andbyt hehypo t he s i so fi nduc t i on,wecanf i ndm 1 ,...,mj-1 s ucht ha t 2j-1 -1 1 ö÷ æç 1 ö÷ … æç 1 ö÷ æ 1+ 1+ = ç1 + . m1 ø è m2 ø è mj-1 ø è t 1+ Thu s,weneedon l ot akemj = 2 t -1. yt I fni seven,t ha ti s, n =2 tfors omepo s i t i vei n t ege rt,no t e t ha t Ma t hema t i c a lOl i adi nCh i na ymp 322 2j -1 2 t +2j -1 2 t +2j -2 · = j 2 t 2 t 2 t +2 -2 1+ 2j-1 -1ö÷ 1 ö÷ æç 1+ j t +2 -2ø è 2 t ø = ç1 + Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. æ è hehypo t he s i sofi nduc t i on,wecanf i nd and2 t +2j -2 >0,byt m 1 ,...,mj-1s ucht ha t 2j-1 -1 1 ö÷ æç 1 ö÷ … æç 1 ö÷ æ 1+ 1+ = ç1 + . m1 ø è m2 ø è mj-1 ø è t 1+ Thu s,weneedon l ot akemj = 2 t +2j -2. yt Me thod2.Cons i de rt heb i na r i ono ft her ema i nde r s yexpans fnand -n: o fmod2k o n -1 ≡ 2a1 +2a2 + … +2ar ( mod2k ), whe r e0 ≤ a1 < a2 < … < ar ≤k -1,and -n ≡ 21 +22 + … +2s ( mod2 ), b b b k whe r e0 ≤b1 <b2 < … <bs ≤k -1. S i nc e 0 1 k 1 -1 ≡ 2 +2 + … +2 - ( mod2k ), wehave { a1 ,a2 ,...,ar }∪ { b1 ,b2 ,...,bs } ={ 0,1,...,k -1},andr +s =k. For1 ≤ p ≤r,1 ≤q ≤s,wedeno t e Sp = 2ap +2ap+1 + … +2ar , Tq = 2b1 +2b2 + … +2bq . Andde f i neSr+1 = T0 = 0.No t et ha t S1 +Ts = 2k -1 andn +Ts ≡ 0( mod2k ).Wehave In t e rna t i ona lMa t hema t i c a lOl i ad ymp 323 2k -1 n +S1 +Ts = n n 1+ = n +S1 +Ts ·n +Ts n +Ts n r Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. = 2ap 2bq æç ö÷ æç ö÷ · + + . 1 1 ∏ ∏ n +Sp+1 +Ts ø q=1 è n +Tq-1 ø p=1 è r = s n +Sp +Ts n +Tq ·∏ ∏ + + +Tq-1 n S T n s p+1 p=1 q=1 s Thu s,i ff or1 ≤ p ≤ r,1 ≤ q ≤ s,de f i ne m p = n +Sp+1 +Ts n +Tq-1 ,t hen wege tt her equ i r ed andmr+q = ap 2bq 2 equa l i t l opr ovet ha ta l lmi a r ei n t ege r s.For1 y.Weneedon yt ≤ p ≤r,weknowt ha t n +Sp+1 +Ts ≡ n +Ts ≡ 0( mod2ap ). Andfor1 ≤q ≤s,wehave mod2bq ), n +Tq-1 ≡ n +Ts ≡ 0( wh i chob t a i n st heconc l u s i on. Me thod3.Foranya(≠ 0,-1),wehave 1ö æ 1+ 1 ö÷ 1öæ æç 1 + ÷ ç1 + = çç a ÷÷ . aøè a +1ø è 2ø è ② Wer ewr i t et hel e f t -hands i de of ① i nt hef orm oft he oduc to f2k -1f r ac t i on s: pr 1 ö÷ æç 1 ö÷ æç 1 ö÷ … æ n +2k -1 1+ 1+ = ç1 + nøè n +1ø è n +2ø è n 1 1 æç ö÷ æç ö÷ 1+ 1+ . è n +2k -3ø è n +2k -2ø ③ Fort heevenn,g r oup i ngs uc c e s s i ve l her i -hands i deof yt ght ③i npa i r sf romt hel e f tt ot her i s i ng ② wege tt he ghtandbyu Ma t hema t i c a lOl i adi nCh i na ymp 324 f ormo fpr oduc to f2k-1f r ac t i ons: 1öæ 1 ö æ 1 ææ öö 1 ö÷ ç ç1 + n ÷ ç1 + n ÷ … ç1 + n ÷ ÷ æç1 + . k 1 çç ÷ç ç +1÷ +2 -2÷ ÷ è n +2k -2ø 2øè 2 2 èè ø è øø Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ④ roup i ngt her i t -hands i deo f③i npa r e Fort heevenn,g gh f r omt her i ot hel e f tandbyu s i ng ② ,wege tt heform of ghtt oduc to f2k-1f r ac t i ons: pr æ 1ö æç 1 + ÷ çç nø è è 1 öæ 1 æ ö ç1 +n +1÷ ç1 +n +1 ÷… ç ÷ç +1÷ 2 øè 2 è ø ö 1 æ ö 1 æç ö÷ ÷ ç1 +n +1 ÷ + 1 k 1 ç ( /2 +2k-1 -2ø ÷ . +2 - -3÷ è n +1) 2 è ø ø ⑤ Repea t i ngt heabove g r oup i ngproc e s st ot hef r ac t i onsi n b i r en t he s e so f④ and ⑤ k -2t ime s,r e s t i ve l l l gpa pec y,wewi tt heformo ft her i t -hands i deof①. ge gh 2 A conf i r a t i on o f 4027 po i n t si nt he p l anei s ca l l ed gu Co l omb i ani fi tcons i s t so f2013r edpo i n t sand2014b l ue i n t s,and no t hr ee po i n t s of t he conf i a t i on a r e po gur co l l i nea r.Bydr awi ngs omel i ne s,t hep l anei sd i v i dedi n t o s eve r a lr eg i on s.An a r r angemen tofl i ne si sgood f ora Co l omb i anconf i a t i oni ft hefo l l owi ngtwocond i t i ons gur a r es a t i s f i ed: ● Nol i nepa s s e st hroughanypo i n to ft heconf i a t i on. gur ● Nor eg i oncon t a i n spo i n t sofbo t hco l our s. F i ndt hel ea s tva l ue o fk s ucht ha tf orany Co l omb i an conf i r a t i ono f4027po i n t s,t he r ei sagooda r r angemen t gu o fkl i ne s.( s edby Au s t r a l i a) po In t e rna t i ona lMa t hema t i c a lOl i ad ymp So l u t i on1.Theansweri sk 325 =2 013. t hanexamp l e.Wema rk Fi r s t l ha tk ≥ 2013wi y,weshowt 2013r edpo i n t sand2013b l uepo i n t sonac i r c l ea l t e rna t i ve l y. Thent he r ea r e4026a r c sont hec i r c l ewi t hd i f f e r en tendpo i n t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. co l ou r s.Ma rkano t he rpo i n ti nb l ueont hep l ane.I fkl i ne si s hen each a r ci n t e r s e c t s wi t hs omel i ne and any l i ne good,t i n t e r s e c t st hec i r c l ea tmo s ttwopo i n t s,hencet he r ea r ea tl ea s t 4026/2=2013l i ne s. In t he fo l l owi ng, we s how t ha tt he r e ex i s t s a good a r r angemen to f2013l i ne s. t ht hes ame F i r s t,no t et ha tf ortwo po i n t sA and B wi co l our,wecans epa r a t et he s etwopo i n t swi t ho t he r sbydr awi ng twos u f f i c i en tnea rpa r a l l e ll i ne st oAB oneachs i deofAB . heconvexhu l lo fa l lco l ou r edpo i n t s.Taketwo Le tP bet ad acen tve r t i ce so fP ,s aypo i n t sA andB .Theno t he rpo i n t s j I foneoft hemi sr ed,s ayA . a r el oca t edonones i deo fl i neAB . Thenwecandr awonel i net os epa r a t eA wi t ho t he rpo i n t s.The r ema i n i ng2012r ed po i n t scanbeg r oupedi n1006 pa i r s;each i ro fr edpo i n t scanbes epa r a t edbytwol i ne s.Soa l lt oge t he r pa 2013l i ne smee tt her equ i r emen t.I fA andB a r ea l lb l ue,t hen t heycanbes epa r a t ed wi t ho t he rve r t i ce sbyu s i ngal i ne.The r ema i n i ng2012b l uepo i n t scanbeg roupedi n1006pa i r s;each i ro fb l uepo i n t scanbes epa r a t edby2l i ne s.Soa l lt oge t he r pa 2013l i ne smee tt her equ i r emen t s. So l u t i on2.Wehaveamoregenera lr e s u l ta sfo l l ows. Suppo s et he r ea r enpo i n t si nr edorb l ueonap l ane,andno t hr eeo ft hem a r eco l l i nea r.Thent he r eex i s t[ n/2]l i ne st ha t mee tt her equ i r emen t s. Proo f.Weprovebyinduc t i on.Theconc l u s i oni sobv i ou si fn ≤ 2.Forn ≥ 3,con s i de ral i ne ℓ pa s s i ngpo i n t sA andB ,s uch Ma t hema t i c a lOl i adi nCh i na ymp 326 t ha tt her ema i n i ngpo i n t sa r eonones i deo fl i ne ℓ.Byt he n/2]- 1 hypo t he s e so fi nduc t i on,t her ema i n i ngpo i n t shave [ l i ne st ha tmee tt her equ i r emen t s. I fA andB havet hes ameco l ou rori nd i f f e r en tco l our sbu t Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. a r es epa r a t edby al i ne,wecans epa r a t eA andB wi t ho t he r n/2]l i ne s mee tt he i n t sby al i ne pa r a l l e lt o ℓ.Thu s,[ po r equ i r emen t s.I fA andB a r ei nd i f f e r en tco l our sbu tbo t hi na thave r eg i onR , t hent her ema i n i ngpo i n t si nt her eg i onR dono aco l ou r,s ay,b l ue,wecandr aw al i net os epa r a t et heb l ue i n to fA orB wi t ho t he rpo i n t s.Thu s,t he [ n/2]l i ne smee t po t her equ i r emen t sandcomp l e t et hei nduc t i on. Rema r k: Wecan a s k a gene r a lprob l em t ha ts ub s t i t u t e s 2013and2014byanypo s i t i vei n t ege r sm andn,r e s t i ve l pec y, m ≤n.Deno t edt hes o l u t i ont ot heg ene r a lp r ob l embyf( m ,n). Wecanob t a i nt ha tm ≤ f( m ,n)≤ m +1byt hei deaof seven,t henf( m ,n)= m .Ont heo t he r So l u t i on1.Andi fmi hand,fort heca s et ha tm i sodd,t hent he r ei sanN ,s ucht ha t foranym ≤n ≤ N ,wehavef( m ,n)= m ,andforanyn > N , m ,n)= m +1. wehavef( 3 Suppo s et ha tt heexc i r c l eo f △ABC oppo s i t et heve r t exA t ouche st he s i de BC a tt he i n tA1 .De f i net he po i n t s po B1 on CA and C1 on AB ana l ogou s l s i ng t he y by u exc i r c l e soppo s i t et oBandC , r e s c t i v e l s et ha tt he pe y.Suppo c i r c umc en t r eo f△A1B1C1 l i e s on t he c i r cumc i r c l e of F i .3 .1 g In t e rna t i ona lMa t hema t i c a lOl i ad ymp 327 △ABC .Pr ovet ha t△ABCi sr i ang l ed.( Theexc i r c l e ght s i t et heve r t exA i st hec i r c l et ha tt ouched o f△ABC oppo t hel i nes egmen tBCt ot her ayAB beyondBandt ot her ay AC beyondC .Theexc i r c l e soppo s i t eB andCa r es imi l a r l y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. de f i ned.)( s edby Ru s s i a) po So l u t i on.Denot et hec i r cumc i r c l e sof△ABCand△A1B1C1by ︵ he mi dpo i n to fa r cBC on Ω Ωand G,r e s c t i ve l tA0 bet pe y.Le con t a i n i ngpo i n tA .Po i n t sB0 andC0 a r ede f i nedana l ogou s l y. hec en t r eo fc i r c l e G,t henQi sonΩbyt hehypo t he s i s Le tQ bet oft hepr ob l em.F i r s t,weg i vet hefo l l owi ngl emma. Lemma.A0B1 = A0C1 .Po i n t sA ,A0 ,B1 and C1 a r e concyc l i c. I f po i n t s A0 and A co i nc i de,t hen △ABC i si s o s ce l e s t r i ang l e,t hu sAB1 = AC1 .Ot he rwi s e,A0B = A0C by t he ti sev i den tt ha t de f i n i t i onofA0 .I æ 1 ö b +c -a)÷ , BC1 = CB1 ç = ( è 2 ø and ∠C1BA0 = ∠ABA0 = ∠ACA0 = ∠B1CA0 . Thu s, △A0BC1 ≌ △A0CB1 . So,wehaveA0B1 = A0C1 . ① Al s o, by ① , we have ∠A0C1B = ∠A0B1C , hence ∠A0C1A = ∠A0B1A .Thu s,po i n t sA ,A0 ,B1 and C1 a r e concyc l i c. Obv i ou s l i n t sA1 ,B1 andC1 a r eonas emi a r co fΓ, y,po t hu s △A1B1C1 i san ob t u s e ang l edt r i ang l e. Wi t hou tl o s sof sob t u s eang l e,t hen r a l i t uppo s et ha t ∠A1B1C1i gene y,wemays i n t sQ andB1 a r eond i f f e r en ts i de sofA1C1 .Obv i ou s l o po y,s Ma t hema t i c a lOl i adi nCh i na ymp 328 a r epo i n t sB andB1 ,henc e,po i n t sQ andB a r eont hes ames i de o fA1C1 . n t e r s ec t swi t h No t et ha tt hepe rpend i cu l a rb i s ec t orofA1C1i Ga ttwo po i n t swh i cha r eond i f f e r en ts i de sofA1C1 .Byt he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. r eamongt hei n t e r s ec t i on po i n t s, abovea r t,B0 andQ a gumen andB0 andQ a r eont hes ames i deofA1C1 .SoB0 = Q ,a s s howni nt hed i ag r am. By t he l emma,l i ne s QA0 and QC0 a r e pe rpend i cu l a r e s c t i ve l r e b i s ec t or sofB1C1 andA1B1 ,r pe y,andA0 andC0 a mi dpo i n t so fa r c sCB andBA ,r e s c t i ve l r e for e, pe y.The ∠C1B0A1 = ∠C1B0B1 + ∠B1B0A1 = 2∠A0B0B1 +2∠B1B0C0 = 2∠A0B0C0 = 1 80 °- ∠ABC . Ont heo t he rhand,byt hel emmaoncemor e,wehave ∠C1B0A1 = ∠C1BA1 = ∠ABC . Henc e,∠ABC = 180 ° - ∠ABC ,s o ∠ABC = 90 °.Th i s comp l e t e st heproo f. S e c ondDay 9:00~13:30,Ju l y24,2013 4 Le t△ABC beanacu t e ang l edt r i ang l e wi t hor t hocen t e r H ,andl e tW bea po i n tont hes i deBC ,l i ngs t r i c t l y y be tweenB andC .Thepo i n t sM andN a r et hef ee to ft he a l t i t ude sf r om B andC ,r e s c t i ve l t ebyω1 t he pe y.Deno c i r cumc i r c l eo f△BWN ,andl e tX bet hepo i n tont heω1 s ucht ha tWX i l ogou s l t e st hed i ame t e rofω1 .Ana y,deno byω2t hec i r cumc i r c l eof△CWM ,andY bet hepo i n ton ω2s sad i ame t e rofω2 .Pr ucht ha tWYi ovet ha tX ,Y and In t e rna t i ona lMa t hema t i c a lOl i ad ymp 329 Ha r eco l l i nea r.( s edby po Tha i l and) So l u t i on. Le t AL be t he a l t i t udet os i deBC ,andZ bet he Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. i n t e r s e c t i on po i n to fc i r c l e sω1 how andω2 o t he rt hanW .Wes t ha tpo i n t sX ,Y ,Z and H a r e co l l i nea r. Po i n t sB ,C ,M and N a r e concyc l i c( deno t et hec i r c l e by ω3)s i nc e ∠BNC = ∠BMC = F i .4 .1 g n t e r s e c ta ta po i n ts i ncet hey a r et he 90 °.WZ ,BN andCM i r ad i ca lax i so fω1 andω2 ,ω1 andω3 ,ω2 andω3 ,r e s t i ve l pec y. Ands i nceBN andCM i n t e r s e c ta tA ,WZ pa s s e st hroughA . ∠WZX = ∠WZY =9 r et hed i ame t e r s 0 °s i nc eWX andWYa o fω1andω2 , r e s c t i ve l s,po i n t sX andY a r eont hel i ne pe y.Thu lpe rpend i cu l a rf r omZt oWZ . So,po i n t sB ,L ,H andN a r econcyc l i cs i nc e ∠BNH = ∠BLH = 9 0 °.Byt heCi r c l e -Powe rTheor em, AL ·AH = AB ·AN = AW ·AZ . ① I fpo i n tH i sont hel i ne AW ,t hen H and Z co i nc i de. Ot he rwi s e,by ① ,wehave AZ AL = . AH AW Thu s,△AHZ ∽ △AWL, c on s e en t l qu y,∠HZA = ∠WLA = 90 °,hence,po i n tH i sa l s oonl i nel. 5 Le tQ+ bet hes e to fa l lpo s i t i ver a t i ona lnumbe r s.Le tf: + unc t i on s a t i s f i ng t he f o l l owi ng t hr ee Q → R be a f y Ma t hema t i c a lOl i adi nCh i na ymp 330 cond i t i on s: ( x) xy); i)fora l lx,y ∈ Q+ ,wehavef( f( y)≥ f( ( i i)f ora l lx,y ∈ Q+ ,wehavef( x +y)≥f( x)+f( y); ( i i i)t he r ee x i s t sar a t i ona lnumb e ra >1s a)=a. u cht ha tf( Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Pr ovet ha tf( x) = x fora l lx ∈ Q+ . ( s ed by po Bu l r i a) ga So l u t i on.Le tx = 1,y = ai n( i),wehave 1)≥ 1. f( By ( i i)andi nduc t i ononn,wehave ① nx)≥ nf( x),"n ∈ Z+ ( t hes e tofa l lpo s i t i vei n t ege r s), f( "x ∈ Q+ . ② 1)≥ n."n ∈ Z+ . n)≥ nf( f( ③ æm ö n)≥ f( m ),"m ,n ∈ Z+ . f ç ÷f( èn ø ④ n ② ,wehave Tak i ngx = 1i By ( i)onc eaga i n,wehave Henc e,by ③ and ④ ,wehave + f( q)> 0,"q ∈ Q . ⑤ x)≥ f(x )≥ x > x -1,"x ∈ Q+ ,x > 1. f( ⑥ n x)≥ f( xn ),"n ∈ Z+ ,"x ∈ Q+ , f ( ⑦ n x)≥ f( xn )> xn -1,"x ∈ Q+ ,x > 1. f ( ⑧ Thenby ( i i)and ⑤ ,fi ss t r i c t l nc r ea s i ng,and yi By ( i)andi nduc t i ononn,wehave Thu s,by ⑥ and ⑦ ,wehave Thenby ⑧ ,wehave In t e rna t i ona lMa t hema t i c a lOl i ad ymp 331 x)> xn -1,"n ∈ Z+ ,"x ∈ Q+ ,x > 1. f( ⑨ x)≥ x "x ∈ Q+ ,x > 1. f( ⑩ an )= an . f( ( 11) n Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. Tak i ngt hel imi ta snt end st ot hei nf i n i t n ⑨ ,wehave yi By ( i i i),⑦ and ⑩ ,wehavean = fn ( a)≥ f( an )≥ an . Thu s, Fora l lx ∈ Q+ ,x > 1,we maychoo s en ∈ Z+ ,s ucht ha t an -x > 1.By ( 11),( i i)and ⑩ ,wehave an = f( an )≥ f( x)+f( an -x)≥ x + ( an -x)= an . The r e f or e, x)= xfora l lx ∈ Q+ ,x > 1. f( ( 12) La s t l l lx ∈ Q+ ,andfora l ln ∈ Z+ ,by ( 12),( i)and y,fora ② ,wehave nf( x)= f( n) x)≥ f( nx)≥ nf( x), f( "n ∈ Z+ , n > 1."x ∈ Q+ , ( 13) nx)= nf( x),"n ∈ Z+ ,n > 1,"x ∈ Q+ , f( ( 14) t ha ti s, Tak i ngx = m/ nfora l lm ≤ n ∈ Z+ , n > 1,wehave m) m æ m ö f( = . fç ÷ = èn ø n n Tha ti s,f( x)= xfora l lx ∈ Q+ ,x ≤ 1. Remark.Thecond i t i onf( a)=a > 1i se s s en t i a l.Inf ac t,f or b ≥ 1,f unc t i onf( x)=bx2s a t i s f i e s( i)and ( i i)f ora l lx,y ∈ 1 + saun i i xedpo i n t ≤ 1. Q ,andf ha quef b Ma t hema t i c a lOl i adi nCh i na ymp 332 6 Suppo s et ha ti n t ege rn ≥ 3,andcons i de rac i r c l ewi t hn + 1 equa l l ed po i n t s ma rked on i t. Cons i de ra l l ys pac l abe l l i ng so ft he s epo i n t swi t ht henumbe r s0,1,...,n s ucht ha teach numbe ri su s ed exac t l e;twos uch y onc Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. l abe l l i ng sa r econ s i de r edt o bet hes amei fonecan be ob t a i nedf r om t heo t he rby aro t a t i on o ft hec i r c l e.A eaut i li f,foranyfou rl abe l sa <b < l abe l l i ngi sca l l edb fu c < d wi t ha + d = b +c,t a,d)doe hechord ( sno t b,c). i n t e r s ec tt hechord ( Le tM bet henumbe ro fbeau t i f u ll abe l l i ng s,andl e tN be x,y)s ucht ha tx < y,x +y ≤n t henumbe ro fchord s( andgcd ( x,y)= 1.Provet ha tM = N + 1.( s edby po Ru s s i a) So l u t i on1.Not et ha tt hed i s t anc ebe tweenma rkedpo i n t sdoe s no tma t t e r.Thei n t e r s e c t i onoft hechord son l sont he ydepend orde roft hepo i n t s.Forac i r cu l a t i onofpe rmu t a t i onof[ 0,n] ={ 0,1,...,n},weca x,y)ak -chordi fx +y = l lachord ( k.I fx =y, t hent hechorddegene r a t e s. Weca l lt hr eed i s o i n tchord sa r ei norde r j i fachordpa r t so t he rtwochord s,e. g., i nF i .6 .1,chord sA ,B and C a r ei n g orde rbu tchord sB ,CandD no t.Weca l l m ≥3d i s o i n tchord sa r ei norde r,i fany j t hr eechord sa r ei norde r. F i .6 .1 g Aux i l i aryLemma.Inabeaut i f u ll abe l l i ng,a l lk -chord sa r ei n orde rforanyi n t ege rk. Proo f.Weprovei tbyi nduc t i ononn.Thel emmai st r i v i a lfor n ≤ 3.Forn ≥ 4,s uppo s et ha tt he r ewe r eabeau t i f u ll abe l l i ng S,s ha tt hr eek -chord sA ,B andC we r eno ti norde r.I fn ucht i sno tt heendpo i n tofchord sA ,B andC ,wecande l e t et he In t e rna t i ona lMa t hema t i c a lOl i ad ymp 333 i n tnandob t a i nabeau t i f u ll abe l l i ngS\{ n}of[ 0,n -1].By po r ei norde r. t hehypo t he s i so fi nduc t i on,chord sA ,B andC a S imi l a r l f0i sno tt heendpo i n tofchord sA ,B andC ,we y,I Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. cande l e t et he po i n t0,and mi nu seachl abe l l i ng by1,s o we ob t a i n a beau t i f u ll abe l l i ng S\{ 0}. By t he hypo t he s i s of r ei n orde r, wh i ch i sa i nduc t i on,chord s A , B and C a con t r ad i c t i on.Thu s,0andn mu s tappea ramongt heendpo i n t s n -chord so fA ,B andC .Suppo s et ha tchord s( 0,x)and( y,n) t hu s a r eamongA ,B andC .Thenn ≥ 0 +x =k =n +y ≥n, x =nandy =0.Tha ti s,( 0,n) i soneofn -chord so fA ,B and C .Wi t hou tl o s so fgene r a l i t uppo s et ha tC = ( 0,n). y,s u,v)bead ac en t Le tchordD = ( j andpa r a l l e lt ochordC ,s eeF i .6 .2, g ft = n,t hen anddeno t et = u +v.I n -chord sA ,B andD a r eno ti norde r i nt he beau t i f u ll abe l l i ng S\{ 0,n}, wh i ch con t r ad i c t st he hypo t he s i s of i nduc t i on.I ft <n, -chord ( 0, t) t hent henchordC i s doe sno ti n t e r s e c tD ,t F i .6 .2 g apa r tpo i n ttandchordD .Andn -chordE = ( t,n -t)doe sno t i n t e r s e c tchordC .So,po i n t s tandn -ta r eont hes ames i d eo fC. Bu tcho r d sA ,B andE a r eno ti no r d e r,wh i chi sac on t r ad i c t i on. La s t l i nc et he mapp i ng o fx t on - x con s e r ve st he y,s beau t i f u l ne s so fac i r cu l a rpe rmu t a t i on,at -chordmapst o( 2 n- t) -chord, t ha ti s, t > n equ i va l en t st ot < n.Thu s,wehave ovedt heaux i l i a r emma. pr yl , Int hefo l l owi ng wepr ovet hepr ob l embyi nduc t i ononn. t i f u l Theca s eo fn = 2i st r i v i a l.Le tn ≥ 3,andS beabeau rmu t a t i onof [ 0,n].T i sob t a i nedby de l e t i ngn i nS.Al l pe Ma t hema t i c a lOl i adi nCh i na ymp 334 n -chord si nT a r ei norde r,t he i rendpo i n t si nc l udenumbe r s[ 0, n -1].Weca so ff i r s tk i ndi f0i sl oca t edbe tweentwo l ls uchTi n -chord s,o t he rwi s e,weca l ls uchTi sofs econdk i nd.Wesha l l showt ha teachf i r s tk i ndofbeau t i f u lpe rmu t a t i onof[ 0,n -1] Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. cor r e s st oexac tonebeau t i f u lpe rmu t a t i onof[ 0,n].And pond eachs e condk i ndo fpe rmu t a t i onof [ 0,n - 1]cor r e s st o pond exac t l t i f u lpe rmu t a t i on sof[ 0,n]. ytwobeau I fTi so ft hef i r s tk i nd,s uppo s et ha t0l oca t e sont hea r c be tweenchord sA andB .S i nc echord sA ,( 0,n)andchord sB i nSa r ei norde r, n mu s tl oca t eont heo t he ra r cbe tweenchord s A andB .Thu s,wecanr e t r i eveS f rom T un i l he que y.Ont o t he rhand,foreachT o ft hef i r s tk i nd,wecanaddni nt he above manne rt o ob t a i n S. We s ha l ls how t ha t cyc l i c rmu t a t i onS mu s tbebeau t i f u l. pe For0 <k <n, t hek -chordofSi sa l s ot hek -chordofT , s o k -chord sa r ei norde r. I fT i so ft hes e condk i nd,t hent he po s i t i onofn t ot he cor r e s i ngSha stwopo s s i b i l i t i e s,t ha ti s, nad ac i en tt o0on pond j e i t he rs i de.S imi l a r l ha tS i s a beau t i f u l y, we can check t rmu t a t i ono f[ 0,n]. pe Deno t et het o t a lnumbe ro fbeau t i f u lpe rmu t a t i ono f[ 0,n] by M n ,t he t o t a l numbe r of t he s econd k i nd of beau t i f u l rmu t a t i ono f[ 0,n -1]byLn .Thenwehave pe Mn = ( M n-1 -Ln-1)+2Ln-1 = M n-1 +Ln-1 . st he numbe r of po s i t i ve I ts u f f i ce st os how t ha tLn-1 i x,y)wi x, i n t ege rpa i r s( t ht hecon s t r a i n t sx +y =nandgcd( y)= 1. f i ne Forn ≥ 3,de In t e rna t i ona lMa t hema t i c a lOl i ad ymp 335 n)= # { x:1 ≤ x ≤ n,gcd( x,n)= 1}. φ( Wes ha l lshowt ha tLn-1 =φ( n).Cons i de ras econdk i ndof ] , , [ , rknumbe r s0 ...,n beau t i f u lpe rmu t a t i ono f 0 n -1 ma Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. 1c l ockwi s eont hec i r c l es ucht ha tnumbe r0co i nc i de swi t ht he s i t i on0.Deno t et henumbe ro fpo s i t i onia sf( i).Suppo s e po t ha tf( a)= n -1. S i nceeachnumbe ro f[ 1,...,n -1]i sa tanendpo i n tof n -chord,anda l ln -chord sa r ei norde r,byt hede f i n i t i ono ft he s e condk i nd,( 1,n - 1)i sann -chord.Thu s,a l ln -chord sa r e i)+f( n -i)= nfori = 1,...,n -1. r a l l e l.Tha ti s,f( pa S imi l a r l i nc ea l l( n -1) -cho r d sa r ei no r de rande a chpo i n ti s y,s anendpo i n to fan ( n -1) -cho r d,a l l( n -1) -cho r d sa r epa r a l l e l. Thu s,f( i)+f( mod( a -i,n))=n -1f o ri = 1,...,n -1. mod( equen t l a -i,n)+1.Cons The r e f or e,f(-i)= f( y, byt ak i ngi =a,2 0)=0, a,...,( n -1) as uc ce s s i ve l y,andf( wehave k,n),k = 0,1,2,...,( n -1). f(-ak)= mod( ① S i nc ef i sape rmu t a t i on,we mu s thavegcd ( a,n) = 1. Tha ti s,Ln-1 ≤ φ( n).Toshowt ha tt heequa l i t l d sfor ① , yho weneedon l os how t ha tt he pe rmu t a t i onf g i ven by ① i s yt beau t i f u l.Tos eet h i s,wecon s i de rf ournumbe r sw ,x,yandz ont hec i r c l es a t i s f i ngw + y = x + z.The i rcor r e s i ng y pond s i t i on sont hec i r c l es a t i s f po y (-aw )+ (-ay)≡ (-ax)+ (az)mod ( n),t ha ti s,t hechord ( w ,y)andchord ( x,z)a r e r a l l e l.Thu s,t hepe rmu t a t i on ① i sbeau t i f u l,andi soft he pa s e condk i ndbycon s t r uc t i on. Ma t hema t i c a lOl i adi nCh i na ymp 336 2014 ( C a eT own,T h e p R e u b l i co fS o u t h p ) A f r i c a Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. F i r s tDay 9:00~13:30,Ju l y8,2014 1 Le ta0 < a1 < a2 < … beani nf i n i t es equenceofpo s i t i ve i n t ege r s.Provet ha tt he r eex i s t saun i s i t i vei n t ege r quepo ns ucht ha t an < a0 +a1 + … +an ≤ an+1 . n ( s t edby Au s t r i a) po So l u t i on.De f i nedn No t et ha t =( a0 +a1 + … +an )-nan ,n =1,2,.... nan+1 - ( a0 +a1 + … +an ) =( n +1) an+1 - ( a0 +a1 + … +an +an+1) = -dn+1 . So,t hepr ob l emi sequ i va l en tt o pr ovet ha tt he r eex i s t sa un i s i t i vei n t ege rns ucht ha tdn > 0 ≥ dn+1 . quepo Wes eet ha td1 = ( a0 +a1)-1·a1 = a0 > 0,and dn+1 -dn = (( a0 +a1 + … +an )-nan+1)(( a0 +a1 + … +an )-nan ) = n( an -an+1)< 0, t ha ti s,{ dn }i sas t r i c t l r ea s i ngs equenceofi n t ege r swi t h y dec r e ex i s t s a un i t hef i r s tt e rm be i ng po s i t i ve. Henc e,t he que s i t i vei n t ege rn, s ucht ha tdn > 0 ≥ dn+1 . po In t e rna t i ona lMa t hema t i c a lOl i ad ymp 337 Remark.I ti se s s en t i a l l n t e rmed i a t et heor emi nd i s c r e t e yani f orm. 2 s s boa rd Le tn ≥ 2beani n t ege r.Con s i de rann × n che Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. cons i s t i ngo fn2 un i ts r e s.Aconf i a t i onofn r ooks qua gur ont h i sboa rdi spe a c e li feve r ow andeve r l umn fu yr yco con t a i n s exac t l ook.Fi nd t he g r ea t e s t po s i t i ve y one r ucht ha tf oreach peac e f u lconf i a t i ono fn i n t ege rk s gur r ook s,t he r ei sanemp t r ewi t hou tanyrookson yk ×ks qua anyofi t sk2 un i ts r e s.( s edbyCroa t i a) qua po So l u t i on.Wesha l ls howt ha tt heanswe ri skmax = [ n -1]by twos t ep s.Le tlbeapo s i t i vei n t ege r. ( 1)I fn >l2 , t hent he r eex i s t sanemp t r ef orany yl×ls qua e f u lconf i r a t i on. peac gu 2 ( 2)I fn ≤ l ,t hent he r eex i s t sapeace f u lconf i a t i on, gur s ucht ha teachl ×ls r ei sno temp t qua y. Proo fo f( 1).Therei sarowR wi t hf i r s tco l umnha sarook. Takes uc c e s s i ve l owscon t a i n i ngrow R ,wh i cha r edeno t ed ylr byU .I fn >l2 , t henl2 +1 ≤n,andf romco l umn2t oco l umn l2 +1i nU ,con t a i n i ngll ×ls r e swh i chhavea tmo s tl -1 qua r ook s.So,a tl ea s tonel ×ls r ei semp t qua y. Proo fo f (2). For n 2 = l , we s ha l lf i nd a peac e f u l conf i r a t i onwh i chha snoemp t r e.Wel abe lt he gu yl ×ls qua rowsf r ombo t t omt ot opandt heco l umnsf roml e f tt or i t h ghtbo by0,1,..., heun i ts r ea tt he l2 -1.Sodeno t eby ( r,c)t qua r owrandco l umnc. Wepu tar ooka t( i l +j,j l +i)fori,j = 0,1,2,..., l -1.Thef i r ebe l owshowst heca s eforl = 3.S i nceeach gu numbe rbe tween0t ol2 -1canbewr i t t enun i l nt heform que yi Ma t hema t i c a lOl i adi nCh i na ymp 338 o fi l +j,( 0 ≤i,j ≤l -1),s ucha conf i r a t i oni speac e f u l. gu Foranyl ×ls r eA ,s uppo s e qua Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. t ha tt hel owe s tr owo fA i sr owpl + s i nc epl +q ≤ q,0 ≤ p,q ≤l -1 ( 2 l -l).The r ei sonerooki nA byt he conf i r a t i on,andt heco l umnl abe l s gu l+ oft herookmayb eq l +p,( q +1) l -1) l +p,p +1, l+ ( p,...,( p+ F i .2 .1 g ( 1),..., l +p +1.Rea r r anget he s enumbe r si ni nc r ea s i ng q -1) orde rt o l +p +1,q l +p, p +1,l + ( p +1),...,( q -1) ( l +p,...,( l -1) l +p. q +1) Thent hef i r s tnumbe ri sl e s st hanorequa lt ol,t hel a s ti s r ea t e rt hanorequa lt o( l -1) landt hed i f f e r encebe tweentwo g r e for e,t he r eex i s t sonerooki nA . ad acen tnumbe r si sl.The j Fort heca s eofn <l2 ,cons i de rt heconf i a t i onabove, gur bu tde l e t el2 -nco l umnsandr owsf romt her i romt he ghtandf bo t t om,r e s c t i ve l tanl ×ls r e.Thu s,weob t a i n pe y,wege qua ann ×ns r e,whe r et he r ei snoemp t r e. Bu ts ome qua yl ×ls qua rowsandco l umn smaybeemp t trooksa tt hec ro s s y.Wecanpu s r e so f emp t l umns t o ob t a i nt he de s i r ed qua y rows and co e f u lconf i r a t i on. peac gu Remark.Theanswercou l da l s obei nt heformof n 3 -1. Aconvexquadr i l a t e r a lABCD ha °. s ∠ABC = ∠CDA =90 st hef oo to ft hepe rpend i cu l a rf rom A t oBD . Po i n tH i Po i n t sSandTl i eons i de sAB andAD ,r e s c t i ve l uch pe y,s t ha tH l i e si ns i de △SCT and ∠CHS - ∠CSB = 90 °, In t e rna t i ona lMa t hema t i c a lOl i ad ymp 339 ∠THC - ∠DTC = 9 0 °. st angen tt ot he c i r cumc i r c l eo f Pr ovet ha tl i ne BD i s edbyI r an) △TSH .( po So l u t i on.Suppos et ha tt hel i nepa s s i ngC andpe rpend i cu l a rt o Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. n t e r s e c t sl i neABa tpo i n tQ , s eet hef i ebe l ow.Then l i neSCi gur 256 ∠SQC = 9 0 °- ∠BSC = 180 °- ∠SHC . r ec on c c l i c wi t h SQ i t s Hen c e,po i n t sC, H ,S and Q a y Ma t hema t i c a lOl i adi nC ymp ,hina f△SHCi sonl i neAB. I n d i ame t e r. The r e f o r et hec i r c umc en t r eK o f△CHTi sonl i neAD . t hes amemanne r,t hec i r c umc en t r eL o , Byt heCo s i neLaw DF2 = BD2 +BF2 -2BD ·BFcosB =2 y 1 2 = 2 2 2 ( -( x +y) x +z) y +z) + ( x +y)( 2( y +z) 4xy2z . ( x +y)( y +z) Thu s 4xyz 4xyz · i .3 .1 g ) KF × HD CF ( x +y) AD ( y +zF = FH × DK DF DF x· z AD CF Tos ho wt ha tl i ne BD i st angen tt ot he c i r cumc i r c l eo f 16 x z y t s u f f i c e st os how t ha tt he i n t e r s ec t i on po i n t of △TSH ,i = = 4. 2 ( ( +y) +z) x D F y sonl i neAH .Bu tt he rpend i cu l a rb i s e c t or sofHS and HT i pe 2 ,i e rp e n d i cu l a rb i s e c t o r sa r e u s tt h eang l eb s e c t or so f ∠AKH DKHF App l i ngPt o l emy sT h e o r e mt oc c l i c uad r i l a t e r a l p j y q and ∠ALH ,r e s c t i ve l he In t e rna l Ang l e Bi s ec t or wehave pe y. By t Theor em,i ts u f f i c e st oshowt ha t KF ·HD = DF ·HK +FH ·DK . AK × AL KF × HD FD = HK. = 4,weob t a i n Comb i n i ngwi t h KH LH =3. FH × DK FH × DK h ef o l l owi ng,wewi l lg i vetwopr oof so f①. So l u t i on 2. Fi r s t we I r ov e a pnt l emma. ① Proo f1.Le tM bet hei n t e r s ec t i onpo i n to fl i ne sKL andHC , Lemma.Asshowni n F i .3 . s e et h ef i u r ebe l ow. g g2. t I ft hec i r c l et ouche sAB andACa B and C ,r sa e s c t i ve l pe y,Q i n t e r s e c t s i n tont hec i r c l e.AQ i po hen t he c i r c l e a t po i n t P, t F i .3 .2 g wehave PQ ·BC = 2BP ·QC = 2BQ ·PC . Proo fo ft hel emma.ByPto l emy sTheor em,wes eet ha t PQ ·BC = BP ·QC +BQ ·PC . Ma t hema t i c a lOl i adi nCh i na ymp 340 S i nc eKH = KCandLH = LC , po r e i n t s H and C a s t r i cove rl i neKL .Thu s, ymme Mi st he mi d i n to fHC .Le t po Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. O be t he c i r cumc en t r eo ft he c e,O i s ad r i l a t e r a lABCD .Hen qu t he mi d -po i n t o f AC and c on s equ en t l t he r e f o r e yOM ‖AH , OM ⊥ BD .Fur t he rby OB = F i .3 .2 g OD , we s ee t ha t OM i st he rpend i cu l a rb i s e c t oro fBD , t hu sBM = DM . pe r ec onc c l i cwi t hKC, S i n c eCM ⊥ KL, i n t sB,C,M andK a y po r ec onc c l i cwi t h t hed i ame t e r.S imi l a r l i n t sL,C,M andD a y y,po LC, t hed i ame t e r.So,byt heS i neTheo r em,wehav e AK s i n∠ALK DM ·CK CK KH , = = = = AL s i n∠AKL CL BM CL LH t ha ti s,①. P r oo f2.I fpo i n t sA ,H andCa r ec o l l i ne a r,t henAK = AL,KH = LH , t hu s① f o l l ows.El s e,l e tω bet hec i r c l epa s s i ngA ,H andC .S i nepo i n t sA ,B ,C andD a r eonac i r c l e ∠BAC = ∠BDC = 9 0 °- ∠ADH = ∠AHD . Le tN ≠ A bet heo t he ri n t e r s ec t i on po i n toft hec i r c l eω andt heb i s e c t oro f ∠CAH ,t hen AN i sa l s ot he b i s ec t orof ∠BAD . S i nc epo i n t sH andC a r es t r i cove rl i neKL ,and ymme eet ha tpo i n tN andt HN = NC ,wes hecen t r eofωa r ebo t hon ti s,t hec i r c l eωi st heApo l l on i u sc i r c l eofpo i n t s l i neKL .Tha K andL , t hu s① f o l l ows. Remark.Prob l em3ha sanex t en s i onbyt heprob l ems e l e c t i on In t e rna t i ona lMa t hema t i c a lOl i ad ymp 341 commi t t eea sfo l l ows. Foraconvexquadr i l a t e r a lABCD ,po i n t sS andT a r eon sl oca t edi nt he s i de sAB andAD ,r e s c t i ve l i n tH i pe y.And po i nne rpa r to f△SCT ,s a t i s f i ng ∠BAC = ∠DAH , ∠CHS y Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. ∠CSB = 9 0 °and ∠THC - ∠DTC = 90 °. Then t he c i r cumcen t r eo f △TSH i s on AH and t he c i r cumcen t r eof△SCTi sonAC . S e c ondDay 9:00~13:30,Ju l y9,2014 4 Po i n t sP andQl i eons i d eBC o fa c u t e ang l ed △ABC s ot ha t ∠PAB = ∠BCAa nd ∠CAQ = ∠ABC.Po i n t sM andNl i eon l i n e sAP andAQ,r e s e c t i v e l u cht ha tP i st h e mi dpo i n to f p y,s andQi AM , st hemi dpo i n to fAN .P r ov et ha tl i n e sBM andCN i n t e r s e c tont hec i r c umc i r c l eo f△ABC.( s e dbyGe o r i a) po g So l u t i on.Le tS bet hei n t e r s e c t i onpo i n to fl i ne sBM andCN , s eet hef i ebe l ow.Deno t eβ = ∠QAC = ∠CBA ,γ = ∠PAB gur = ∠ACB , t henwecans eet ha t△ABP ∽ △CAQ , t hu s BP BP AQ NQ = = = . PM PA QC QC By ∠BPM = β + γ = ∠CQN ,△BPM ∽ △NQC ,t hu s ∠BMP = ∠NCQ .Con s equen t l hu s y, △BPM ∽ △BSC ,t ∠CSB = ∠BPM =β +γ = 1 80 °- ∠BAC . 5 Foreachpo s i t i vei n t ege rn, t hebankofCapeTowni s s ue s 1 co i nsofdenomi na t i on .Gi venaf i n i t eco l l ec t i ono fs u ch n c o i n s( o fno tne c e s s a r i f f e r en tdenomi na t i on s)wi t ht o t a l yd 342 Ma t hema t i c a lOl i adi nCh i na ymp v a l u ea tmo s t99 + 1, ove pr 2 t ha ti ti spo s s i b l et os l i tt h i s p co l l e c t i oni n t o100 orf ewe r Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. r oup s,s ucht ha teachg r oup g ha st o t a l va l ue a t mo s t 1. ( s edbyLuxembou r po g) So l u t i on. We sha l l pr ove a r a l conc l u s i on: f or any gene s i t i vei n t ege r N ,g i ven af i n i t e po F i .5 .1 g co l l ec t i ono fs uchco i n swi t hat o t a lva l uea tmo s tN - 1, prove 2 t ha ti ti spo s s i b l et os l i tt h i sco l l e c t i oni n t oN orf ewe rg r oups, p s ucht ha teachg roupha sat o t a lva l ueo fa tmo s t1. k ( ki s a po s i t i ve I fs ome co i ns have t o t a lva l ue of1/ i n t ege r),wer ep l ac et he s eco i nsbyoneco i nofva l ue1/ k,wh i ch doe sno ta f f e c tt hepr ob l em.Int h i sway,foreacheveni n t ege r k,a tmo s toneco i nha sva l ueo f1/ k( o t he rwi s e,twos uchco i ns mayber ep l ac edbyoneco i no fva l ue2/ k); foreachoddi n t ege r k,a tmo s t( k -1)co i n so fva l ue1/ k,( o t he rwi s e,ks uchco i ns canber ep l ac edbyoneco i no fva l ue1).So,we mays uppo s e t ha tt he r ei snomor er ep l ac emen tcanbemakefort heco i ns. F i r s twet akeeachco i no fva l ue1a sag roup.Suppo s et he r e a r ed < N s uchg r oup s.I ft he r ea r enoo t he rco i n s,t hent he ob l emi ss o l ved.Ot he rwi s e,t akeco i nofva l ue1/2a sag roup pr ( ft he r ei sany.Le tm = N - d ≥ 1.Thenforeach d +1)i l ue i n t ege rki n2,...,m , t akeco i nso fva l ue1/( 2 k -1)andva 1/( 2 k) i ng r oup ( d + k)i ft he r ea r eany,i n wh i cht het o t a l /( va l uedoe sno texc eed( 2 k -2) 2 k -1)+1/( 2 k)<1.Forco i ns ofva l uel e s st han1/( 2m ), i ft he r ea r eany,wecanpu tt hemi n In t e rna t i ona lMa t hema t i c a lOl i ad ymp 343 s omeg r oup ( d +j)s ucht ha tt het o t a lva l uei sl e s st han1( s i nce d +j)ha i feachg r oup ( sva l ueg r ea t e rt han1 -1/( 2m ),t hen 1 -1/( 2m ))= m + t het o t a lva l uewi l lbeg r ea t e rt hand + m ( d -1/2 =N - 1/2).Repea tt h i s procedu r ef i n i t et ime s,a l l Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. co i nsa r epu ti nN orf ewe rg r oup swi t heachg roupo fva l uea t mo s tone. 6 As e to fl i ne si nt hep l anei si ng ene r a lpo s i t i oni fnotwo o ft hema r epa r a l l e landnot hr eeoft hempa s st hrought he s amepo i n t.A s e to fl i ne si n gene r a lpo s i t i on cu t st he l anei n t or eg i ons,s omeo f wh i ch havef i n i t ea r ea;we p ca l lt hem f i n i t er eg i on s.Provet ha tf ora l ls u f f i c i en t l y l a r n any s e to fn l i ne si n gene r a lpo s i t i on,i ti s gen,i s s i b l et oco l ou ra tl ea s t no ft hel i ne si nb l ues ucht ha t po noneo fi t sf i n i t er eg i on sha sacomp l e t e l l uebounda r yb y. ( s edby Au s t r i a) po So l u t i on.Le hemax ima ls e to fl i ne ss ucht ha tnof i n i t e tB bet r eg i onha scomp l e t e l l uebounda r t|B |=k,andco l our yb y.Le , t heo t he rn -kl i ne si nr ed.Foreachr edl i ne ℓ t he r ei sa s l ea s tonef i n i t er eg i onA who s eun i eds i del i e son ℓ.We quer s ayapo i n ti sr edi fi ti st hei n t e r s e c t i onpo i n to fr edandb l ue l i ne s.Andwes ayapo i n ti sb l uei fi ti st hei n t e r s ec t i onpo i n tof twob l uel i ne s.Deno t et heve r t i c e so fA c l ockwi s eby ( p1 ,p2 , ...,ps )wi t hpo i n tp1i nr edandp2i nb l ue.Then wes ayt he r edl i ne ℓ cor r e s st ot her edpo i n tp1 andb l uepo i n tp2 . pond Now wea r et oshowt ha tanyb l uepo i n tcanbecor r e s o pondedt æk ö a tmo s ttwor edl i ne s.( I ft h i si st r ue,t henn -k ≤2ç ÷ ⇔n ≤ è2ø k2 .) Wepr ovei tbycon t r ad i c t i on.I fno t,s uppo s et ha tt he r e 344 Ma t hema t i c a lOl i adi nCh i na ymp a r et hr eer edl i ne sℓ1 ,ℓ2and ℓ3cor r e s oab l uepo i n tb, pondt andt hecor r e s i ngr edpo i n t sonr edl i ne s ℓ1 ,ℓ2 and ℓ3 pond a r er1 , r2andr3 , hei n t e r s ec t i onpo i n tof r e s c t i ve l tbbet pe y.Le twob l uel i ne s.Wi t hou tl o s so fgene r a l i t uppo s et ha t y,wemays Mathematical Olympiad in China (2011–2014) Downloaded from www.worldscientific.com by NATIONAL UNIVERSITY OF SINGAPORE on 05/06/18. For personal use only. s i de s( r eononeoft hetwob l uel i ne s,and r2 ,b)and ( r3 ,b)a ( r1 ,b) i st hes i deo fr eg i onA ont heo t he rb l uel i ne.S i nceA ha son l eds i de,A mu s tbeat r i ang l e △r1b r2 .Bu tt hen yoner s sr2 , r edl i ne sℓ1andℓ2pa andab l uel i nea l s opa s s e sr2 ,wh i ch i sacon t r ad i c t i on.