IMO2018 Shortlisted Problems with Solutions

1 ShortlistedProblems(with Solutions )59thInternational Mathematical OlympiadCluj-Napoca Romania, 3 14 July 2018 NoteofCon dentialityTheShortlisthastobekeptstri tly on dentialuntilthe on lusionofthefollowingInternationalMathema ti tionCommitteeofIMO2018thankthefollowing4 9 ountriesfor ontributing168problemproposals:Armenia,A ustralia,Austria,Azerbaijan,Belarus,Belg ium,BosniaandHerzegovina,Brazil,Bulgaria ,Canada,China,Croatia,Cyprus,Cze hRepubli ,Denmark,Estonia,Germany,Gree e,HongKong,I eland,India,Indonesia,Iran,Ireland,Israe l,Japan,Kosovo,Luxembourg,Mexi o,Moldova,Mongolia,Netherlands,Ni aragua,Poland,Russia,Serbia,Singapore,Sl ovakia,Slovenia,SouthAfri a,SouthKorea,Switzerland,Taiwan,Tanzania ,Thailand,Turkey,Ukraine,UnitedKingdom, tionCommitteeCalinPopes u,RaduGologan,MarianAndrona he,MihailBaluna,Ni olaeBeli,IlyaBogdanov,PavelKozhevnikov,G zaK s, tionsf:Q 0 Q 0satisfyingf`x2fpyq2 fpxq2fpyqforallx, yPQ 0.

2 (Switzerland) 3forwhi hthereexistrealnumbersa1, a2, .. , an,an`1 a1,an`2 a2su hthataiai`1`1 ai`2foralli 1,2, .. , n.(Slovakia) ,showthatatleastoneofthefollowingtwoasse rtionsholds:(1)Thereexistdistin t nitesubsetsFandGofSsu hthat xPF1{x xPG1{x;(2)Thereexistsapositiverationalnu mberr 1su hthat xPF1{x rforall nitesubsetsFofS.(Luxembourg) , a1, a2, ..beasequen eofrealnumberssu hthata0 0,a1 1,andforeveryn 2thereexists1 k nsatisfyingan an 1` `an a2017.(Belgium) tionsf:p0,8q Rsatisfying x`1x fpyq fpxyq `f yx forallx, y 0.(SouthKorea) , n , .. , xnqbeapolynomialwithreal oe ientssu hthatfpx1, .. , xnq Yx1`..`xnm]foreveryx1.}}}

3 , xnP 0,1, .. , m 1(.Provethatthetotaldegreeoffisatleastn. (Brazil) 3cab`7`3cbc`7`3ccd`7`3cda`7,wherea, b, c, darenonnegativerealnumberswhi hsatisfya`b`c`d 100.(Taiwan)4 Cluj-Napo a Romania,3 14 July2018 Combinatori :Foreverym 2,3, .. , nthesetS anbepartitionedintotwosubsetswithequalsu msofelements,withoneofsubsetsof ardinalitym.(I eland) 20 ,Horstpla esabla kknightonanemptysquareinsu hawaythathisnewknightdoesnotatta nishedwhensomebody hthat,regardlessofthestrategyofQueenie,H orst anputatleastKknightsontheboard.(Armenia) eofturnsonaboard onsistingofn`1squaresinarow, ,nstonesareputintosquare0, ,Sisyphus hoosesanynonemptysquare,saywithkstones,t akesoneofthosestonesandmovesittotheright byatmostksquares(thestoneshouldstaywithi ntheboard).)

4 Sisyphus' annotrea htheaiminlessthanQn1U`Qn2U`Qn3U` `QnnUturns.(Asusual,rxsstandsfortheleast integernotsmallerthanx.)(Netherlands) alpyramidisa nitesetofnumbers,pla edinatriangle-shapedarraysothatthe rstrowofthearray ontainsonenumber,these ondrow ontainstwonumbers,thethirdrow ontainsthreenumbersandsoon;and,ex eptforthenumbersinthebottomrow,ea hnumberequalstheabsolutevalueofthedi eren e,thetrianglebelowisananti-Pas alpyramidwithfourrows,inwhi heveryintegerfrom1to1`2`3`4 10o ursexa tlyon alpyramidwith2018rows,usingeveryintegerf rom1to1`2` `2018exa tlyon e?(Iran) ommitteeofatennistournamentistos hedulethemat hesfor2kplayerssothateverytwoplayersplay on e,ea hdayexa tlyonemat hisplayed,andea hplayerarrivestothetournamentsitethedayo fhis rstmat h,anddepartsthedayofhislastmat ,the ommitteehastopay1 hedulesoastominimisethetotal ostofallplayers' ost.

5 (Russia) nitepro esstakespla ,we hoosesu hapairandin reaseoneofits hpairexists, ,nomatterhowwemakethe hoi esinpiq,operationpiiqwillbeperformedonly nitelymanytimes.(Serbia) rossing ir lesnothreeofwhi hare on ir lessubdividetheplaneintoregionsboundedby ir ularedgesthatmeetatverti ethatthereareanevennumberofverti esonea h ir ir le,alternately olourtheverti esonthat ir h ir le,everyvertexis olouredtwi e on eforea hofthetwo ir lesthat olouringsagreeatavertex,thenitisassigned that olour;otherwise,itbe ,ifsome ir le ontainsatleast2061yellowpoints,thentheve rti esofsomeregionareallyellow.(India)6 Cluj-Napo a Romania,3 ute-angledtrianglewith ir um ir le.

6 LetDandEbepointsonthesegmentsABandAC,res pe tively,su hthatAD ularbise torsofthesegmentsBDandCEinterse tthesmallar s ABand ACatpointsFandGrespe G.(Gree e) AC, hthatP B P CandP BandP C,respe tively,sothatBliesonthesegmentP X,CliesonthesegmentP Y,and=P XM =P Y XYis y li .(Australia) ir le olle tionToftrianglesis alledgood,ifthefollowing onditionshold:piqea htrianglefromTisins ribedin ;piiqnotwotrianglesfromThavea hthat,forea hpositiveintegern,thereexistsagood olle tionofntriangles,ea hofperimetergreaterthant.(SouthAfri a) ,B1,andC1bethere e tionsofTinBC,CA,andAB,respe bethe ir um ir ,B1T,andC1 Tmeet againatA2,B2,andC2,respe ,BB2,andCC2are on urrenton.

7 (Mongolia) ir um ir le andin interse tsthelinesAI,BI,andCIatpointsD,E,andF,re spe tively,distin tfromthepointsA,B,C, ularbise torsx,y,andzofthesegmentsAD,BE,andCF,res pe tivelydetermineatriangle .Showthatthe ir um ir leofthetriangle istangentto .(Denmark) onvexquadrilateralABCD satis esAB CD BC hoseninsidethequadrilateralsothat=XAB =XCDand=XBC = `=CXD 180 .(Poland) ir um entre,and bethe ir um ir leofana ,distin tfromA,B,C,andtheirantipodesin .Denotethe ir um entresofthetrianglesAOP,BOP,andCOPbyOA,O B,andOC,respe A, B,and Cperpendi ulartoBC,CA,andABpassthroughOA,OB,andOC, respe ir um ir leofthetriangleformedby A, B,and CistangenttothelineOP.

8 (Russia) , kqofdistin tpositiveintegerssu hthatthereexistsapositiveintegersforwhi hthenumbersofdivisorsofsnandofskareequal .(Ukraine) h ellofann ntable onditionsaresatis ed:piqEa hnumberinthetableis ongruentto1modulon;piiqThesumofnumbersin anyrow,aswellasthesumofnumbersinany olumn,is tofthenumbersintheithrow,andCjbetheprodu tofthenumbersinthejth ` `RnandC1` `Cnare ongruentmodulon4.(Indonesia) nethesequen ea0, a1, a2, ..byan 2n`2tn{ nitelymanytermsofthesequen ewhi h anbeexpressedasasumof(twoormore)distin ttermsofthesequen e,aswellasin nitelymanyofthosewhi h annotbeexpressedinsu haway.(Serbia) ,a2,..,an,..beasequen eofpositiveintegerssu hthata1a2`a2a3` `an 1an`ana1isanintegerforalln k, hthatan an`1foralln m.}

9 (Mongolia) ,y,z,andtsatisfytherelationsxy zt x`y z` tsquares?(Russia) :t1,2,3, ..u t2,3, ..ubeafun tionsu hthatfpm`nq |fpmq `fpnqforallpairsm, 1whi hdividesallvaluesoff.(Mexi o) 2018beaninteger,andleta1, a2, .. , an, b1, b2, .. , bnbepairwisedistin tpositiveintegersnotex ea1b1,a2b2, .. ,anbnformsanarithmeti eareequal.(Thailand)8 Cluj-Napo a Romania,3 14 July2018 Shortlistedproblems tionsf:Q 0 Q 0satisfyingf`x2fpyq2 fpxq2fpyqp qforallx, yPQ 0.(Switzerland)Answer:fpxq 1forallxPQ , bPQ fpaq,y bandx fpbq,y aintop qwegetf`fpaq 2fpbq f`fpaq2fpbq2 f`fpbq 2fpaq,whi hyieldsf`fpaq 2fpaq f`fpbq 2fpbqforalla, bPQ ,thisshowsthatthereexistsa onstantCPQ 0su hthatf`fpaq 2 Cfpaq,or f`fpaq C 2 fpaqCforallaPQ 0.

10 (1)Denotebyfnpxq fpfp..pfloooomoooonnpxqq.. (1)yieldsfpaqC f2paqC 2 f3paqC 4 fn`1paqC ,fpaq{ {C 1,sin eotherwisetheexponentofsomeprimeinthepri mede ompositionoffpaq{Cisnotdivisiblebysu ,fpaq CforallaPQ ,aftersubstitutingf Cintop qwegetC C3,when eC 1istheuniquefun tionsatisfyingp e,onemaystartwith ndingfp1q ,letd y 1andx d2,y 1intop qwegetfpd2q d3andfpd6q fpd2q2 d 1,y d2weobtainfpd6q d2 d3 ,d7 fpd6q d5,when ed ,therestofthesolutionsimpli esabit,sin ewealreadyknowthatC fpfp1qq2fp1q eequationp1qbe omesmerelyfpfpaqq2 fpaq,whi hyieldsfpaq onstantfun tionsf:R` R`satisfyingp qforallrealx, y 0 ,fpxq ?}}}