IMO2020 Shortlisted Problems with Solutions

1 ShortlistedProblems(with Solutions )61stInternational Mathematical OlympiadSaint-Petersburg Russia, 18th 28th September 2020 Note of ConfidentialityThe Shortlist has to be kept strictly confidentialuntil the conclusion of the followingInternational Mathematical General Regulations CountriesThe Organising Committee and the Problem Selection Committee of IMO 2020 thank thefollowing 39 countries for contributing 149 problem proposals:Armenia, Australia, Austria, Belgium, Brazil, Canada, Croatia, Cuba,Cyprus, Czech Republic, Denmark, Estonia, France, Georgia, Germany,Hong Kong, Hungary, India, Iran, Ireland, Israel, Kosovo, Latvia,Luxembourg, Mongolia, Netherlands, North Macedonia, Philippines, Poland,Slovakia, Slovenia, South Africa, Taiwan, Thailand, Trinidad and Tobago,Ukraine, United Kingdom, USA, VenezuelaProblem Selection CommitteeIlya I.

2 Bogdanov, Sergey Berlov, Alexander Gaifullin, Alexander S. Golovanov, G za K s,Pavel Kozhevnikov, Dmitry Krachun, Ivan Mitrofanov, FedorPetrov, Paul Vaderlind4 Saint-Petersburg Russia, 18th 28th September a positive integer, and setN 2n. Determine the smallest realnumberansuch that, for all realx,Ncx2N`12 anpx 1q2` every positive integerN, determine the smallest real numberbNsuch that,for all realx,Ncx2N`12 bNpx 1q2`x.(Ireland) the set of all polynomials in three variablesx,y,zwith integercoefficients. LetBdenote the subset ofAformed by all polynomials which can be expressed aspx`y`zqPpx,y,zq ` pxy`yz`zxqQpx,y,zq `xyzRpx,y,zqwithP,Q,RPA. Find the smallest non-negative integernsuch thatxiyjzkPBfor all non-negative integersi,j,ksatisfyingi`j`k n.(Venezuela) thata,b,c,dare positive real numbers satisfyingpa`cqpb`dq ac` the smallest possible value ofab`bc`cd`da.

3 (Israel) ,b,c,dbe four real numbers such thata b c d 0anda`b`c`d thatpa`2b`3c`4dqaabbccdd 1.(Belgium) magician intends to perform the following trick. She announces a positive integern, along with2nreal numbersx1 .. x2n, to the audience. A member of the audience thensecretly chooses a polynomialPpxqof degreenwith real coefficients, computes the2nvaluesPpx1q,..,Ppx2nq, and writes down these2nvalues on the blackboard in non-decreasing that the magician announces the secret polynomial to the the magician find a strategy to perform such a trick?(Luxembourg) all functionsf:Z Zsuch thatfa2`b2pa`bq afpaq `bfpbqfor everya, ,fndenotes thenthiteration off, ,f0pxq xandfn`1pxq fpfnpxqqfor alln 0.(Slovakia) Shortlisted positive integers. Prove that fora1,..,anP r1,2ksone hasn i 1aiaa21`.

4 `a2i 4?kn.(Iran) `be the set of positive real numbers. Determine all functionsf:R` R`such that, for all positive real numbersxandy,f`x`fpxyq `y fpxqfpyq `1.(Ukraine)6 Saint-Petersburg Russia, 18th 28th September a positive integer. Find the number of permutationsa1,a2,..,anof thesequence1,2,..,nsatisfyinga1 2a2 3a3 .. nan.(United Kingdom) a regular100-gon,41vertices are colored black and the remaining59vertices arecolored white. Prove that there exist24convex quadrilateralsQ1, .. ,Q24whose corners arevertices of the100-gon, so that the quadrilateralsQ1, .. ,Q24are pairwise disjoint, and every quadrilateralQihas three corners of one color and one corner of the other color.(Austria) an integer withn 2. On a slope of a mountain,n2checkpoints aremarked, numbered from1ton2from the bottom to the top.

5 Each of two cable car companies,AandB, operateskcable cars numbered from1tok; each cable car provides a transfer fromsome checkpoint to a higher one. For each company, and for anyiandjwith1 i j k,the starting point of carjis higher than the starting point of cari; similarly, the finishing pointof carjis higher than the finishing point of cari. Say that two checkpoints arelinkedby somecompany if one can start from the lower checkpoint and reach the higher one by using one ormore cars of that company (no movement on foot is allowed).Determine the smallestkfor which one can guarantee that there are two checkpoints thatare linked by each of the two companies.(India) Fibonacci numbersF0,F1,F2,..are defined inductively byF0 0,F1 1, andFn`1 Fn`Fn 1forn 1. Given an integern 2, determine the smallest size of a setSofintegers such that for everyk 2,3.

6 ,nthere exist somex,yPSsuch thatx y Fk.(Croatia) an odd prime, and putN 14pp3 pq 1. The numbers1,2,..,Narepainted arbitrarily in two colors, red and blue. For any positive integern N, denote byrpnqthe fraction of integers int1,2,..,nuthat are that there exists a positive integeraP t1,2,..,p 1usuch thatrpnq a{pfor alln 1,2,..,N.(Netherlands) of weights1,2,3,..,4nare given. Each coin is colored in one ofncolorsand there are four coins of each color. Show that all these coins can be partitioned into twosets with the same total weight, such that each set contains two coins of each color.(Hungary) Shortlisted any rectangular table having finitely many rows andcolumns, with a realnumberapr,cqin the cell in rowrand columnc. A pairpR,Cq, whereRis a set of rows andCa set of columns, is called asaddle pairif the following two conditions are satisfied:piqFor each rowr1, there isrPRsuch thatapr,cq apr1,cqfor allcPC;piiqFor each columnc1, there iscPCsuch thatapr,cq apr,c1qfor saddle pairpR,Cqis called aminimal pairif for each saddle pairpR1,C1qwithR1 RandC1 C, we haveR1 RandC1 that any two minimal pairs contain the same number of rows.}

7 (Thailand) a game on a blackboard that initially contains2020copiesof the number 1. In every round, playerAerases two numbersxandyfrom the blackboard,and then playerBwrites one of the numbersx`yand|x y|on the blackboard. The gameterminates as soon as, at the end of some round, one of the following holds:(1) one of the numbers on the blackboard is larger than the sumof all other numbers;(2) there are only zeros on the then give as many cookies to playerAas there are numbers on the to get as many cookies as possible, whereas playerBwants to give as few aspossible. Determine the number of cookies thatAreceives if both players play optimally.(Austria)8 Saint-Petersburg Russia, 18th 28th September an isosceles triangle withBC CA, and letDbe a point inside sideABsuch thatAD DB.

8 LetPandQbe two points inside sidesBCandCA, respectively,such that=DPB =DQA 90 . Let the perpendicular bisector ofPQmeet line segmentCQatE, and let the circumcircles of trianglesABCandCPQmeet again at pointF, thatP,E,Fare collinear. Prove that=ACB 90 .(Luxembourg) a convex quadrilateral. Suppose thatPis a point in the interior ofABCD such that=PAD:=PBA:=DPA 1 : 2 : 3 =CBP:=BAP:= internal bisectors of anglesADPandPCBmeet at a pointQinside the thatAQ BQ.(Poland) a convex quadrilateral with=ABC 90 ,=CDA 90 , and=DAB =BCD. Denote byEandFthe reflections ofAin linesBCandCD, that the segmentsAEandAFmeet the lineBDatKandL, respectively. Prove thatthe circumcircles of trianglesBEKandDFLare tangent to each other.(Slovakia) the plane, there aren 6pairwise disjoint disksD1,D2.

9 ,Dnwith radiiR1 R2 .. Rn. For everyi 1,2,..,n, a pointPiis chosen in diskDi. LetObe anarbitrary point in the plane. Prove thatOP1`OP2`..`OPn R6`R7`..`Rn.(A disk is assumed to contain its boundary.)(Iran) a cyclic quadrilateral with no two sides parallel. LetK,L,M, andNbe points lying on sidesAB,BC,CD, andDA, respectively, such thatKLMNis a rhombuswithKLkACandLMkBD. Let 1, 2, 3, and 4be the incircles of trianglesANK,BKL,CLM, andDMN, respectively. Prove that the internal common tangents to 1and 3and the internal common tangents to 2and 4are concurrent.(Poland) the incenter and theA-excenter of an acute-angled triangleABCwithAB AC. Let the incircle meetBCatD. The lineADmeetsBIAandCIAatEandF, respectively. Prove that the circumcircles of trianglesAIDandIAEFare tangent toeach other.

10 (Slovakia) Shortlisted a point on the circumcircle of an acute-angled triangleABC. LetD,E, andFbe the reflections ofPin the midlines of triangleABCparallel toBC,CA, andAB,respectively. Denote by A, B, and Cthe circumcircles of trianglesADP,BEP, andCFP,respectively. Denote by the circumcircle of the triangle formed by the perpendicular bisectorsof segmentsAD, that A, B, C, and have a common point.(Denmark) andIbe the circumcircle and the incenter of an acute-angled circles Band Cpassing throughBandC, respectively, are tangent atI. Let Bmeetthe shorter arcABof and segmentABagain atPandM, respectively. Similarly, let Cmeet the shorter arcACof and segmentACagain atQandN, respectively. The raysPMandQNmeet atX, and the tangents to Band CatBandC, respectively, meet that the pointsA,X, andYare collinear.