# Solution to The IMO Shortlist

#### by Evan Chen

The first step is to solve the math problems in each of the four sections (except for the geometry section, see below). This is easier to do than it might seem, because we fully expect solvers to abuse computer algebra systems, brute-force programming, Internet searching, engineer's induction, geometry-plotting software, and so on, to figure out the correct answers without necessarily having a complete proof. Nonetheless, for entertainment value, you can check this PDF file (or its TeX source) for complete, mathematically rigorous solutions to the posed problems.

Each of the “proposing” countries for the fake math problems is one of the hosts of a recent International Math Olympiad between the years 2000 and 2020, each exactly once. These countries are listed on the first page of the puzzle in chronological order, as an additional hint. The main gimmick for the problem is that each of the four categories gives a way to transform the IMO corresponding to that host to one of the other hosts:

• For the algebra section, inputting the number of the IMO (e.g. 43rd IMO) into the answer to the problem (which is a function in x) yields a different number which also the number of an IMO.
• For the combinatorics section, inputting the year of the IMO (e.g. 2003) into the answer to the problem (which is a function in n) yields a different number which also the year of an IMO.
• For the geometry section, the problem statement asks to prove three points are collinear, and the names of those points correspond to the given host country. However, each problem has another hidden triple of collinear points which correspond to the names of a different host country.
• For the number theory section, taking the city corresponding to that IMO as a latitude-longitude pair and adding it to the answer to the problem (which is an ordered pair) gives the latitude-longitude pair which corresponds to the city of a different host country.

These are in fact the first four columns of the IMO timeline. This is subtly hinted both by “timeline ... official IMO” in the flavor text as well as the first page (in which the bold portions are these four parameters in order).

Problem Proposer Host Year (Input) Input Data Answer to Problem Output Data Host Year (Output)
A1 UNK 2002 43 (4x+2)/3 58 2017
A2 HEL 2004 45 √(80x) 60 2019
A3 UNK 2019 60 3φ(x)−1 47 2006
A4 ESP 2008 49 x + √x 56 2015
A5 THA 2015 56 ⌈8x2 46 2005
A6 VNM 2007 48 (9x/24+1)/2 41 2000
C1 KAZ 2010 2010 n + s2(n) 2018 2018
C2 NLD 2011 2011 2⌈n/2⌉ 2012 2012
C3 ARG 2012 2012 (n+2014) / gcd(n,2014) 2013 2013
C4 JPN 2003 2003 n+⌈log2(n)⌉ 2014 2014
C5 SAF 2014 2014 n+6 for even n > 4 2020 2020
C6 KOR 2000 2000 n+1 for n > 2 2001 2001
G1 USA 2001 USA n/a UNK 2002
G2 ROU 2018 ROU n/a GER 2009
G3 HKG 2016 HKG n/a HEL 2004
G4 RUS 2020 RUS n/a ESP 2008
G5 MEX 2005 MEX n/a VNM 2007
NT1 BRA 2017 Rio de Janeiro (−22.91, −43.20) (+74, +115) Astana (+51.17, +71.43) 2010
NT2 GER 2009 Bremen (+53.08, +8.8) (−31, +105) Hong Kong (+22.3, +114.2) 2016
NT3 SVN 2006 Ljubljana (+46.06, +14.51) (+6, −10) Amsterdam (+52.34, +4.9) 2011
NT4 COL 2013 Santa Marta (+11.24, −74.21) (+24, +214) Tokyo (+35.69, +139.69) 2003

This bijection between the 21 hosts gives a full cycle. There is a bit of redundancy in this cycle for error checking: the problems of each section are in increasing order (e.g. A1 appears before A2, and so on). This also gives the starting point of the cycle: it should start at G1 and end at C6.

The final insight is that the four letters A, C, G, T correspond to the genetic code. This is clued by the use of “NT” instead of the more traditional “N”, as well as more subtly by the presence of “stranded” in the flavortext. One thus arrives at the following sequence. Indeed, there are 21 letters, and we can map each group of three to an amino acid, which has a canonical letter; hence we get seven letters.

GAT [D] CGT [R] GAA [E] TCC [S] TCC [S] GAA [E] G1 → A1 → NT1 → C1 → G2 → NT2 → G3 → A2 → A3 → NT3 → C2 → C3 → NT4 → C4 → C5 → G4 → A4 → A5 → G5 → A6 → C6