# Solution to The IMO Shortlist

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 | ⌈8x/π^{2}⌉ |
46 | 2005 |

A6 | VNM | 2007 | 48 | (9^{x/24}+1)/2 |
41 | 2000 |

C1 | KAZ | 2010 | 2010 | n + s_{2}(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+⌈log_{2}(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.

G1 | → | A1 | → | NT1 | GAT [D] | |
---|---|---|---|---|---|---|

→ | C1 | → | G2 | → | NT2 | CGT [R] |

→ | G3 | → | A2 | → | A3 | GAA [E] |

→ | NT3 | → | C2 | → | C3 | TCC [S] |

→ | NT4 | → | C4 | → | C5 | TCC [S] |

→ | G4 | → | A4 | → | A5 | GAA [E] |

→ | G5 | → | A6 | → | C6 | GAC [D] |

Reading the seven letters gives the answer DRESSED.