I have been involved in mathematical olympiads, whether as a marker, proposer, coach, or as a contestant. In the past, I have also proposed problems and do quality checks for "Kontes Terbuka Olimpiade Matematika" (KTOM), an Indonesian nonprofit mathematical olympiad organization. My main topic of interest in contest math is functional equations.
Below is a list of some nice problems, "generalized" from the IMO Shortlist; I do not propose any of these problems. Some of these have essentially the same solutions as the official solution or the solutions you can find in AoPS thread for the original version. Enjoy!
Consider a configuration of finitely many pebbles stacked in finitely many piles. At every point, one can choose two piles, take out the same amount of pebbles from both piles, say \(k\) pebbles, then put the \(2k\) pebbles into a new pile. For each initial configuration, determine the smallest number of non-empty piles one can obtain after performing moves.Difference with the original formulation
The original formulation restricts the initial configuration to only have one pebble in each pile.
Let \(\mathbb{N}\) denote the set of positive integers. Find all functions \(f : \mathbb{N} \to \mathbb{Z}\) such that:
- \(f(mn) = f(m) + f(n)\) for any positive integers \(m\) and \(n\); and
- there exists infinitely many positive integers \(n\) such that \(f(n - k) = f(k)\) for all positive integers \(k < n\).
Difference with the original formulation
The original formulation has codomain \(\mathbb{N} \cup \{0\}\) instead of \(\mathbb{Z}\).
Spoiler
The official solution does not work anymore, but the functions that work are still defined by \(p\)-adic valuation. The functional equation can be generalized further, but the set of solutions become more mysterious.
Let \(F\) be a field. Find all functions \(f : F \to F\) such that for any \(x, y \in F\), we have \[ f(f(x) f(y)) + f(x + y) = f(xy). \]Difference with the original formulation
The original formulation uses \(F = \mathbb{R}\).
Spoiler
The official solution does not work, but a particular magical solution for the original problem (can be found on AoPS) works as long as \(F\) does not have characteristic equal to \(2\). Readers can try working out the case where \(F\) has characteristic \(2\), which has an entirely different proof of injectivity.
Let \(F\) be a finite field of cardinality not equal to \(11\). Prove that every element of \(F\) can be expressed as \(a^2 + b^5\) for some \(a, b \in F\).Difference with the original formulation
The original formulation only considers \(F = \mathbb{F}_p\) for some prime \(p\), and furthermore it assumes \(p > 100\).
Spoiler
There are no pairs \((a, b)\) of integers such that \(a^2 + b^5 \equiv 7 \pmod{11}\). The official solution works as long as \(\#F \geq 41\).