IMO 2026 Problems

The 67th International Mathematical Olympiad was held in Shanghai, China, from 10–20 July 2026. As usual, the contest ran over two days, with three problems per day and four and a half hours per day to solve them; each problem is worth 7 points, for a maximum score of 42. Below are all six problems, quoted verbatim, with the answer to each left blank for now — I want to sit with them myself before writing anything up.

Problem 1 (Day 1, 7 points)

There are 2026 integers greater than 1 written on a blackboard, not necessarily different. In a move, Confucius chooses two integers $m > 1$ and $n > 1$ from different places on the blackboard and replaces these two integers with

He continues to make moves while it is possible to do so.

(a) Prove that, regardless of the choices of Confucius, after finitely many moves, exactly one integer $M$ on the blackboard is greater than 1.

(b) Prove that the value of $M$ does not depend on the choices of Confucius.

My solution

General idea for (a): The product of the terms is non-increasing after the move. If gcd(m,n) = 1, then one of the terms now becomes 1. Otherwise, the product of the terms is strictly decreasing.

If any term becomes 1, then it stays 1 as it can no longer be chosen by Confucius. In any case, this implies that at any move we can either increase our count of 1s by one or decrease our product of terms.

Consider for the sake of contradiction that it is not possible for exactly one integer $M$ to be left on the blackboard that is greater than 1.

Then, we have strictly less than 2025 applications of a move being made on (m,n) where gcd(m,n) = 1. In turn, all other moves strictly decrease the product.

Since it is not possible for only one non-one integer to be left on the board, we have that there are always at least two non-one integers left on the board.

We now after all our moves with gcd(m,n) = 1 only can apply moves the decrease the product and so we have a strictly decreasing product with at least two non-one integers left on the board. Therefore, this product is eventually 4 corresponding to 2 x 2, which is the smallest product possible given at least two non-one integers are left on the board. Then, we can apply the move to (2,2) and are left with (1, 2).

And so we have a contradiction, as all but one term is 1 in this situation.

General idea for (b):

Problem 2 (Day 1, 7 points)

Let $ABC$ be a triangle and let points $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Let points $K$ and $L$ be chosen inside triangles $BMC$ and $BNC$, respectively, such that $K$ lies strictly inside triangle $LBA$, and $L$ lies strictly inside triangle $ACK$. Suppose that

Let $O$ be the circumcenter of triangle $AKL$. Prove that $OM = ON$.

My solution

(solution not yet written — come back and fill this in)

Problem 3 (Day 1, 7 points)

Let $n$ be a positive integer. Liu Bang and Xiang Yu have a stick of length 1 and want to divide it between themselves. Liu Bang marks at most $n$ points on the stick, and then Xiang Yu marks at most $n$ points on the stick. The marked points are distinct. Then, the stick is cut at all marked points, creating a number of pieces. Afterwards, they take turns claiming any unclaimed piece of the stick, with Liu Bang going first. Each player’s goal is to maximise the total length of their own pieces. For each $n$, determine the largest value $c$ such that Liu Bang may guarantee a total length of at least $c$, regardless of Xiang Yu’s play.

My solution

(solution not yet written — come back and fill this in)

Problem 4 (Day 2, 7 points)

Shan-Yu and Mulan are playing a game. Let $\theta$ be an angle with $0^\circ < \theta < 180^\circ$ known to both players. Initially, Shan-Yu makes a paper triangle $\mathcal{T}$ with measurements of his choice. Then, they repeatedly perform the following steps:

• If $\mathcal{T}$ has at least one angle measuring exactly $\theta$, then the game stops and Mulan wins.
• Otherwise, Mulan chooses a point $P$ on the perimeter of $\mathcal{T}$, different from its three vertices. She then makes a straight cut from $P$ to the opposite vertex of $\mathcal{T}$, splitting it into two triangles.
• Shan-Yu discards one of the two triangles. The remaining triangle becomes the new $\mathcal{T}$.

For which real values of $\theta$ can Mulan guarantee her victory in finitely many steps, no matter how Shan-Yu plays?

My solution

(solution not yet written — come back and fill this in)

Problem 5 (Day 2, 7 points)

Solve over $\mathbb{R}_{>0}$ the functional equation

My solution

(solution not yet written — come back and fill this in)

Problem 6 (Day 2, 7 points)

Let $a_1, a_2, a_3, \ldots$ be an infinite sequence of positive integers greater than 1. Suppose that for all positive integers $n$, the number $a_{n+1}$ is the smallest positive integer greater than $a_n$ such that $\gcd(a_{n+1}, a_i) > 1$ for every $i = 1, 2, \ldots, n$. Prove that there exist positive integers $T$ and $L$ such that

for every positive integer $n$.

My solution

(solution not yet written — come back and fill this in)

Source note: Problem statements above were cross-checked against the official IMO problem archive at imo-official.org/problems/2026, Evan Chen’s IMO 2026 Solution Notes (dated 23 July 2026), and discussion threads on the Art of Problem Solving forums.