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Solutions to the 67th International Mathematical Olympiad (2026)

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P1. 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.

(Note that $\gcd(x,y)$ denotes the greatest common divisor of the positive integers $x$ and $y$, and $\operatorname{lcm}(x,y)$ denotes their least common multiple.)

S1.

After prime factorization, the given operation can be considered independently for each prime, allowing us to observe the following facts.

Fact 1. If only one number contains a certain prime factor, that prime can neither disappear nor be newly created through the given operation.

Fact 2. If one number contains exactly one factor of a certain prime $p$, while the remaining numbers are either not divisible by $p$ or contain only powers of $p$ as factors, the number of occurrences of $p$ strictly decreases under the above operation. By Fact 1, the number of occurrences of $p$ continues to decrease until only one remains.

Fact 3. If the $p$-parts of all the numbers are $1,p^k,p^{2k},p^{3k},\ldots$, then eventually only one $p^k$ remains. The term $p^k$ cannot be divided any further.

For example, $(3,9)\to(3,3)\to(1,3)$, verifying Fact 2. For example, $(4,12,16)\to(4,3,16)\to(4,3,4)\to(1,4,3)), verifying Fact 3. For two coprime numbers $a$ and $b$, the operation produces $1$ and $ab$, and by Fact 1, $M\ne1$. Therefore, statement ⒜ has been proved. Define $d_p$ to be the greatest common divisor of the $p$-adic exponents. In other words, define $p^{d_p}$ to be the largest power of $p$ such that the $p$-parts of all the numbers can be expressed as integral powers of it. Then, regardless of the order in which the operations are performed, the board is reduced to 2025 copies of $1$ and one number

Therefore, statement ⒝ has been proved.

P2. 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 strictly inside triangles $BMC$ and $BNC$, respectively, such that $K$ lies strictly inside triangle $ABL$ and $L$ lies strictly inside triangle $AKC$. Suppose that

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

S2.

Let $P$ be the intersection of $BK$ and $AC$, and let $Q$ be the intersection of $CL$ and $AB$. Since $K$ and $L$ lie inside $\triangle ABC$, the points occur in the orders $B-K-P, C-L-Q.$ Since $\angle PBQ=\angle PCQ,$ the points $B,C,P,Q$ are concyclic. Also, $\angle LBP+\angle LNP=180^\circ,$ so $B,P,N,L$ are concyclic. Similarly, $C,K,M,Q$ are concyclic.

Take $A$ as the origin and identify each point with its position vector. Write

Since $K\in BP$, there exists $t\in(0,1)$ such that

Because $B,C,P,Q$ are concyclic, the power of $A$ with respect to their circumcircle gives $AB\cdot AQ=AC\cdot AP.$ Hence

In general, a circle with center $h$ and radius $r$ has equation $\lVert x-h\rVert^2=r^2,$ which may be rewritten as

for some vector $u$ and scalar $d$. Apply this equation to the circle through $C,K,M,Q$. Substituting $x=sB$, we obtain

Since the intersections of this circle with $AB$ are $M=\frac12B$ and $Q=\mu B$, the roots are $s=\frac12,\qquad s=\mu.$ Therefore,

and

Since $C$ also lies on this circle,

Since $K$ lies on the same circle,

Using $K=(1-t)B+t\lambda C,$ together with earlier equations, we obtain

This simplifies to

Let $H$ be the center of the circumcircle of $B,C,P,Q$, and let its radius be $R$. By the power of $A$,

Since $K$ lies inside the chord $BP$, the power of $K$ gives

Therefore,

Equivalently by (★),

Hence $AH^2-KH^2=AK^2,$ or $AH^2=AK^2+KH^2.$ Thus, by the converse of the Pythagorean theorem, $AK\perp KH.$ Similarly, $AL\perp LH.$ Therefore, $\angle AKH=\angle ALH=90^\circ,$ so $A,K,H,L$ are concyclic, with $AH$ as a diameter. Consequently, the circumcenter $O$ of $\triangle AKL$ is the midpoint of $AH$. Since $M$ is the midpoint of $AB$ and $O$ is the midpoint of $AH$, the midpoint theorem in $\triangle ABH$ gives $OM=\frac12 BH.$ Similarly, since $N$ is the midpoint of $AC$, $ON=\frac12 CH.$ Finally, $H$ is the center of the circumcircle of $B,C,P,Q$, so $BH=CH.$ Therefore,

P3. 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 marks at most $n$ points on the stick, and then Xiang 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 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 may guarantee a total length of at least $c$, regardless of Xiang’s play.

S3.

Since Liu Bang moves first, we have $c \ge \frac12$. We should also keep in mind that $c \to \frac12$ as $n \to \infty$.

For $n=2$, if Liu Bang marks the midpoint, Xiang Yu will also place his mark very close to the midpoint, and the two players will end up receiving approximately half of the stick each. However, suppose Liu Bang places his mark at a point one-third of the way from the left endpoint. If Xiang Yu places his mark near the same position, one of the remaining pieces becomes too large. Therefore, Xiang Yu is effectively forced to place his mark at the point two-thirds of the way from the left endpoint. Consequently, Liu Bang obtains a total length of $\frac23$.

This problem is analogous to optimizing a symmetric inequality, so the maximum value of $c$ is attained under a symmetric equality configuration. This is reminiscent of the uvw principle of inequality. After Liu Bang chooses his points and Xiang Yu responds, the remaining subdivision of the stick exhibits a self-similar structure, much like a fractal. This makes it relatively easy to identify the equality configuration. By “self-similar,” we mean that, as suggested by the case $n=2$, Liu Bang first receives a piece of length $x$, Xiang Yu then receives another piece of length $x$, Liu Bang next receives a piece of length $\frac{x}{2}$, Xiang Yu receives another piece of length $\frac{x}{2}$, and so on. The final smallest piece is taken by Liu Bang, and this piece represents precisely the advantage of moving first. Therefore, $c_n$ is given by

P4. 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?

S4.

It is easy to see that victory cannot always be guaranteed when $\theta>90^\circ$, because Shan-Yu may discard the obtuse triangle. We know that $\theta=90^\circ$ is a real value for which victory can always be guaranteed, because when a triangle is divided, an angle of measure $\theta$ must occur in one of the two resulting triangles. It follows that every value of the form

also guarantees victory. We know that $\theta=60^\circ$ is a real value for which victory can always be guaranteed, because when a triangle is divided, angles of measures $\theta$ and $2\theta$ can necessarily be produced. It follows that every value of the form

also guarantees victory. Generalising this, if

then $\theta$ is a real value for which victory can always be guaranteed, and this includes all the preceding cases. For example, if $13\theta=180^\circ$, then $13=1101_2$. First, split it into $8\theta$, corresponding to $1000_2\theta$, and $5\theta$, corresponding to $101_2\theta$. If the triangle with angle $5\theta$ is selected, split it into $4\theta$, corresponding to $100_2\theta$, and $\theta$. If $8\theta$ or $4\theta$ is selected, continue dividing the angle in half.

On the other hand, when attempting to create $72^\circ$, the following alternate-angle configuration is impossible. Constructing it from $36^\circ$ is also impossible, leaving only the method of splitting from $144^\circ$. However, as mentioned earlier, a strategy that first creates $144^\circ$ is impossible.

Since $72^\circ=\frac25\times180^\circ,$ and this value is impossible, every rational multiple that is not of the form $180^\circ/m$, where $m\in\mathbb N$, is also impossible. This is because rational numbers not of the form $1/m$ have the same status as $2/5$, and therefore, by group theory, they must follow the same rules. For irrational numbers, the operation itself is not closed, so such values of $\theta$ are impossible. Therefore, the answer is

P5. Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f:\mathbb R_{>0}\to\mathbb R_{>0}$ such that

for every $x,y\in\mathbb R_{>0}$.

S5.

Substituting $x=f(t)$ and $y=t$ gives $f(f(t))=2f(t)-t.$ From this equation, mathematical induction gives $f^{(n)}(x)=x+n\bigl(f(x)-x\bigr).$ Since every $f^{(n)}(x)>0$, we have $f(x)-x\ge0.$ Using the given inequality again,

Since the above inequality holds for all $x,y>0$, bringing $x$ arbitrarily close to $y$ gives $\left|g’(x)\right|=0.$ Therefore,

For completeness, the existence of the solutions must also be established by substituting them into the original inequality and verifying that it holds. Note that the differentiability of $g(x)$ is not guaranteed, so a more rigorous approach is required. Nevertheless, the main idea is similar.

P6. Let $a_1,a_2,a_3,\ldots$ be an infinite sequence of positive integers greater than 1. Suppose that for every positive integer $n$, 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 $n$. Note that $\gcd(x,y)$ denotes the greatest common divisor of the positive integers $x$ and $y$.

S6.

I’ll use the argument based on the periodicity and density of primes.

To facilitate understanding of the problem, let us first consider several examples. If $a_1$ is even, then $T=1$ and $L=2$, so the claim is easily proved. If $a_1$ is prime, then $T=1$ and $L=a_1$, so the claim is again straightforward. If $a_1$ is an odd composite number, however, the situation becomes more complicated. For example, if $a_1=15$, then $a_2=18, a_3=20.$

It can be shown that the sequence ${a_n}$ contains $a_1,2a_1,3a_1,\ldots.$ Likewise, it can be shown that it contains $a_2,2a_2,3a_2,\ldots.$ We cannot simply take $L=a_1$, because some terms of the sequence lying between $a_1$ and $2a_1$ may fail to satisfy $a_{n+T}=a_n+L.$ If such a situation occurs, it may also affect the position of $a_1$, so that there may be no fixed period $T$ satisfying $a_{1+kT}=a_1+kL$ for all relevant $k$. Instead, suppose that we choose $L$ to be a common multiple of $a_1,a_2,\ldots,a_k,$ where $a_k<2a_1\leq a_{k+1},$ so that every term between $a_1$ and $2a_1$ satisfies $a_{n+T}=a_n+L.$ This means that a configuration of $k$ points with the same relative spacings as $a_1,a_2,\ldots,a_k$ appears again later in the sequence.

On the other hand, as $n$ increases, the terms $a_n$ are subject to an increasing number of constraints. Consequently, their density decreases monotonically; in other words, the sequence becomes progressively sparser. The reappearance of the same configuration of $k$ points would therefore imply that the density is eventually preserved uniformly across all intervals. Since the density is preserved, while the periodic structure induced by the primes greatly restricts the possible positions of the terms of ${a_n}$, the sequence must ultimately reduce to a relation of the form $a_{n+T}=a_n+L.$ At this point, one might also attempt to replace the periodicity argument for primes with an application of the pigeonhole principle.

Posted: 2026.07.16 20:01