IMO 2026 · Problem 4

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?

Official text of the problem. Official problem paper (PDF)

Answer

\(\displaystyle \theta=\frac{180^\circ}{n}\) for any integer \(n\ge 2\).

Solution

We prove that Mulan can win for these values and that Shan-Yu can avoid defeat for every other value.

Mulan’s strategy when \(180^\circ=n\theta\).

First, suppose the current triangle has an angle \(k\theta\), where \(1\le k\le n-1\). Mulan can win in at most \(k-1\) further cuts. Indeed, if \(k=1\), she has already won. Otherwise, she cuts from that vertex, splitting its angle into \(\theta\) and \((k-1)\theta\). Whichever triangle Shan-Yu retains either has an angle \(\theta\), or has an angle \((k-1)\theta\). Repeating this procedure proves the claim.

It therefore suffices to force an angle that is an integer multiple of \(\theta\).

Let the current triangle be \(ABC\), with angles \(a,b,c\), respectively, and suppose none is an integer multiple of \(\theta\). Relabel the vertices so that \(a\) is a largest angle. We claim that there is an integer \(m\) such that \[ b<m\theta<180^\circ-c=b+a. \]

  • If \(n=2\), both \(b\) and \(c\) are strictly less than \(90^\circ\): otherwise, since \(a\) is largest, two angles would sum to at least \(180^\circ\). Thus \(m=1\) works.
  • If \(n\ge3\), then \(a\ge60^\circ\ge\theta\). Since \(a\) is not an integer multiple of \(\theta\), we have \(a>\theta\). Taking \[ m=\left\lfloor\frac b\theta\right\rfloor+1 \] gives \[ b<m\theta\le b+\theta<b+a. \]

Now Mulan cuts along \(AP\), where \(P\) lies on \(BC\) and \[ \angle BAP=m\theta-b. \] The inequalities just proved show that this angle lies strictly between \(0^\circ\) and \(a\), so the ray meets the interior of \(BC\), making the cut legal. The two angles at \(P\) are \[ \angle APB=180^\circ-m\theta=(n-m)\theta, \qquad \angle APC=m\theta. \] Moreover, \(1\le m\le n-1\). Hence either retained triangle has an angle that is a positive integer multiple of \(\theta\). Mulan then applies the decreasing-multiple strategy above.

This guarantees victory in finitely many steps; in fact, at most \(n-1\) cuts suffice.

Shan-Yu’s strategy when \(180^\circ/\theta\) is not an integer.

Call an angle marked if it is a positive integer multiple of \(\theta\), and call a triangle safe if none of its angles is marked.

Shan-Yu initially chooses an equilateral triangle. It is safe, since an equality \(60^\circ=k\theta\) would imply \[ \frac{180^\circ}{\theta}=3k, \] contrary to our assumption.

We show that whenever Mulan cuts a safe triangle, at least one of the resulting triangles is safe. Shan-Yu can therefore always retain a safe triangle.

Let \(ABC\) be safe, with angles \(a,b,c\), and suppose the cut is \(AP\), where \(P\) is in the interior of \(BC\). Write \[ x=\angle BAP,\qquad y=\angle PAC,\qquad u=\angle APB,\qquad v=\angle APC. \] The angle sums give \[ a=x+y,\qquad b=v-x,\qquad c=u-y,\qquad 180^\circ=u+v. \]

Suppose, for a contradiction, that neither resulting triangle is safe. Since the inherited angles \(b\) and \(c\) are not marked, at least one of \(x,u\) is marked, and at least one of \(y,v\) is marked. There are four possibilities:

  • If \(x,y\) are marked, then \(a=x+y\) is marked.
  • If \(x,v\) are marked, then \(b=v-x>0^\circ\) is marked.
  • If \(u,y\) are marked, then \(c=u-y>0^\circ\) is marked.
  • If \(u,v\) are marked, then \(180^\circ=u+v\) is an integer multiple of \(\theta\).

The first three contradict the safety of \(ABC\), and the fourth contradicts the assumption on \(\theta\). Thus at least one resulting triangle is safe.

By retaining such a triangle after every cut, Shan-Yu ensures that no angle ever equals \(\theta\). Consequently, Mulan cannot guarantee victory for any other value of \(\theta\).

The solution was checked against Evan Chen's published IMO solution notes.