Problem 4

Problem

Let $\theta$ satisfy $0^\circ<\theta<180^\circ$, with its value known to both players. Shan-Yu draws an arbitrary initial triangle on paper. If the triangle has an angle exactly equal to $\theta$, Mulan wins immediately. Otherwise Mulan chooses a non-vertex point $P$ on the perimeter and cuts from $P$ to the opposite vertex, splitting the triangle into two triangles. Shan-Yu discards one of them, and the remaining triangle becomes the new current triangle.

Determine all values of $\theta$ for which Mulan can guarantee a win after finitely many moves, regardless of Shan-Yu's choices.

Solution

Mulan can force a win exactly for

We describe a triangle only by the multiset of its three positive angle measures, whose sum is $180^\circ$. If a cut splits an angle $x+y$ and the other two angles are $a,b$, the two possible remaining triangles have angle multisets

Conversely, every decomposition $x+y$ with $x,y>0$ is realized by a ray inside that vertex angle; the ray meets the opposite side at a non-vertex point. Thus this angle model describes exactly the legal cuts used below.

Necessity

Any angle that can be forced must lie strictly between $0^\circ$ and $180^\circ$, since it eventually occurs as an angle of a nondegenerate triangle.

Let

be the additive subgroup of all integral multiples of $\theta$. Suppose, for a contradiction, that $180^\circ\notin H$.

Call a triangle $H$-avoiding if none of its angles belongs to $H$. The key invariant is that after any cut of an $H$-avoiding triangle, at least one of the two resulting triangles is still $H$-avoiding. Indeed, suppose both children contained an angle in $H$. If the chosen angles were either of the inherited angles $a$ or $b$, the original triangle would already fail to be $H$-avoiding. Otherwise the first child's chosen angle is $x$ or $b+y$, while the second's is $y$ or $a+x$. The four possible pairs force, respectively,

to lie in $H$ by closure under addition and subtraction. Each possibility contradicts either the original avoidance or $180^\circ\notin H$.

Also, $\theta/2\notin H$: otherwise an integer multiple of $\theta$ would equal $\theta/2$, forcing an integer to equal $1/2$. Since $180^\circ\notin H$, we also have $180^\circ-\theta\notin H$. Therefore

is an $H$-avoiding triangle. After each cut, Shan-Yu can retain an $H$-avoiding child. Such a triangle never contains $\theta\in H$, so Mulan cannot force a win in finitely many moves. This contradiction proves

Thus $180^\circ=n\theta$ for some integer $n$. Because $0<\theta<180^\circ$, necessarily $n\ge2$, and hence

Sufficiency

Now assume $n\ge2$ and $n\theta=180^\circ$.

First observe that if a triangle contains an angle $k\theta$, where $1\le k<n$, Mulan can force $\theta$. For $k=1$ she has already won. For $k>1$, she cuts that angle as

One child contains $\theta$, while the other contains $(k-1)\theta$. Whichever child Shan-Yu keeps, induction on $k$ finishes the game.

It remains to produce such a multiple. The following balanced-cut fact is used: in every triangle whose angle sum is $n\theta$, either

1. one angle already equals $k\theta$ for some $1\le k<n$; or
2. Mulan can make one cut such that one child contains $k\theta$ and the other contains $(n-k)\theta$, with both multipliers strictly between $0$ and $n$.

To see this, write the angles as $p,q,r$. Suppose first that $n\ge3$ and no angle is already a proper multiple of $\theta$. At least one angle, say $r$, is greater than $\theta$; otherwise all three would be smaller than $\theta$, contradicting $p+q+r=n\theta$. Let $k\theta$ be the first positive multiple of $\theta$ strictly above $q$. Then

Split the angle $r$ into

which are both positive. One child has the new angle $q+y=k\theta$. In the other child the complementary new angle is

The unused third angle is positive, so $q+r<180^\circ=n\theta$; hence $1\le k<n$, and likewise $1\le n-k<n$. This is the required balanced cut.

For $n=2$, the only proper multiple is $\theta=90^\circ$. If no angle equals $\theta$, choose an angle $q<\theta$. At least one of the other angles, say $r$, satisfies $q+r>\theta$: otherwise adding the two inequalities involving $q$ would force $q\le0$. Set $y=\theta-q$ and $x=r-y$. Then $x,y>0$, and the same construction with $k=1$ gives the angle $\theta$ in both children.

In the first case, apply the preceding induction directly. In the second case, make the balanced cut and apply it in both children. Therefore Mulan has a finite winning strategy for every initial triangle.