Problem 4

Assume $\theta=180^\circ/n$ for an integer $n\ge2$. Let the largest angle of the initial triangle be at $V$; its altitude to the opposite side meets that side in its interior (the other two angles are acute), so this is a legal first cut. Either retained piece is a right triangle. Relabel the retained triangle as $ABC$ with $\angle A=90^\circ$ and, after interchanging $B,C$, $\angle B\le45^\circ$. If $n=2$, then $\theta=90^\circ=\angle A$ and Mulan wins immediately. For $n\ge3$, choose an integer $k$ with $n/4<k\le n/2$: for $n=3$ use $k=1$, while for $n\ge4$ the half-open interval $(n/4,n/2]$ contains an integer because its length is at least one. Hence $45^\circ<k\theta\le90^\circ$. Since $\angle B\le45^\circ$, the number $k\theta-\angle B$ is strictly between $0^\circ$ and $90^\circ$; therefore the ray from $A$ making this angle with $AB$ meets the interior of $BC$ at a legal point $D$. In $\triangle ABD$, the exterior-angle theorem gives $\angle ADC=\angle B+\angle BAD=k\theta$. The adjacent angle is $\angle ADB=180^\circ-k\theta=(n-k)\theta$, and $n-k\ge1$. Thus one child, $ABD$, contains the positive integer multiple $(n-k)\theta$, while the other, $ACD$, contains $k\theta$; step 1 wins from whichever child Shan-Yu retains, after only these two cuts.