22. G8 (ISR) ${ }^{\mathrm{IMO6}}$ Let $A B C$ be a triangle with $\angle B A C=60^{\circ}$. Let $A P$ bisect $\angle B A C$ and let $B Q$ bisect $\angle A B C$, with $P$ on $B C$ and $Q$ on $A C$. If $A B+$ $B P=A Q+Q B$, what are the angles of the triangle?
22. Let $S$ and $T$ respectively be the points on the extensions of $A B$ and $A Q$ over $B$ and $Q$ such that $B S=B P$ and $Q T=Q B$. It is given that $A S=$ $A B+B P=A Q+Q B=A T$. Since $\angle P A S=\angle P A T$, the triangles $A P S$
and $A P T$ are congruent, from which we deduce that $\angle A T P=\angle A S P=$ $B / 2=\angle Q B P$. Hence $\angle Q T P=\angle Q B P$.
If $P$ does not lie on $B T$, then the last equality implies that $\triangle Q B P$ and $\triangle Q T P$ are congruent, so $P$ lies on the internal bisector of $\angle B Q T$. But $P$ also lies on the internal bisector of $\angle Q A B$; consequently, $P$ is an excenter of $\triangle Q A B$, thus lying on the internal bisector of $\angle Q B S$ as well. It follows that $\angle P B Q=\beta / 2=\angle P B S=180^{\circ}-\beta$, so $\beta=120^{\circ}$, which is impossible. Therefore $P \in B T$, which means that $T \equiv C$. Now from $Q C=Q B$ we conclude that $120^{\circ}-\beta=\gamma=\beta / 2$, i.e., $\beta=80^{\circ}$ and $\gamma=40^{\circ}$.