22. G8 (ISR) ${ }^{\mathrm{IMO}}$ Let $A B C$ be a triangle with $\measuredangle B A C=60^{\circ}$. 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 \( AB \) and \( AQ \) over \( B \) and \( Q \) such that \( BS = BP \) and \( QT = QB \). It is given that \( AS = AB + BP = AQ + QB = AT \). Since \( \angle PAS = \angle PAT \), the triangles \( APS \) and \( APT \) are congruent, from which we deduce that \( \angle ATP = \angle ASP = \beta / 2 = \angle QBP \). Hence \( \angle QTP = \angle QBP \). If \( P \) does not lie on \( BT \), then the last equality implies that \( \triangle QBP \) and \( \triangle QTP \) are congruent, so \( P \) lies on the internal bisector of \( \angle BQT \). But \( P \) also lies on the internal bisector of \( \angle QAB \); consequently, \( P \) is an excenter of \( \triangle QAB \), thus lying on the internal bisector of \( \angle QBS \) as well. It follows that \( \angle PBQ = \beta / 2 = \angle PBS = 180^\circ - \beta \), so \( \beta = 120^\circ \), which is impossible. Therefore \( P \in BT \), which means that \( T \equiv C \). Now from \( QC = QB \) we conclude that \( 120^\circ - \beta = \gamma = \beta / 2 \), i.e., \( \beta = 80^\circ \) and \( \gamma = 40^\circ \).