In triangle $A B C$, let $A P$ be the angle bisector of $\angle B A C$, where $P$ is on side $B C$, and $B Q$ be the angle bisector of $\angle A B C$, where $Q$ is on side $C A$. We know that $\angle B A C=60^{\circ}$ and that $A B+B P=A Q+Q B$.

What are the possible values of the angles of triangle $A B C$?
Let $M$ be the point on the extension of segment $AB$ beyond $B$ such that $BM = BP$. Similarly, let $N$ be the point on the extension of segment $AQ$ beyond $Q$ such that $BQ = QN$. The angle bisector of $\angle BQC$ intersects $BC$ at $S$.

Let $\angle BPM = \varphi$. Since $BM = BP$, $\angle BMP = \varphi$ and $\angle PBA = 2\varphi$. Therefore, $\angle QBC = \varphi$. The triangles $\triangle QBS$ and $\triangle QNS$ are congruent because $QB = QN$, the side $QS$ is common, and $\angle SQB = \angle NQS$. Thus, $\angle QBS = \angle QNS = \varphi$. Therefore, the triangles $\triangle AMP$ and $\triangle ANP$ are also congruent because:

$$
AM = AB + BM = AB + BP = AQ + BQ = AQ + QN = AN
$$

The sides $AP$ are common, and the angles at $A$ are equal. Hence, $\angle AMP = \angle ANP = \varphi$.

Thus, $\angle QNP = \angle QNS = \varphi$. Two cases are possible: $P = S$ or $P \neq S$, and thus $P, S, N$ are collinear.

If $P = S$, then from the congruence of triangles $\triangle BQP$ and $\triangle QPN$, as well as $\triangle APM$ and $\triangle APN$, we get that $BP = PN = PM$, which means $BP = PM = BM$. Therefore, the angle at $B$ in $\triangle ABC$ is $180^\circ - 60^\circ = 120^\circ$, which is impossible because then $\angle ABC + \angle BAC = 180^\circ$.

Therefore, $P, S, N$ are collinear, which means $N = C$. Thus, $\angle ABC = \angle ANS = \varphi$, and $180^\circ - \angle BAC + \angle ABC + \angle ACB = 60^\circ + 2\varphi + \varphi$. From this, $\varphi = 40^\circ$ and $2\varphi = 80^\circ$. Therefore, the only possible solution is $\angle ABC = 80^\circ$, $\angle ACB = 40^\circ$, and $\angle BAC = 60^\circ$.

This solution is valid because from the relationship $\angle QBC = \angle QCB$, we get that $BQ = QC$. Furthermore, the triangles $\triangle APM$ and $\triangle APC$ are congruent because two of their angles, at $A$ and at $M$ and $C$, are equal, and they have a common side $AP$. Therefore, $AB + BP = AB + BM = AM = AC = AQ + QC = AQ + QB$.

Thus, the only solution is the triangle with angles $60^\circ$, $80^\circ$, and $40^\circ$.

Péter Csikvári (Fazekas M. Főv. Gyak. Gimn., 11th grade)

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