## Task 16/82

In a plane, let two (non-coincident) points $A$ and $B$ be fixed. Given is the set of all triangles $A B C$ in this plane (where the vertices are labeled in the mathematically positive direction), for which $\measuredangle B C A=\gamma=$ constant. The set of the incenters $M$ is sought.
If $M$ is the incenter of a triangle, then by the angle sum theorem

$$
\measuredangle B M A+\measuredangle M A B+\measuredangle A B M=180^{\circ}
$$

Since the incenter is the intersection of the angle bisectors, we have

$$
\measuredangle M A B=\frac{1}{2} \measuredangle C A B=\frac{1}{2} \alpha \quad ; \quad \measuredangle A B M=\frac{1}{2} \measuredangle A B C=\frac{1}{2} \beta
$$

thus

$$
\measuredangle B M A+\frac{1}{2} \alpha+\frac{1}{2} \beta=180^{\circ} \quad \text { or } \quad \measuredangle B M A=180^{\circ}-\frac{1}{2}(\alpha+\beta)
$$

By the angle sum theorem, $\alpha+\beta=180^{\circ}-\gamma$,

$$
\measuredangle B M A=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\gamma\right)=90^{\circ}-\frac{1}{2} \gamma=\text { constant }
$$

Thus, $M$ lies on the arc through $A$ and $B$ with the inscribed angle $\varphi=90^{\circ}+\frac{1}{2} \gamma$, which lies in the same (half-plane generated by the line through $A$ and $B$) as $C$. The desired set is therefore this arc. For the radius $r$ of this arc, we have

$$
\frac{A B}{2 r}=\frac{c}{2 r}=\sin \left(90^{\circ}-\frac{1}{2} \gamma\right)=\cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos \gamma}{2}} \rightarrow r=\frac{c}{2} \sqrt{\frac{2}{1+\cos \gamma}}
$$

Additional: If $Q$ is the center of this arc, then

$$
\measuredangle B Q A=360^{\circ}-\measuredangle A Q B=360^{\circ}-2 \measuredangle B M A=360^{\circ}-2\left(90^{\circ}+\frac{\gamma}{2}\right)=180^{\circ}-\gamma=\alpha+\beta
$$

Thus, the quadrilateral $A Q B C$ is a cyclic quadrilateral (the sum of opposite angles is $180^{\circ}$). It follows that $Q$ lies on the common circumcircle of the triangles $A B C$. By symmetry, $Q$ bisects the arc $B A$.