Problem 12.2. The incircle of $\triangle A B C$ is tangent to the sides $A C$ and $B C$, $A C \neq B C$, at points $P$ and $Q$, respectively. The excircles to the sides $A C$ и $B C$ are tangent to the line $A B$ at points $M$ and $N$. Find $\Varangle A C B$ if the points $M, N, P$ and $Q$ are concyclic.

Oleg Mushkarov, Nikolai Nikolov
12.2. The perpendicular bisector of the segment $A B$ and the bisector of $\Varangle A C B$ meet at the midpoint $D$ of the arc $\overparen{A B}$ of the circumcircle of $\triangle A B C$ which does not contain $C$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-098.jpg?height=783&width=849&top_left_y=1181&top_left_x=496)

Next we use the standard notation for the elements of $\triangle A B C$. Since $A M=$ $B N=C P=C Q=p-c$, the condition of the problem is equivalent to the equality $D M=D P$. The Cosine theorem gives

$$
\begin{aligned}
D P^{2} & =D C^{2}+C P^{2}-2 D C \cdot C P \cos \frac{\gamma}{2} \\
D M^{2} & =D A^{2}+A M^{2}+2 D A \cdot A M \cos \frac{\gamma}{2}
\end{aligned}
$$

Subtracting these equalities, we get

$$
D C-D A=2(p-c) \cos \frac{\gamma}{2}
$$

On the other hand, we have $D C=\frac{(a+b) D A}{c}$ by the Ptolemy's theorem. Since $D A=\frac{c}{2 \cos \frac{\gamma}{2}}$, we get $D C=\frac{a+b}{2 \cos \frac{\gamma}{2}}$. Now (1) implies that $\cos ^{2} \frac{\gamma}{2}=\frac{1}{2}$, i.e. $\gamma=90^{\circ}$.

Remark. The solution above shows that the points $M, N, P$ and $Q$ are concyclic if and only if $A C=B C$ or $\Varangle A C B=90^{\circ}$.