## Problem I - 3

In triangle $A B C$, the internal bisector $C D$ is drawn. It is known that the center of the circle inscribed in triangle $B C D$ coincides with the center of the circle circumscribed around triangle $A B C$. Calculate the angles of triangle $A B C$.
Solution

Let $O$ be the common center of the circles mentioned in the statement. By the hypothesis, $BO$ and $CO$ are the bisectors of the angles $\angle ABC$ and $\angle BCD$ ( $O$ is the center of the inscribed circle in $BCD$).

Moreover, $AO=BO=CO$ ( $O$ is the center of the circumscribed circle of $ABC$).

From here, if we call $\alpha=\angle ABO$, then we have

$$
\angle OAB=\alpha, \angle OCB=\angle OBC=\alpha, \angle BCD=2\alpha \quad(CD \text{ is the bisector })
$$

From here, $\angle OAC=\angle OCA=3\alpha$

Since the sum of the angles of triangle $ABC$ is $180^{\circ}$, it follows that $\alpha=18^{\circ}$, and $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$