Austrian Mathematical Olympiad

Figure 2: Problem 10

As usual, we denote the angles of the triangle at $A$, $B$ and $C$ with $\alpha$, $\beta$ and $\gamma$. It is sufficient to show that the center $I_c$ of the excircle touching the line $AB$ lies on the two circles. To do this, we look at the respective inscribed angles. Since the triangle $AYB$ is isosceles, the following holds: $$\angle CYA=\angle BYA=90^{\circ }-\angle YBA/2=90^{\circ }-90^{\circ }+\beta /2=\beta /2.$$

But it is also true that $$\angle C{I}_{c}A=180^{\{}\circ \}-(180^{\{}\circ \}-\alpha )/2-\alpha -\gamma /2=90^{\{}\circ \}-\alpha /2-\gamma /2=\beta /2.$$ So $I_c$ lies on the circumcircle of $ACY$ by the inverse of the inscribed angle theorem. In the same way, one also obtains that $I_c$ lies on the circumcircle of $BCX$. So $I_c$ is the second point of intersection, which therefore lies on the angle bisector through $C$ as required.