Austrian Mathematical Olympiad

Let $w_\gamma$ be the angle bisector of $\gamma$, $k$ the circumcircle of $ABC$ and $D = k \cap w_\gamma$. Since $\angle DCB = \angle DCA$ we certainly have $|DA| = |DB|$. Considering the triangle $DBI$, we note that $\angle CDB = \angle CAB = \alpha$. Also, $\angle DBI = \angle DBA + \angle ABI = \angle DCA + \angle ABI = \frac{\gamma}{2} + \frac{\beta}{2}$, and therefore $\angle DIB = 180^\circ - \alpha - (\frac{\gamma}{2} + \frac{\beta}{2}) = \frac{\gamma}{2} + \frac{\beta}{2}$. We see that $DBI$ is isosceles with $|DI| = |DB|$. Furthermore, since $D$ lies on the bisector of $UI$, we also have $|DU| = |DI|$. It follows that $D$ is the mid-point of a circle through all four points $A$, $B$, $I$ and $U$.

Since $U$ is the mid-point of the circumcircle of $ABC$, we have $\angle AUB = 2 \cdot \angle ACB = 2\gamma$. On the other hand, since $\angle IAB = \frac{\alpha}{2}$ and $\angle IBA = \frac{\beta}{2}$, we have $\angle AIB = 180^\circ - \frac{\alpha}{2} - \frac{\beta}{2}$. Since $A$, $B$, $I$ and $U$ lie on a common circle, we have $\angle AUB = \angle AIB$, and therefore $2 \cdot \angle ACB = 2\gamma = 180^\circ - \frac{\alpha}{2} - \frac{\beta}{2}$, which is equivalent to $\gamma = \frac{1}{2}(\alpha + \beta)$. Since the value of $\gamma$ is the arithmetic mean of the values of the angles $\alpha$ and $\beta$, it is certainly the second largest angle in the triangle $ABC$ as claimed. $\square$