\section*{Problem 13 - V01213}

In triangle \(A B C\), the angle \(\gamma\) is to be calculated, given that:

\[
\sin \gamma=\sqrt{2} \cdot \cos \frac{\alpha+\beta}{2}
\]
}

Let \(\alpha, \beta\) and \(\gamma\) be the interior angles of a triangle. According to the interior angle sum theorem,

\[
\alpha+\beta+\gamma=180^{\circ} \Longleftrightarrow \frac{\alpha+\beta}{2}=90^{\circ}-\frac{\gamma}{2}
\]

and thus

\[
\sin (\gamma)=\sqrt{2} \cdot \cos \left(90^{\circ}-\frac{\gamma}{2}\right)
\]

Since \(\cos \left(90^{\circ}-\frac{\gamma}{2}\right)=\sin \left(\frac{\gamma}{2}\right)\) holds for any angle \(\gamma\), it follows that

\[
\sin (\gamma)=\sqrt{2} \cdot \sin \left(\frac{\gamma}{2}\right)
\]

According to the addition theorem,

\[
\sin (x+y)=\sin (x) \cos (y)+\sin (y) \cos (x)
\]

for all real numbers \(x\) and \(y\), so

\[
\sin (\gamma)=2 \cdot \sin \left(\frac{\gamma}{2}\right) \cdot \cos \left(\frac{\gamma}{2}\right)=\sqrt{2} \cdot \sin \left(\frac{\gamma}{2}\right)
\]

holds. Since \(\gamma \in\left(0^{\circ}, 180^{\circ}\right)\) must hold, it follows that

\[
2 \cos \left(\frac{\gamma}{2}\right)=\sqrt{2}
\]

and thus

\[
\gamma=2 \cdot \arccos \left(\frac{1}{\sqrt{2}}\right)=90^{\circ}
\]

Therefore, \(\gamma\) is a right angle.