65th Czech and Slovak Mathematical Olympiad

Let us use the standard notation of the inner angles of the triangle $ABC$, further let $I$ be the incenter and $I_a$ be the excenter (of the excircle opposite $A$), and let $D$ and $E$ be in order the touching points of the thought circles. Since the bisectors $BI$ and $B{I}_a$ of the supplementary angles are perpendicular to each other (as well as $CI$ and $C{I}_a$), the points $B$, $C$, $I$, and $I_a$ lie on the circle with the diameter $I{I}_a$.

Thus $D$ and $E$, the orthogonal projections of $I$ and $I_a$ onto the secant $BC$, are point reflections of each other with respect to the center of $BC$.

The right triangles $BID$ and $I_{a}BE$ are obviously similar and $$|BD|:|ID|=|I_{a}E|:|BE|\text{or}|BD|\cdot |BE|=|ID|\cdot |I_{a}E|$$ considering the mentioned point reflection also $$|BD|+|BE|=|BD|+|CD|=|BC|=r+r_a=|ID|+|I_{a}E|.$$ Fig. 1

The two equations imply that the pairs ($|ID|$, $|E{I}_a|$) and ($|BD|$, $|BE|$) are roots of the same quadratic equation, that is $|ID|=|BD|$ or $|ID|=|BE|$.

$|ID|=|BD|$ means the right-angled triangle $BID$ is isosceles, which is $\beta =90^{\circ }$.

Similarly, if $|ID|=|BE|$ that is $|ID|=|CD|$ ($D$ and $E$ are point reflections in the mentioned reflection) means the right triangle $CID$ is isosceles, that is $\gamma =90^{\circ }$.

In both cases the triangle $ABC$ is right-angled.