62nd Czech and Slovak Mathematical Olympiad

Let $\alpha ,\beta ,\gamma$ denote the measures of the interior angles at the vertices $A,B,C$, respectively, of the triangle $ABC$, and let $J$ be the intersection point of the line $AI$ and the side $BC$. (Fig. 2). The inscribed angle $PBI$ corresponds to the chord $PI$, while the inscribed angle $QBI$ corresponds to the chord $IQ$, and since the measure of both of these angles is $\frac{1}{2}\beta$, the chords $PI$ and $IQ$ share the same length as well.

Fig. 2

Since the inscribed angle $JIQ$ also corresponds to the chord $IQ$, its measure is also $\frac{1}{2}\beta$. It follows from the congruence of vertical angles that $|\angle RIA|=\frac{1}{2}\beta$. This measure is also shared by the inscribed angle $PIA$ as it corresponds to the chord $PI$ of equal length as $IQ$. Further, $|\angle RAI|=|\angle PAI|=\frac{1}{2}\alpha$. The triangles $RIA$ and $PIA$ are thus congruent by ASA, hence $|RI|=|PI|$.

Therefore, the measure of the angle $QIB$ is $$\begin{array}{rl}|\angle QIB|& =180^{\circ }-|\angle AIB|-|\angle JIQ|=180^{\circ }-(90^{\circ }+\frac{\gamma }{2})-\frac{\beta }{2}\\ & =90^{\circ }-(\frac{\beta }{2}+\frac{\gamma }{2})=\frac{\alpha }{2}=|\angle RAI|.\end{array}$$

The measure of the angle $QIB$ could also be determined as follows: Since the inscribed angles $AIP$ and $IPQ$ correspond to chords of equal length, we have $|\angle AIP|=\frac{1}{2}\beta =|\angle IPQ|$. The congruence of alternate angles implies that $AI\parallel PQ$. Hence $|\angle QPB|=|\angle IAB|=\frac{1}{2}\alpha$, and so $|\angle QIB|=\frac{1}{2}\alpha$ as well since they are both inscribed angles corresponding to the chord $QB$.

Since $|\angle QIB|=|\angle RAI|$ and $|\angle QBI|=|\angle RIA|$, it follows that $AIR\sim IBQ$. Therefore, $|AR|/|RI|=|IQ|/|QB|$, so $$|AR|\cdot |QB|=|RI|\cdot |IQ|=|PI{|}^2.$$