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
$$\begin{aligned} |\angle QIB| &= 180^\circ - |\angle AIB| - |\angle JIQ| = 180^\circ - (90^\circ + \frac{\gamma}{2}) - \frac{\beta}{2} \\ &= 90^\circ - (\frac{\beta}{2} + \frac{\gamma}{2}) = \frac{\alpha}{2} = |\angle RAI|. \end{aligned}$$

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
$$|AR| \cdot |QB| = |RI| \cdot |IQ| = |PI|^2.$$