Philippine Mathematical Olympiad

Solution:

Let $C$ be the intersection of the circle through $C_1, C_2$ and $A$, and the bisector of $\angle C_1 A C_2$. It suffices to show that $C S = C A$. Let $\angle C_1 A C_2 = C_1 B C_2 = 2 \alpha$ (fixed), $\angle C_1 P B = \angle C_1 B P = 2 \beta$ and $\angle C_2 Q B = \angle C_2 B Q = 2 \gamma$. Since
$$\begin{aligned} \angle C_1 R C_2 & = 180^\circ - \angle C_1 P B - \angle C_2 Q B = 180^\circ - \angle C_1 B P - \angle C_2 B Q \\ & = \angle C_1 B C_2 = 2 \alpha \end{aligned}$$
then $R$ is concyclic with $A, C_1, C_2$ and $C$. Also, since $R I$ bisects $\angle C_1 R C_2$, then $\angle C_1 R I = \alpha = \angle C_1 R C$, so $C$ is collinear with $R$ and $I$. We note too that by considering the angles of $\triangle P Q R$, $\alpha + \beta + \gamma = 90^\circ$.

Let the perpendicular bisectors of $P I$ and $Q I$ meet $P Q$ at $M$ and $N$, respectively. Since $S P = S I = S Q$, then $\angle S I M = \angle S P M = \angle S Q N = \angle S I N$ and so $S I$ bisects $\angle M I N$, where
$$\begin{aligned} \angle M I N & = 180^\circ - (\angle M I P + \angle M P I) - (\angle N I Q + \angle N Q I) \\ & = 180^\circ - 2 \beta - 2 \gamma = 2 \alpha . \end{aligned}$$

Thus, $\alpha = \angle S I M = \angle S I N$. Consequently, $S$ is collinear with $R, I$ and $C$ since $\angle P I S = \angle P I M + \angle S I M = \beta + \alpha = \angle I P R + \angle I R P = 180^\circ - \angle R I P$. Furthermore, since $\angle S Q P = \angle S I N = \alpha = \angle S R P$, then $P R Q S$ is cyclic.

By Ptolemy's Theorem applied to $C_1 A C_2 C$,
$$\begin{aligned} C A \cdot C_1 C_2 & = C C_1 \cdot A C_2 + C C_2 \cdot A C_1 = C C_1 (A C_1 + A C_2) \\ & = C C_1 (P C_1 + Q C_2) = C C_1 (R P - R C_1 + R Q - R C_2) \\ & = C C_1 (R P + R Q) - C C_1 (R C_1 + R C_2) \end{aligned}$$
We use Ptolemy's Theorem next on $C_1 R C_2 C$ and $P R Q S$. Since $\triangle C_1 C C_2 \sim \triangle P S Q$ (both are isosceles with base angle $\alpha$ ),
$$\begin{aligned} C A & = \frac{S P}{P Q}(R P + R Q) - \frac{C C_1}{C_1 C_2}(R C_1 + R C_2) \\ & = \frac{S Q \cdot R P + S P \cdot R Q}{P Q} - \frac{C C_2 \cdot R C_1 + C C_1 \cdot R C_2}{C_1 C_2} \\ & = \frac{R S \cdot P Q}{P Q} - \frac{R C \cdot C_1 C_2}{C_1 C_2} = R S - R C = C S . \end{aligned}$$