25. In the triangle $A B C, A B=585, B C=520, C A=455$. Let $P, Q$ be points on the side $B C$, and $R \neq A$ the intersection of the line $A Q$ with the circumcircle $\omega$ of the triangle $A B C$. Suppose $P R$ is parallel to $A C$ and the circumcircle of the triangle $P Q R$ is tangent to $\omega$ at $R$. Find $P Q$.
25. Answer: 64
Solution.

First by cosine rule, $\cos C=2 / 7$. Reflect $A$ about the perpendicular bisector of $B C$ to get the point $A^{\prime}$ on $\omega$. Then $A A^{\prime} B C$ is an isosceles trapezoid with $A^{\prime} A$ parallel to $B C$. Thus $A^{\prime} A=B C-2 A C \cos C=520-2 \times 455 \times 2 / 7=260$. Consider the homothety $h$ centred at $R$ mapping the circumcircle of $P Q R$ to $\omega$. We have $h(Q)=A$, and $h(P)=A^{\prime}$ because $P Q$ is parallel to $A^{\prime} A$. Thus $A^{\prime}, P, R$ are collinear and $A A^{\prime} P C$ is a parallelogram. Hence $P C=A A^{\prime}=260$, and $P$ is the midpoint of $B C$. Also $P A^{\prime}=C A=455$. As $P A^{\prime} \times P R=B P \times P C$, we have $455 \times P R=260^{2}$ giving $P R=1040 / 7$. Since the triangles $P Q R$ and $A^{\prime} A R$ are similar, we heve $P Q / A^{\prime} A=R P / R A^{\prime}$. Therefore, $P Q=$ $260 \times(1040 / 7) /(455+1040 / 7)=64$