12. $A B C D$ is a square of side length 1 . $P$ is a point on $A C$ and the circumcircle of $\triangle B P C$ cuts $C D$ at $Q$. If the area of $\triangle C P Q$ is $\frac{6}{25}$, find $C Q$.
$A B C D$ 是個邊長為 1 的正方形。 $P$ 是 $A C$ 上的一點, 且 $\triangle B P C$ 的外接圓交 $C D$於 $Q$ 。若 $\triangle C P Q$ 的面積為 $\frac{6}{25}$, 求 $C Q$ 。
12. $\frac{3}{5}$
12. Since $B P Q C$ is a cyclic quadrilateral with $\angle C=90^{\circ}$, we have $\angle B P Q=90^{\circ}$. Also, $P B=P Q$ since $\angle P C B=$ $\angle P C Q=45^{\circ}$. By Ptolemy's Theorem, we have
$$
P B \times Q C+P Q \times B C=P C \times B Q .
$$

Since $B Q=\sqrt{2} P B=\sqrt{2} P Q$ and $B C=1$, this simplifies to $Q C+1=\sqrt{2} P C$. Let $C Q=y$. Then $P C=\frac{y+1}{\sqrt{2}}$ and considering the area of $\triangle C P Q$ gives $\frac{1}{2}\left(\frac{y+1}{\sqrt{2}}\right) y \sin 45^{\circ}=\frac{6}{25}$, or $(5 y+8)(5 y-3)=0$, so that $y=\frac{3}{5}$.