11. Let $A B C D$ be a square of side length 1. $P$ and $Q$ are two points on the plane such that $Q$ is the circumcentre of $\triangle B P C$ and $D$ is the circumcentre of $\triangle P Q A$. Find the largest possible value of $P Q^{2}$. Express the answer in the form $a+\sqrt{b}$ or $a-\sqrt{b}$, where $a$ and $b$ are rational numbers.
(2 marks)
設 $A B C D$ 為正方形, 邊長為 $1 \circ P 、 Q$ 為平面上的兩點, 使得 $Q$ 是 $\triangle B P C$ 的外心、 $D$ 是 $\triangle P Q A$ 的外心。求 $P Q^{2}$ 的最大可能值, 答案以 $a+\sqrt{b}$ 或 $a-\sqrt{b}$形式表示, 其中 $a \vee b$ 為有理數。
11. $2+\sqrt{3}$
11. As $Q$ is the circumcentre of $\triangle B P C$, it lies on the perpendicular bisector of $B C$, and we have $P Q=C Q$. As $D$ is the circumcentre of $\triangle P Q A$, we have $D Q=D A=1$ and hence $Q$ lies on the circle with centre $D$ and radius 1 . These two loci intersect at two points, giving two possible positions of $Q$. We choose the one outside $\triangle A B C$ in order to maximise $P Q(=C Q)$.

Let $E$ and $F$ be the midpoints of $A D$ and $B C$ respectively. Since $\triangle A P Q$ is equilateral with side length 1, we have $E Q=\frac{\sqrt{3}}{2}$. Appling Pythagoras Theorem in $\triangle C F Q$, we have
$$
P Q^{2}=C Q^{2}=C F^{2}+F Q^{2}=\left(\frac{1}{2}\right)^{2}+\left(1+\frac{\sqrt{3}}{2}\right)^{2}=2+\sqrt{3} .
$$