In $\triangle A B C, \angle A=90^{\circ}$ and $\angle B=\angle C=45^{\circ}$. $P$ is a point on $B C$ and $Q, R$ are the circumcentres of $\triangle A P B$ and $\triangle A P C$ respectively. If $B P=\sqrt{2}$ and $Q R=2$, find $P C$.
(2 marks)
In $\triangle A B C, \angle A=90^{\circ}$ and $\angle B=\angle C=45^{\circ}$. $P$ is a point on $B C$ and $Q, R$ are the circumcentres of $\triangle A P B$ and $\triangle A P C$ respectively. If $B P=\sqrt{2}$ and $Q R=2$, find $P C$.
18. $\sqrt{6}$
18. Let $P^{\prime}$ be the image of $P$ by an anticlockwise rotation about $A$ through $90^{\circ}$ so that $B$ goes to C. As $\angle A P^{\prime} C + \angle A P C = \angle A P B + \angle A P C = 180^{\circ}$, $A P C P^{\prime}$ is a cyclic quadrilateral. So the circumcentre $R$ is the image of the circumcentre $Q$ under the rotation. Thus $A Q R$ is also a right-angled isosceles triangle, and so $A Q P R$ is a square and $A P = Q R = 2$. Let $A B = x$. Applying the cosine law in $\triangle A B P$, we have
$$
x^{2} + (\sqrt{2})^{2} - 2 x(\sqrt{2}) \cos 45^{\circ} = 2^{2} \text{.}
$$

This gives $x^{2} - 2 x - 2 = 0$ and hence $x = 1 + \sqrt{3}$. It follows that $P C = \sqrt{2}$ and $A B - B P = \sqrt{2} + \sqrt{6} - \sqrt{2} = \sqrt{6}$