HongKong 2022-23 IMO Selection Tests

Answer: $\frac{5-2\sqrt{2}}{6}$

Set the square in the first quadrant of the coordinate plane with $B$ at the origin and $C$ at $(1,0)$. Then the coordinates of $P$ and $Q$ are $(0,\frac{1}{\sqrt{2}})$ and $(\frac{1}{\sqrt{2}},0)$ respectively.

As $PC$ has slope $-\frac{1}{\sqrt{2}}$, the slope of $BN$ is $\sqrt{2}$. Let $N=(n,\sqrt{2}n)$. Considering the slope of $CN$, we have $\frac{\sqrt{2}n}{n-1}=-\frac{1}{\sqrt{2}}$, which gives $n=\frac{1}{3}$.



Hence $N=(\frac{1}{3},\frac{\sqrt{2}}{3})$ and so $$N{Q}^2={\left(\frac{1}{3}-\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2=\frac{5-2\sqrt{2}}{6}$$