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Geometry Difficulty 7.4 National olympiad, round 2 Prove it Hong Kong

$ABCD$ is a square with side length $1$ . $P$ and $Q$ are points on $AB$ and $BC$ respectively such that $BP = BQ = \frac{1}{\sqrt{2}}$ . $N$ is the foot of perpendicular from $B$ to $CP$ . Find $NQ^2$ .

Solution

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
$NQ^2 = \left(\frac{1}{3} - \frac{1}{\sqrt{2}}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2 = \frac{5 - 2\sqrt{2}}{6}$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.