19. The diagram shows a circle of radius 1 touching three sides of a $2 \times 4$ rectangle. A diagonal of the rectangle intersects the circle at $P$ and $Q$, as shown.
What is the length of the chord $P Q$ ?

A $\sqrt{5}$
B $\frac{4}{\sqrt{5}}$
C $\sqrt{5}-\frac{2}{\sqrt{5}}$
D $\frac{5 \sqrt{5}}{6}$
E 2
Solution
B

We let the the vertices of the rectangle be $K, L, M$, and $N$ as shown in the figure. We let $O$ be the centre of the circle, and $R$ be the point where the perpendicular from $O$ to the chord $P Q$ meets the chord.

We note first that, by Pythagoras' Theorem applied to the rightangled triangle $K L M, K M^{2}=K L^{2}+L M^{2}=4^{2}+2^{2}=20$.

Therefore $K M=\sqrt{20}=2 \sqrt{5}$.
The radius $O Q$ is parallel to $K L$. (You are asked to show this in Problem 19.1.) Therefore the alternate angles $\angle O Q R$ and $\angle M K L$ are equal. It follows that the right-angled triangles $O Q R$ and $M K L$ are similar. In particular,
$$
\frac{R Q}{O Q}=\frac{K L}{K M} .
$$

It follows that
$$
R Q=\frac{K L}{K M} \times O Q=\frac{4}{2 \sqrt{5}} \times 1=\frac{2}{\sqrt{5}} .
$$

The right-angled triangles $O R Q$ and $O R P$ are congruent as they share the side $O R$, and their hypotenuses $O Q$ and $O P$ are equal because they are radii of the same circle. It follows that $P R=R Q=\frac{2}{\sqrt{5}}$. We therefore conclude that
$$
P Q=P R+R Q=\frac{2}{\sqrt{5}}+\frac{2}{\sqrt{5}}=\frac{4}{\sqrt{5}} .
$$