## Problem 4

$\mathrm{P}$ lies between the rays $\mathrm{OA}$ and $\mathrm{OB}$. Find $\mathrm{Q}$ on $\mathrm{OA}$ and $\mathrm{R}$ on $\mathrm{OB}$ collinear with $\mathrm{P}$ so that $1 / \mathrm{PQ}+1 / \mathrm{PR}$ is as large as possible.


## Solution

Let the line through $\mathrm{P}$ parallel to $\mathrm{OA}$ meet $\mathrm{OB}$ at $\mathrm{C}$. Note that $\mathrm{C}$ is fixed. Let the line through $\mathrm{C}$ parallel to $\mathrm{QR}$ meet $\mathrm{OP}$ at $\mathrm{D}$. $\mathrm{D}$ varies as $\mathrm{Q}$ varies. Triangles $\mathrm{ODC}, \mathrm{OPR}$ are similar, so $\mathrm{CD} / \mathrm{PR}=\mathrm{OD} / \mathrm{OP}$. Also triangles OPQ and $\mathrm{PDC}$ are similar, so $\mathrm{CD} / \mathrm{PQ}=\mathrm{DP} / \mathrm{OP}$. Adding, $\mathrm{CD} / \mathrm{PR}+\mathrm{CD} / \mathrm{PQ}=1$. Hence we maximise $1 / \mathrm{PQ}+1 / \mathrm{PR}$ by making $\mathrm{CD}$ as small as possible. That is achieved by making angle $\mathrm{CDP}=90^{\circ}$ and hence QR perpendicular to OP.

