# Difference between revisions of "1969 Canadian MO Problems/Problem 10"

## Problem

Let $\displaystyle ABC$ be the right-angled isosceles triangle whose equal sides have length 1. $\displaystyle P$ is a point on the hypotenuse, and the feet of the perpendiculars from $\displaystyle P$ to the other sides are $\displaystyle Q$ and $\displaystyle R$. Consider the areas of the triangles $\displaystyle APQ$ and $\displaystyle PBR$, and the area of the rectangle $\displaystyle QCRP$. Prove that regardless of how $\displaystyle P$ is chosen, the largest of these three areas is at least $\displaystyle 2/9$.

## Solution

Let $\displaystyle AQ=x.$ Because triangles $\displaystyle APQ$ and $\displaystyle BPR$ both contain a right angle and a $\displaystyle 45^\circ$ angle, they are isosceles right triangles. Hence, $\displaystyle PQ=RC=x$ and $\displaystyle QC=PR=BR=1-x.$

Now let's consider when $\displaystyle \frac13 or else one of triangles $\displaystyle APQ$ and $\displaystyle PBR$ will automatically have area greater than $\displaystyle \frac29.$ In this case, $\displaystyle [QCRP]>[ABC]-[APQ]-[PBR]>\frac29.$ Therefore, one of these three figures will always have area greater than $\displaystyle \frac29,$ regardless of where $\displaystyle P$ is chosen.