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

## Problem

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

## Solution

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

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

 1969 Canadian MO (Problems) Preceded byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed byLast question