# 1969 Canadian MO Problems/Problem 4

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## Problem

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, $\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}$.

## Solution 1

Let a side of the triangle be $s$ and let $[ABC]$ denote the area of $ABC.$ Note that because $2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],$ $\frac{\sqrt3}{2}s^2=s(PD+PE+PF).$ Dividing both sides by $s$, the sum of the perpendiculars from $P$ equals $PD+PE+PF=\frac{\sqrt3}{2}s.$ (It is independant of point $P$) Because the sum of the sides is $3s$, the ratio is always $\cfrac{s\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.$

## Solution 2

Let a be the side of the triangle. By Viviani's theorem (which has a similar proof to solution 1) PD + PE + PF = h where h is the height of the equilateral triangle. Now it is not hard to see that the ratio h : 3 * a = 1 / 2 root 3 from 30 - 60 - 90 triangles or trigonometry. ~AK2006

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