1969 Canadian MO Problems/Problem 4

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Problem

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

Solution

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


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