# 1971 Canadian MO Problems/Problem 8

## Problem

A regular pentagon is inscribed in a circle of radius $r$. $P$ is any point inside the pentagon. Perpendiculars are dropped from $P$ to the sides, or the sides produced, of the pentagon.

a) Prove that the sum of the lengths of these perpendiculars is constant.

b) Express this constant in terms of the radius $r$.

## Solution

Let the pentagon be $ABCDE$, and the perpendiculars from $P$ to sides $AB$, $BC$, $CD$, $DE$, and $EA$ have lengths $p_1$, $p_2$, $p_3$, $p_4$, and $p_5$, respectively.

Since $P$ is inside the pentagon, we have that the area of the pentagon is the sum of the areas of the triangles $ABP$, $BCP$, $CDP$, $DEP$, and $EAP$. This sum is simply equal to $$[ABCDE]=\dfrac{1}{2}(AB\cdot p_1+BC\cdot p_2+CD\cdot p_3+DE\cdot p_4+EA\cdot p_5).$$

This is constant. However, since $ABCDE$ is a regular pentagon, we have that all of its sides are equal to some positive $s$, and $$p_1+p_2+p_3+p_4+p_5=2[ABCDE]/s.$$

Therefore the sum of the lengths of the perpendiculars is constant.

Let $O$ be the center of the pentagon. We know that $\angle AOB=72^{\circ}$, so $[AOB]=(1/2)r^2\sin 72^{\circ}$. We were given that $ABCDE$ is a regular pentagon, so it follows that $$p_1+p_2+p_3+p_4+p_5=(5r^2/s)\sin 72^{\circ}.$$

We now find $s$ in terms of $r$. We know that $\angle OAB=54^{\circ}$, so $[AOB]=(1/2)rs\sin 54^{\circ}$. Therefore $s=r\left(\sin 72^{\circ}/\sin 54^{\circ}\right)$. This shows that $$p_1+p_2+p_3+p_4+p_5=5r\sin 54^{\circ}.$$

Note: The value of $\sin 54^{\circ}$ can be proven to be $\frac{\sqrt{5}+1}{4}$.