# 1950 AHSME Problems/Problem 48

## Problem

A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is: $\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$

## Solution

Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be $ABC$ with $AB=BC=AC=s$. We will call the aforementioned point $P$. Call altitude from $P$ to $BC$ $PA'$. Similarly, we will name the other two altitudes $PB'$ and $PC'$. We can see that $$\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh$$ Where h is the altitude. Multiplying both sides by $2$ and dividing both sides by $s$ gives us $$PA'+PB'+PC'=h$$ The answer is $\textbf{(C)}$

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