# Difference between revisions of "1967 AHSME Problems/Problem 37"

## Problem

Segments $AD=10$, $BE=6$, $CF=24$ are drawn from the vertices of triangle $ABC$, each perpendicular to a straight line $RS$, not intersecting the triangle. Points $D$, $E$, $F$ are the intersection points of $RS$ with the perpendiculars. If $x$ is the length of the perpendicular segment $GH$ drawn to $RS$ from the intersection point $G$ of the medians of the triangle, then $x$ is:

$\textbf{(A)}\ \frac{40}{3}\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ \frac{56}{3}\qquad \textbf{(D)}\ \frac{80}{3}\qquad \textbf{(E)}\ \text{undetermined}$

## Solution

$\fbox{A}$

WLOG let $RS$ be the x-axis, or at least horizontal. The three lengths represent the y-coordinates of points $A,B,C$. As $G$ is by definition the average of $A,B,C$, it's coordinates are the average of the coordinates of $A,B,$ and $C$. Hence, the y-coordinate of $G$, which is also the distance from $G$ to $RS$, is the average of the y-coordinates of $A,B,$ and $C$, or $\frac{10+6+24}{3}$, hence the result.