# 1967 AHSME Problems/Problem 37

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## 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.

## See also

 1967 AHSME (Problems • Answer Key • Resources) Preceded byProblem 36 Followed byProblem 38 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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