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.

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 36	Followed by Problem 38
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1967 AHSME Problems/Problem 37

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