1962 AHSME Problems/Problem 15

Problem

Given triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line, the intersection point of the three medians moves on:

$\textbf{(A)}\ \text{a circle}\qquad\textbf{(B)}\ \text{a parabola}\qquad\textbf{(C)}\ \text{an ellipse}\qquad\textbf{(D)}\ \text{a straight line}\qquad\textbf{(E)}\ \text{a curve here not listed}$

Solution

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Let $CM$ be the median through vertex $C$ , and let $G$ be the point of intersection of the triangle's medians.

Let $CH$ be the altitude of the triangle through vertex $C$ and $GP$ be the distance from $G$ to $AB$ , with the point $P$ laying on $AB$ .

Using Thales' intercept theorem, we derive the proportion:

$\frac{GP}{CH} = \frac{GM}{CM}$

The fraction $\frac{GM}{CM}$ in any triangle is

equal to $\frac{1}{3}$ . Therefore $GP = \frac{CH}{3}$ .

Since the problem states that the vertex $C$ is moving in a straight line, the length of $CH$ is a constant value. That means that the length of $GP$ is also a constant. Therefore the point $G$ is moving in a straight line.

Answer: D

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1962 AHSME Problems/Problem 15

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