# 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

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 on 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 on a straight line.

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