# 1950 AHSME Problems/Problem 49

## Problem

A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $1\frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is: $\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches} \qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches} \qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along } BA \qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus}$

## Solution

The locus of the median's endpoint on $BC$ is the circle about $A$ and of radius $1\frac{1}{2}$ inches. The locus of the vertex $C$ is then the circle twice as big and twice as far from $B$, i.e. of radius $3$ inches and with center $4$ inches from $B$ along $BA$ which means that our answer is: $\textbf{(D)}$.

## Solution 2

Let $A(a,y_A)$, $B(b,y_B)$ and $C(x,y)$.

Hence, $D\left(\frac{x+b}{2},\frac{y+y_B}{2}\right)$ is a midpoint of $BC$.

Thus, the equation of needed locus is $$\left(\frac{x+b}{2}-a\right)^2+\left(\frac{y+y_B}{2}-y_A\right)^2=\left(\frac{3}{2}\right)^2,$$ which is equation of the circle: $$(x-(2a-b))^2+(y-(2y_A-y_B))^2=3^2.$$

Thus, D) is valid because $$\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.$$

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