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: $\textbf{(D)}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 48	Followed by Problem 50
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1950 AHSME Problems/Problem 49

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