1953 AHSME Problems/Problem 41

Problem

A girls' camp is located $300$ rods from a straight road. On this road, a boys' camp is located $500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is:

$\textbf{(A)}\ 400\text{ rods} \qquad \textbf{(B)}\ 250\text{ rods} \qquad \textbf{(C)}\ 87.5\text{ rods} \qquad \textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these}$

Solution

[asy] draw((0,0)--(6,0)); draw((1,0)--(1,3)--(5,0)); draw((1,3)--(1.875,0)); label("$A$",(1,0),S); label("$B$",(5,0),S); label("$C$",(1.875,0),S); label("$G$",(1,3),NW); label("$r$",(0,0),SW); [/asy] Let $r$ be the straight road, $G$ be the girls' camp, $B$ be the boys' camp, and $C$ be the water canteen. $\overline{AG}$ is the perpendicular from $G$ to $r$. Suppose each rod is one unit long. $AG=300$ and $BG=500$. Since $\angle GAB$ is a right angle, $\triangle GAB$ is a $3-4-5$ right triangle, so $AB=400$.

Let $x$ be the distance from the canteen to the girls' and boys' camps. We have $BC=CG=x$ and $AC=400-x$. Using the Pythagorean Theorem on $\triangle GAC$, we have \[300^2+(400-x)^2=x^2\] Simplifying the left side gives \[90000+160000-800x+x^2=x^2\] Subtracting $x^2$ from both sides and rearranging gives \[800x=250000\] Therefore, $x=\frac{250000}{800}=312.5$. The answer is $\boxed{\textbf{(E)}\ \text{none of these}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 40
Followed by
Problem 42
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