# 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 (Problems • Answer Key • Resources) Preceded byProblem 40 Followed byProblem 42 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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