# 1954 AHSME Problems/Problem 35

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## Problem 35

In the right triangle shown the sum of the distances $BM$ and $MA$ is equal to the sum of the distances $BC$ and $CA$. If $MB = x, CB = h$, and $CA = d$, then $x$ equals:

$[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; draw((0,0)--(8,0)--(0,5)--cycle); label("C",(0,0),SW); label("A",(8,0),SE); label("M",(0,5),N); dot((0,3.5)); label("B",(0,3.5),W); label("x",(0,4.25),W); label("h",(0,1),W); label("d",(4,0),S);[/asy]$

$\textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h$

## Solution 1

The question states that $$h+d = x+\sqrt{(x+h)^2+d^2}$$

We move $x$ to the left: $$h+d-x = \sqrt{(x+h)^2+d^2}$$

We square both sides: $$h^2 + d^2 + x^2 - 2xh - 2xd + 2hd = x^2 + 2xh + h^2 + d^2$$

Cancelling and moving terms, we get: $$4xh + 2xd = 2hd$$

Factoring $x$: $$x(4h+2d) = 2hd$$

Isolating for $x$: $$x=\frac{2hd}{4h+2d}=\frac{hd}{2h+d}$$

Therefore, the answer is $\fbox{A}$

## Solution 2

Realize that a 3 - 4 - 5 triangle satisfies these requirements. Checking the answer choices, $\fbox{A}$ is the correct solution.