Problem

1954 AHSME 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


Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.


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