1954 AHSME Problems/Problem 35

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.

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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