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 $\fbox{A}$

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