1954 AMC 12 Problem 35

Question

In the right triangle shown the sum of the distances

\overline{BM}

and

\overline{MA}

is equal to the sum of the distances

\overline{BC}

and

\overline{CA}.

If

\overline{MB}=x,

\overline{CB}=h,

and

\overline{CA}=d,

then

x

equals:

\dfrac{hd}{2h+d}

d-h

\dfrac12d

h+d-\sqrt{2d}

\sqrt{h^2+d^2}-h

Use the Pythagorean theorem to express

\overline{MA}

in terms of

d,h,x

From

x+\sqrt{d^2+(h+x)^2}=h+d,

isolate the radical and square

The condition and the Pythagorean theorem give

x+\sqrt{d^2+(h+x)^2}=h+d.

After isolating the radical and squaring,

d^2+(h+x)^2=(h+d-x)^2.

Cancelling common terms leaves

4hx+2dx=2hd,

so

x=\frac{hd}{2h+d}.

Thus, the correct answer is A.

Problem 35 in other years