1956 AMC 12 Problem 38

Question

In a right triangle with sides

a

and

b,

and hypotenuse

c,

the altitude drawn on the hypotenuse is

x.

Then:

ab=x^2

\dfrac1a+\dfrac1b=\dfrac1x

a^2+b^2=2x^2

\dfrac1{x^2}=\dfrac1{a^2}+\dfrac1{b^2}

\dfrac1x=\dfrac ba

Compute the triangle’s area using either the legs or the hypotenuse and its altitude

From

ab=cx,

substitute

c^2=a^2+b^2

and divide by

a^2b^2x^2

Equating two area formulas gives

\frac12ab=\frac12cx,

so

ab=cx.

Squaring and using

c^2=a^2+b^2

yields

a^2b^2=x^2(a^2+b^2).

Dividing by

a^2b^2x^2

gives

\frac1{x^2}=\frac1{a^2}+\frac1{b^2}.

Thus, the correct answer is D.

Problem 38 in other years