1980 AMC 12 Problem 23

Question

Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths

\sin x

and

\cos x,

where

x

is a real number such that

0\lt x\lt\frac\pi2.

The length of the hypotenuse is

\frac43

\frac32

\frac{3\sqrt5}{5}

\frac{2\sqrt5}{3}

not uniquely determined by the given information

Put the hypotenuse on the

x

-axis and assign coordinates to the right-angle vertex

Add the squares of the two trisector-segment lengths so the unknown horizontal position cancels

Let the hypotenuse have endpoints

(0,0)

and

(h,0),

and let the right-angle vertex be

(u,v).

The right-angle condition gives

u^2+v^2=hu.

The squared distances to

(\frac{h}{3},0)

and

(\frac{2h}{3},0)

sum to

\begin{aligned} &(u-\frac{h}{3})^2+v^2\\ &\quad+(u-\frac{2h}{3})^2+v^2 =\frac{5h^2}{9}. \end{aligned}

This also equals

\sin^2x+\cos^2x=1.

Hence

h=\frac{3}{\sqrt5}=\frac{3\sqrt5}{5}.

Therefore, the correct answer is C.

Problem 23 in other years