1960 AMC 12 Problem 40

Question

Given right triangle

ABC

with legs

\overline{BC}=3,

\overline{AC}=4.

Find the length of the shorter angle trisector from

C

to the hypotenuse:

\dfrac{32\sqrt3-24}{13}

\dfrac{12\sqrt3-9}{13}

6\sqrt3-8

\dfrac{5\sqrt{10}}6

\dfrac{25}{12}

Place

C=(0,0),

B=(3,0),

A=(0,4)

The two trisector rays make angles

30^\circ

and

60^\circ

with

CB

; intersect each with

\frac{x}{3}+\frac{y}{4}=1

Place

C=(0,0),

B=(3,0),

A=(0,4).

The hypotenuse has equation

\frac{x}{3}+\frac{y}{4}=1.

The

30^\circ

trisector ray is

y=\frac{x}{\sqrt3}.

Substitution gives

x=\frac{12\sqrt3}{4\sqrt3+3}.

Its length is

\frac{x}{\cos30^\circ}=\frac{2x}{\sqrt3},

hence

\frac{24}{4\sqrt3+3} =\frac{32\sqrt3-24}{13}.

The

60^\circ

ray gives the longer trisector.

Thus, the correct answer is A.

Problem 40 in other years