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:

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

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

$6\sqrt3-8$

$\dfrac{5\sqrt{10}}6$

$\dfrac{25}{12}$

Answer: A

Concepts:coordinate geometry trigonometry right triangle

Difficulty rating: 2000

Small Hint:

Place $C=(0,0),$ $B=(3,0),$ $A=(0,4)$

Big Hint:

The two trisector rays make angles $30^\circ$ and $60^\circ$ with $CB$ ; intersect each with $\frac{x}{3}+\frac{y}{4}=1$

Solution:

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.

1960 AMC 12 Problem 40

Answer: A

Difficulty rating: 2000

Small Hint:

Big Hint:

Solution:

Problem 40 in Other Years