1987 AMC 12 Problem 30

Question

In the figure,

\triangle ABC

has

\angle A=45^\circ

and

\angle B=30^\circ.

A line

DE,

with

D

on

AB

and

\angle ADE=60^\circ,

divides

\triangle ABC

into two pieces of equal area. (Note: the figure may not be accurate; perhaps

E

is on

CB

instead of

AC.

) The ratio

\frac{AD}{AB}

is

\frac1{\sqrt2}

\frac2{2+\sqrt2}

\frac1{\sqrt3}

\frac1{\sqrt[3]{6}}

\frac1{\sqrt[4]{12}}

Answer: E

Concepts:coordinate geometry triangle area similar scaling trigonometry

Difficulty rating: 2400

Small hint:

Scale so

AB=1

and place

A=(0,0),\ B=(1,0)

Big hint:

Find the height of

C

and the height of

E

in terms of

d=AD

, then equate half the total area to the area of

\triangle ADE

Solution:

Set

AB=1,\ A=(0,0),

and

B=(1,0).

Since

\angle A=45^\circ,

side

AC

lies on

y=x.

The line through

B

making angle

30^\circ

with

BA

meets it at height

h_C=\frac1{1+\sqrt3}.

Let

d=AD.

The ray

DE

has slope

-\sqrt3,

so its intersection with

y=x

has height

h_E=\frac{\sqrt3\,d}{1+\sqrt3}.

Equal areas require

\frac12d h_E=\frac12\left(\frac12h_C\right).

Therefore

\begin{aligned} \frac{\sqrt3\,d^2}{2(1+\sqrt3)} &=\frac1{4(1+\sqrt3)},\\ d^2&=\frac1{2\sqrt3} =\frac1{\sqrt{12}}. \end{aligned}

Hence

\frac{AD}{AB}=d=\frac{1}{\sqrt[4]{12}}.

Thus the correct answer is E.