1981 AMC 12 Problem 27

Question

In the adjoining figure triangle

ABC

is inscribed in a circle. Point

D

lies on

\overset{\frown}{AC}

with

\overset{\frown}{DC}=30^\circ,

and point

G

lies on

\overset{\frown}{BA}

with

\overset{\frown}{BG}\gt\overset{\frown}{GA}.

Side

AB

and side

AC

each have length equal to the length of chord

DG,

and

\angle CAB=30^\circ.

Chord

DG

intersects sides

AC

and

AB

at

E

and

F,

respectively. The ratio of the area of

\triangle AFE

to the area of

\triangle ABC

is

\frac{2-\sqrt3}{3}

\frac{2\sqrt3-3}{3}

7\sqrt3-12

3\sqrt3-5

\frac{9-5\sqrt3}{3}

Answer: C

Concepts:arc chord area ratio

Difficulty rating: 2260

Small hint:

Use equal chords to identify the relevant equal arcs and isosceles triangles

Big hint:

Normalize

AB=AC=DG=1

and find

AE

from a

30^\circ

-

60^\circ

-

90^\circ

triangle

Solution:

Scale so

AB=AC=DG=1.

The equal-chord arc relations show that triangle

DEC

is

30^\circ

-

60^\circ

-

90^\circ

and

AE=DE.

Put

AE=DE=x.

Then

CE=1-x=\frac{2x}{\sqrt3},

so

x=2\sqrt3-3.

Equal chords also give

AF=FG=EF=\frac{1-x}{2}.

Hence

\begin{aligned} \frac{[AFE]}{[ABC]} &=\frac{x(1-x)}2\\ &=7\sqrt3-12. \end{aligned}

Therefore, the correct answer is C.