1980 AMC 12 Problem 21

Question

In triangle

ABC,

\angle CBA=72^\circ,

E

is the midpoint of side

AC,

and

D

is a point on side

BC

such that

2BD=DC;

AD

and

BE

intersect at

F.

The ratio of the area of

\triangle BDF

to the area of quadrilateral

FDCE

is

\frac15

\frac14

\frac13

\frac25

none of these

Area ratios are unchanged by an affine transformation, so convenient coordinates may be used

Find the ratio

AF:FD

, then compare the small triangles with

\triangle ABC

Scale the total area of

\triangle ABC

to

1.

Since

BD:DC=1:2,

we have

[ABD]=\frac{1}{3}

and

[ACD]=\frac{2}{3}.

The midpoint and trisection conditions give

AF:FD=3:1.

Hence

[BDF]=\frac14[ABD]=\frac1{12}.

Also

\begin{aligned} [AEF] &=\frac{AE}{AC}\frac{AF}{AD}[ACD]\\ &=\frac12\cdot\frac34\cdot\frac23\\ &=\frac14. \end{aligned}

Therefore

[FDCE]=\frac{2}{3}-\frac{1}{4}=\frac{5}{12},

and the requested ratio is

\frac{\frac{1}{12}}{\frac{5}{12}}=\frac{1}{5}.

Therefore, the correct answer is A.

Problem 21 in other years