1959 AMC 12 Problem 10

Question

In triangle

ABC,

with

\overline{AB}=\overline{AC}=3.6,

a point

D

is taken on

\overline{AB}

at a distance

1.2

from

A.

Point

D

is joined to point

E

in the prolongation of

\overline{AC}

so that triangle

AED

is equal in area to triangle

ABC.

Then

\overline{AE}

equals:

4.8

5.4

7.2

10.8

12.6

Compare the altitudes from

B

and

D

to the line

AC

Because

\frac{AD}{AB}=\frac{1}{3},

the altitude from

D

is one-third the altitude from

B

Triangles

AED

and

ABC

use bases on the same line

AC.

Since

\frac{AD}{AB}=\frac{1.2}{3.6}=\frac13,

the perpendicular distance from

D

to line

AC

is one-third the corresponding distance from

B.

Equality of areas therefore requires

AE\left(\frac h3\right)=AC(h),

so

AE=3AC=10.8.

Thus, the correct answer is D.

Problem 10 in other years