2000 AMC 12 Problem 19

Question

In triangle

ABC,

AB = 13,

BC = 14,

and

AC = 15.

Let

D

denote the midpoint of

\overline{BC}

and let

E

denote the intersection of

\overline{BC}

with the bisector of angle

BAC.

Which of the following is closest to the area of triangle

ADE?

2

2.5

3

3.5

4

Answer: C

Concepts:Heron’s Formula angle bisector theorem triangle area

Difficulty rating: 1810

Small hint:

By Heron’s formula the area of

\triangle ABC

is

84,

so the altitude from

A

to

BC

is

12

Big hint:

D

is at distance

7

from

B

; the bisector foot

E

satisfies

BE : EC = 13 : 15

Solution:

By Heron’s formula, the area of

\triangle ABC

is

\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,

so the altitude from

A

to

BC

is

\dfrac{2 \cdot 84}{14} = 12.

The midpoint

D

is

7

from

B.

The bisector from

A

meets

BC

at

E

with

BE : EC = AB : AC = 13 : 15,

so

BE = 14 \cdot \dfrac{13}{28} = 6.5.

Both

D

and

E

lie on

BC,

so

\triangle ADE

has base

DE = 7 - 6.5 = 0.5

and altitude

12,

giving area

\tfrac12 \cdot 0.5 \cdot 12 = 3.

Thus, the correct answer is C.