2010 AMC 10A Problem 16

Question

Nondegenerate

\triangle ABC

has integer side lengths,

\overline{BD}

is an angle bisector,

AD = 3,

and

DC = 8.

What is the smallest possible value of the perimeter?

30

33

35

36

37

Use the Angle Bisector Theorem

AB:BC=3:8

, and

AC=11

Using the Angle Bisector Theorem, we have that

\dfrac{AB}{3} = \dfrac{BC}{8}

AB = \dfrac{3}{8} BC.

For

AB

and

BC

to be integers, we must have that

BC

is a multiple of

8.

To minimize the perimeter, we can set

BC = 8

and

AB = 3.

This, however, makes the triangle degenerate.

BC

must then be

16

and

AB = 6.

Since

AC = AD + DC = 11,

the perimeter is

16 + 6 + 11 = 33.

Thus, B is the correct answer.

Problem 16 in other years