2011 AMC 12A Problem 13

Question

Triangle

ABC

has side-lengths

AB = 12,

BC = 24,

and

AC = 18.

The line through the incenter of

\triangle ABC

parallel to

\overline{BC}

intersects

\overline{AB}

at

M

and

\overline{AC}

at

N.

What is the perimeter of

\triangle AMN?

27

30

33

36

42

Let

I

be the incenter. Since

MN \parallel BC,

the angle

\angle MIB

equals

\angle IBC

That makes

\triangle MBI

isosceles with

MB = MI,

and similarly

NC = NI

Let

I

be the incenter. Because

\overline{BI}

bisects

\angle B

and

MN \parallel BC,

alternate angles give

\angle MIB = \angle IBC = \angle MBI,

so

\triangle MBI

is isosceles with

MB = MI.

Similarly

NC = NI.

Therefore the perimeter of

\triangle AMN

is

\begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30. \end{gathered}

Thus, the correct answer is B.

Problem 13 in other years