2011 AMC 12A Problem 13

Problem 13 of 25IntermediateGeometry

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?

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Solution

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.

More practice

Concepts: incircle, incenter, and inradius · parallel lines · isosceles triangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.