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 $\triangle AMN?$

$27$

$30$

$33$

$36$

$42$

Answer: B

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

Difficulty rating: 1600

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.

2011 AMC 12A Problem 13

Answer: B

Difficulty rating: 1600

Solution:

Problem 13 in Other Years