AMC12 2011 A

AMC12 2011 A · Q13

AMC12 2011 A · Q13. It mainly tests Angle chasing, Triangles (properties).

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 $\triangle AMN?$

三角形 $ABC$ 的边长为 $AB = 12, BC = 24,$ 且 $AC = 18$。过 $\triangle ABC$ 的内心且平行于 $\overline{BC}$ 的直线与 $\overline{AB}$ 交于 $M$，与 $\overline{AC}$ 交于 $N$。求 $\triangle AMN$ 的周长。

(A) 27 27

(B) 30 30

(D) 36 36

(E) 42 42

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $O$ be the incenter of $\triangle{ABC}$. Because $\overline{MO} \parallel \overline{BC}$ and $\overline{BO}$ is the angle bisector of $\angle{ABC}$, we have \[\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}\] It then follows due to alternate interior angles and base angles of isosceles triangles that $MO = MB$. Similarly, $NO = NC$. The perimeter of $\triangle{AMN}$ then becomes \begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{(B)} \end{align*}

设 $O$ 为 $\triangle{ABC}$ 的内心。因为 $\overline{MO} \parallel \overline{BC}$ 且 $\overline{BO}$ 是 $\angle{ABC}$ 的角平分线，所以 \[\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}\] 由内错角相等与等腰三角形的底角性质可得 $MO = MB$。同理，$NO = NC$。因此 $\triangle{AMN}$ 的周长为 \begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{(B)} \end{align*}

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