HMMT November, 2015

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Let $\triangle ABC$ be a right triangle with right angle $C$ . Let $I$ be the incenter of $ABC$ , and let $M$ lie on $AC$ and $N$ on $BC$ , respectively, such that $M$ , $I$ , $N$ are collinear and $\overline{MN}$ is parallel to $AB$ . If $AB = 36$ and the perimeter of $CMN$ is $48$ , find the area of $ABC$ .

Solution

Solution:

Answer: $252$

Note that $\angle MIA = \angle BAI = \angle CAI$ , so $MI = MA$ . Similarly, $NI = NB$ . As a result, $CM + MN + NC = CM + MI + NI + NC = CM + MA + NB + NC = AC + BC = 48$ .

Furthermore, $AC^2 + BC^2 = 36^2$ . As a result, we have $AC^2 + 2AC \cdot BC + BC^2 = 48^2$ , so $2AC \cdot BC = 48^2 - 36^2 = 12 \cdot 84$ , and so $\dfrac{AC \cdot BC}{2} = 3 \cdot 84 = 252$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.