Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Let ABC\triangle ABC be a right triangle with right angle CC. Let II be the incenter of ABCABC, and let MM lie on ACAC and NN on BCBC, respectively, such that MM, II, NN are collinear and MN\overline{MN} is parallel to ABAB. If AB=36AB = 36 and the perimeter of CMNCMN is 4848, find the area of ABCABC.

Solution

Solution:

Answer: 252252

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

Furthermore, AC2+BC2=362AC^2 + BC^2 = 36^2. As a result, we have AC2+2ACBC+BC2=482AC^2 + 2AC \cdot BC + BC^2 = 48^2, so 2ACBC=482362=12842AC \cdot BC = 48^2 - 36^2 = 12 \cdot 84, and so ACBC2=384=252\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.