Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Ukraine

Let $I$ be the incenter of triangle $ABC$ . Points $M$ and $N$ are chosen on sides $AB$ and $AC$ respectively such that $M \neq B$ , $N \neq C$ and the points $A$ , $I$ , $M$ , $N$ are cyclic. Prove that $BM + CN = BC$ .

Solution

Візьмемо на стороні $BC$ таку точку $K$ , що $BM = BK$ . Легко бачити, що $\Delta BMI = \Delta BKI$ . Оскільки навколо чотирикутника $ANIM$ можна описати коло, і $\angle MAI = \angle NAI$ , то $MI = NI = KI$ . Зауважимо, що $\angle BKI = \angle BMI = 180^\circ - \angle AMI = \angle ANI$ , $\angle CNI = \angle CKI$ . З того, що $\angle NCI = \angle KCI$ , випливає рівність кутів $CIN$ і $CIK$ . Отже, $\Delta CNI = \Delta CKI$ , $CN = CK$ , $BM + CN = BK + KC = BC$ .

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