Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Silk Road Mathematics Competition

In triangle $ABC$ let $A_0, B_0, C_0$ are the midpoints of the sides $BC, CA, AB$ , respectively, and $A_1, B_1, C_1$ are the midpoints (by length) of the broken lines $BAC, CBA, BCA$ , respectively. Prove that the lines $A_0A_1, B_0B_1, C_0C_1$ are concurrent.

Solution

W.l.o.g. assume that $AC \le AB$ . Then $A_1$ belongs to the segment $AB$ , moreover, it lies between $A$ and $C_0$ . Then notice that $AC + AA_1 = BA_1 \iff AC + AC_0 - A_1C_0 = C_0B + A_1C_0 \iff A_1C_0 = \frac{1}{2}AC \Rightarrow A_1C_0 = A_0C_0$ i.e. $\angle C_0A_1A_0 = \angle A_1A_0C_0 \Rightarrow \angle A_1A_0C_0 = \angle A_1A_0B_0$ since $AB \parallel A_0B_0$ . So, $A_0A_1$ is bisector of the angle $\angle C_0A_0B_0$ .

Similarly, $B_0B_1$ is bisector of $\angle C_0B_0A_0$ and $C_0C_1$ is bisector of $\angle A_0C_0B_0$ . So the lines $A_0A_1, B_0B_1$ and $C_0C_1$ intersect in the incenter of the triangle $A_0B_0C_0$ .

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