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Problem 760

AMC 12 late, AIME early

Geometry Difficulty 4.1 Prove it Junior Macedonian Mathematical Olympiad · North Macedonia

Let $\triangle ABC$ be an equilateral triangle. Let $C_1$ and $C_2$ be on $AB$ , $B_1$ and $B_2$ on $AC$ and $A_1$ and $A_2$ on $BC$ such that $\overline{A_1A_2} = \overline{B_1B_2} = \overline{C_1C_2}$ . Let $A_2B_1$ and $B_2C_1$ , $B_2C_1$ and $C_2A_1$ , $C_2A_1$ and $A_2B_1$ intersect at $E$ , $F$ , $G$ correspondently. Prove that the triangle formed by the segments $B_1A_2$ , $A_1C_2$ and $C_1B_2$ is similar to $\triangle EFG$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let us denote the triangle formed by the segments $B_1A_2$ , $A_1C_2$ and $C_1B_2$ with $\triangle A_3B_3C_3$ . Let $P$ be a point of the interior of the triangle $\triangle EFG$ such that $C_1C_2PB_2$ is a parallelogram. Then $\triangle B_2PB_1$ is equilateral, hence $PA_1A_2B_1$ is a parallelogram. From the above observations we get that $PC_2 \parallel EF$ , $PA_1 \parallel EG$ . Now it's obvious that $\triangle PC_2A_1 \sim \triangle EFG$ and because $\triangle PC_2A_1 \cong \triangle A_3B_3C_3$ we conclude that $\triangle A_3B_3C_3 \sim \triangle EFG$ .

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