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Croatian Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it Croatia

A point $T$ is chosen inside the triangle $ABC$ . Let $A_1, B_1$ and $C_1$ be the reflections of $T$ across the lines $BC, CA$ and $AB$ , respectively. The lines $A_1T, B_1T$ and $C_1T$ intersect the circle $k$ circumscribed to the triangle $A_1B_1C_1$ again at $A_2, B_2$ and $C_2$ , respectively.
Prove that the lines $AA_2, BB_2$ and $CC_2$ are concurrent on $k$ .
(IMO Shortlist 2018)

Solution

Let $K$ be the intersection of $CC_2$ and $k$ .

Since $CB$ and $CA$ are the bisectors of $\overline{TA_1}$ and $\overline{TB_1}$ , respectively, the point $C$ is the circumcentre of the triangle $A_1TB_1$ . Hence,
$\triangle(CA_1, CB) = \triangle(CB, CT) = \triangle(B_1A_1, B_1T) = \triangle(B_1A_1, B_1B_2).$
Observing the circle $k$ , we have $\triangle(B_1A_1, B_1B_2) = \triangle(C_2A_1, C_2B_2)$ and $\triangle(CA_1, CB) = \triangle(B_1A_1, B_1B_2) = \triangle(C_2A_1, C_2B_2)$ . Similarly, we get $\triangle(BA_1, BC) = \triangle(B_2A_1, B_2C_2)$ , hence the triangles $A_1BC$ and $A_1B_2C_2$ are similar, so as the triangles $A_1BB_2$ and $A_1CC_2$ , from which it follows that $\triangle(C_2C, C_2A_1) = \triangle(B_2B, B_2A_1)$ and the point $K$ lies on $BB_2$ .

Analogously, we show that the point $K$ lies on $AA_2$ , and the proof is finished.

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