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Geometry Difficulty 8.5 Shortlist Prove it Romania

Let $ABC$ be a non-isosceles triangle, and let $I$ be its incenter. Let $\gamma_A$ be the circle through $I$ , tangent to $AB$ and $AC$ and crossing the segment $AI$ , let $\gamma_B$ be the circle through $I$ tangent to $BC$ and $BA$ and crossing the segment $BI$ , and let $\gamma_C$ be the circle through $I$ , tangent to $CA$ and $CB$ and crossing the segment $CI$ . The circles $\gamma_B$ and $\gamma_C$ cross again at $A'$ , the circles $\gamma_C$ and $\gamma_A$ cross again at $B'$ , and the circles $\gamma_A$ and $\gamma_B$ cross again at $C'$ . Prove that the circles $AA'I, BB'I$ and $CC'I$ cross again at a point different from $I$ .
IMO 1997, Longlist

Solution

Let the internal bisectrices of the angles $BAC$ , $CBA$ and $ACB$ cross the circle $ABC$ again at $M_A$ , $M_B$ and $M_C$ , respectively, and let $\gamma_A$ , $\gamma_B$ and $\gamma_C$ be centred at $O_A$ , $O_B$ and $O_C$ , respectively; clearly, $O_A$ , $O_B$ and $O_C$ lie on the segments $AI$ , $BI$ and $CI$ , respectively.

The centre of the circle $AA'I$ is the point where the perpendicular bisectrix of the segment $AI$ crosses the perpendicular bisectrix of the segment $A'I$ . It is a fact that the former is the line $M_B M_C$ ; the latter is, of course, the line $O_B O_C$ , so the lines $M_B M_C$ and $O_B O_C$ cross at the centre of the circle $AA'I$ . Similarly, the lines $M_C M_A$ and $O_C O_A$ cross at the centre of the circle $BB'I$ , and the lines $M_A M_B$ and $O_A O_B$ cross at the centre of the circle $CC'I$ .

Finally, since the triangles $M_A M_B M_C$ and $O_A O_B O_C$ are in perspective from $I$ , the conclusion follows by Desargues' theorem.

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