Selection tests for the International Mathematical Olympiad 2013

Question

Triangle ABC is inscribed in circle . Point P lies inside triangle ABC. Lines AP, BP and CP intersect again at points A_1, B_1 and C_1 (other than A, B, C), respectively. The tangent lines to at A_1 and B_1 intersect at C_2. The tangent lines to at B_1 and C_1 intersect at A_2. The tangent lines to at C_1 and A_1 intersect at B_2. Prove that the lines AA_2, BB_2 and CC_2 are concurrent. FIGURE_0

Accepted Answer

Let R be the circumradius of triangle ABC. We have, using sine law, C_1A = 2R ACP and AB_1 = 2R PBA. FIGURE_0 On the other hand, because lines A_2C_1 and A_2B_1 are tangent to , we have AC_1A_2 = ACP and A_2B_1A = PBA. Applying sine law to triangles AA_2C_1 and AB_1A_2, we obtain AA_2C_1AC_1 = AC_1A_2AA_2 and AA_2B_1AB_1 = A_2B_1AAA_2. Therefore, we have C_1A_2A AA_2B_1 = AC_1 AC_1A_2AB_1 A_2B_1A = ^2 ACP^2 PBA We have, in a similar way A_1B_2B BB_2C_1 = ^2 BAP^2 PCB and B_1C_2C CC_2A_1 = ^2 CBP^2 PAC But, because AP, BP and CP are concurrent, we have by trigonometric Ceva BAP PAC CBP PBA ACP PCB = 1. We deduce that C_1A_2A AA_2B_1 A_1B_2B BB_2C_1 B_1C_2C CC_2A_1 = 1. This proves, by using the reciprocal of trigonometric Ceva, that lines AA_2, BB_2 and CC_2 are concurrent.