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Selection tests for the International Mathematical Olympiad, 2013

Geometry Difficulty 7.4 National Olympiad, round 2 Prove it Saudi Arabia

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

Solution

Let $R$ be the circumradius of triangle $ABC$ . We have, using sine law,
$C_1A = 2R \sin \angle ACP \quad \text{ and } \quad AB_1 = 2R \sin \angle PBA.$

On the other hand, because lines $A_2C_1$ and $A_2B_1$ are tangent to $\omega$ , we have
$\angle AC_1A_2 = \angle ACP \quad \text{ and } \quad \angle A_2B_1A = \angle PBA.$
Applying sine law to triangles $AA_2C_1$ and $AB_1A_2$ , we obtain
$\frac{\sin \angle AA_2C_1}{AC_1} = \frac{\sin \angle AC_1A_2}{AA_2} \quad \text{ and } \quad \frac{\sin \angle AA_2B_1}{AB_1} = \frac{\sin \angle A_2B_1A}{AA_2}.$
Therefore, we have
$\frac{\sin \angle C_1A_2A}{\sin \angle AA_2B_1} = \frac{AC_1 \cdot \sin \angle AC_1A_2}{AB_1 \cdot \sin \angle A_2B_1A} = \frac{\sin^2 \angle ACP}{\sin^2 \angle PBA}$
We have, in a similar way
$\frac{\sin \angle A_1B_2B}{\sin \angle BB_2C_1} = \frac{\sin^2 \angle BAP}{\sin^2 \angle PCB} \quad \text{ and } \quad \frac{\sin \angle B_1C_2C}{\sin \angle CC_2A_1} = \frac{\sin^2 \angle CBP}{\sin^2 \angle PAC}$
But, because $AP$ , $BP$ and $CP$ are concurrent, we have by trigonometric Ceva
$\frac{\sin \angle BAP}{\sin \angle PAC} \cdot \frac{\sin \angle CBP}{\sin \angle PBA} \cdot \frac{\sin \angle ACP}{\sin \angle PCB} = 1.$
We deduce that
$\frac{\sin \angle C_1A_2A}{\sin \angle AA_2B_1} \cdot \frac{\sin \angle A_1B_2B}{\sin \angle BB_2C_1} \cdot \frac{\sin \angle B_1C_2C}{\sin \angle CC_2A_1} = 1.$
This proves, by using the reciprocal of trigonometric Ceva, that lines $AA_2$ , $BB_2$ and $CC_2$ are concurrent.

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