Maths Olympiad Prep

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, 2013

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

Triangle ABCABC is inscribed in circle ω\omega. Point PP lies inside triangle ABCABC. Lines APAP, BPBP and CPCP intersect ω\omega again at points A1A_1, B1B_1 and C1C_1 (other than AA, BB, CC), respectively. The tangent lines to ω\omega at A1A_1 and B1B_1 intersect at C2C_2. The tangent lines to ω\omega at B1B_1 and C1C_1 intersect at A2A_2. The tangent lines to ω\omega at C1C_1 and A1A_1 intersect at B2B_2. Prove that the lines AA2AA_2, BB2BB_2 and CC2CC_2 are concurrent.

Solution

Let RR be the circumradius of triangle ABCABC. We have, using sine law,
C1A=2Rsin⁡∠ACP and AB1=2Rsin⁡∠PBA. C_1A = 2R \sin \angle ACP \quad \text{ and } \quad AB_1 = 2R \sin \angle PBA.
Figure 1
On the other hand, because lines A2C1A_2C_1 and A2B1A_2B_1 are tangent to ω\omega, we have
∠AC1A2=∠ACP and ∠A2B1A=∠PBA. \angle AC_1A_2 = \angle ACP \quad \text{ and } \quad \angle A_2B_1A = \angle PBA.
Applying sine law to triangles AA2C1AA_2C_1 and AB1A2AB_1A_2, we obtain
sin⁡∠AA2C1AC1=sin⁡∠AC1A2AA2 and sin⁡∠AA2B1AB1=sin⁡∠A2B1AAA2. \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
sin⁡∠C1A2Asin⁡∠AA2B1=AC1⋅sin⁡∠AC1A2AB1⋅sin⁡∠A2B1A=sin⁡2∠ACPsin⁡2∠PBA \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
sin⁡∠A1B2Bsin⁡∠BB2C1=sin⁡2∠BAPsin⁡2∠PCB and sin⁡∠B1C2Csin⁡∠CC2A1=sin⁡2∠CBPsin⁡2∠PAC \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 APAP, BPBP and CPCP are concurrent, we have by trigonometric Ceva
sin⁡∠BAPsin⁡∠PAC⋅sin⁡∠CBPsin⁡∠PBA⋅sin⁡∠ACPsin⁡∠PCB=1. \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
sin⁡∠C1A2Asin⁡∠AA2B1⋅sin⁡∠A1B2Bsin⁡∠BB2C1⋅sin⁡∠B1C2Csin⁡∠CC2A1=1. \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 AA2AA_2, BB2BB_2 and CC2CC_2 are concurrent.

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