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Turkey — Team Selection Test

Geometry Difficulty 9.0 IMO level Prove it Turkey

A circle ω\omega with center PP intersects the sides BC,CA,ABBC, CA, AB of a triangle ABCABC at points A1A_1 and A2,B1A_2, B_1 and B2,C1B_2, C_1 and C2C_2, respectively. Let A′A' be the circumcenter of the triangle A1A2PA_1A_2P, B′B' be the circumcenter of the triangle B1B2PB_1B_2P and C′C' be the circumcenter of the triangle C1C2PC_1C_2P. Prove that the lines AA′,BB′AA', BB' and CC′CC' are concurrent.

Solution

Figure 1
Let A3,B3,C3A_3, B_3, C_3 be the feet of the perpendiculars from PP to the sides BC,CA,ABBC, CA, AB respectively. Clearly, A′,B′,C′A', B', C' are on the lines PA3,PB3,PC3PA_3, PB_3, PC_3, respectively. By the Pythagoras Theorem, we have
A′P2−A′A32=A′A12−A′A32=A1P2−A3P2=A1P2−(A3A′+A′P)2 A'P^2 - A'A_3^2 = A'A_1^2 - A'A_3^2 = A_1P^2 - A_3P^2 = A_1P^2 - (A_3A' + A'P)^2
and hence PA′⋅PA3=PA12/2PA' \cdot PA_3 = PA_1^2/2. Similarly we get that PB′⋅PB3=PB12/2PB' \cdot PB_3 = PB_1^2/2 and PC′⋅PC3=PC12/2PC' \cdot PC_3 = PC_1^2/2. Therefore we conclude that PA′⋅PA3=PB′⋅PB3=PC′⋅PC3PA' \cdot PA_3 = PB' \cdot PB_3 = PC' \cdot PC_3. Hence the quadrilaterals A3C3C′A′,B3A3A′B′,C3B3B′C′A_3C_3C'A', B_3A_3A'B', C_3B_3B'C' are cyclic. By the sine law in triangles A3B3A′A_3B_3A' and AC3A′AC_3A', we have

A′B3AA′=sin⁡∠A′AB3cos⁡∠A′B3PandA′C3AA′=sin⁡∠A′AC3cos⁡∠A′C3P \frac{A'B_3}{AA'} = \frac{\sin \angle A'AB_3}{\cos \angle A'B_3P} \quad \text{and} \quad \frac{A'C_3}{AA'} = \frac{\sin \angle A'AC_3}{\cos \angle A'C_3P}
Hence we get
sin⁡∠A′AB3sin⁡∠A′AC3=A′B3A′C3⋅cos⁡∠A′B3Pcos⁡∠A′C3P \frac{\sin \angle A'AB_3}{\sin \angle A'AC_3} = \frac{A'B_3}{A'C_3} \cdot \frac{\cos \angle A'B_3P}{\cos \angle A'C_3P}
Similarly, we get that
sin⁡∠B′BC3sin⁡∠B′BA3=B′C3B′A3⋅cos⁡∠B′C3Pcos⁡∠B′A3P \frac{\sin \angle B'BC_3}{\sin \angle B'BA_3} = \frac{B'C_3}{B'A_3} \cdot \frac{\cos \angle B'C_3P}{\cos \angle B'A_3P}
and
sin⁡∠C′CA3sin⁡∠C′CB3=C′A3C′B3⋅cos⁡∠C′A3Pcos⁡∠C′B3P \frac{\sin \angle C'CA_3}{\sin \angle C'CB_3} = \frac{C'A_3}{C'B_3} \cdot \frac{\cos \angle C'A_3P}{\cos \angle C'B_3P}
By the converse of the Trigonometric Ceva Theorem, in order to show that AA′,BB′AA', BB' and CC′CC' are concurrent it is sufficient to show that
A′B3A′C3⋅B′C3B′A3⋅C′A3C′B3⋅cos⁡∠A′B3Pcos⁡∠A′C3P⋅cos⁡∠B′C3Pcos⁡∠B′A3P⋅cos⁡∠C′A3Pcos⁡∠C′B3P=1 \frac{A'B_3}{A'C_3} \cdot \frac{B'C_3}{B'A_3} \cdot \frac{C'A_3}{C'B_3} \cdot \frac{\cos \angle A'B_3P}{\cos \angle A'C_3P} \cdot \frac{\cos \angle B'C_3P}{\cos \angle B'A_3P} \cdot \frac{\cos \angle C'A_3P}{\cos \angle C'B_3P} = 1
By the concyclicity, ∠A′C3P=C′A3P\angle A'C_3P = C'A_3P, B′A3P=A′B3PB'A_3P = A'B_3P, C′B3P=B′C3PC'B_3P = B'C_3P and hence
cos⁡∠A′B3Pcos⁡∠A′C3P⋅cos⁡∠B′C3Pcos⁡∠B′A3P⋅cos⁡∠C′A3Pcos⁡∠C′B3P=1 \frac{\cos \angle A'B_3P}{\cos \angle A'C_3P} \cdot \frac{\cos \angle B'C_3P}{\cos \angle B'A_3P} \cdot \frac{\cos \angle C'A_3P}{\cos \angle C'B_3P} = 1
Now, we have to show that
A′B3A′C3⋅B′C3B′A3⋅C′A3C′B3=1 \frac{A'B_3}{A'C_3} \cdot \frac{B'C_3}{B'A_3} \cdot \frac{C'A_3}{C'B_3} = 1
By concyclicity again, the triangles PA3C′PA_3C' and PC3A′PC_3A' are similar and hence
C′A3A′C3=PA3PC3 \frac{C'A_3}{A'C_3} = \frac{PA_3}{PC_3}
and similarly we get that
A′B3B′A3=PB3PA3andB′C3C′B3=PC3PB3 \frac{A'B_3}{B'A_3} = \frac{PB_3}{PA_3} \quad \text{and} \quad \frac{B'C_3}{C'B_3} = \frac{PC_3}{PB_3}
Side by side product of last three equalities yields
A′B3A′C3⋅B′C3B′A3⋅C′A3C′B3=PA3PC3⋅PB3PA3⋅PC3PB3=1 \frac{A'B_3}{A'C_3} \cdot \frac{B'C_3}{B'A_3} \cdot \frac{C'A_3}{C'B_3} = \frac{PA_3}{PC_3} \cdot \frac{PB_3}{PA_3} \cdot \frac{PC_3}{PB_3} = 1
as desired.

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