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Geometry Difficulty 6.9 National Olympiad Prove it Romania

Two circles $\gamma_1$ and $\gamma_2$ meet at two points; let $A$ be one of these points. The tangent to $\gamma_1$ at $A$ meets again $\gamma_2$ at $B$ , the tangent to $\gamma_2$ at $A$ meets again $\gamma_1$ at $C$ , and the line $BC$ meets again $\gamma_1$ and $\gamma_2$ at $D_1$ and $D_2$ , respectively. Let $E_1$ and $E_2$ be interior points of the segments $AD_1$ and $AD_2$ , respectively, such that $AE_1 = AE_2$ . The lines $BE_1$ and $AC$ meet at $M$ , the lines $CE_2$ and $AB$ meet at $N$ , and the lines $MN$ and $BC$ meet at $P$ . Show that the line $PA$ is tangent to the circle $ABC$ .

Solution

To begin, apply Menelaus' theorem to triangles $ABD_2$ , $ACD_1$ , $ABC$ , to write
$\frac{NB}{NA} \cdot \frac{CD_2}{CB} \cdot \frac{E_2A}{E_2D_2} = 1, \quad \frac{MA}{MC} \cdot \frac{E_1D_1}{E_1A} \cdot \frac{BC}{BD_1} = 1, \quad \frac{MC}{MA} \cdot \frac{NA}{NB} \cdot \frac{PB}{PC} = 1,$
so, multiplying the three, $\frac{E_1 D_1}{E_2 D_2} \cdot \frac{C D_2}{B D_1} \cdot \frac{P B}{P C} = 1$ , (*) on account of $AE_1 = AE_2$ . Since $\angle A D_1 B = \angle B A C = \angle A D_2 C$ , it follows that $AD_1 = AD_2$ , so $E_1 D_1 = E_2 D_2$ , with reference again to $AE_1 = AE_2$ . Consequently, $P B / P C = B D_1 / C D_2$ , by (*).

Finally, similarity of the triangles $ABC$ and $D_1BA$ yields $BD_1 = AB^2/BC$ . Similarly, $CD_2 = AC^2/BC$ , so $PB \cdot AC^2 = PC \cdot AB^2$ , by the preceding, q.e.d.

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