Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Ukraine

$A$ is one of the points of intersection of circles $\omega_1$ and $\omega_2$ , $T_1T_2$ is the external common tangent to these circles, which tangents $\omega_1$ at $T_1$ and tangents $\omega_2$ at $T_2$ . $B_1$ is any point on the circle $\omega_1$ , $B_2$ is any point on the circle $\omega_2$ , such that $A$ , $B_1$ and $B_2$ are not collinear. Circumcircle of the $\triangle AB_1B_2$ intersects lines $T_1B_1$ , $T_2B_2$ at points $C_1, C_2$ respectively. Prove that $C_1T_2$ and $C_2T_1$ intersect on $\omega$ .

Solution

Let $X$ be the second intersection point of $\omega$ and $(AT_1T_2)$ . Observe that $\angle (AX, XC_1) = \angle (AB_1, B_1C_1) = \angle (AB_1, T_1B_1) = \angle (AT_1, T_1T_2) = \angle (AX, XT_2)$ and then $X \in C_1T_2$ . Similarly, $X \in C_2T_1$ and hence $C_1T_2$ and $C_2T_1$ intersect on $\omega$ .

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