Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

Two circles touch each other externally at point $C$ . Consider two diameters $A_1A_2$ , $B_1B_2$ of the same direction. Circle with the center on the common internal tangent passes through the point of intersection of $A_1B_2$ , $A_2B_1$ , and meets these lines at points $M$ , $N$ . Prove that $MN$ is perpendicular to $A_1A_2$ , $B_1B_2$ .

Solution

Since point $C$ is a center of homothety that transforms one circle into another, then $C = A_1B_2 \cap A_2B_1$ , $A_1B_2 \perp A_2B_1$ (fig. 21).

Let $D = MN \cap B_1B_2$ . Then, $\angle DB_2C = \angle B_2A_1A_2 = \angle A_2CO = \angle CND$ . Therefore $DB_2NC$ is cyclic and $\angle B_2DN = \angle B_2CA_2 = \frac{\pi}{2}$ , which implies that $MN \perp B_1B_2$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.