Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 AIME, harder Prove it

Example 23 As shown in Figure $11-33$ , given that $A$ is an intersection point of two circles $\odot O_{1}$ and $\odot O_{2}$ with unequal radii, the two external common tangents $P_{1} P_{2}$ and $Q_{1} Q_{2}$ touch the two circles at $P_{1}, P_{2}, Q_{1}, Q_{2}$ , respectively, and $M_{1}, M_{2}$ are the midpoints of $P_{1} Q_{1}$ and $P_{2} O_{2}$ . Prove that: $\angle O_{1} A O_{2}=\angle M_{1} A M_{2}$ .
(IMO - 24 Problem)

Solution

Proof: Let the line $P_{1} P_{2}$ intersect $Q_{1} Q_{2}$ at point $O$ . Then $O, O_{1}, O_{2}$ are collinear. Suppose this line intersects $\odot O_{1}$ at points $D$ and $E$ . It is known that $M_{1}, O$ harmonically divide $DE$ , thus the midpoint $O_{1}$ of $DE$ satisfies
$O_{1} M_{1} \cdot O_{1} O = O_{1} E^{2} = O_{1} A^{2}$ ,
which implies $\frac{O_{1} A}{O_{1} M_{1}} = \frac{O_{1} O}{O_{1} A}$ . Given that $\angle A O_{1} M_{1}$ is common, we have $\triangle O_{1} A M_{1} \sim \triangle O_{1} O A$ .
Thus, $\angle O_{1} A M_{1} = \angle O_{1} O A$ .
Similarly, $\angle O_{2} A M_{2} = \angle O_{2} O A$ .
Therefore, $\angle O_{1} A O_{2} = M_{1} A M_{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.