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Problem 1306

AIME late

Geometry Difficulty 5.5 Prove it Iranian Mathematical Olympiad · Iran

Given circles $\omega_1$ and $\omega_2$ with two intersection points $A$ and $B$ . Points $X$ on $\omega_1$ and $Y$ on $\omega_2$ are chosen such that $XY$ is tangent to both circles and $XY$ is closer to $B$ than $A$ . If $C$ and $D$ are the reflections of $X$ and $Y$ with respect to $B$ , prove that $\angle CAD < 90^\circ$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Suppose that $AB$ intersects $XY$ and $CD$ at $M$ and $N$ , respectively. By comparing the power of $M$ with respect to $\omega_1$ and $\omega_2$ we have
$MX^2 = MB \cdot MA = MY^2 \implies MX = MY$

$\angle CAD < 90^\circ$ it suffices to prove that $A$ is outside the circle with diagonal $CD$ ; or equivalently to prove $NA > NC$ . It is clear that
$NA = NM + MA = MB + MA \quad \text{and} \quad NC = 2MX.$
So we have to show that $MB + MA > 2MX$ . Then by AM-GM inequality
$MX^2 = MA \cdot MB \le \left( \frac{MA + MB}{2} \right)^2 \implies MA + MB \ge 2MX.$
Note that the equality case only occurs whenever $MB = MA$ , however we know that $MB > MA$ , so we have $MB + MA > 2MX$ .

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