2023 AIME II Problems/Problem 9

Solution

Denote by $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ , respectively. Let $XY$ and $AO_1$ intersect at point $C$ . Let $XY$ and $BO_2$ intersect at point $D$ .

Because $AB$ is tangent to circle $\omega_1$ , $O_1 A \perp AB$ . Because $XY \parallel AB$ , $O_1 A \perp XY$ . Because $X$ and $P$ are on $\omega_1$ , $O_1A$ is the perpendicular bisector of $XY$ . Thus, $PC = \frac{PX}{2} = 5$ .

Analogously, we can show that $PD = \frac{PY}{2} = 7$ .

Thus, $CD = CP + PD = 12$ . Because $O_1 A \perp CD$ , $O_1 A \perp AB$ , $O_2 B \perp CD$ , $O_2 B \perp AB$ , $ABDC$ is a rectangle. Hence, $AB = CD = 12$ .

Let $QP$ and $AB$ meet at point $M$ . Thus, $M$ is the midpoint of $AB$ . Thus, $AM = \frac{AB}{2} = 6$ .

In $\omega_1$ , for the tangent $MA$ and the secant $MPQ$ , following from the power of a point, we have $MA^2 = MP \cdot MQ$ . By solving this equation, we get $MP = 4$ .

We notice that $AMPC$ is a right trapezoid. Hence, $\begin{align*} AC & = \sqrt{MP^2 - \left( AM - CP \right)^2} \\ & = \sqrt{15} . \end{align*}$

Therefore, $\begin{align*} {\rm Area} \ XABY & = \frac{1}{2} \left( AB + XY \right) AC \\ & = \frac{1}{2} \left( 12 + 24 \right) \sqrt{15} \\ & = 18 \sqrt{15}. \end{align*}$

Therefore, the answer is $18 + 15 = \boxed{\textbf{(033) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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2023 AIME II Problems/Problem 9

Solution