2023 AIME II Problems/Problem 9

Problem

Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q,$ and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B,$ respectively. The line parallel to $AB$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y,$ respectively. Suppose $PX=10,$ $PY=14,$ and $PQ=5.$ Then the area of trapezoid $XABY$ is $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n.$

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=x-5VYR1Dfw4

Solution 1

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, $AM = \frac{AB}{2} = 6$ . This is the case because $PQ$ is the radical axis of the two circles, and the powers with respect to each circle must be equal.

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

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

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

Solution 2

Notice that line $\overline{PQ}$ is the radical axis of circles $\omega_1$ and $\omega_2$ . By the radical axis theorem, we know that the tangents of any point on line $\overline{PQ}$ to circles $\omega_1$ and $\omega_2$ are equal. Therefore, line $\overline{PQ}$ must pass through the midpoint of $\overline{AB}$ , call this point M. In addition, we know that $AM=MB=6$ by circle properties and midpoint definition.

Then, by Power of Point,

$\begin{align*} AM^2&=MP*MQ \\ 36&=MP(MP+5) \\ MP&=4 \\ \end{align*}$

Call the intersection point of line $\overline{A\omega_1}$ and $\overline{XY}$ be C, and the intersection point of line $\overline{B\omega_2}$ and $\overline{XY}$ be D. $ABCD$ is a rectangle with segment $MP=4$ drawn through it so that $AM=MB=6$ , $CP=5$ , and $PD=7$ . Dropping the altitude from $M$ to $\overline{XY}$ , we get that the height of trapezoid $XABY$ is $\sqrt{15}$ . Therefore the area of trapezoid $XABY$ is

$\begin{align*} \frac{1}{2}\cdot(12+24)\cdot(\sqrt{15})=18\sqrt{15} \end{align*}$

Giving us an answer of $\boxed{033}$ .

~Danielzh

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=RUv6qNY_agI

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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