Difference between revisions of "2023 AIME II Problems/Problem 9"

Latest revision as of 15:14, 7 June 2024

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

@@ Line 2: / Line 2: @@
 Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at two points <math>P</math> and <math>Q,</math> and their common tangent line closer to <math>P</math> intersects <math>\omega_1</math> and <math>\omega_2</math> at points <math>A</math> and <math>B,</math> respectively. The line parallel to <math>AB</math> that passes through <math>P</math> intersects <math>\omega_1</math> and <math>\omega_2</math> for the second time at points <math>X</math> and <math>Y,</math> respectively. Suppose <math>PX=10,</math> <math>PY=14,</math> and <math>PQ=5.</math> Then the area of trapezoid <math>XABY</math> is <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math>
+==Video Solution & More by MegaMath==
+https://www.youtube.com/watch?v=x-5VYR1Dfw4
 ==Solution 1==
@@ Line 21: / Line 24: @@
 Let <math>QP</math> and <math>AB</math> meet at point <math>M</math>.
 Thus, <math>M</math> is the midpoint of <math>AB</math>.
-Thus, <math>AM = \frac{AB}{2} = 6</math>.
+Thus, <math>AM = \frac{AB}{2} = 6</math>. This is the case because <math>PQ</math> is the radical axis of the two circles, and the powers with respect to each circle must be equal.
 In <math>\omega_1</math>, for the tangent <math>MA</math> and the secant <math>MPQ</math>, following from the power of a point, we have <math>MA^2 = MP \cdot MQ</math>.
@@ Line 44: / Line 47: @@
 </cmath>
-Therefore, the answer is <math>18 + 15 = \boxed{\textbf{(033) }}</math>.
+Therefore, the answer is <math>18 + 15 = \boxed{\textbf{(033)}}</math>.
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==Solution 2==
-Notice that line <math>overline{PQ}</math> is the radical axis of circles <math>w1</math> and <math>w2</math>. By the radical axis theorem, we know that the tangents of any point on line <math>overline{PQ}</math> to circles <math>w1</math> and <math>w2</math> are equal. Therefore, line <math>overline{PQ}</math> must pass through the midpoint of <math>overline{AB}</math>, call this point M. In addition, we know that <math>AM=MB=6</math> by circle properties and midpoint definition.
+Notice that line <math>\overline{PQ}</math> is the [[radical axis]] of circles <math>\omega_1</math> and <math>\omega_2</math>. By the radical axis theorem, we know that the tangents of any point on line <math>\overline{PQ}</math> to circles <math>\omega_1</math> and <math>\omega_2</math> are equal. Therefore, line <math>\overline{PQ}</math> must pass through the midpoint of <math>\overline{AB}</math>, call this point M. In addition, we know that <math>AM=MB=6</math> by circle properties and midpoint definition.
 Then, by Power of Point,
@@ Line 55: / Line 58: @@
 <cmath>
 \begin{align*}
-AP^2=MP*MQ //
+AM^2&=MP*MQ \\
-=MP(MP+5) //
+&=MP(MP+5) \\
-MP=4 //
+MP&=4 \\
+\end{align*}
+</cmath>
+Call the intersection point of line <math>\overline{A\omega_1}</math> and <math>\overline{XY}</math> be C, and the intersection point of line <math>\overline{B\omega_2}</math> and <math>\overline{XY}</math> be D. <math>ABCD</math> is a rectangle with segment <math>MP=4</math> drawn through it so that <math>AM=MB=6</math>, <math>CP=5</math>, and <math>PD=7</math>. Dropping the altitude from <math>M</math> to <math>\overline{XY}</math>, we get that the height of trapezoid <math>XABY</math> is <math>\sqrt{15}</math>. Therefore the area of trapezoid <math>XABY</math> is
+<cmath>
+\begin{align*}
+\frac{1}{2}\cdot(12+24)\cdot(\sqrt{15})=18\sqrt{15}
 \end{align*}
 </cmath>
-Call the intersection point of line <math>overline{AW1}</math> and <math>overline{XY}</math> be C, and the intersection point of line <math>overline{BW2}</math> and <math>overline{XY}</math> be D. <math>ABCD</math> is a rectangle with segment <math>MP=4</math> drawn through it so that <math>AM=MB=6</math>, <math>CP=5</math>, and <math>PD=7</math>. Dropping the altitude from <math>M</math> to <math>overline{XY}</math>, we get that the height of the rectangle is <math>\sqrt{15}</math>. Therefore the area of trapezoid <math>XABY</math> is <math>frac{1}{2}\cdot(12+24)\cdot(\sqrt{15})</math>
+Giving us an answer of <math>\boxed{033}</math>.
+~[[Daniel Zhou's Profile|Danielzh]]
+==Video Solution 1 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=RUv6qNY_agI
 == See also ==
 {{AIME box|year=2023|num-b=8|num-a=10|n=II}}
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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