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

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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>
 
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>
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 +
==Video Solution & More by MegaMath==
 +
https://www.youtube.com/watch?v=x-5VYR1Dfw4
  
 
==Solution 1==
 
==Solution 1==
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Let <math>QP</math> and <math>AB</math> meet at point <math>M</math>.
 
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>M</math> is the midpoint of <math>AB</math>.
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Thus, <math>AM = \frac{AB}{2} = 6</math>.
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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>.
 
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>.
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</cmath>
 
</cmath>
  
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Therefore, the answer is <math>18 + 15 = \boxed{\textbf{(033) }}</math>.
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Therefore, the answer is <math>18 + 15 = \boxed{\textbf{(033)}}</math>.
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==Solution 2==
 
==Solution 2==
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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.
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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,
 
Then, by Power of Point,
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<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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AP^2&=MP*MQ \\
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AM^2&=MP*MQ \\
 
36&=MP(MP+5) \\
 
36&=MP(MP+5) \\
 
MP&=4 \\
 
MP&=4 \\
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</cmath>
 
</cmath>
  
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~Danielzh
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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 ==
 
== See also ==
 
{{AIME box|year=2023|num-b=8|num-a=10|n=II}}
 
{{AIME box|year=2023|num-b=8|num-a=10|n=II}}
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[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See also

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
All AIME Problems and Solutions

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