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

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(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
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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,
 
Then, by Power of Point,
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</cmath>
 
</cmath>
  
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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>
+
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 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>
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:10, 17 February 2023

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 $m+n.$

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, $AM = \frac{AB}{2} = 6$.

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*} AP^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 the rectangle is $\sqrt{15}$. Therefore the area of trapezoid $XABY$ is $frac{1}{2}\cdot(12+24)\cdot(\sqrt{15})$

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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