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

(Created page with "==Solution== Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. Let <math>XY</math> and <math>AO_1<...")
 
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<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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{\rm Area} \ XABY & = \frac{1}{2} \left( AB + XY \right) AC \\
+
[XABY] & = \frac{1}{2} \left( AB + XY \right) AC \\
 
& = \frac{1}{2} \left( 12 + 24 \right) \sqrt{15} \\
 
& = \frac{1}{2} \left( 12 + 24 \right) \sqrt{15} \\
 
& = 18 \sqrt{15}.
 
& = 18 \sqrt{15}.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
== See also ==
 +
{{AIME box|year=2023|num-b=8|num-a=10|n=II}}
 +
{{MAA Notice}}

Revision as of 20:02, 16 February 2023

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*} [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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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