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

m (→‎Solution 1)
(→‎Solution 2)
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~[[Daniel Zhou's Profile|Danielzh]]
 
~[[Daniel Zhou's Profile|Danielzh]]
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==Solution 3==
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Refer to Solution 1.
 +
 +
We let <math>AC=BD=x</math> and the extension of <math>AC</math> to the circle <math>\neq A</math> as <math>E.</math> By Power of a Point on point <math>C</math> of circle <math>w_1</math> we find <cmath>x\cdot{CE}=5\cdot5.</cmath> <cmath>CE=\frac{25}{x}.</cmath> We have diameter <math>AE = AC+CE=x+\frac{25}{x}.</math> Therefore the radius of <math>w_1</math> is <math>\frac{x+\frac{25}{x}}{2}=\frac{25+x^2}{2x} = O_1A = O_1P.</math>
 +
 +
Similarly repeating this procedure on <math>w_2</math> we find the radius of <math>w_2</math> is <math>\frac{49+x^2}{2x} = O_2P = O_2B.</math>
 +
 +
Next we solve for <math>O_1O_2</math> in two ways. Let the perpendicular from <math>O_1</math> to <math>BO_2</math> intersect at <math>K</math> we have <math>O_1K =AB = 12.</math> We also have <cmath>O_2K = BO_2 - AO_1 =\frac{49+x^2}{2x}- \frac{25+x^2}{2x}=\frac{12}{x}.</cmath>
 +
Therefore since <math>\triangle{O_1KO_2}</math> is right, we have <math>(O_1O_2)^2 = (O_1K)^2+(O_2K)^2 = 12^2 + \frac{12}{x}^2 =144 + \frac{144}{x^2}.</math>
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For our second way, we let the midpoint of <math>PQ</math> be <math>M.</math> Note that <math>PM</math> forms the right triangles <math>PO_1M</math> and <math>PO_2M</math> both of which share an leg of <math>PM</math> or <math>\frac{5}{2}.</math> Using Pythag we can solve for <math>O_1O_2.</math>
 +
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<cmath>O_1O_2 = O_1M+O_2M = \sqrt{(O_1P)^2 - (PM)^2}+\sqrt{(O_2P)^2 - (PM)^2}</cmath>
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<cmath>\sqrt{144 + \frac{144}{x^2}} = \sqrt{(\frac{25+x^2}{2x})^2 - (\frac{5}{2})^2}+\sqrt{(\frac{49+x^2}{2x})^2 - (\frac{5}{2})^2}</cmath>
 +
<cmath>\sqrt{144 + \frac{144}{x^2}} = \sqrt{\frac{(25+x^2)^2}{4x^2} - \frac{25}{4}}+\sqrt{\frac{(49+x^2)^2}{4x^2} - \frac{25}{4}}</cmath>
 +
We let <math>x^2 = n</math> to slightly simplify the equation,
 +
<cmath>\sqrt{144 + \frac{144}{a}} = \sqrt{\frac{(25+a)^2}{4a} - \frac{25}{4}}+\sqrt{\frac{(49+a)^2}{4a} - \frac{25}{4}}</cmath>
 +
<cmath>
 +
12\sqrt{1+\frac{1}{a}} = \sqrt{\frac{a^2 + 25a + 625}{4a}} + \sqrt{\frac{a^2 +73a + 2401}{4a}}.
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</cmath>
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<cmath>
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24\sqrt{1+\frac{1}{a}} = \sqrt{\frac{a^2 + 25a + 625}{a}} + \sqrt{\frac{a^2 +73a + 2401}{a}}.
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</cmath>
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<cmath>
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24\sqrt{a+1} = \sqrt{a^2 + 25a + 625} + \sqrt{a^2 +73a + 2401}.
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</cmath>
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<cmath>
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\sqrt{a^2 + 25a + 625}=24\sqrt{a+1}-\sqrt{a^2 +73a + 2401}.
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</cmath>
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<cmath>
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a^2 + 25a + 625=576a+576+a^2+73a+2401-48\sqrt{(a+1)(a^2 +73a + 2401)}.
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</cmath>
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<cmath>
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48*\sqrt{(a+1)(a^2 +73a + 2401)}=624a+2352.
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</cmath>
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<cmath>
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\sqrt{(a+1)(a^2 +73a + 2401)}=13a+49.
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</cmath>
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<cmath>a^3 + 74 a^2 + 2474 a + 2401=169 a^2 + 1274 a + 2401.</cmath>
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<cmath>a^3 -95a^2 + 1200a = 0</cmath>
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<cmath>a(a-15)(a-80)=0</cmath>
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Thus the solutions are <math>a=0,15,80.</math> Checking bounds <math>a=15</math> is the only valid solution, which means <math>x=\sqrt{15}.</math>
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Finally to find the area of <math>XABY,</math> we have the bases <math>XY=24</math> and <math>AB=12</math> and the height <math>x=\sqrt{15}</math> therefore <cmath>[XABY]=\frac{1}{2}\cdot(12+24)\cdot\sqrt{15}=18\sqrt{15}.</cmath>
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Giving us an answer of <math>18+15 = \boxed{033}.</math>
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~[[Mathkiddus's Profile|mathkiddus]]
  
 
==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==

Revision as of 16:52, 19 January 2025

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 $XP$. 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

Solution 3

Refer to Solution 1.

We let $AC=BD=x$ and the extension of $AC$ to the circle $\neq A$ as $E.$ By Power of a Point on point $C$ of circle $w_1$ we find \[x\cdot{CE}=5\cdot5.\] \[CE=\frac{25}{x}.\] We have diameter $AE = AC+CE=x+\frac{25}{x}.$ Therefore the radius of $w_1$ is $\frac{x+\frac{25}{x}}{2}=\frac{25+x^2}{2x} = O_1A = O_1P.$

Similarly repeating this procedure on $w_2$ we find the radius of $w_2$ is $\frac{49+x^2}{2x} = O_2P = O_2B.$

Next we solve for $O_1O_2$ in two ways. Let the perpendicular from $O_1$ to $BO_2$ intersect at $K$ we have $O_1K =AB = 12.$ We also have \[O_2K = BO_2 - AO_1 =\frac{49+x^2}{2x}- \frac{25+x^2}{2x}=\frac{12}{x}.\] Therefore since $\triangle{O_1KO_2}$ is right, we have $(O_1O_2)^2 = (O_1K)^2+(O_2K)^2 = 12^2 + \frac{12}{x}^2 =144 + \frac{144}{x^2}.$

For our second way, we let the midpoint of $PQ$ be $M.$ Note that $PM$ forms the right triangles $PO_1M$ and $PO_2M$ both of which share an leg of $PM$ or $\frac{5}{2}.$ Using Pythag we can solve for $O_1O_2.$

\[O_1O_2 = O_1M+O_2M = \sqrt{(O_1P)^2 - (PM)^2}+\sqrt{(O_2P)^2 - (PM)^2}\] \[\sqrt{144 + \frac{144}{x^2}} = \sqrt{(\frac{25+x^2}{2x})^2 - (\frac{5}{2})^2}+\sqrt{(\frac{49+x^2}{2x})^2 - (\frac{5}{2})^2}\] \[\sqrt{144 + \frac{144}{x^2}} = \sqrt{\frac{(25+x^2)^2}{4x^2} - \frac{25}{4}}+\sqrt{\frac{(49+x^2)^2}{4x^2} - \frac{25}{4}}\] We let $x^2 = n$ to slightly simplify the equation, \[\sqrt{144 + \frac{144}{a}} = \sqrt{\frac{(25+a)^2}{4a} - \frac{25}{4}}+\sqrt{\frac{(49+a)^2}{4a} - \frac{25}{4}}\] \[12\sqrt{1+\frac{1}{a}} = \sqrt{\frac{a^2 + 25a + 625}{4a}} + \sqrt{\frac{a^2 +73a + 2401}{4a}}.\] \[24\sqrt{1+\frac{1}{a}} = \sqrt{\frac{a^2 + 25a + 625}{a}} + \sqrt{\frac{a^2 +73a + 2401}{a}}.\] \[24\sqrt{a+1} = \sqrt{a^2 + 25a + 625} + \sqrt{a^2 +73a + 2401}.\] \[\sqrt{a^2 + 25a + 625}=24\sqrt{a+1}-\sqrt{a^2 +73a + 2401}.\] \[a^2 + 25a + 625=576a+576+a^2+73a+2401-48\sqrt{(a+1)(a^2 +73a + 2401)}.\] \[48*\sqrt{(a+1)(a^2 +73a + 2401)}=624a+2352.\] \[\sqrt{(a+1)(a^2 +73a + 2401)}=13a+49.\] \[a^3 + 74 a^2 + 2474 a + 2401=169 a^2 + 1274 a + 2401.\] \[a^3 -95a^2 + 1200a = 0\] \[a(a-15)(a-80)=0\] Thus the solutions are $a=0,15,80.$ Checking bounds $a=15$ is the only valid solution, which means $x=\sqrt{15}.$ Finally to find the area of $XABY,$ we have the bases $XY=24$ and $AB=12$ and the height $x=\sqrt{15}$ therefore \[[XABY]=\frac{1}{2}\cdot(12+24)\cdot\sqrt{15}=18\sqrt{15}.\] Giving us an answer of $18+15 = \boxed{033}.$ ~mathkiddus

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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