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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 6"

(Created page with "==Problem == <math>C_1</math> is a circle with radius <math>164</math> and <math>C_2</math> is a circle internally tangent to <math>C_1</math> that passes through the center o...")
 
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==Solution==
 
==Solution==
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{{Solution}}
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Let the center of <math>C_1</math> be <math>O_1</math>, and let the center of <math>C_2</math> be <math>O_2</math>. Extend <math>O_1O_2</math> through <math>O_2</math> to meet <math>C_1</math> at <math>X</math>. Then <math>O_1X=164</math> since it is a radius. Drop perpendiculars from <math>X,O_2,</math> and <math>C_1</math> to <math>AB</math>. Since <math>D</math> is the tangent point, we must have <math>D</math> be the foot of the perpendicular from <math>O_2</math>. Let the perpendiculars from <math>X</math> and <math>C_1</math> have feet <math>E,F</math> respectively. Then since <math>XO_2=O_2O_1=82</math> due to being a radius (note that the radius of <math>C_2</math> is half that of <math>C_1</math> since two radii of the first make one radii of the second from construction), we must have <math>XE+O_1F=2O_2D=164</math>.
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Let <math>XE=y</math>. Then <math>O_1F=164-y</math>. Notice that since <math>O_1F</math> is perpendicular to the chord, we must have <math>AF=BF=160</math>. Additionally, <math>O_1A=164</math> due to being a radius. Then, by the [[Pythagorean Theorem]], we must have <math>O_1F^2+AF^2=AO_1^2</math>. Substituting values results in <math>y=128</math>.
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Let <math>EF=x</math>. From earlier, <math>O_1X=164,XE=y=128,</math> and <math>O_1F=164-y=36</math>. Then, by the Pythagorean Theorem:
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<cmath>EF^2+(XE-O_1F)^2=XO_1^2</cmath>
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<cmath>\Rightarrow x^2+(128-36)^2=164^2</cmath>
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<cmath>\Rightarrow x^2=164^2-92^2=256\cdot72=16^2\cdot6^2\cdot2</cmath>
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<cmath>\Rightarrow x=96\sqrt{2}</cmath>
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Since <math>BF=160</math>, <math>BE=160-x</math>. Additionally, since <math>XO_2=O_2O_1</math>, then <math>ED=DF</math>, so they are equal to <math>\frac{1}{2}x</math>. Then <math>BD=BE+ED=160-\frac{1}{2}x</math>. Therefore, <math>BD=160-48\sqrt{2}</math>, so the answer is <math>160+48+2=\boxed{210}</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]

Latest revision as of 18:13, 23 March 2025

Problem

$C_1$ is a circle with radius $164$ and $C_2$ is a circle internally tangent to $C_1$ that passes through the center of $C_1$. $\overline{AB}$ is a chord in $C_1$ of length $320$ tangent to $C_2$ at $D$ where $AD>BD$. Given that $BD=a-b\sqrt{c}$ where $a,b,c$ are positive integers and $c$ is not divisible by the square of any prime, what is $a+b+c$?

Solution

Let the center of $C_1$ be $O_1$, and let the center of $C_2$ be $O_2$. Extend $O_1O_2$ through $O_2$ to meet $C_1$ at $X$. Then $O_1X=164$ since it is a radius. Drop perpendiculars from $X,O_2,$ and $C_1$ to $AB$. Since $D$ is the tangent point, we must have $D$ be the foot of the perpendicular from $O_2$. Let the perpendiculars from $X$ and $C_1$ have feet $E,F$ respectively. Then since $XO_2=O_2O_1=82$ due to being a radius (note that the radius of $C_2$ is half that of $C_1$ since two radii of the first make one radii of the second from construction), we must have $XE+O_1F=2O_2D=164$.

Let $XE=y$. Then $O_1F=164-y$. Notice that since $O_1F$ is perpendicular to the chord, we must have $AF=BF=160$. Additionally, $O_1A=164$ due to being a radius. Then, by the Pythagorean Theorem, we must have $O_1F^2+AF^2=AO_1^2$. Substituting values results in $y=128$.

Let $EF=x$. From earlier, $O_1X=164,XE=y=128,$ and $O_1F=164-y=36$. Then, by the Pythagorean Theorem: \[EF^2+(XE-O_1F)^2=XO_1^2\] \[\Rightarrow x^2+(128-36)^2=164^2\] \[\Rightarrow x^2=164^2-92^2=256\cdot72=16^2\cdot6^2\cdot2\] \[\Rightarrow x=96\sqrt{2}\]

Since $BF=160$, $BE=160-x$. Additionally, since $XO_2=O_2O_1$, then $ED=DF$, so they are equal to $\frac{1}{2}x$. Then $BD=BE+ED=160-\frac{1}{2}x$. Therefore, $BD=160-48\sqrt{2}$, so the answer is $160+48+2=\boxed{210}$.

~ eevee9406

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