Mock AIME 6 2006-2007 Problems/Problem 6

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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Mock AIME 6 2006-2007 Problems/Problem 6

Problem

Solution