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Difference between revisions of "2006 iTest Problems/Problem 38"

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[[Category: Intermediate Geometry Problems]]

Latest revision as of 19:13, 13 October 2025

Problem

Segment $AB$ is a diameter of circle $\Gamma_1$. Point $C$ lies in the interior of segment $AB$ such that $BC=7$, and $D$ is a point on $\Gamma_1$ such that $BD=CD=10$. Segment $AC$ is a diameter of the circle $\Gamma_2$. A third circle, $\omega$, is drawn internally tangent to $\Gamma_1$, externally tangent to $\Gamma_2$, and tangent to segment $CD$. If $\omega$ is centered on the opposite side of $CD$ as $B$, then the radius of $\omega$ can be expressed as $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Solution 1

Place circle $\Gamma_1$ with center $O_1=(0,0)$ and radius $R_1$ so that $\overline{AB}$ is the diameter on the x-axis with $A=(-R_1,0)$, $B=(R_1,0)$. Then $C$ lies at $(R_1-7,0)$. Let point D lie on $\Gamma_1$ and let $BD=CD=10$. Then $(x-R_1)^2+y^2=100$, and $(X-(R_1-7))^2+y^2=100$. Subtracting gives $14x-14R_1+49=0$, so $x=R_1-3.5$. Then $(x-R_1)^2+y^2=100$ means that $y^2=100-3.5^2=\frac{351}{4}$, so $D=(R_1-3.5,\frac{\sqrt{351}}{2})$. Since $D$ lies on $\Gamma_1$, $x^2+y^2=R_1^2$ means that $(R_1-3.5)^2+\frac{351}{4}=R_1^2$. Solving for $R_1$ yields that $R_1=\frac{100}{7}$. Therefore, $C=(\frac{51}{7},0)$ and $D=(\frac{151}{14},\frac{\sqrt{351}}{2})$. Since segment $\overline{AC}$ is a diameter of $\Gamma_2$, knowing that $A=(\frac{-100}{7}, 0)$ and $C=(\frac{51}{7},0)$, $AC=\frac{151}{7}$, so the center of $\Gamma_2$ is the midpoint of $\overline{AC}$, which is $(-3.5,0)$, and it has radius $\frac{AC}{2}=\frac{151}{14}$. We then find that the equation of $CD$ is $-\frac{\sqrt{351}}{7}x+y+\frac{\sqrt{351}*51}{49}=0$. To find the center of circle $\omega$, we can use the tangencies in the problem. Internal tangency to $\Gamma_1$ and external tangency to $\Gamma_2$ give $h^2+k^2=(R_1-r)^2=(\frac{100}{7}-r)^2$ and $(h+3.5)^2+k^2=(R_2+r)^2=(\frac{151}{14}+r)^2$, where $h,k$ are the coordinates of the center of circle $\omega$ and $r$ is the radius. Subtracting the two equations gives $\frac{7h}{2}+\frac{49}{4}=(\frac{151}{14}+r)^2-(\frac{100}{7}-r)^2=-\frac{17199}{196}+\frac{351r}{7}$, so $h=\frac{-200}{7}+\frac{702r}{49}$. Using the point to line distance formula, we find that $k=\frac{20r}{7}+\frac{h\sqrt{351}}{7}-\frac{51\sqrt{351}}{49}$. From internal tangency to $\Gamma_1$, $k^2=(\frac{100}{7}-r)^2-h^2$. Plugging the expressions for $h$ and $k$ yields that $r=\frac{1757}{702}$. It is in lowest terms, so $m+n=1757+402=\boxed{2459}$.

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 37 		Followed by:
Problem 39
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