2006 iTest Problems/Problem 38

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 $m+n=1757+402=\boxed{2459}$ .

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2006 iTest Problems/Problem 38

Problem

Solution 1