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

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==Solution 1==
 
==Solution 1==
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Place circle <math>\Gamma_1</math> with center <math>O_1=(0,0)</math> and radius <math>R_1</math> so that <math>\overline{AB}</math> is the diameter on the x-axis with <math>A=(-R_1,0)</math>, <math>B=(R_1,0)</math>. Then <math>C</math> lies at <math>(R_1-7,0)</math>. Let point D lie on <math>\Gamma_1</math> and let <math>BD=CD=10</math>. Then <math>(x-R_1)^2+y^2=100</math>, and <math>(X-(R_1-7))^2+y^2=100</math>. Subtracting gives <math>14x-14R_1+49=0</math>, so <math>x=R_1-3.5</math>. Then <math>(x-R_1)^2+y^2=100</math> means that <math>y^2=100-3.5^2=\frac{351}{4}</math>, so <math>D=(R_1-3.5,\frac{\sqrt{351}}{2}</math>. Since <math>D</math> lies on <math>\Gamma_1</math>, <math>x^2+y^2=R_1^2</math> means that <math>(R_1-3.5)^2+\frac{351}{4}=R_1^2</math>. Solving for <math>R_1</math> yields that <math>R_1=\frac{100}{7}</math>. Therefore, <math>C=(\frac{51}{7},0)</math> and <math>D=(\frac{151}{14},\frac{\sqrt{351}}{2})</math>. Since segment <math>\overline{AC}</math> is a diameter of <math>\Gamma_2</math>, knowing that A=(-\frac{100}{7}, 0) and <math>C=(\frac{51}{7},0)</math>, <math>AC=\frac{151}{7}</math>, so the center of <math>\Gamma_2</math> is the midpoint of <math>\overline{AC}</math>, which is <math>(-3.5,0)</math>, and it has radius <math>\frac{AC}{2}=\frac{151}{14}</math>. We then find that the equation of <math>CD</math> is <math>-\frac{\sqrt{351}}{7}x+y+\frac{\sqrt{351}*51}{49}=0</math>. To find the center of circle <math>\omega</math>, we can use the tangencies in the problem. Internal tangency to <math>\Gamma_1</math> and external tangency to <math>\Gamma_2</math> give <math>h^2+k^2=(R_1-r)^2=(\frac{100}{7}-r)^2</math> and <math>(h+3.5)^2+k^2=(R_2+r)^2=(\frac{151}{14}+r)^2</math>, where <math>h,k</math> are the coordinates of the center of circle <math>\omega</math> and <math>r</math> is the radius. Subtracting the two equations gives <math>\frac{7h}{2}+\frac{49}{4}=(\frac{151}{14}+r)^2-(\frac{100}{7}-r)^2=-\frac{17199}{196}+\frac{351r}{7}</math>, so <math>h=\frac{-200}{7}+\frac{702r}{49}</math>. Using the point to line distance formula, we find that <math>k=\frac{20r}{7}+\frac{h\sqrt{351}}{7}-\frac{51\sqrt{351}}{49}</math>. From internal tangency to <math>\Gamma_1</math>, <math>k^2=(\frac{100}{7}-r)^2-h^2</math>. Plugging the expressions for <math>h</math> and <math>k</math> yields that <math>r=\frac{1757}{702}</math>. It is in lowest terms, so <math>m+n=1757+402=\boxed{2459}</math>.
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Place circle <math>\Gamma_1</math> with center <math>O_1=(0,0)</math> and radius <math>R_1</math> so that <math>\overline{AB}</math> is the diameter on the x-axis with <math>A=(-R_1,0)</math>, <math>B=(R_1,0)</math>. Then <math>C</math> lies at <math>(R_1-7,0)</math>. Let point D lie on <math>\Gamma_1</math> and let <math>BD=CD=10</math>. Then <math>(x-R_1)^2+y^2=100</math>, and <math>(X-(R_1-7))^2+y^2=100</math>. Subtracting gives <math>14x-14R_1+49=0</math>, so <math>x=R_1-3.5</math>. Then <math>(x-R_1)^2+y^2=100</math> means that <math>y^2=100-3.5^2=\frac{351}{4}</math>, so <math>D=(R_1-3.5,\frac{\sqrt{351}}{2})</math>. Since <math>D</math> lies on <math>\Gamma_1</math>, <math>x^2+y^2=R_1^2</math> means that <math>(R_1-3.5)^2+\frac{351}{4}=R_1^2</math>. Solving for <math>R_1</math> yields that <math>R_1=\frac{100}{7}</math>. Therefore, <math>C=(\frac{51}{7},0)</math> and <math>D=(\frac{151}{14},\frac{\sqrt{351}}{2})</math>. Since segment <math>\overline{AC}</math> is a diameter of <math>\Gamma_2</math>, knowing that <math>A=(\frac{-100}{7}, 0)</math> and <math>C=(\frac{51}{7},0)</math>, <math>AC=\frac{151}{7}</math>, so the center of <math>\Gamma_2</math> is the midpoint of <math>\overline{AC}</math>, which is <math>(-3.5,0)</math>, and it has radius <math>\frac{AC}{2}=\frac{151}{14}</math>. We then find that the equation of <math>CD</math> is <math>-\frac{\sqrt{351}}{7}x+y+\frac{\sqrt{351}*51}{49}=0</math>. To find the center of circle <math>\omega</math>, we can use the tangencies in the problem. Internal tangency to <math>\Gamma_1</math> and external tangency to <math>\Gamma_2</math> give <math>h^2+k^2=(R_1-r)^2=(\frac{100}{7}-r)^2</math> and <math>(h+3.5)^2+k^2=(R_2+r)^2=(\frac{151}{14}+r)^2</math>, where <math>h,k</math> are the coordinates of the center of circle <math>\omega</math> and <math>r</math> is the radius. Subtracting the two equations gives <math>\frac{7h}{2}+\frac{49}{4}=(\frac{151}{14}+r)^2-(\frac{100}{7}-r)^2=-\frac{17199}{196}+\frac{351r}{7}</math>, so <math>h=\frac{-200}{7}+\frac{702r}{49}</math>. Using the point to line distance formula, we find that <math>k=\frac{20r}{7}+\frac{h\sqrt{351}}{7}-\frac{51\sqrt{351}}{49}</math>. From internal tangency to <math>\Gamma_1</math>, <math>k^2=(\frac{100}{7}-r)^2-h^2</math>. Plugging the expressions for <math>h</math> and <math>k</math> yields that <math>r=\frac{1757}{702}</math>. It is in lowest terms, so <math>m+n=1757+402=\boxed{2459}</math>.
  
 
==See Also==
 
==See Also==
 
{{iTest box|year=2006|num-b=37|num-a=39|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
 
{{iTest box|year=2006|num-b=37|num-a=39|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
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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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