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Difference between revisions of "2005 AIME II Problems/Problem 8"

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== Problem ==
 
== Problem ==
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[[Circle]]s <math> C_1 </math> and <math> C_2 </math> are externally [[tangent]], and they are both internally tangent to circle <math> C_3. </math> The radii of <math> C_1 </math>  and <math> C_2 </math> are 4 and 10, respectively, and the [[center]]s of the three circles are all [[collinear]]. A [[chord]] of <math> C_3 </math> is also a common external tangent of <math> C_1 </math> and <math> C_2. </math> Given that the length of the chord is <math> \frac{m\sqrt{n}}p </math> where <math> m,n, </math> and <math> p </math> are positive integers, <math> m </math> and <math> p </math> are [[relatively prime]], and <math> n </math> is not divisible by the [[square]] of any [[prime]], find <math> m+n+p. </math>
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[[Circle]]s <math> C_1 </math> and <math> C_2 </math> are externally [[tangent]], and they are both internally tangent to circle <math> C_3. </math> The radii of <math> C_1 </math>  and <math> C_2 </math> are 4 and 10, respectively, and the [[center]]s of the three circles are all [[collinear]]. A [[chord]] of <math> C_3 </math> is also a common external tangent of <math> C_1 </math> and <math> C_2. </math> Given that the length of the chord is <math> \frac{m\sqrt{n}}p </math> where <math> m,n, </math> and <math> p </math> are positive integers, <math> m </math> and <math> p </math> are [[relatively prime]], and <math> n </math> is not divisible by the square of any [[prime]], find <math> m+n+p. </math>
  
 
== Solution ==
 
== Solution ==
−
Let <math>O_1, O_2</math> be the centers and <math>r_1 = 4, r_2 = 10</math> the radii of the circles <math>C_1, C_2</math>, <math>O_1O_2 = r_1 + r_2 = 14</math>. Let the common external tangent  of <math>C_1, C_2</math> meet the center line <math>O_1O_2</math> at a point H. Let <math>\displaystyle T_1, T_2</math> be the tangency points of the circles <math>C_1, C_2</math> with their common external tangent. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2</math>,
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<center>
 +
<asy>
 +
pointpen = black;
 +
pathpen = black + linewidth(0.7);
 +
size(200);
  
−
<math>\frac{r_2}{r_1} = \frac{O_2T_2}{O_1T_1} = \frac{HO_2}{HO_1} = \frac{HO_1 + O_1O_2}{HO_1} = 1 + \frac{O_1O_2}{HO_1}</math>
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pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));
 +
path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H);
 +
pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t);
  
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<math>HO_1 = O_1O_2 \cdot \frac{r_1}{r_2 - r_1} = r_1 \cdot \frac{r_2 + r_1}{r_2 - r_1}</math>
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draw(cir1); draw(cir2); draw(cir3);
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draw((14,0)--(-14,0));
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draw(A--B);
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MP("H",H,W);
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draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4"));
  
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and by Pythagorean theorem for the right angle triangle <math>\triangle HO_1T_1</math>,
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draw(MP("O_1",C1));
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draw(MP("O_2",C2));
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draw(MP("O_3",C3));
  
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<math>HT_1 = \sqrt{HO_1^2 - O_1T_1^2} = r_1 \sqrt{\left(\frac{r_2 + r_1}{r_2 - r_1}\right)^2 - 1} = 2r_1\ \frac{\sqrt{r_1r_2}}{r_2 - r_1}</math>
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draw(MP("T",T,N));
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draw(MP("A",A,NW));
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Let O be the center and <math>r = r_1 + r_2 = 14</math> the radius of the circle <math>C_3</math>. Since <math>O_1O = r - r_1 = r_2</math>,
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draw(MP("B",B,NE));
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draw(C1--MP("T_1",T1,N));
  
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<math>HO = HO_1 + O_1O = r_1 \cdot \frac{r_2 + r_1}{r_2 - r_1} + r_2 = \frac{r_1^2 + r_2^2}{r_2 - r_1}</math>
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draw(C2--MP("T_2",T2,N));
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draw(C3--T);
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draw(rightanglemark(C3,T,H));
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</asy></center>
  
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Let the common external tangent <math>T_1T_2</math> of the circles <math>C_1, C_2</math> intersects the circle <math>C_3</math> at points A, B, so that the points H, A, B follow on the tangent in this order. Let T be the midpoint of the chord AB. Then <math>HT = \frac{HA + HB}{2}</math>. Since the angle <math>\angle HTO = 90^\circ</math> is right, the right angle triangles <math>\triangle HOT \sim \triangle HO_1T_1</math> are similar, <math>\frac{HT}{HT_1} = \frac{HO}{HO_1}</math> and
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Let <math>O_1, O_2, O_3</math> be the centers and <math>r_1 = 4, r_2 = 10,r_3 = 14</math> the radii of the circles <math>C_1, C_2, C_3</math>. Let <math>T_1, T_2</math> be the points of tangency from the common external tangent of <math>C_1, C_2</math>, respectively, and let the extension of <math>\overline{T_1T_2}</math> intersect the extension of <math>\overline{O_1O_2}</math> at a point <math>H</math>. Let the endpoints of the chord/tangent be <math>A,B</math>, and the foot of the perpendicular from <math>O_3</math> to <math>\overline{AB}</math> be <math>T</math>. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T </math>,
  
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<math>HA + HB = 2 HT = 2 HT_1 \cdot \frac{HO}{HT_1} = \frac{4 \sqrt{r_1r_2}}{r_2 - r_1} \cdot \frac{r_1^2 + r_2^2}{r_1 + r_2} =</math>
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<cmath> \frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}. </cmath>
  
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<math>= 4 \sqrt{r_1r_2}\ \frac{r_1^2 + r_2^2}{r_2^2 - r_1^2} = \frac{232 \sqrt{10}}{21}</math>
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It follows that <math>HO_1 = \frac{28}{3}</math>, and that <math>O_3T = \frac{58}{7}\dagger</math>. By the [[Pythagorean Theorem]] on <math>\triangle ATO_3</math>, we find that
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The power of the point H to the circle <math>C_3</math> is equal to
 
  
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<math>HA \cdot HB = (HO - r)(HO + r) = HO^2 - r^2 =</math>
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<cmath>AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}</cmath>
  
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<math>= \frac{(r_1^2 + r_2^2)^2 - (r_1 + r_2)^2(r_2 - r_1)^2}{(r_2 - r_1)^2} = \frac{4 r_1^2 r_2^2}{(r_2 - r_1)^2} = \frac{1600}{9}</math>
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and the answer is <math>m+n+p=\boxed{405}</math>.
  
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Substituting <math>HB = \frac{1600}{9 HA}</math> or <math>HA = \frac{1600}{9 HB}</math> into the formula for HA + HB leads to the same quadratic equation for HA or HB:
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<math>\dagger</math> Alternatively, drop an altitude from <math>O_1</math> to <math>O_3T</math> at <math>O_3'</math>, and to <math>O_2T_2</math> at <math>O_2'</math>. Then, <math>O_2O_2'=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}</math>. From rectangles, <math>O_3'T=O_1T_1=4</math> so <math>O_3T=4+\frac{30}{7}=\frac{58}{7}</math>.
  
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<math>63 x^2 - 696 x \sqrt{10} + 11200 = ax^2 + bx + c = 0</math>
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==Solution 2==
  
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Since HA, HB are the 2 roots of this quadratic equation, their difference is
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Call our desired length <math>x</math>. Note for any <math>X</math> on <math>\overline{AB}</math> and <math>Y</math> on <math>\overline{O_1O_2}</math> such that <math>\overline{XY}\perp\overline{AB}</math> that the function <math>f</math> such that <math>f(\overline{O_1Y})=\overline{XY}</math> is linear. Since <math>(0,4)</math> and <math>(14,10)</math>, we can quickly interpolate that <math>f(10)=\overline{O_3T}=\frac{58}{7}</math>. Then, extend <math>\overline{O_3T}</math> until it reaches the circle on both sides; call them <math>P,Q</math>. By Power of a Point,
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<math>\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}</math>.
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<math>AB = HB - HA = \frac{\sqrt{b^2 - 4ac}}{a} = \frac{\sqrt{696^2 \cdot 10 - 4 \cdot 63 \cdot 11200}}{63} =</math>
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Since <math>\overline{AT}=\overline{TB}=\frac{1}{2}x</math>,
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<cmath>(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2</cmath>
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<math>= \frac{8\sqrt{29^2 \cdot 10 - 7 \cdot 700}}{21} = \frac{8\sqrt{3510}}{21} = \frac{8\sqrt{390}}{7}</math>
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<cmath>\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2</cmath>
 +
After solving for <math>x</math>, we get <math>x=\frac{8\sqrt{390}}{7}</math>, so our answer is <math>8+390+7=\boxed{405}</math>
  
 
== See also ==
 
== See also ==
Line 43: Line 58:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:03, 28 December 2021

Problem

Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

Solution

[asy] pointpen = black; pathpen = black + linewidth(0.7); size(200); pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); draw(cir1); draw(cir2); draw(cir3); draw((14,0)--(-14,0)); draw(A--B); MP("H",H,W); draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4")); draw(MP("O_1",C1)); draw(MP("O_2",C2)); draw(MP("O_3",C3)); draw(MP("T",T,N)); draw(MP("A",A,NW)); draw(MP("B",B,NE)); draw(C1--MP("T_1",T1,N)); draw(C2--MP("T_2",T2,N)); draw(C3--T); draw(rightanglemark(C3,T,H)); [/asy]

Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$. Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$, respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$. Let the endpoints of the chord/tangent be $A,B$, and the foot of the perpendicular from $O_3$ to $\overline{AB}$ be $T$. From the similar right triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$,

\[\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.\]

It follows that $HO_1 = \frac{28}{3}$, and that $O_3T = \frac{58}{7}\dagger$. By the Pythagorean Theorem on $\triangle ATO_3$, we find that

\[AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}\]

and the answer is $m+n+p=\boxed{405}$.

$\dagger$ Alternatively, drop an altitude from $O_1$ to $O_3T$ at $O_3'$, and to $O_2T_2$ at $O_2'$. Then, $O_2O_2'=10-4=6$, and $O_1O_2=14$. But $O_1O_3O_3'$ is similar to $O_1O_2O_2'$ so $O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}$. From rectangles, $O_3'T=O_1T_1=4$ so $O_3T=4+\frac{30}{7}=\frac{58}{7}$.

Solution 2

Call our desired length $x$. Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$, we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$. Then, extend $\overline{O_3T}$ until it reaches the circle on both sides; call them $P,Q$. By Power of a Point, $\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}$. Since $\overline{AT}=\overline{TB}=\frac{1}{2}x$, \[(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2\] \[\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2\] After solving for $x$, we get $x=\frac{8\sqrt{390}}{7}$, so our answer is $8+390+7=\boxed{405}$

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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