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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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Circles <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 centers 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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== 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>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 angle triangles <math>\triangle HO_1T_1 \sim \triangle HO_2T_2</math>,
 +
 
 +
<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>
 +
 
 +
<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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== Solution ==
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and by Pythagorean theorem for the right angle triangle <math>\triangle HO_1T_1</math>,
 +
 
 +
<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>
 +
 +
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>,
 +
 
 +
<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>
 +
 
 +
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
 +
 
 +
<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>
 +
 
 +
<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>
 +
 +
The power of the point H to the circle <math>C_3</math> is equal to
 +
 
 +
<math>HA \cdot HB = (HO - r)(HO + r) = HO^2 - r^2 =</math>
 +
 
 +
<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>
 +
 
 +
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:
 +
 
 +
<math>63 x^2 - 696 x \sqrt{10} + 11200 = ax^2 + bx + c = 0</math>
  
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{{solution}}
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Since HA, HB are the 2 roots of this quadratic equation, their difference is
  
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== See Also ==
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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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*[[2005 AIME II Problems/Problem 7| Previous problem]]
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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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*[[2005 AIME II Problems/Problem 9| Next problem]]
 
−
*[[2005 AIME II Problems]]
 
  
 +
== See also ==
 +
{{AIME box|year=2005|n=II|num-b=7|num-a=9}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 19:28, 21 March 2007

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

Let $O_1, O_2$ be the centers and $r_1 = 4, r_2 = 10$ the radii of the circles $C_1, C_2$, $O_1O_2 = r_1 + r_2 = 14$. Let the common external tangent of $C_1, C_2$ meet the center line $O_1O_2$ at a point H. Let $T_1, T_2$ be the tangency points of the circles $C_1, C_2$ with their common external tangent. From the similar right angle triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2$,

$\frac{r_2}{r_1} = \frac{O_2T_2}{O_1T_1} = \frac{HO_2}{HO_1} = \frac{HO_1 + O_1O_2}{HO_1} = 1 + \frac{O_1O_2}{HO_1}$

$HO_1 = O_1O_2 \cdot \frac{r_1}{r_2 - r_1} = r_1 \cdot \frac{r_2 + r_1}{r_2 - r_1}$

and by Pythagorean theorem for the right angle triangle $\triangle HO_1T_1$,

$HT_1 = \sqrt{HO_1^2 - O_1T_1^2} = r_1 \sqrt{\left(\frac{r_2 + r_1}{r_2 - r_1}\right)^2 - 1} = 2r_1\ \frac{\sqrt{r_1r_2}}{r_2 - r_1}$

Let O be the center and $r = r_1 + r_2 = 14$ the radius of the circle $C_3$. Since $O_1O = r - r_1 = r_2$,

$HO = HO_1 + O_1O = r_1 \cdot \frac{r_2 + r_1}{r_2 - r_1} + r_2 = \frac{r_1^2 + r_2^2}{r_2 - r_1}$

Let the common external tangent $T_1T_2$ of the circles $C_1, C_2$ intersects the circle $C_3$ 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 $HT = \frac{HA + HB}{2}$. Since the angle $\angle HTO = 90^\circ$ is right, the right angle triangles $\triangle HOT \sim \triangle HO_1T_1$ are similar, $\frac{HT}{HT_1} = \frac{HO}{HO_1}$ and

$HA + HB = 2 HT = 2 HT_1 \cdot \frac{HO}{HT_1} = \frac{4 \sqrt{r_1r_2}}{r_2 - r_1} \cdot \frac{r_1^2 + r_2^2}{r_1 + r_2} =$

$= 4 \sqrt{r_1r_2}\ \frac{r_1^2 + r_2^2}{r_2^2 - r_1^2} = \frac{232 \sqrt{10}}{21}$

The power of the point H to the circle $C_3$ is equal to

$HA \cdot HB = (HO - r)(HO + r) = HO^2 - r^2 =$

$= \frac{(r_1^2 + r_2^2)^2 - (r_1 + r_2)^2(r_2 - r_1)^2}{(r_2 - r_1)^2} = \frac{4 r_1^2 r_2^2}{(r_2 - r_1)^2} = \frac{1600}{9}$

Substituting $HB = \frac{1600}{9 HA}$ or $HA = \frac{1600}{9 HB}$ into the formula for HA + HB leads to the same quadratic equation for HA or HB:

$63 x^2 - 696 x \sqrt{10} + 11200 = ax^2 + bx + c = 0$

Since HA, HB are the 2 roots of this quadratic equation, their difference is

$AB = HB - HA = \frac{\sqrt{b^2 - 4ac}}{a} = \frac{\sqrt{696^2 \cdot 10 - 4 \cdot 63 \cdot 11200}}{63} =$

$= \frac{8\sqrt{29^2 \cdot 10 - 7 \cdot 700}}{21} = \frac{8\sqrt{3510}}{21} = \frac{8\sqrt{390}}{7}$

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AIME Problems and Solutions
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