Difference between revisions of "2009 AIME II Problems/Problem 5"

m (Solution)
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Let <math>X</math> be the intersection of the circles with centers <math>B</math> and <math>E</math>, and <math>Y</math> be the intersection of the circles with centers <math>C</math> and <math>E</math>. Since the radius of <math>B</math> is <math>3</math>, <math>AX</math> = <math>4</math>. Assume <math>AE</math> = <math>m</math>. Then <math>EX</math> and <math>EY</math> are radii of circle <math>E</math> and have length <math>4+m</math>. <math>AC</math> = <math>8</math>, and it can easily be shown that angle <math>CAE</math> = <math>60</math> degrees. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain
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Let <math>X</math> be the intersection of the circles with centers <math>B</math> and <math>E</math>, and <math>Y</math> be the intersection of the circles with centers <math>C</math> and <math>E</math>. Since the radius of <math>B</math> is <math>3</math>, <math>AX =4</math>. Assume <math>AE</math> = <math>m</math>. Then <math>EX</math> and <math>EY</math> are radii of circle <math>E</math> and have length <math>4+m</math>. <math>AC = 8</math>, and it can easily be shown that angle <math>CAE = 60</math> degrees. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain
  
<math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m)</math> cos <math>60</math>.
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<math>(6+m)^2 =m^2 + 64 - 2(8)(m) \cos 60</math>.
  
The <math>2</math> and the cos <math>60</math> cancel out:
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The <math>2</math> and the <math>\cos 60</math> terms cancel out:
  
<math>m^2</math> + <math>12m</math> + <math>36</math> = <math>m^2</math> + <math>64</math> - <math>8m</math>
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<math>m^2 + 12m +36 = m^2 + 64 - 8m</math>
  
<math>12m</math> + <math>36</math> = <math>64</math> - <math>8m</math>
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<math>12m+ 36 = 64 - 8m</math>
  
<math>m</math> = <math>\frac {28}{20}</math> = <math>\frac {7}{5}</math>. The radius of circle <math>E</math> is <math>4</math> + <math>\frac {7}{5}</math> = <math>\frac {27}{5}</math>, so the answer is <math>27</math> + <math>5</math> = <math>\boxed{032}</math>.
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<math>m =\frac {28}{20} = \frac {7}{5}</math>. The radius of circle <math>E</math> is <math>4 + \frac {7}{5} = \frac {27}{5}</math>, so the answer is <math>27 + 5 = \boxed{032}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 20:14, 13 March 2015

Problem 5

Equilateral triangle $T$ is inscribed in circle $A$, which has radius $10$. Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$. Circles $C$ and $D$, both with radius $2$, are internally tangent to circle $A$ at the other two vertices of $T$. Circles $B$, $C$, and $D$ are all externally tangent to circle $E$, which has radius $\dfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4;  pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep};  draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5));  dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy]


Solution

Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX =4$. Assume $AE$ = $m$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+m$. $AC = 8$, and it can easily be shown that angle $CAE = 60$ degrees. Using the Law of Cosines on triangle $CAE$, we obtain

$(6+m)^2 =m^2 + 64 - 2(8)(m) \cos 60$.

The $2$ and the $\cos 60$ terms cancel out:

$m^2 + 12m +36 = m^2 + 64 - 8m$

$12m+ 36 = 64 - 8m$

$m =\frac {28}{20} = \frac {7}{5}$. The radius of circle $E$ is $4 + \frac {7}{5} = \frac {27}{5}$, so the answer is $27 + 5 = \boxed{032}$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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