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

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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
 
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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<math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m) cos 60</math>.
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<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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The <math>2</math> and the <math>cos  60</math> cancel out:
+
The <math>2</math> and the cos <math>60</math> cancel out:
  
 
<math>m^2</math> + <math>12m</math> + <math>36</math> = <math>m^2</math> + <math>64</math> - <math>8m</math>
 
<math>m^2</math> + <math>12m</math> + <math>36</math> = <math>m^2</math> + <math>64</math> - <math>8m</math>

Revision as of 19:42, 17 April 2009

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 $60$ cancel out:

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

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

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

See Also

2009 AIME II (Problems • Answer Key • Resources)
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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