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

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<math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m) cos 60</math>.
 
<math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m) cos 60</math>.
  
−
The <math>2 and the </math>cos 60<math> cancel out:
+
The <math>2</math> and the <math>cos 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>
  
−
</math>12m<math> + </math>36<math> = </math>64<math> - </math>8m<math>
+
<math>12m</math> + <math>36</math> = <math>64</math> - <math>8m</math>
  
−
</math>m<math> = </math>28/20<math> = </math>7/5<math>. The radius of circle </math>E<math> is </math>4<math> + </math>7/5<math> = </math>27/5<math>, so the answer is </math>27<math> + </math>5<math> = </math>\boxed{032}$.
+
<math>m</math> = <math>28/20</math> = <math>7/5</math>. The radius of circle <math>E</math> is <math>4</math> + <math>7/5</math> = <math>27/5</math>, so the answer is <math>27</math> + <math>5</math> = <math>\boxed{032}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2009|n=II|num-b=4|num-a=6}}

Revision as of 19:40, 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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