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

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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
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Let <math>X</math> 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.
 
(6+m)^2 = m^2 + 64 - 2(8)(m) cos 60.

Revision as of 19:37, 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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