2009 AIME II Problems/Problem 5

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 = $\boxed{032}$ .

2009 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2009 AIME II Problems/Problem 5

Problem 5

Solution

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