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Revision as of 23:36, 4 July 2013

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:

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

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

$m$ = $\frac {28}{20}$ = $\frac {7}{5}$. 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 (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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