2009 AIME II Problems/Problem 5

Revision as of 20:14, 13 March 2015 by Mathgeek2006 (talk | contribs) (Solution)

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]


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 $\cos 60$ terms 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 (ProblemsAnswer KeyResources)
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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