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 $AE$ = $p$ . Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$ . $AC = 8$ , and it can easily be shown that angle $CAE = 60$ degrees. Using the Law of Cosines on triangle $CAE$ , we obtain

$(6+p)^2 =p^2 + 64 - 2(8)(p) \cos 60$ .

The $2$ and the $\cos 60$ terms cancel out:

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

$12p+ 36 = 64 - 8p$

$p =\frac {28}{20} = \frac {7}{5}$ . The radius of circle $E$ is $4 + \frac {7}{5} = \frac {27}{5}$ , so the answer is $27 + 5 = \boxed{032}$ .

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

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