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 1

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 angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. 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 $27 + 5 = \boxed{032}$.

Solution 2 (NO TRIGONOMETRY!!!)

Draw $\overline{CD}$ from circle center $C$ to center $D$. Let its midpoint be point $F$.

Draw $\overline{BF}$ perpendicular to line segment $CD$ and intersecting $CD$ at point $F$.

Let $G$ be the common external tangent point of circle $B$ and circle $E$. $\overline{GA} = 4$

Circle centers $A$ and $E$ lie on $\overline{BF}$.

Draw $\overline{AC}$ from circle center $A$ to center $C$. $\overline{AC}$ has length $8$.

Triangle $ACF$ is a right triangle with $\angle ACF = 30^{\circ}$.

(If you take the original equilateral triangle with centroid A and draw the medians, dividing the triangle into 6 30 degree triangles, you find that triangle $ACF$ is similar to one of the triangles).

$\overline{AF} = 4$ and $\overline{CF} = 4\sqrt{3}$.

Let the radius of circle $E$ have length $r$. Draw $\overline{EC}$ from circle center $E$ to center $C$.

$\overline{EC}$ has length $r + 2$. $\overline{EF}$ has length $GA + AF - GE = 8 - r$.

Since $CEF$ is a right triangle, by the Pythagorean theorem $EF^2 + CF^2 = EC^2$.

$(8 - r)^2 + (4\sqrt{3})^2 = (r + 2)^2$. Solving, $r = \frac {27}{5}$ and the answer is $27 + 5 = \boxed{032}$.

-unhappyfarmer

Video Solution

https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh

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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