Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.9 AMC 10/12 Find the answer

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 $m+n$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

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 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 $27 + 5 = \boxed{032}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.