Question

Circle $C$ with radius $2$ has diameter $\overline{AB}.$ Circle $D$ is internally tangent to circle $C$ at $A.$ Circle $E$ is internally tangent to circle $C,$ externally tangent to circle $D,$ and tangent to $\overline{AB}.$ The radius of circle $D$ is three times the radius of circle $E$ and can be written in the form $\sqrt{m} - n,$ where $m$ and $n$ are positive integers. Find $m + n.$

Answer

Answer: 254

Concepts:tangent circles Pythagorean Theorem radical

Difficulty rating: 2710

Solution:

Let $C,$ $D,$ $E$ also name the circles' centers, let $s$ be the radius of circle $E,$ so circle $D$ has radius $3s,$ and let $F$ be the foot of $E$ on $\overline{AB}.$ Tangency gives $CE = 2 - s,$ $DE = 3s + s = 4s,$ and $EF = s,$ while $D$ lies on $\overline{AB}$ with $DC = 2 - 3s.$

Right triangles $CEF$ and $DEF$ give $CF = \sqrt{(2-s)^2 - s^2}$ $= \sqrt{4 - 4s}$ and $DF = \sqrt{(4s)^2 - s^2} = s\sqrt{15}.$ Since $F$ is on the opposite side of $C$ from $A,$ we have $DF = DC + CF,$ so $s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s}.$

Moving $2 - 3s$ to the left and squaring gives $24s^2 - 8s = 2\sqrt{15}\,s\,(2 - 3s),$ i.e. $12s - 4 = \sqrt{15}\,(2 - 3s);$ squaring again yields $9s^2 + 84s - 44 = 0,$ so $s = \frac{-14 + 4\sqrt{15}}{3}.$ The radius of circle $D$ is $3s = 4\sqrt{15} - 14 = \sqrt{240} - 14,$ and $m + n = 240 + 14 = 254.$

2014 AIME II Problem 8

Answer: 254

Difficulty rating: 2710

Solution:

Problem 8 in Other Years