Bunuel

Circles A, B and C are externally tangent to each other, and internally tangent to circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the center of D. What is the radius of circle B?
(A) 2/3
(B) \(\frac{\sqrt{3}}{2}\)
(C) 7/8
(D) 8/9
(E) \(\frac{1 + \sqrt{3}}{3}\)
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This is beyond the difficulty level that I'd expect on an actual test (find me an official geometry question that's this difficulty), but who can resist a "Problem Solving Weekend Challenge?"
Let's start by looking at triangle COP.
The radius of D is 2, so OC=2-r.
CP=r
\(CP^2 + PO^2 = OC^2\)
\(r^2 + PO^2 = (2-r)^2\)
\(r^2 + PO^2 = r^2 - 4r + 4\)
\(PO^2 = 4-4r\)
\(PO = \sqrt{4-4r}\)
Let's also look at triangle CAP.
CP=r
AC is the radius of A plus the radius of C, so AC=1+r
AP is the radius of A plus PO, so AP = 1+PO
\(CP^2 + PA^2 = AC^2\)
\(r^2 + (1+PO)^2 = (1+r)^2\)
\(r^2+1+2PO+PO^2 = 1+2r+r^2\)
\(PO^2+2PO=2r\)
Substituting in from the red equations for PO^2 and PO, we get
\((4-4r)+2(\sqrt{4-4r})=2r\)
\(2(\sqrt{4-4r})=6r-4\)
Square both sides
\(4(4-4r)=36r^2-48r+16\)
\(16-16r=36r^2-48r+16\)
\(0=36r^2-32r\)
\(r=32/36\)
\(r=8/9\)
Answer choice D.
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