
Take a circle and inscribe 4 interiordisjoint congruent circles A, B, C, D
that are cyclically tangent (A is tangent to B is tangent to C is tangent to D
is tangent to A), and so that each is tangent to the large circle. A, B,
C, and D are colored white in the diagram, below. Then, in each of these four,
inscribe four congruent tangent circles in exactly the same way, to get 16 small
circles, colored red in the diagram, below. Finally, inscribe a small circle
C17, colored yellow in the diagram, in the space around the center of the large
circle and tangent to each of A, B, C, D. Which is larger, the yellow C17 or one
of the sixteen red circles inside A, B, C, D?
(Source:
http://mathforum.org/wagon/fall01/p944.html)
Let's draw this whole thing on an xy coordinate system. First, the outer circle. It will be the unit circle, centered at (0,0). Next, we'll draw circle A. Each white circle occupies a 90� sector of the larger circle. For convenience, I'll draw it in the upper right sector. Let r be the radius of the smaller circle. Its center, then, is (r,r) because it touches the axes. The distance from the origin to the center (r,r) is r*sqrt(2). Because the smaller circle is tangent to the larger one, r*sqrt(2)+r=1.

Now the radius of the circles inscribed in A bears the same
relationship to circle A as circle A bears to the unit circle. The ratio of
their radii is the same. So the radius of the inscribed circles is
The radius of the small circle centered at the origin, C17, is 1 minus twice the radius of circle A, which is
It's the same radius as the 16 red circles. 
Descartes' Circle Theorem  Given four mutually tangent circles whose curvatures (curvature is the reciprocal of radius) are k1, k2, k3, and k4, (k1+k2+k3+k4)^2 = 2(k1+k2+k3+k4)
Three Circles Puzzle  only a quarter of the largest circle is visible  four mutually tangent circles
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