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 Math Help > Math Puzzles > Circles Puzzles > Circles Puzzle 1 -- hierarchy of circles > Circles Puzzle 1

# Answer to the Circles Puzzle

Take a circle and inscribe 4 interior-disjoint 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 x-y 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. r*sqrt(2)+r=1 r*(sqrt(2)+1)=1 r=1/(sqrt(2)+1) r=(sqrt(2)-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 r^2 = (sqrt(2)-1)^2 = 3 - 2sqrt(2) The radius of the small circle centered at the origin, C17, is 1 minus twice the radius of circle A, which is 1-2(sqrt(2)-1) 1-2sqrt(2)+2 3-2sqrt(2) It's the same radius as the 16 red circles.

### Related pages in this website

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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