A correct solution came from Bradley University students Scott Small, Limin Liang. Other correct solutions were submitted by Ahmed AboAdham, Cee Ann Franklin, Dan Dima, Bill Webb, Lisa Schechner, Keith Anker, Paul Botham, Bruce Branscom, Lou Cairoli, Juan Carlos Marivela, Iñigo Picaza, Lorenzo Pini, Angel Plaza, Sunil Sangwal, Ives Thiry.  A special commendation to Larry Bell for a particularly nice solution.  

Let's first consider a square C of side 2a with center at the origin and the next smaller square D in the upper right hand corner. Circle C has radius a and equation x² + y² = a² while C and D intersect on the line x = y which occurs when x = y = aÖ2/2.  Let (d,d) be the center of circle D with radius R.  It's easy to see that R = a - d, so D has equation (x - d)² + (y - d)² = (a - d)².  Substituting (aÖ2/2,aÖ2/2), which lies on D, into this equation and solving for d gives d = 2a(Ö2 - 1); therefore a - d = a(Ö3 - 2Ö2).

Now we return to the specific situation.  The first circle has radius a₀ = 1/2.  The next smaller circle therefore has radius a₁ = a₀(3 - 2Ö2).  The third circle has radius a₂ = a₁(3 - 2Ö2) and, in general, a_k = a_k-1(3 - 2Ö2) = a₀(3 - 2Ö2)^k = (1/2)(3 - 2Ö2)^k.

The total area occupied by the circle is given by , which is a geometric series converging to .  Similarly, the circumference is given by .

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©2005 Alberto L. Delgado