a) The following recurrence between one radius and the next one: r sub (k + 1) = lambda times r sub k, where lambda = (sqrt 2 – 1)/(sqrt 2 + 1) < 1.

b) The total length of the circles is 2.pi.sum of r sub k = 2.pi.r sub 1/(1 – lambda) = 1,06 times length of greatest circle. Here r sub 1 = a/2, and a = side of the first square.

c) The total area is pi times a squared times (9/16)(3sqrt 2 + 2)/14).

In any case, both the total length and the total area converge. each of the sum is the sum of an infinite geometric series; the ratios are lambda and lambda squared, respectively.

The total area can also be obtained approximately by using the above recurrence—squared— and adding up. After multiplication by pi—and neglecting the term [r sub (n + 1)]*2—, the above result follows. ¡Remarkable! ]]>

My approach was to suppose the big circle has radius 1, find the radius of the 2nd circle, and square it. That gives you the ratio for an infinite geometric series, with the first term being pi/4 for the area of the first circle relative to the square. The figure quoted above it its sum. That 2nd radius (call it r) can be found by observing that the distance along the y-axis between the circles’ tangent point and the top of the square (call it h) is 1- sqrt(2)/2. h is also equal to r*(1+sqrt(2)/2), and then it’s trivial to solve for r.

]]>Kind regards. ]]>

If the first radius R is 1, then the sum of areas tends to 3,2368… (in other words: pi*R2*(3+2*SQRT(2))/(4*SQRT(2)) and the sum of the circles tends to infinity.

Its a finite area contained in an infinite limit.

Nice problem and nice page!

]]>#2: round peg in square hole is easy, pi/4. square peg in round hole is 2/pi since the inscribed square is half the area of the circumscribed. Then pi/4 is more than 3/4 and 2/pi is less than 3/4, so the round peg fits better.

#3: I think I’ll come back to this one sometime when I have scratch paper. I don’t see any beautiful shortcut here since there’s only one square, but maybe I’m missing something nice.

]]>