A vicious circle
The ends of the diameter of a unit circle are joined to make a smaller circle. The ends of the diameter of this smaller circle are joined to make an even smaller circle. And so on indefinitely.
If the radii of all the circles (including the initial unit circle) are laid end to end in a line, and the ends of that line are joined to make a circle, what is the radius of the circle?
The diameter of each circle becomes the circumference of the next smaller circle down.
Let C1 be the circumference of the unit circle.
i.e. C1 = 2πR1, where R1 is the radius of the unit circle, so R1 = 1.
The diameter, D1 = 2R1
And this is the circumference of the 2nd circle, C2. i.e. C2 = D1.
C2 = 2πR2 = D1 = 2R1. i.e. R2 = R1/π
Continuing ...
C3 = 2πR3 = D2 = 2R2. i.e. R3 = R2/π
C4 = 2πR4 = D3 = 2R3. i.e. R4 = R3/π
C5 = 2πR5 = D4 = 2R4. i.e. R5 = R4/π
Cn = 2πRn = D_(n-1) = 2R_(n-1). i.e. Rn = R_(n-1)/π
If the radii of all these circles were laid end to end, the total length would be,
L = R1 + R2 + R3 + ... + Rn, n = 1 to ∞,
L = R1 + R1/π + R2/π + R3/π + ... + Rn/π, n = 1 to ∞,
L = R1 + (1/π){R1 + R2 + R3 + ... + Rn}, n = 1 to ∞,
L = R1 + (1/π)*L
L – L/π = R1
L(π – 1) = πR1
L = πR1/(π-1)
This length, of all the radii joined together, now becomes the circumference of a final circle.
The radius of this final circle now is, Rf = L/(2π)
Rf = (R1/2)/(π – 1)
Since R1 is the radius of a unit circle, then R1 = 1 and Rf = 0.5/(π – 1) = 0.23347