In the diagram below, in between two circles, each with a radius of 1 meter, three smaller circles are squeezed in, centred at C1 , C2 and C3. If you imagine it was possible to continue drawing and squeezing in more circles until you had a billion of them, how much space would be left between the top of the smallest circle and point A? And what ubiquitous thing’s diameter is very close to that distance? What could fit into the tiny space between point A and the smallest of a billion circles squeezed into the space between the two large circles, one of which is centered at B? Diagram by the author.

This is a variation of an old problem, and although the situation is often used as an example of a non-geometric series in calculus books, it could also be solved with some basic algebra and induction. And for that reason, versions of the puzzle have appeared in math competitions for bright kids who have yet to venture into calculus.

Let’s begin by solving for the radius of the first circle (the red one centered at C1). If we come with expressions or numbers for all three sides,  we could apply the Pythagorean theorem to triangle ABC1.

Notice that AC1 ‘s length can simply be obtained by subtracting the radius of the red circle from the 1-meter length of the square’s side, which we obtain from the radius of the large, identical circles.

AC1 = 1 – r1.

Since AB² + AC1 ² = BC1 ²,  where the hypotenuse, BC1 , is simply the sum of the radii, we obtain:

1² + (1 – r1) ² = (1 + r1) ²

Solving for r1 yields  r1 = 1/4 so the red circle’s diameter = d1  = 2 * 1/4 =  1/2 .

Not to bore you, we will walk through the steps only one more time to obtain the diameter, d2, of the blue circle centered at C2.

We add the red circle’s obtained diameter of  1/2 to the radius of  the blue circle, r2 , and then subtract the sum from 1 meter to obtain AC2.

AC2 = 1 – (1/2 + r2 )=   1/2 – r2

Then we apply the Pythagorean theorem to the second triangle, ABC2,

AB² + AC2 ² = BC2 ²,  or:

1² + ( 1/2 – r2) ² = (1 + r2) ²

1 + 1/4 – r2 + r2 ² = 1 + 2r2 + r2 ²

1/4 = 3r2

r2 = 1/12

so the blue circle’s diameter = 2 * 1/12 = d2 = 1/6.

As promised, without going through the similar details, d3 = 1/12 and for a fourth circle that we squeeze in, d4 will be 1/20.

Now a pattern becomes apparent. For the diameter of the nth circle, dn,  dn = 1/n(n+1). But if you did not realize that, you could still solve the problem. After all, we’re interested in the leftover distance between the top of the billionth circle and point A.

After we squeezed in one circle , we had a diameter of  1/2 . After squeezing in a pair, the sum of the two circles’ diameters 1/21/6 = 2/3. Squeeze in three circles and the sum is  1/21/6 + 1/12 = 3/4 .  A fourth insertion yields 1/21/6 + 1/12 + 1/20 = 48/60 = 4/5.

So for squeezing in an n-number of  circles, the sum of the diameters is n/n+1 and since the side of the square is 1 meter in length, the remaining distance will be 1 – n/n+1. Using a common denominator, that expression is equal to n+1 -n/n+1 = 1/n+1.  For a billion circles then, the leftover distance would be

1/1 000 000 000 +1 = 1/1 000 000 001 meters,

which is extremely close to being to a nanometer.