In the diagram below, in between two circles, each with a radius of 1 meter, three smaller circles are squeezed in, centred at C_{1} , C_{2 }and C_{3. }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?

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 C_{1}). If we come with expressions or numbers for all three sides, we could apply the Pythagorean theorem to triangle ABC_{1.}

Notice that AC_{1} ‘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.

AC_{1} = 1 – r_{1}.

Since AB² + AC_{1} ² = BC_{1} ², where the hypotenuse, BC_{1} , is simply the sum of the radii, we obtain:

1² + (1 – r_{1}) ² = (1 + r_{1}) ²

Solving for r_{1} yields r_{1} = ^{1}/_{4} so the red circle’s diameter = d_{1} = 2 * ^{1}/_{4} = ^{1}/_{2} .

Not to bore you, we will walk through the steps only one more time to obtain the diameter, d_{2}, of the blue circle centered at C_{2}.

We add the red circle’s obtained diameter of ^{1}/_{2} to the radius of the blue circle, r_{2} , and then subtract the sum from 1 meter to obtain AC_{2.}

AC_{2} = 1 – (^{1}/_{2} + r_{2} )= ^{1}/_{2} – r_{2}

Then we apply the Pythagorean theorem to the second triangle, ABC_{2},

AB² + AC_{2} ² = BC_{2} ², or:

1² + ( ^{1}/_{2} – r_{2}) ² = (1 + r_{2}) ²

1 + ^{1}/_{4} – r_{2} + r_{2} ² = 1 + 2r_{2} + r_{2} ²

^{1}/_{4} = 3r_{2}

r_{2} = ^{1}/_{12}

so the blue circle’s diameter = 2 * ^{1}/_{12} = d_{2} = ^{1}/_{6}.

As promised, without going through the similar details, d_{3} = ^{1}/_{12} and for a fourth circle that we squeeze in, d_{4 } will be ^{1}/_{20}.

Now a pattern becomes apparent. For the diameter of the nth circle, d* _{n}*, d

*=*

_{n}^{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}/_{2} + ^{1}/_{6} = ^{2}/_{3}. Squeeze in three circles and the sum is ^{1}/_{2} + ^{1}/_{6} + ^{1}/_{12} = ^{3}/_{4} . A fourth insertion yields ^{1}/_{2} + ^{1}/_{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.

^{1}/_{1 000 000 000} meters or 1 nanometer, is a little more than three times the size of a single water molecule, and just slightly wider than the ring of one glucose molecule, the fuel for all brains, including those that devised and solved this puzzle. And use up just an extra iota of glucose to realize that the leftover space would be long enough but not wide enough to accommodate the entire molecule. Finally, imagine a circle for every person alive on Earth in 2050, about 9.4 billion, and the distance remaining between the smallest circle and point A will only as big as the diameter of the smallest and most common atom in the universe.

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