# Stacking circles with $r=\frac{1}{p}$ inside a circle with $r = 1$

Let's start with a circle with radius $1$. Now suppose we would continuously insert circles from above with radii $\frac{1}{p}$ (first a circle with $r = \frac{1}{2}$, then a circle with $r = \frac{1}{3}$, etc...) so that they would rest at the lowest possible position, meaning that the circles can't cross or overlap each other. To illustrate what I mean, here's an image: My question is if we can keep doing this infinitely without ever exceeding the outer circle.

Now it is known that the sum of squared prime reciprocals converges to approximately $0.452247$, so we know that the total area of all the inner circles is less than half of the area of the outer circle. Looking at the picture then leads me to believe that we can easily fill the circle infinitely from above, but I'm not entirely sure and I'd like to find out if this can be proven mathematically.

• – Lucian Oct 22 '15 at 8:52

## 1 Answer

See my question which asks the same sort of thing, but considers circles with radii $1/n, \ \forall n \in \Bbb{N}$ Can all circles of radius $1/n$ be packed in a unit disk, excluding the circle of radius $1/1$?

Because if that question is true, then obviously circles with radii taken over just the primes would fit within the unit circle.

• Your question neglects gravity, though, doesn't it? – Brian Tung Oct 27 '15 at 0:34
• Yes, but the circles must first fit inside the unit disk in order to occupy their respective lowest possible positions. My question considers only if they can fit or not. – Rob Bland Oct 27 '15 at 0:36
• Maybe I didn't communicate my intention properly: What you say is exactly the observation I meant to make. – Brian Tung Oct 27 '15 at 2:06
• Then yes, you are correct – Rob Bland Oct 27 '15 at 2:07
• Thanks for the link. Indeed your question deals with packing instead of stacking, but it's an interesting read. Perhaps the stacking problem could also be stated with circles of radius 1/n – Marijn Oct 27 '15 at 13:22