# Diameter of a 10-ball in a 10-box is larger than the side length of box?

I came across this idea in a lecture on elementary topology. While it makes sense algebraically, I'm hoping someone could shed some light on the way this is possible.

So you begin with a square of side length $1$ and $4$ disks lying within the square, one in each corner, of diameter $1/2$. In the space at the center, there is another disk, of maximum possible diameter. To find the diameter we recognize that the diagonal length of the square is $\sqrt{2}$, and if one extends the construction in a similar way along the diagonal direction, there is half the center disk's diameter in each corner, making the total $\sqrt{2} = 1+2d$. (The two disks along the diagonal contribute twice their diameter, which comes to 1).

From here we can transpose to get:

$$d = \frac{\sqrt{2}-1}{2}$$

It is not much of a step to see that this generalises to higher dimensions, as the only thing that changes is the term under the square root, due to Pythagoras' theorem.

$$d_m = \frac{\sqrt{m} - 1}{2}$$

Where m is the dimension of the ambient space and all the balls.

However, while the equation is sensible for low dimensionalities, it gives a strange result for $m > 9$, ie.

$$d_{10} = \frac{\sqrt{10} - 1}{2} > 1$$

Here the diameter of the ball exceeds the side length of the container! How is this possible?

• When you say 4 disks in each corner, do you mean the disks are centered at the 4 corners of the square? – TomGrubb Jul 6 '15 at 1:32
• No, I mean four disks which only touch the edges of the square. A picture speaks a thousand words but I don't have one. – user02139502945 Jul 6 '15 at 1:36