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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).

four disks packed in square

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?

hardmath
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user02139502945
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1 Answers1

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Your computation never did anything to make sure the ball would fit in the container.

In high dimensions, the corners of a hypercube get progressively far away from the inscribed hypersphere. This happens particularly to the length-1/2 hypercubes that contain one of the corner spheres each. The center of each hyperface of the large hypercube is where the corners of a number of such small hypercubes, so it gets farther and farther away from the surfaces of the corner spheres.

In high enough dimension, the space between the corner spheres is simply large enough that there's room for the central sphere to poke out between them!

hmakholm left over Monica
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    Thanks for your answer. Would I be right in saying that as the number of dimensions increases, the ratio of the 'volume' of each corner ball to the total hypercube gets smaller and smaller until the total volume occupied by all the corner balls is small enough in comparison that a hypersphere that fits between them has a longer diameter than the side length of the hypercube? I just want to make sure I'm understanding correctly. – user02139502945 Jul 06 '15 at 01:53