You have $n+1$ points in $n$-dimensional Euclidean space, such that the mutual distance between each pair of points is the same. Now imagine an $n$-dimensional generalization of a sphere, such that all of these $n+1$ points lie on this sphere. Let's call this circum-hypersphere. Similarly, we can define an in-hypersphere, such that the midpoints of each $n$-point subset lie on the in-hypersphere.

How man dimensions $n$ are needed to make the $n$-dimensional hypervolume of the circum-hypersphere more than a billion times larger than the one of the in-hypersphere?

Probably this way of asking the question is a bit abstract, but here are some examples in low-dimensional spaces to explain what I mean:

  • In $n = 2$ dimensions we have $n+1=3$ points. If we arrange them such that the mutual distance between each pair is the same, it results in an equilateral triangle. The "circum-hypersphere" is also called circumscribed circle is this case and the "hypersphere that contains all midpoints of $n$-point subsets" is the incribed circle, as it contains all midpoints of the triangle edges.
  • In $n = 3$ dimensions the $n+1=4$ points form a regular tetrahedron. The circumsphere is again trivial and the insphere is constructed such that it touches the midpoints of all the triangular faces.

This problem seems to be hard in terms of the length of the required calculations, but there is an easy way. Hence, the answer with the simplest explanation will be accepted.


1 Answer 1


Let $R$ denote the radius of the circum-hypersphere and $r$ the radius of the in-hypersphere. We claim that


To prove this, let $x_1,x_2,\cdots ,x_{n+1}$ be the $n+1$ points (these are really vectors in $\mathbb R^n$). They form a $n$ dimensional simplex. Now it's clear from symmetry that

the centers of both hyperspheres are at $O=\tfrac{x_1+\cdots +x_{n+1}}{n+1}$, and $r$ is the distance from the center to the centroid of the set $\{x_2,\cdots ,x_{n+1}\}$ (call this point $P_1$), and $R$ is the distance from the centre to $x_1$ (call this point $X_1$). But $$\frac{x_1+n\cdot\tfrac{x_2+\cdots + x_{n+1}}{n}}{n+1}=O.$$ So by section formula, $O$ divides $X_1P_1$ in the ratio $n:1$, so $$\frac{R}{r}=\frac{X_1O}{OP_1}=n.$$

This proves our claim. Now of course,

$n-$hypervolume of a hypersphere is proportional to the $n-$th power of the radius, so the ratio of hypervolumes of the hyperspheres here is $n^n:1$. We need to find the smallest $n$ so that $n^n$ is bigger than a billion, which is $10^9$. Clearly $9^9<10^9$, whereas $10^{10}>10^9$, so $10$ is our answer. $\blacksquare$

  • $\begingroup$ And there was me trying to do a pretty geometric proof using the natural right-angled triangle with $R$ as hypotenuse and $r$ as one of the other two sides ... :-( $\endgroup$ Commented Dec 17, 2017 at 14:59

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