Unintuitive

Geometry Level 5

Consider a square partitioned into 4 smaller squares of equal size, and in each square, draw a circle that touches its sides. Now draw a circle at the center that touches all 4 circles:

We could take this up into the 3 $^\text{rd}$ dimension. Take a cube, partition it into 8 smaller cubes of equal size, and in each cube, inscribe a sphere that touches the smaller cube's sides. Now draw a sphere at the center that touches all 8 spheres.

Now let's generalize it to $n$ -dimensions. We take an $n$ -cube $\big(n^\text{th}$ dimensional hypercube $\big)$ , partition it into $2^n$ smaller $n$ -cubes of equal size, and in each smaller $n$ -cube, inscribe an $n$ -sphere $\big(n^\text{th}$ dimensional hypersphere $\big)$ that touches all the smaller $n$ -cube's sides. Now generate another $n$ -sphere at the center such that it touches all $2^n$ $n$ -spheres.

Here's the question: What is the lowest dimension $n \geq 1$ where the center $n$ -sphere (that touches the other $2^n$ spheres) intersects the sides of the larger $n$ -cube as well? (The "larger $n$ -cube" refers to the $n$ -cube made up of the $2^n$ smaller cubes.)

If you think that this never happens, enter the answer of 0.

The answer is 9.

1 solution

Mark Hennings
Dec 28, 2016

Consider the original cube $[-2,2]^N$ (of side length $4$ ) subdivided into $2^N$ sub-cubes (each of side $2$ ). These sub-cubes are inscribed by the spheres $S_{\mathbf{\epsilon}}$ , for $\mathbf{\epsilon} \in \{-1,1\}^N$ , where $S_{\mathbf{\epsilon}} \; =\; \big\{ \mathbf{x} \in \mathbb{R}^N \, \big|\, \Vert\mathbf{x} - \mathbf{\epsilon}\Vert^2 \;\le 1 \big\}$ Each of these spheres are of radius $1$ . Diametrically opposite spheres are $S_{\mathbf{\epsilon}}$ and $S_{\mathbf{-\epsilon}}$ , and the centres of these spheres are a distance $\Vert \mathbf{\epsilon} - (\mathbf{-\epsilon})\Vert \; = \; 2\Vert\mathbf{\epsilon}\Vert \; = \; 2\sqrt{N}$ apart, and hence the centre sphere has diameter $2\sqrt{N}-2$ , and so has radius $\sqrt{N}-1$ . Thus we want to know the value of $N$ for which $\sqrt{N}-1 = 2$ , namely $N = \boxed{9}$ .

Beware of "geometric intuition" at larger dimensions. Some of them are no longer true.

At larger dimensions, the geometric idea that "spheres are round" is no longer true. Sphere end up being very pointy, which is what allows the central sphere to be tangential to the small spheres, but still touch the sides of the large cube.

Calvin Lin Staff - 4 years, 5 months ago

An object which is $O(N)$ -invariant is "pointy"? Pull the other one! It is only "pointy" if you apply the intuition of three dimensions to higher dimensions, the very fault you are warning people about.

The bottom line is that we need to be very careful about what geometrical principles to apply; most of our familiar ones are grounded in three-dimensional intuition

Mark Hennings - 4 years, 5 months ago

Unintuitive

The answer is 9.

1 solution

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