Messing around in higher dimensions 1

For the figure above, the following are the coordinates of the vertices of the pentachoron, where all the edge lengths are $\sqrt{20}$ .

$\left( 0,0,2,0 \right)$
$\left( 0,4,4,0 \right)$
$\left( 0,1,5,\sqrt { 10 } \right)$
$\left( 0,3,1,\sqrt { 10 } \right)$
$\left( \dfrac { 5 }{ \sqrt { 2 } } ,2,3,\sqrt { \dfrac { 5 }{ 2 } } \right)$

The last vertex at the bottom of this list is at the center of this figure.

Parallel planes are shown passing through the $5$ vertices having a spacing of $1$

This is solvable using coordinate geometry. Let the five parallel planes be perpendicular to the $w$ -axis, and let the five vertices of the 5-cell be $A,B,C,D,E$ , with coordinates as below:

Rather than make all the points arbitrary, we have put $A$ at the origin, then assigned the other vertices to the parallel hyperplanes with their $w$ -coordinates. We can further orient the 5-cell so that $B$ lies in the $wx$ -plane, and so on. There's no loss of generality in doing this, but it reduces the number of unknowns to the $9$ shown above, together with the edge-length $s$ .

We have ten unknowns; the ten equations we need are simply that the distance between any two vertices must be $s$ , that is, $AB=AC=AD=\cdots=CD=s$

This is a system of quadratic equations; with a bit of manipulation, it's possible to show that one solution is

Other solutions are possible in terms of the vertex positions, but these just amount to reflections and rotations; $s=\sqrt{20}$ is the only valid edge-length, so the required answer is $s^2=\boxed{20}$ .

We work in a 4-dimensional euclidian space with a standard orthonormal standard basis.

We could rephrase the problem as: Given a 5-cell with vertices $\vec{v_1}, ..., \vec{v_5}$ ,  could we find a vector $\vec{p}=(p_1,..p_4)$ such that the dot products $\vec{p} \cdot \vec{v_1}, \vec{p} \cdot \vec{v_2}, \vec{p} \cdot \vec{v_3}, \vec{p} \cdot \vec{v_4} \text { and }\vec{p} \cdot \vec{v_5}$ form an arithmetic series?

We are going to project the vertices onto $\vec{p}$ . Note that all points that lie on a hyperplane perpendicular to a $\vec{p}$ are projected onto the same point. We want these projection points to be at regular distances. Once we have found an example, we will scale the 5-cell so that the distance between adjacent projection points is 1.

Projections of the vertices $\vec{v_i}$ are given by $\frac{(\vec{p} \cdot\vec{v_i})\vec{p}}{|p|^2}$ . The distance between the two projections then is $\frac{p \cdot (\vec{v_1}-\vec{v_2})}{|p|}$ .

We will use the simplest 5-cell vertex coordinates: $(2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(φ,φ,φ,φ)$ , so that the respective dot products with $\vec{p}$ are $2p_1, 2p_2, 2p_3, 2p_4, φ(p_1+p_2+p_3+p_4)$

If these form an arithmetic series, we can set $d=p_2-p_1$ and the dot products become $2p_1, 2p_1+2d, 2p_1+4d, 2p_1+6d, φ(4p_1+6d)$ .

The last dot product has to be equal to $2p_1+8d$ , so we infer

$φ(4p_1+6d)=2p_1+8d$

Working this out via $(4φ-2) p_1 =d(8-6φ)$ and $\frac{d}{p_1}=\frac{2φ-1}{ 4-3φ} = \frac{2\sqrt{5}}{ 5-3\sqrt{5}}$ we arrive at

$\frac{d}{p_1}= s=  -\frac{ \sqrt{5}+3}{ 2}$

Now the vector $\vec{p}$ can be written as $\vec{p}=p_1(1, 1+s, 1+2s, 1+3s)$

and hence we find that $|p|^2= p_1^2(4+ 12s + 14s^2)= (35+15\sqrt{5})p_1^2$

The distance between the two projections of $\vec{v_1}$ and $\vec{v_2}$ then is the projection of the difference vector $\vec{v_2}-\vec{v_1}=(-2,2,0,0)$ which has length $\frac{-2p1+2p2}{|p|}$ $= \frac{2d}{|p|} =  \frac{2sp_1}{|p|}$ $= \frac{\sqrt{5}+3}{ \sqrt{35+15\sqrt{5} }}$ $= \frac{1}{5}\sqrt{10}$ . This is also the distance between the five parallel planes containing the vertices.

In order to get the planes a unit distance apart we now have to scale up the 5-cell by the reciprocal of this, which is a factor $\frac{1}{2}\sqrt{10}$ . The squared side lengths were $2^2+2^2=8$ before scaling. These now become $8×\frac{10}{4}=\boxed{20}$ .