I am trying to learn some coordinates in more than $3$ dimensions. And suddenly stumbled into this:
If we say that in a $4$D regular figure (a regular 5-cell) $ABCDE$, a hyperplane cuts it such that it's parallel and equidistant to edge $AB$ and plane $CDE$, then at what points does the hyperplane cut the $4$D figure? Can we say that the points formed will be $\frac{AC}{2}, \frac{AD}{2}, \frac{AE}{2}, \frac{BC}{2}, \frac{BD}{2}, \frac{BE}{2}$ where $\frac{XY}{2}$ means the midpoint of $X, Y$. I am not sure about this but this is just from a pattern I could observe. I cannot think about the $4$ dimensions, so I just followed a pattern from what happens in the analogous $3$D, $2$D, cases. Now, we can take some equations and matrices and solve it.
To begin with, we can choose the coordinates in the following manner : $A(2,0,0,0), B(0,2,0,0), C(0,0,2,0), D(0,0,0,2), E(\tau, \tau, \tau, \tau)$ where $\tau$ is the golden ratio(from Wikipedia). Since it's a regular figure, we can always consider a homothety/dilation to get the figure of the desired shape. So, the above coordinates work. Now, we can carry on from here. But I'm a not sure how to do that
But I'm not really getting much idea about that.
We are working in four dimensional Euclidean space, where points are identified by four real coordinates, say $(x,y,z,t)$. It may help with "visualization" to think of the fourth coordinate at time, so that increasing it amounts to shifting a figure forward in time, etc.
Let's choose the coordinates in a way that makes the intersecting hyperplane as simple as the hyperplane $t=0$, a three dimensional subspace.
To do this place the edge $AB$ and the triangle $CDE$ at equal but opposing distances from the hyperplane, so that $t=0$ is halfway between them. For simplicity we set the $t$-coordinates of $A,B$ to $1$ and set the $t$-coordinates of $C,D,E$ to $-1$. The six edges that connect vertices $A,B$ to vertices $C,D,E$ will thus be bisected by hyperplane $t=0$.
The regularity of the figure means that its five vertex points are equidistant from one another. The triangle $CDE$ will be equilateral, and its three edges will have the same lengths as the edge $AB$.
Center the equilateral triangle $CDE$ around $(0,0,0,-1)$ in the plane $z=0,t=-1$, and center the edge $AB$ on the point $(0,0,0,1)$ so it can be rotated in the $x,y$-coordinates without changing distances to the points $C,D,E$. To be specific, let:
$$ A = (0,0,+b,1) $$
$$ B = (0,0,-b,1) $$
$$ C = (a,0,0,-1) $$
$$ D = (-a/2,a\sqrt{3}/2,0,-1) $$
$$ E = (-a/2,-a\sqrt{3}/2,0,-1) $$
Now the length of any of the edges from $A$ or $B$ to one of $C,D,E$ is:
$$ \sqrt{a^2 + b^2 + 4} $$
while the length of $AB$ is:
$$ \sqrt{4b^2} $$
and the length of any side of $CDE$ is:
$$ \sqrt{3a^2} $$
Solving for $a,b$ that make the three quantities above equal gives:
$$ a = \frac{4}{5} \sqrt{5} $$
$$ b = \frac{2}{5} \sqrt{15} $$
It remains only to find the midpoints of the edges $AC,AD,AE,BC,BD,BE$ (where $t=0$) and connect them (as corners of the three dimensional figure formed by the intersection of the 5-cell and the hyperplane) accordingly as midpoints are connected in triangles $ACD,ADE,AEC,BCD,BDE,BEC$ and $ABC,ABD,ABE$.
Although details of the vertex coordinates remain to be worked out, the outline of the figure already comes into focus. With nine edges and six vertices, Euler's polyhedron formula tells us there are five faces. This means the polyhedron is a triangular prism.