# Ask Uncle Colin: What's the Bisector of a Tesseract

Dear Uncle Colin

If you hang a square up by one corner and cut it in the plane perpendicular to the vertical diagonal, you get a regular hexagon. What’s the corresponding result for a tesseract?

- Have You Proved Everything, Really?

Hi, HYPER, and thanks for your message! Let’s start in 3D.

If you dangle a 3D cube from one vertex (call it O), you can mark each other vertex with a height – those that share an edge with O have a height of 1, the three vertices in the next layer have a height of 2, and the final opposite vertex has a height of 3.

The mid-plane is therefore at height 1.5, and cuts all of the edges connecting 1-vertices to 2-vertices. There are six of those, and by symmetry they have to form a regular hexagon.

When you move to 4D, the first thing you get is a headache.

However, once you get past that, a similar argument holds sway: pick a vertex and call it at ‘height’ 0. It’s not really a height, it’s whatever you want to call the fourth dimension. Time, maybe.

The four vertices connected to the initial vertex have time 1; the six vertices connected to those have time 2; then four of the remaining vertices have time 3 and the final one has time 4.

The mid-hyperplane passes through the six vertices with height 2. So it’s a hexagon as well, right? Wrong.

In a unit tesseract((the best kind)), the six points have coordinates:

- A: (1,1,0,0)
- B: (1,0,1,0)
- C: (1,0,0,1)
- D: (0,1,0,1)
- E: (0,1,1,0)
- F: (0,0,1,1)

(Hey look! Their coordinates each sum to 2. That’s interesting.)

Point A is $\sqrt{2}$ units away from B, C, D and E, and two units away from F. It’s similar for each of the points: they’re $\sqrt{2}$ units away from four others, and two from the last. Also, the mid-hyperplane of a 4D shape should give us a 3D object.

Six vertices, each with four neighbours? We’re looking at a *regular octahedron*.

Hope that helps!

- Uncle Colin