The “moral” of the famous Nine Points problem that we’ve been dealing with last week (and not for the first or last time) is that sometimes, without realizing it, we impose more conditions on ourselves than we originally intended are . In the present case, when we see nine points arranged in a grid, we tend to assume that a dashed line connecting them is also included in the grid, or in other words that all of their vertices are connected to some of them its vertices coincide. the nine points.

In the problem of splitting an obtuse triangle into acute triangles, it is common to impose on oneself a similar constraint, although it seems unrelated to the previous one: in this case, that all vertices of the acute triangles are points of the perimeter of the obtuse angle , and with this self-imposed condition, division is not possible; But without this unnecessary restriction, an obtuse triangle can be subdivided into… how many acute angles at least?

Referring to the classic of the four equidistant trees, our featured user Salva Fuster says, “The problem of the four trees, where each of them is an equal distance from the other three, seems to me like that.” Dependent depending on the terrain, the shape of the trees and the definition of the distance between them, we have different alternatives. For example, if the terrain is flat and we take the distance between two trees as the distance between their centers of mass, we could have a small (tall) tree surrounded by three taller (shorter) trees such that the centers of mass of the four trees form a tetrahedron. If you consider the distance between two trees as the minimum distance between any point of them, we can have four trees equidistant at different heights or even just in contact with each other.”

The classic solution is that there is a hill on the ground: one of the trees is planted on it, and the other three around it and at a lower height, so that all four stand on the vertices of a regular tetrahedron.

And to form four equilateral triangles with six matches, we must also jump from plane to space, resorting to the tetrahedron; hence the analogy to the problem of trees, with which it apparently has nothing to do.

However, there is a flat solution if the matches are allowed to cross each other. Which is it?

### Platonic Hyperbodies

What if we had to plant five equally spaced trees instead of four? In this case we would have to go into the fourth dimension and place the trees (or hypertrees) at the vertices of a pentachoron, the simplest of the four-dimensional regular polytopes (called polychores).

As three-dimensional beings that we are (without the temporal dimension or possible additional residual dimensions as suggested by string theory), we cannot conceive of four-dimensional bodies, but we can infer their properties. We’ll see if it’s true:

How many edges does the pentachoron have? What are their faces and how many are there?

Dalí’s cross is a tesseract, a figure made up of eight three-dimensional cubes in a four-dimensional space.

Even more difficult:

A polytope not only has vertices (dimensionless), edges (one-dimensional), and faces (two-dimensional); it also has three-dimensional ‘cells’, such as the eight cubes that (in other words) surround the hypercube (also called the octachoron or tesseract) whose development Dalí made famous in one of his paintings (Corpus Hypercubus, 1954).

How many cells does a pentachoron have and how are they structured?

Besides the pentachoron and the hypercube, there are four other four-dimensional regular polytopes that are almost as difficult to name as to imagine: hexadecachoron, icositetrachoron, hecatonicosachoron, and hexacosichoron. The first five can be viewed as four-dimensional analogues of the Platonic solids, while the hexacosichoron, with its 1,200 triangular faces and 600 tetrahedral cells, has no three-dimensional equivalent. Don’t try to draw it.

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