The “moral” of the famous problem of the nine points, which we dealt with (and not for the first or last time) last week, is that sometimes, without realizing it, we impose more conditions on ourselves than those established in the approach . In the case at hand, when we see nine points arranged in a grid, we tend to assume that a broken line that joins them will also be enclosed in the grid, or, in other words, that all its vertices will coincide with some of its vertices. the nine points.
In the problem of dividing an obtuse triangle into acute triangles, although it apparently has nothing to do with the previous one, it is common to impose a similar restriction on oneself: in this case, that all the vertices of the acute triangles be points on the perimeter of the obtuse angle, and with this self-imposed condition the division is not possible; but without this unnecessary restriction an obtuse triangle can be divided into… how many acute angles at least?
With respect to the classic of the four equidistant trees, this is what our “featured user” Salva Fuster says: “In the problem of the four trees in which each of them is the same distance from the other three, it seems to me that we have different alternatives depending on the terrain, the shape of the trees and the definition of distance between them. 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 short (high) tree surrounded by three taller (shorter) ones so that the centers of mass mass of the four trees form a tetrahedron. If the distance between two trees is considered to be the minimum distance between any point of them, we can have four trees at the same distance at different heights, or even simply being in contact with each other”.
The classical solution is that there is a mound on the ground: one of the trees is planted on top of it and the other three around it and at a lower height, so that all four are at the vertices of a regular tetrahedron.
And to form four equilateral triangles with six matches we also have to jump from the plane to space and resort to the tetrahedron; hence its analogy with the problem of trees, with which it apparently has nothing to do.
Although, if the matches are allowed to cross, there is a flat solution. Which is it?
Platonic hypersolids
What if we had to plant five equidistant trees instead of four? In that case we would have to move to 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 (not counting the temporal dimension or possible extra vestigial dimensions, such as those proposed by string theory), we cannot conceive of four-dimensional bodies, but we can deduce their characteristics. We’ll see if it’s true:
How many edges does the pentachoron have? How are their faces and how many are they?
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 surround (in other words) the hypercube (also called an octachoron or tesseract), whose development Dalí made famous with one of his paintings (Corpus Hypercubus1954).
How many cells does a pentachoron have and what are they like?
In addition to the pentachoron and the hypercube, there are four other four-dimensional regular polytopes, almost as difficult to name as they are to conceive: hexadecachoron, icositetrachoron, hecatonicosachoron, and hexacosichoron. The first five can be considered the 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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