# Regular polyhedrons. Part 2. Four-dimensional

- Tutorial

Previous publication: Correct polyhedrons. Part 1. Three-dimensional

I see that serious people gathered on Habré. The article about three-dimensionality at the expense of "times" was dismantled. However, spaces of constant curvature will not surprise anyone in our time. Nevertheless, there are always those who want to look higher in four-dimensionality. Well, it is with such inquisitive colleagues that we continue the conversation and move on to the next level in terms of dimension.

My task is not just to talk about partitions of spaces of constant curvature of any dimension into regular polyhedra, but to make it so that even yesterday’s school students who graduated from 11 classes understood the material. I love articles on Habré precisely for their intelligibility, comprehensibility, simplicity, despite the complexity of the material, and in the same quality I try to submit information in publications. In universities and in domestic publications, the proposed material may be considered, but, as it seems to me, not in this form. I think that the information will be useful for students. In foreign literature, this material is, respectively, not in Russian, in a highly compressed form and using higher mathematics. Here I “chew” everything for schoolchildren, without higher mathematics, in fact, on the same geometric intuition. We will see in the next article, how will the transition from 4D to 5D be made using geometry, clearly, without higher algebra. This will be the most difficult step, but whoever understands it will understand all other dimensions from 6 and above. I'm not sure that I managed to thoroughly “chew” everything, so if there are additional questions, ask, this will help me improve the article.

In this publication, the idea of calculations is completely the same as in the previous article, only one dimension higher, so if someone else has not had time to get acquainted with it , it is advisable to do this in order to understand what is happening.

First, we give all the definitions again, only for 4-dimensional polyhedra and the corresponding Schleaf symbol. I do not want to immediately give general formulations, so as not to confuse unprepared readers who, with this subject, may be dealing for the first time. Then we give the statement of the problem. In this publication, it seems that she took a more austere and harmonious look. If I am mistaken in some details, then it’s okay, this is not a paper publication, I’ll edit it and correct it to make everything beautiful. The main calculations and the result are correct, don’t worry about it, then I’ll even give links to authoritative authors, if that. Looking at these authors, you will realize that my work, in the form of these several publications, is not in vain.

There are many different angles in polyhedra; we call a dihedral angle a dihedral adjacent angle, between adjacent faces, i.e. faces having a common edge.

I give a definition of the correct polyhedron recursively, in my own words, because Habr is not a place for copy pasteors.

A regular 4-dimensional polyhedron is a convex polyhedron, in which all 3-dimensional faces are regular polyhedra, equal to each other and all angles between 3-dimensional faces are equal to each other.

There is an assertion that a partition of an (n-1) -dimensional sphere corresponds uniquely to a regular n-dimensional polyhedron, for n> 1. That is, the dimension of the polyhedron is one higher than the dimension of the sphere that we divide. For example, if the vertices of the partition of a 3-dimensional sphere are connected by edges corresponding to the arcs of the partition, then we get a 4-dimensional polyhedron. I do not know how this statement is proved and whether it is proved at all. Therefore, it is taken for an axiom and we consider it true in all dimensions of spaces. For a two-dimensional sphere, using the example of an icosahedron, this was shown in a video clip in a previous publication. We can say that in three-dimensionality (i.e., for partitioning a two-dimensional sphere) this fact is established experimentally, and in four-dimensionality and higher, we believe that everything is similar. Everything seems to be clean, intuitively clear that it is so.

I also give the definition of the Shlefly symbol in my own words.

Shlefli’s symbol is a sequence of numbers {p1, p2, p3} that defines an algorithm for constructing a regular polyhedron as follows:

- take regular {p1} squares, combine them along edges, so that p2 pieces of such {p1} converge at each vertex, we get { {p1}, p2} or briefly {p1, p2}

- take the {p1, p2} obtained in the previous step and combine them along flat faces so that p3 pieces of such {p1, p2} three-dimensional polyhedra converge in each edge.

Induction in the definition is already asking, isn't it? The general definition for an n-dimensional regular polyhedron will be given in the next publication. For now, it is better to consider a special case so that it is easier to understand and not be distracted by general formulations.

And one more thought (statement of the problem), which, expressed in a rough form, a loophole, turns out to be incorrect, but it is not used in calculations, therefore it does not affect the result, but for that it is the driving force of research :) The idea is , that for any Schleafly symbol in any dimension, there is a regular polyhedron that partitions one of the spaces of constant curvature of the corresponding dimension. We saw that when dividing two-dimensional spaces this idea is true, there any Schlefli symbol broke something. The word "Exists" is marked with a capital letter, since in fact this is not true for dimensions 4 and above. To allow existence always is of course liberties, but one can formulate the statement of the problem in such a way that this thought turns out to be legitimate, namely:

the number of different values of the Shlefli symbol is a countable set, divide this set into disjoint subsets (classes) finite or infinite in type:

- a class of characters defining a partition of the Spherical space into regular polytopes of the corresponding dimension;

- --- // --- partition of the Euclidean space --- // ---

- --- // --- partition of the Lobachevsky space into regular polyhedra of Finite volume --- // ---

- --- / / --- partition of the Lobachevsky space into regular limit polytopes of finite volume --- // --- (those whose vertices fall directly on the absolute - the boundary of the Poincare disk)

- --- // --- partition of the Lobachevsky space into regular polyhedra of infinite volume --- // --- (if the resulting figures are conventionally considered to be regular polyhedra)

- a class of "bad" symbols that cannot be associated with any polyhedron or some kind of partition of a space of constant curvature.

When listing these classes, we ran a little ahead, and in the partition of the Lobachevsky spaces above, we identified three subsets (classes), instead of one, as for the Sphere and Euclid.

In other words, our task is to study all the possible values of the Shlefli symbol in all dimensions, in all three spaces of constant curvature. This is motivation, goal and task. In this publication, the dimension of the partitioned spaces of constant curvature = 3, the dimension of the resulting polytopes = 4.

The idea of the solution itself is simple, as in the previous article on three-dimensionality, on the one hand, according to the parameters of the Schleuly symbol {p1, p2}, calculate the dihedral angle ( ) of the polyhedron and compare this angle with ie so that the complete revolution ( ) contains an integer number of polyhedra converging in an edge. Here p3 is the third parameter of the Schleufly symbol {p1, p2, p3}, which defines the partition of 4-dimensional space into polytopes {p1, p2}, which means the number of polytopes {p1, p2} converging in an edge.

1. If it turns out that = it means there is no need to bend the space, everything came together beautifully, that is, this is Euclidean space. As you can see, the deuce begs to contract, therefore, in the calculations it appears not the dihedral angle of the polyhedron itself, but its half =

2. If it turns out that < ie the dihedral angle is small, then the polyhedron, together with its angles, needs to be “inflated” so that the angles increase to the desired values, so you need to place it on a three-dimensional sphere, i.e. This is a spherical case.

3. If it turns out that > ie the dihedral angle is large, then the polyhedron, together with its angles, needs to be “blown away" so that the angles decrease to the desired values, so you need to place it in the three-dimensional Lobachevsky space, i.e. this is a hyperbolic case.

About what it means to “inflate” and “deflate” a polyhedron (polygon) was described in a previous article, where we saw that the sum of the angles of a triangle (polygon) increases when we place it on a sphere, the triangle seems to inflate and that the sum of the angles the triangle decreases when we put it in hyperbolic space, the triangle is blown off, as it were. There we saw it in two-dimensionality, everything similarly happens with corners in 3-dimensional spaces and in spaces of higher dimensions.

However, if we know the corners of regular polygons by heart, then the dihedral angles of three-dimensional regular polyhedra are already harder to remember by heart, although, of course, all these angles are known. We will see in the next publication when we climb into five-dimensional space that the derivation of the formula of the dihedral angle of a polyhedron becomes not a trivial task and the formula itself comes to the fore. One could even name the article: the dihedral angle formula of a regular, convex polyhedron in n-dimensional Euclidean space. But in fact, this formula is not only the goal of the proposed publications, but also a tool to help identify polyhedra in spaces of higher dimensions. Those. all questions here are closely related and interesting, each of them individually and all taken together,

Now all that remains is to make calculations for calculating the dihedral angle according to the parameters of the Shlefli symbol {p1, p2} , and the student will be able to compare the obtained values with and write out the answer in the form of a table. However, it is also not difficult to calculate the dihedral angle from the parameters p1, p2. Here in the next dimension, it’s more complicated, there is already a highlight :) but more on that in the next publication.

I apologize, in the previous article I forgot about the auxiliary angles to say, which we will need for the calculations. These are corners and . Who carefully read the previous article, he noticed that there, on the drawing, the angle was indicated, calculated and turned out . There we measured the angle of a regular polygon at vertex C. Now, taking this vertex and two vertices adjacent to it, we get an isosceles triangle, here is the angle at the base of this triangle, we now need it. Also for a 3-dimensional polyhedron we need an angle- this is the angle at the base of the isosceles triangle, at the regular polygon {p2}, lying at the base of the isosceles pyramid. The pyramid is obtained by cutting off {p1, p2} one vertex under study along the plane of nearby vertices. Pay attention to the simple but important fact that since p2 pieces of regular p1-gons converge at the top of the pyramid, the regular p2-gon is obtained at the base of this pyramid. This is what we use to calculate the angle of

the figure is considered an example of a icosahedron, then , in the general case for arbitrary {p1, p2}, we obtain , . So far, we are in Euclidean space, so the sum of the angles of the triangle is still equal .

Now consider this isosceles pyramid closer. To calculate the dihedral angle, it’s enough to consider not the whole pyramid, but only one edge and two faces containing this edge.

In the general case, for {p1, p2} we consider an edge and find this and this is half of the angle we are looking for, that is, half of the dihedral angle of the polyhedron {p1, p2}. I will give explanations about additional constructions: and - heights in the corresponding isosceles triangles. The plane of an isosceles triangle is orthogonal to the edge by construction, so the plane angleof this triangle is equal in magnitude to the desired dihedral angle of the regular polyhedron {p1, p2}. Not complicated calculations are written to the left of the drawing, please read them carefully. If you know the definition of sine and cosine, then difficulties should not arise. So:

You can, of course, calculate this angle through the fundamental tetrahedron (with vertices in the center of the 3D face, the center of the 2D face, the middle of the edge and the vertex of the polyhedron), so the classics do, but then the transition in dimensions becomes difficult, at least for schoolchildren this is unlikely to be clarified. Therefore, the approach proposed in this publication, to the calculation of the dihedral angle, was deliberately chosen. At least this is another way to calculate this angle. Someone can understand one way, someone else.

As they said, we compare the angle with and write down the results of the comparison in the form of a table. Comparisons are more convenient to perform not for the corners themselves, but for the sines in the square of these angles, so as not to clutter up the records with arcsines, since such a formula for betta turned out.

The essence of the comparison is in the form of a formula:

where X ^ 3 is the general designation of spaces of constant curvature, S ^ 3 is a three-dimensional sphere, E ^ 3 is a Euclidean three-dimensional space, Λ ^ 3 is a three-dimensional Lobachevsky space.

The result of the comparison in the form of a table:

Where:

Once again, a little explanation - the second column is compared, for and the third row, for . For instance:

1. the column {4,3} is compared with the row p3 = 4, we get 0.5 = 0.5 - that means {4,3,4} - splits the Euclidean 3-dimensional space, these are the usual cubes converging 4 in the edge. :) Elementary.

2. For the column {3,3} and row p3 = 3 we have 0.5 <0.75 - which means {3,3,3} - splits the 3-dimensional sphere (and it corresponds to the 4-dimensional polyhedron).

3. For column {5,3} and row p3 = 6 we have 0.723> 0.25 — which means {5,3,6} —divides the 3-dimensional Lobachevsky space into a regular limit polyhedron. The limit can be understood by the fact that he has the vertex figure {p2, p3} = {3,6} - splits the two-dimensional Euclidean space, i.e. Euclidean plane.

In the general case, if {p1, p2, p3} - splits the Lobachevsky space, then the vertex figure plays the role:

- if {p2, p3} - splits Euclidean, then {p1, p2, p3} is a limit polyhedron,

- if {p2, p3} - splits a sphere, then {p1, p2, p3} - has finite volume and dimensions,

- if {p2, p3} - splits the Lobachevsky plane, then {p1, p2, p3} - has infinite volume.

This statement is without proof, I think that it is true and I know why, but I can not prove and show it strictly and simply. It includes the horosphere, on the surface of which Euclidean geometry. And the vertex figure just cuts off the horosphere for the limit polyhedron.

So, we learned how to calculate the dihedral angle of a regular polyhedron by its Shlefli symbol {p1, p2}, just in case I will write it again:

And we found 6 partitions of the three-dimensional sphere into regular three-dimensional polyhedra:

{3,3,3} - tetrahedra, converging 3 in the edge,

{4,3,3} - cubes converging in 3 in the edge,

{3,3,4} - tetrahedrons, converging in 4 in the edge,

{3,4,3} - octahedrons in converging in 3 in an edge,

{5,3,3} - dodecahedrons converging in 3 in an edge,

{3,3,5} - tetrahedrons in converging 5 in an edge.

And that means 6 corresponding regular 4-dimensional polyhedra with the same Shlefli symbols. It is true that counting in general terms the number of vertices, edges, faces, and facets of these polyhedra is not as simple as it seems at first glance. In special cases for a 4-tetrahedron, 4-cube, 4-octahedron (orthahedron) this can be done by induction, but for the other polyhedra this trick does not work.

Unfortunately, I do not have my own pictures of these polyhedrons, so I refer to Wikipedia , there are beautiful pictures for four-dimensionality.

Just as when we look at a notebook in a cell (partition {4, 4}) we see a structure, as if the Euclidean plane itself has a structure {4, 4}, we already know that there are only two more structures of the same plane { 3, 6} and {6, 3}. Now we have learned that the partition of 3-dimensional Euclidean space is unique {4, 3, 4} - into cubes converging 4 pieces in an edge. We imagine this partition in our minds and feel the structure of this space. Now, imagine mentally {5, 3} - dodecahedrons, we apply them together on flat faces (pentagons), 3 such dodecahedrons converge in the edge and there will still be a gap. Now mentally uniformly inflate these dodecahedrons until the gap disappears. When it disappears, exactly 3 dodecahedrons converge in the edge. Now mentally add three dodecahedrons in each edge. In all the ribs it’s more difficult to imagine at the same time, but necessary. If we take 120 of such bloated dodecahedrons and attach them all without gaps between them along the faces, we get a closed partition of a 3-dimensional sphere. You can also imagine with the other 5 tilings of a 3-dimensional sphere. Such judicious experiments with polyhedra help to mentally understand the structure of the 3-dimensional sphere and touch the 4-dimensional space of Euclid. Similarly, one can speculate with a hyperbolic 3-dimensional space. The calculations presented in the publication help to understand what space we are in for a given Schlefly symbol, i.e. for a given mental construction. Such judicious experiments with polyhedra help to mentally understand the structure of the 3-dimensional sphere and touch the 4-dimensional space of Euclid. Similarly, one can speculate with a hyperbolic 3-dimensional space. The calculations presented in the publication help to understand what space we are in for a given Schlefly symbol, i.e. for a given mental construction. Such judicious experiments with polyhedra help to mentally understand the structure of the 3-dimensional sphere and touch the 4-dimensional space of Euclid. Similarly, one can speculate with a hyperbolic 3-dimensional space. The calculations presented in the publication help to understand what space we are in for a given Schlefly symbol, i.e. for a given mental construction.

In the next publication, we will do the same thing only one dimension higher. I think that I will give there a video clip explaining the main idea of the transition (rise) in dimensions. If someone else does not understand what is happening, then perhaps this video will shed light on all the calculations done here and in the previous article.

If everything is clear, then go to the next level in dimension.

References:

Correct polyhedrons. Part 1. Three-dimensional

regular polyhedrons. Part 2. Four-dimensional

regular polyhedrons. Part 2.5 (auxiliary)

The symbol Shafly. Part 2.6

## Introduction

I see that serious people gathered on Habré. The article about three-dimensionality at the expense of "times" was dismantled. However, spaces of constant curvature will not surprise anyone in our time. Nevertheless, there are always those who want to look higher in four-dimensionality. Well, it is with such inquisitive colleagues that we continue the conversation and move on to the next level in terms of dimension.

My task is not just to talk about partitions of spaces of constant curvature of any dimension into regular polyhedra, but to make it so that even yesterday’s school students who graduated from 11 classes understood the material. I love articles on Habré precisely for their intelligibility, comprehensibility, simplicity, despite the complexity of the material, and in the same quality I try to submit information in publications. In universities and in domestic publications, the proposed material may be considered, but, as it seems to me, not in this form. I think that the information will be useful for students. In foreign literature, this material is, respectively, not in Russian, in a highly compressed form and using higher mathematics. Here I “chew” everything for schoolchildren, without higher mathematics, in fact, on the same geometric intuition. We will see in the next article, how will the transition from 4D to 5D be made using geometry, clearly, without higher algebra. This will be the most difficult step, but whoever understands it will understand all other dimensions from 6 and above. I'm not sure that I managed to thoroughly “chew” everything, so if there are additional questions, ask, this will help me improve the article.

In this publication, the idea of calculations is completely the same as in the previous article, only one dimension higher, so if someone else has not had time to get acquainted with it , it is advisable to do this in order to understand what is happening.

First, we give all the definitions again, only for 4-dimensional polyhedra and the corresponding Schleaf symbol. I do not want to immediately give general formulations, so as not to confuse unprepared readers who, with this subject, may be dealing for the first time. Then we give the statement of the problem. In this publication, it seems that she took a more austere and harmonious look. If I am mistaken in some details, then it’s okay, this is not a paper publication, I’ll edit it and correct it to make everything beautiful. The main calculations and the result are correct, don’t worry about it, then I’ll even give links to authoritative authors, if that. Looking at these authors, you will realize that my work, in the form of these several publications, is not in vain.

## Definitions. Axioms. Formulation of the problem

There are many different angles in polyhedra; we call a dihedral angle a dihedral adjacent angle, between adjacent faces, i.e. faces having a common edge.

I give a definition of the correct polyhedron recursively, in my own words, because Habr is not a place for copy pasteors.

A regular 4-dimensional polyhedron is a convex polyhedron, in which all 3-dimensional faces are regular polyhedra, equal to each other and all angles between 3-dimensional faces are equal to each other.

There is an assertion that a partition of an (n-1) -dimensional sphere corresponds uniquely to a regular n-dimensional polyhedron, for n> 1. That is, the dimension of the polyhedron is one higher than the dimension of the sphere that we divide. For example, if the vertices of the partition of a 3-dimensional sphere are connected by edges corresponding to the arcs of the partition, then we get a 4-dimensional polyhedron. I do not know how this statement is proved and whether it is proved at all. Therefore, it is taken for an axiom and we consider it true in all dimensions of spaces. For a two-dimensional sphere, using the example of an icosahedron, this was shown in a video clip in a previous publication. We can say that in three-dimensionality (i.e., for partitioning a two-dimensional sphere) this fact is established experimentally, and in four-dimensionality and higher, we believe that everything is similar. Everything seems to be clean, intuitively clear that it is so.

I also give the definition of the Shlefly symbol in my own words.

Shlefli’s symbol is a sequence of numbers {p1, p2, p3} that defines an algorithm for constructing a regular polyhedron as follows:

- take regular {p1} squares, combine them along edges, so that p2 pieces of such {p1} converge at each vertex, we get { {p1}, p2} or briefly {p1, p2}

- take the {p1, p2} obtained in the previous step and combine them along flat faces so that p3 pieces of such {p1, p2} three-dimensional polyhedra converge in each edge.

Induction in the definition is already asking, isn't it? The general definition for an n-dimensional regular polyhedron will be given in the next publication. For now, it is better to consider a special case so that it is easier to understand and not be distracted by general formulations.

And one more thought (statement of the problem), which, expressed in a rough form, a loophole, turns out to be incorrect, but it is not used in calculations, therefore it does not affect the result, but for that it is the driving force of research :) The idea is , that for any Schleafly symbol in any dimension, there is a regular polyhedron that partitions one of the spaces of constant curvature of the corresponding dimension. We saw that when dividing two-dimensional spaces this idea is true, there any Schlefli symbol broke something. The word "Exists" is marked with a capital letter, since in fact this is not true for dimensions 4 and above. To allow existence always is of course liberties, but one can formulate the statement of the problem in such a way that this thought turns out to be legitimate, namely:

the number of different values of the Shlefli symbol is a countable set, divide this set into disjoint subsets (classes) finite or infinite in type:

- a class of characters defining a partition of the Spherical space into regular polytopes of the corresponding dimension;

- --- // --- partition of the Euclidean space --- // ---

- --- // --- partition of the Lobachevsky space into regular polyhedra of Finite volume --- // ---

- --- / / --- partition of the Lobachevsky space into regular limit polytopes of finite volume --- // --- (those whose vertices fall directly on the absolute - the boundary of the Poincare disk)

- --- // --- partition of the Lobachevsky space into regular polyhedra of infinite volume --- // --- (if the resulting figures are conventionally considered to be regular polyhedra)

- a class of "bad" symbols that cannot be associated with any polyhedron or some kind of partition of a space of constant curvature.

When listing these classes, we ran a little ahead, and in the partition of the Lobachevsky spaces above, we identified three subsets (classes), instead of one, as for the Sphere and Euclid.

In other words, our task is to study all the possible values of the Shlefli symbol in all dimensions, in all three spaces of constant curvature. This is motivation, goal and task. In this publication, the dimension of the partitioned spaces of constant curvature = 3, the dimension of the resulting polytopes = 4.

## The idea of solving the problem

The idea of the solution itself is simple, as in the previous article on three-dimensionality, on the one hand, according to the parameters of the Schleuly symbol {p1, p2}, calculate the dihedral angle ( ) of the polyhedron and compare this angle with ie so that the complete revolution ( ) contains an integer number of polyhedra converging in an edge. Here p3 is the third parameter of the Schleufly symbol {p1, p2, p3}, which defines the partition of 4-dimensional space into polytopes {p1, p2}, which means the number of polytopes {p1, p2} converging in an edge.

1. If it turns out that = it means there is no need to bend the space, everything came together beautifully, that is, this is Euclidean space. As you can see, the deuce begs to contract, therefore, in the calculations it appears not the dihedral angle of the polyhedron itself, but its half =

2. If it turns out that < ie the dihedral angle is small, then the polyhedron, together with its angles, needs to be “inflated” so that the angles increase to the desired values, so you need to place it on a three-dimensional sphere, i.e. This is a spherical case.

3. If it turns out that > ie the dihedral angle is large, then the polyhedron, together with its angles, needs to be “blown away" so that the angles decrease to the desired values, so you need to place it in the three-dimensional Lobachevsky space, i.e. this is a hyperbolic case.

About what it means to “inflate” and “deflate” a polyhedron (polygon) was described in a previous article, where we saw that the sum of the angles of a triangle (polygon) increases when we place it on a sphere, the triangle seems to inflate and that the sum of the angles the triangle decreases when we put it in hyperbolic space, the triangle is blown off, as it were. There we saw it in two-dimensionality, everything similarly happens with corners in 3-dimensional spaces and in spaces of higher dimensions.

However, if we know the corners of regular polygons by heart, then the dihedral angles of three-dimensional regular polyhedra are already harder to remember by heart, although, of course, all these angles are known. We will see in the next publication when we climb into five-dimensional space that the derivation of the formula of the dihedral angle of a polyhedron becomes not a trivial task and the formula itself comes to the fore. One could even name the article: the dihedral angle formula of a regular, convex polyhedron in n-dimensional Euclidean space. But in fact, this formula is not only the goal of the proposed publications, but also a tool to help identify polyhedra in spaces of higher dimensions. Those. all questions here are closely related and interesting, each of them individually and all taken together,

Now all that remains is to make calculations for calculating the dihedral angle according to the parameters of the Shlefli symbol {p1, p2} , and the student will be able to compare the obtained values with and write out the answer in the form of a table. However, it is also not difficult to calculate the dihedral angle from the parameters p1, p2. Here in the next dimension, it’s more complicated, there is already a highlight :) but more on that in the next publication.

## Calculation of the dihedral angle of a regular polyhedron {p1, p2}

I apologize, in the previous article I forgot about the auxiliary angles to say, which we will need for the calculations. These are corners and . Who carefully read the previous article, he noticed that there, on the drawing, the angle was indicated, calculated and turned out . There we measured the angle of a regular polygon at vertex C. Now, taking this vertex and two vertices adjacent to it, we get an isosceles triangle, here is the angle at the base of this triangle, we now need it. Also for a 3-dimensional polyhedron we need an angle- this is the angle at the base of the isosceles triangle, at the regular polygon {p2}, lying at the base of the isosceles pyramid. The pyramid is obtained by cutting off {p1, p2} one vertex under study along the plane of nearby vertices. Pay attention to the simple but important fact that since p2 pieces of regular p1-gons converge at the top of the pyramid, the regular p2-gon is obtained at the base of this pyramid. This is what we use to calculate the angle of

the figure is considered an example of a icosahedron, then , in the general case for arbitrary {p1, p2}, we obtain , . So far, we are in Euclidean space, so the sum of the angles of the triangle is still equal .

Now consider this isosceles pyramid closer. To calculate the dihedral angle, it’s enough to consider not the whole pyramid, but only one edge and two faces containing this edge.

In the general case, for {p1, p2} we consider an edge and find this and this is half of the angle we are looking for, that is, half of the dihedral angle of the polyhedron {p1, p2}. I will give explanations about additional constructions: and - heights in the corresponding isosceles triangles. The plane of an isosceles triangle is orthogonal to the edge by construction, so the plane angleof this triangle is equal in magnitude to the desired dihedral angle of the regular polyhedron {p1, p2}. Not complicated calculations are written to the left of the drawing, please read them carefully. If you know the definition of sine and cosine, then difficulties should not arise. So:

You can, of course, calculate this angle through the fundamental tetrahedron (with vertices in the center of the 3D face, the center of the 2D face, the middle of the edge and the vertex of the polyhedron), so the classics do, but then the transition in dimensions becomes difficult, at least for schoolchildren this is unlikely to be clarified. Therefore, the approach proposed in this publication, to the calculation of the dihedral angle, was deliberately chosen. At least this is another way to calculate this angle. Someone can understand one way, someone else.

## Angle calculations and comparisons

As they said, we compare the angle with and write down the results of the comparison in the form of a table. Comparisons are more convenient to perform not for the corners themselves, but for the sines in the square of these angles, so as not to clutter up the records with arcsines, since such a formula for betta turned out.

The essence of the comparison is in the form of a formula:

where X ^ 3 is the general designation of spaces of constant curvature, S ^ 3 is a three-dimensional sphere, E ^ 3 is a Euclidean three-dimensional space, Λ ^ 3 is a three-dimensional Lobachevsky space.

The result of the comparison in the form of a table:

Where:

Once again, a little explanation - the second column is compared, for and the third row, for . For instance:

1. the column {4,3} is compared with the row p3 = 4, we get 0.5 = 0.5 - that means {4,3,4} - splits the Euclidean 3-dimensional space, these are the usual cubes converging 4 in the edge. :) Elementary.

2. For the column {3,3} and row p3 = 3 we have 0.5 <0.75 - which means {3,3,3} - splits the 3-dimensional sphere (and it corresponds to the 4-dimensional polyhedron).

3. For column {5,3} and row p3 = 6 we have 0.723> 0.25 — which means {5,3,6} —divides the 3-dimensional Lobachevsky space into a regular limit polyhedron. The limit can be understood by the fact that he has the vertex figure {p2, p3} = {3,6} - splits the two-dimensional Euclidean space, i.e. Euclidean plane.

In the general case, if {p1, p2, p3} - splits the Lobachevsky space, then the vertex figure plays the role:

- if {p2, p3} - splits Euclidean, then {p1, p2, p3} is a limit polyhedron,

- if {p2, p3} - splits a sphere, then {p1, p2, p3} - has finite volume and dimensions,

- if {p2, p3} - splits the Lobachevsky plane, then {p1, p2, p3} - has infinite volume.

This statement is without proof, I think that it is true and I know why, but I can not prove and show it strictly and simply. It includes the horosphere, on the surface of which Euclidean geometry. And the vertex figure just cuts off the horosphere for the limit polyhedron.

## Summarizing

So, we learned how to calculate the dihedral angle of a regular polyhedron by its Shlefli symbol {p1, p2}, just in case I will write it again:

And we found 6 partitions of the three-dimensional sphere into regular three-dimensional polyhedra:

{3,3,3} - tetrahedra, converging 3 in the edge,

{4,3,3} - cubes converging in 3 in the edge,

{3,3,4} - tetrahedrons, converging in 4 in the edge,

{3,4,3} - octahedrons in converging in 3 in an edge,

{5,3,3} - dodecahedrons converging in 3 in an edge,

{3,3,5} - tetrahedrons in converging 5 in an edge.

And that means 6 corresponding regular 4-dimensional polyhedra with the same Shlefli symbols. It is true that counting in general terms the number of vertices, edges, faces, and facets of these polyhedra is not as simple as it seems at first glance. In special cases for a 4-tetrahedron, 4-cube, 4-octahedron (orthahedron) this can be done by induction, but for the other polyhedra this trick does not work.

Unfortunately, I do not have my own pictures of these polyhedrons, so I refer to Wikipedia , there are beautiful pictures for four-dimensionality.

Just as when we look at a notebook in a cell (partition {4, 4}) we see a structure, as if the Euclidean plane itself has a structure {4, 4}, we already know that there are only two more structures of the same plane { 3, 6} and {6, 3}. Now we have learned that the partition of 3-dimensional Euclidean space is unique {4, 3, 4} - into cubes converging 4 pieces in an edge. We imagine this partition in our minds and feel the structure of this space. Now, imagine mentally {5, 3} - dodecahedrons, we apply them together on flat faces (pentagons), 3 such dodecahedrons converge in the edge and there will still be a gap. Now mentally uniformly inflate these dodecahedrons until the gap disappears. When it disappears, exactly 3 dodecahedrons converge in the edge. Now mentally add three dodecahedrons in each edge. In all the ribs it’s more difficult to imagine at the same time, but necessary. If we take 120 of such bloated dodecahedrons and attach them all without gaps between them along the faces, we get a closed partition of a 3-dimensional sphere. You can also imagine with the other 5 tilings of a 3-dimensional sphere. Such judicious experiments with polyhedra help to mentally understand the structure of the 3-dimensional sphere and touch the 4-dimensional space of Euclid. Similarly, one can speculate with a hyperbolic 3-dimensional space. The calculations presented in the publication help to understand what space we are in for a given Schlefly symbol, i.e. for a given mental construction. Such judicious experiments with polyhedra help to mentally understand the structure of the 3-dimensional sphere and touch the 4-dimensional space of Euclid. Similarly, one can speculate with a hyperbolic 3-dimensional space. The calculations presented in the publication help to understand what space we are in for a given Schlefly symbol, i.e. for a given mental construction. Such judicious experiments with polyhedra help to mentally understand the structure of the 3-dimensional sphere and touch the 4-dimensional space of Euclid. Similarly, one can speculate with a hyperbolic 3-dimensional space. The calculations presented in the publication help to understand what space we are in for a given Schlefly symbol, i.e. for a given mental construction.

In the next publication, we will do the same thing only one dimension higher. I think that I will give there a video clip explaining the main idea of the transition (rise) in dimensions. If someone else does not understand what is happening, then perhaps this video will shed light on all the calculations done here and in the previous article.

If everything is clear, then go to the next level in dimension.

References:

Correct polyhedrons. Part 1. Three-dimensional

regular polyhedrons. Part 2. Four-dimensional

regular polyhedrons. Part 2.5 (auxiliary)

The symbol Shafly. Part 2.6