Regular polyhedra harmony

Regular poly­he­drons have been the subject of interest for many great scien­tists. And this interest went far beyond math­e­matics. Plato (427 BC — 347 BC) saw them as the basis of the struc­ture of the Universe, Kepler (1571—1630) tried to link the regular poly­he­drons with the move­ment of the planets of the solar system (of which there) were known five at his time. Perhaps it was the beauty and harmony of the regular poly­he­drons that made the great scien­tists of the past suggest some kind of deeper purpose for them than just geometric objects.

A regular poly­he­dron is a poly­he­dron all of whose faces are regular poly­gons, all of whose planar angles are equal to each other and whose dihe­dral angles are equal to each other. (The planar angles of a poly­he­dron are the angles of polygon-faces, the dihe­dral angles of a poly­he­dron are the angles between faces that have a common edge.)

Note that the convexity of a regular poly­he­dron follows auto­mat­i­cally from this defi­n­i­tion, which in some books is included in the defi­n­i­tion.

In three-dimen­sional space, there are exactly five regular poly­he­drons: tetra­he­dron, octa­he­dron, cube (hexa­he­dron), icosa­he­dron, dodec­a­he­dron. The fact that there are no other regular poly­he­drons was proved by Euclid (about 300 BC) in his great “Ele­ments”.

The tetra­he­dron (from Greek τετρά, in compound words — four, and έδρα — facet) is a regular poly­he­dron with four trian­gular faces. It has four vertices, six edges. Since the faces of the tetra­he­dron are regular trian­gles, its planar angles are $\pi/3$. The dihe­dral angles of the tetra­he­dron are $\arccos(1/3) ≈ 70{,}53^\circ$.

Let’s take the midpoints of the each tetra­he­dron face and connect them to each other with segments. These segments are equal in length and form equi­lat­eral trian­gles. The points are the vertices, the segments are the edges, and the trian­gles are the faces of another tetra­he­dron.

A similar construc­tion is applic­able in the more general case. Consider an arbi­trary convex poly­he­dron and take the midpoints of its faces. Connect points of adja­cent faces with segments. Thus, the points are vertices, the segments are edges and the poly­gons that are bounded by these segments are the faces of another convex poly­he­dron. This poly­he­dron is called dual to the orig­inal poly­he­dron.

As shown above, the dual to a tetra­he­dron is a tetra­he­dron.

Let’s increase the size of the tetra­he­dron, whose vertices are the midpoints of the orig­inal tetra­he­dron faces, to the size of the latter. Eight vertices of such a tetra­he­drons are the vertices of the cube.

The inter­sec­tion of these tetra­he­drons is another regular poly­he­dron, the octa­he­dron (from Greek οκτώ — eight). An octa­he­dron has 8 trian­gular faces, 6 vertices and 12 edges. The flat angles of the octa­he­dron are $\pi/3$ because its faces are right trian­gles, the dihe­dral angles are $\arccos(-1/3) ≈ 109{,}47^\circ$.

Let’s mark the midpoints of the octa­he­dron faces and proceed to the poly­he­dron that is dual to the octa­he­dron. This is a cube or hexa­he­dron (from Greek εξά, six). A cube has edges which are squares. It has 6 faces, 8 vertices and 12 edges. The flat angles of the cube are $\pi/2$, the dihe­dral angles are also $\pi/2$.

If we take the midpoints of the cube faces and consider its dual poly­he­dron, it turns out to be an octa­he­dron. The more general state­ment is correct: if we construct a dual poly­he­dron for a convex poly­he­dron and then a dual to the convex poly­he­dron, it will be the orig­inal poly­he­dron (with exact congru­ence).

Let’s take the points on an edges of an octa­he­dron with the condi­tion that each edge is divided by this point in the ratio $1:(\sqrt5+1)/2$ (the Golden Section) and the points belonging to one edge are the vertices of a regular triangle. The resulting 12 points are the vertices of another regular poly­he­dron — icosa­he­dron (from Greek είκοσι — twenty). An icosa­he­dron is a regular poly­he­dron with 20 trian­gular faces. It has 12 vertices and 30 edges. The flat angles of the icosa­he­dron are $\pi/3$, the dihe­dral angles are $\arccos(-1/3\cdot\sqrt5) ≈ 138{,}19^\circ$.

The icosa­he­dron can be inscribed into a cube. Each face of the cube will have two vertices of the icosa­he­dron.

Let’s rotate the icosa­he­dron, “putting” it on a vertex, to get its more “familiar” appear­ance: two hats of five trian­gles at the south and north poles and a middle layer made of ten trian­gles.

The icosa­he­dron faces’ midpoints are the vertices of another regular poly­he­dron, the dodec­a­he­dron (from the Greek δώδεκα — twelve). The faces of the dodec­a­he­dron are regular pentagons. Thus its planar angles are $3\pi/5$. The dodec­a­he­dron has 12 faces, 20 vertices and 30 edges. The dihe­dral angles of the dodec­a­he­dron are $\arccos(-1/5\cdot\sqrt5) ≈116{,}57^\circ$.

Taking the midpoints of the deca­he­dron faces and going to its dual poly­he­dron, we get the icosa­he­dron again. So the icosa­he­dron and the dodec­a­he­dron are dual to each other. Yet again this illus­trates the fact that the dual to the dual will be the orig­inal poly­he­dron.

When going to a dual poly­he­dron, note that the vertices of the orig­inal poly­he­dron corre­spond to the faces of the dual poly­he­dron, the edges corre­spond to the edges of the dual poly­he­dron, and the faces corre­spond to vertices of the dual poly­he­dron. If an icosa­he­dron has 20 faces then its dodec­a­he­dron has 20 vertices, and they have the same number of edges, If a cube has 8 vertices then its dual octa­he­dron has 8 faces.

There are different methods for regular poly­he­drons to be inscribed into each other, which lead to many remark­able construc­tions. Inter­esting and beau­tiful poly­he­drons are also obtained by combining and inter­secting them.

In a dodec­a­he­dron, we inscribe a cube so that all 8 vertices of the cube coin­cide with vertices of the dodec­a­he­dron. Then we circum­scribe the icosa­he­dron around the dodec­a­he­dron so that its vertices are in the midpoints of the icosa­he­dron edges. Then circum­scribe an octa­he­dron around the icosa­he­dron so that the vertices of the icosa­he­dron rest on the edges of the octa­he­dron. Finally, circum­circle the octa­he­dron around the tetra­he­dron so that the vertices of the octa­he­dron rest on the midpoints of the tetra­he­dron edges.

This construc­tion was made from pieces of broken wooden ski sticks by the great math­e­mati­cian of the 20th century, V. Arnold, in his child­hood. Vladimir Igore­vich kept it for many years, and then gifted it to the labo­ra­tory for popu­lar­iza­tion and propa­ganda of math­e­matics at the Steklov Math­e­mat­ical Insti­tute.

Other etudes in “Polyhedra’s outer geometry”