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There can only be five regular polyedrons.

For regular polyedrons were defined as those all of whose faces are equal regular polygons, and all whose solid angles are equal. These conditions cannot be fulfilled except in a small number of cases.

First. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three those triangles, by four, or by five : hence arise three regular bodies, the tetraedron, the octaedron, and the icosaedron. No other can be formed with equilateral triangles, for six angles of these triangles are equal to four right angles, and cannot form a solid angle (Book V., Prop. XXXIII.).

Secondly.-If the faces are squares, their angles may be arranged by threes; hence results the hexaedron, or cube.

Four angles of a square are equal to four right angles, and cannot form a solid angle.

Thirdlı. —In fine, if the faces are regular pentagons, their angles may likewise be arranged in threes; and the regular dodecaedron will result.

We can go no farther ; for three angles of a regular hexagon are equal to four right angles ; three of a heptagon are greater.

Hence, there can be only five regular polyedrons ; three formed with equilateral triangles, one with squares, and one with pentagons.

SCHOLIUM. It will be proved in the following proposition that these five polyedrons actually exist, and that all their dimensions may be determined when one of their faces is known.


PROBLEM. One of the faces of a regular polyedron being given, or only its side, to construct the polyedron.

This problem embraces five, which will be solved in succession,

1. CONSTRUCTION OF THE TETRAEDRON. Let ABC be the equilateral triangle which is to be one face of the tetraedron. At the point O, the centre of this triangle, erect


OS perpendicular to the plane ABC; terminate this perpendicular in S, so that AS = AB; join SB, SC, and the pyramid S-ABC will be the required tetraedron.

For, by reason of the equal distances OA, OB, OC, the oblique lines SA, SB, SC are equal (Prop. IX., Book V.). One of them, SA AB ; hence, the four faces of the pyramid S-ABC are triangles equal to the given triangle ABC. And the solid angles of this pyramid, are all equal, because each of them is formed by three equal plane angles; hence this pyramid is a regular tetraedron.



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Let ABCD be a given square ; on the base, ABCD, construct a right prism whose altitude, AE, shall be equal to the side AB. It is evident that the faces of this prism are equal squares, and that its solid angles are equal, since they are each formed by three right angles ; hence, this prism is a regular hexaedron, or cube.



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3. CONSTRUCTION OF THE OCTAEDRON. Let AB be the given edge : on AB describe the square ABCD; at the point O, the centre of this square, erect TS perpendicular to its plane, and terminating on both sides in T and S, so that OT = OS AO; then join SA, SB, TA, etc. ; you will have a solid, SABCDT, composed of two quadrangular pyramids, S-ABCD, T-ABCD, united together by their common base, ABCD; this solid will be the required octaedron.

For the triangle AOS is right angled at 0, as is the triangle AOD; the sides AO, OS, OD are equal; hence, those triangles are equal ; hence, AS = AD. It may be shown in like manner that all the other right angled triangles, AOT, BOS, COT, etc., are equal to the triangle AOD; hence all the sides, AB, AS, AT, etc., are equal, and consequently the solid SABCDT is bounded by eight equilateral triangles whose sides are equal to AB. Moreover, the solid angles of this polyedron are all equal ; for instance, the angle S is equal to the angle B.

For it is evident that the triangle SAC is equal to the triangle DAC, and therefore the triangle ASC is right; hence the figure SATC is a square, equal to the square ABCD. But, comparing the pyramid B-ASCT with the pyramid S-ABCD, the base of the first may be placed on the base, ABCD, of the second ; then the point O being a common centre, the altitude OB of the first will coincide with the altitude OS of the second, and the two pyramids will exactly apply to each other in all points ; hence the solid angle S is equal to the solid angle B; hence the solid SABCDT is a regular octaedron.

SCHOLIUM. If three equal straight lines, AC, BD, ST, are perpendicular to each other, and bisect each other, the extremities of these straight lines will be the vertices of a regular octaedron.

4. CONSTRUCTION OF THE DODECAEDRON. Let ABCDE be a given regular pentagon ; let ABP, CBP be two plane angles equal to the angle ABC; with these plane angles form the solid angle B. The mutual inclination of two of these planes, * we will call K. In like manner, at the points C, D, E, A, form solid angles equal to solid angle B, and similarly situated : the plane CBP will be the same as the plane BCG, since they are both inclined by the same quantity, K, to the plane ABCD. Hence, the pentagon BCGFP, equal to the pentagon ABCDE, may be described in the

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plane PBCG. If the same thing is done in each of the other planes, CDI, DEL, etc., we shall have a convex surface, PFGH, etc., composed of six equal regular pentagons, and each inclined to its adjacent plane by the same quantity, K. Let pfgh, etc., be a second surface equal to PFGH, etc., then may these two surfaces be united so as to form a single continuous convex surface. For, the angle opf, for example, may be joined to the two angles OPB, BPF, to make a solid angle, P, equal to the angle B; and in this junction no change will take place in the inclination of the planes BPF, BPO, that inclination being already such as is required to form the solid angle B. But, whilst the solid angle P is formed, the side pf will fall along its equal PF, and at the point F will be found the three planes, PFG, pfe, efg, united and forming a solid angle equal to each of the solid angles already formed : and this junction will take place without changing either the state of the angle P, or that of the surface efgh, etc. ; for the planes PFG, efp, already united at P, have the proper inclination K, as have the planes efg, efp. Continuing the comparison thus, step by step, we see that the two surfaces will mutually adjust themselves to each other so as to form a single continuous convex surface ; which will be that of a regular dodecaedron, since it is composed of twelve equal regular pentagons, and has all its solid angles equal.

* K is determined by methods not given in these elements.

5. CONSTRUCTION OF THE ICOSAEDRON. Let ABC be one of its faces : we must first form a solid angle with five planes equal to ABC, and each equally inclined to its adjacent plane. To do this, on the side B'C' equal to BC, construct the reguiar pentagon B'C'H'I'D' ; at the centre of this pentagon erect a perpendicular to its plane, terminating in A', so that B'A' = B'C'; join A'C', A'H', A'I', A'D' ; the solid angle A', formed by the five planes B'A'C', C'A'H', etc., will be the solid angle required. For, the oblique lines A'B', A'C', etc., are equal, and one of them, A'B', is equal to the side B'C' ; hence, all the triangles, B’A'C', C'A'H', etc., are equal to each other, and to the given triangle, ABC.

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It is further manifest that the planes B'A'C', C'A'H', etc., are each equally inclined to their adjacent planes ; for, the solid angles B', C', etc., are all equal, being each formed by two angles of equilateral triangles, and one of a regular pentagon. Let the inclination of the two planes in which are the equal angles be called K ; the angle K will, at the same time, be the inclination of each of the planes composing the solid angle A' to its adjacent plane.

This being granted, if at each of the points A, B, C, a solid angle be formed equal to the angle A', we will have a convex surface, DEFG, etc., composed of ten equilateral triangles, each one of which will be inclined to its adjacent triangle by the quantity K ; and the angles D, E, F, etc., of its contour will alternately combine three angles and two angles of equilateral triangles. Conceive a second surface equal to the surface DEFG, etc. ; these two surfaces may be mutually adapted to each other, if each triple angle of one is joined to a double angle of the other ; and, since the planes of these angles have already to each other the inclination, K, requisite to form a quintuple solid angle equal to the angle A, there will be nothing changed, by this junction, in the state of each particular surface, and the two together will form a single continuous surface, composed of twenty equilateral triangles. This surface will be that of the regular icosaedron, since all its solid angles are also equal.

SCHOLIUM 1. The following table of the numbers of the different elements of the regular polyedrons enables us to notice particularly some of the properties of each :

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1. In the regular tetraedron the number of vertices is the same as the number of faces.

2. The number of vertices of the regular hexaedron is equal to the number of faces of the octaedron, and reciprocally.

3. The number of the vertices of the dodecaedron is equal to the number of faces of the icosaedron, and reciprocally.

4. The number of edges of the hexaedron and octaedron is the same.

5. The number of edges of the dodecaedron and icosaedron is the same.

The following properties may also be observed in considering these bodies :

6. In a tetraedron the edges are opposite two and two, and each vertex is opposite to a face.

7. In the hexaedron and octaedron the vertices are opposite two and two. The same is true of the edges and faces.

8. In the dodecaedron the faces are opposite two and two, but neither the edges nor the vertices; but each edge is opposite to a vertex.

9. In the regular icosaedron the vertices are opposite two and two, but neither the edges nor the faces; but each edge is opposite to a face.

SCHOLIUM 2. The figures below represent the developments of the surfaces of the five regular polyedrons in their order. We can form

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