1. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five: hence arise three regular bodies, the tetraedron, the octaedron, the icosaedron. No other can be formed with equilateral triangles; for six angles of such a triangle are equal to four right angles, and (356) cannot form a solid angle.

2. 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.

3. In fine, if the faces are regular pentagons, their angles may likewise be arranged by threes; the regular dodecaedron will thus be formed.

We can proceed no farther; 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.

570. Scholium. In the following problem, we shall show that these five polyedrons actually exist; and that all their dimensions may be determined, when one of their faces is known.

PROBLEM.

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

Solution. This problem admits of five cases, which we proceed to solve in succession.

Construction of the Tetraedron.

572. Let ABC (fig. 280) be the equilateral triangle which is Fig. 280. to form one of the faces of the tetraedron. At the point O, the centre of this triangle, erect OS perpendicular to the plane ABC; let this perpendicular terminate in S, so that AS= AB; join SB, SC; the pyramid S-ABC will be the tetraedron required.

For, on account of the equal distances OA, OB, OC, the oblique lines SA, SB, SC are equally removed from the perpendicular SO, and consequently equal to each other. One of them SA=AB; hence the four faces of the pyramid S-ABC are tri

Fig. 281.

Fig. 282.

angles, 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: this pyramid therefore is a regular tetraedron.

Construction of the Hexaedron.

573. Let ABCD (fig. 281) be a given square. On the base ABCD, construct a right prism whose altitude AE shall be equal to the side AB. The faces of this prism will evidently be equal squares; and its solid angles all equal to each other, each being formed by three right angles; this prism therefore is a regular hexaedron or cube.

Construction of the Octaedron.

574. Let AMB (fig. 282) be a given equilateral triangle. On the side AB, describe a square ABCD; through the point O, the centre of this square, let the perpendicular TS be drawn, terminating on the one hand and on the other in T and S, so that OTOS = OA; then join SA, SB, TA, &c.; we shall 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 O, and likewise the triangle AOD, the sides AO, OS, OD, are equal to each other; hence those triangles are equal, and AS = AD. In the same manner we could shew, that, all the other right-angled triangles AOT, BOS, COT, &c., are equal each to the triangle AOD; hence all the sides AB, AS, AT, &c., are equal to each other, and therefore the solid SABCDT is contained by eight triangles, each equal to the given equilateral triangle ABM. We have yet to shew that the solid angles of this polyedron are equal to each other; that the angle S, for example, is equal to the angle B.

Now, the triangle SAC is evidently equal to the triangle DAC, and therefore the angle ASC is a right angle; hence the figure. SATC is a square equal to the square ABCD. But if we compare the pyramid B-ASCT with the pyramid S-ABCD, we shall see that the base ASCT of the first may be placed on the base ABCD of the second; then, the point O being their common centre, the altitude OB of the first will coincide with the altitude

OS of the second; and the two pyramids will exactly coincide with each other in all points; hence the solid angle S is equal to the solid angle B; and therefore the solid SABCDT is a regular octaedron.

575. 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.

Construction of the Dodecaedron.

576. Let ABCDE (fig. 283) be a given regular pentagon; let Fig. 283. ABP, CBP, be two plane angles each equal to the angle ABC. With these plane angles form the solid angle B; and by art. 361 determine the mutual inclination of two of these planes; which inclination we shall call K. In like manner, at the points C, D, E, A, form solid angles, equal to the solid angle B, and which shall be similarly situated; the plane CBP will be the same as the plane BCG, since both of them are inclined at an equal angle K to the plane ABCD; hence in the plane PBCG, we may describe the pentagon BCGFP, equal to the pentagon ABCDE. If the same thing is done in each of the other planes CDI, DEL, &c., we shall have a convex surface PEGH, &c., composed of six regular pentagons, all equal to each other, and each inclined. to its adjacent plane by the same quantity K. Let pfgh, &c. be a second surface equal to PFGH, &c.; we say that these two surfaces may be joined so as to form only a single continuous convex surface. For the angle opf, for example, may be joined to the two angles OPB, BPF, so as to make a solid angle P equal to the angle B; and by this joining together 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. But whilst the solid angle P is forming, the side pf will apply itself to its equal PF, and at the point F will be found three plane angles PFG, pfe, efg, united so as to form a solid angle equal to each of the solid angles already formed; and this junction, like the former, will take place without producing any change either in the state of the angle P or in that of the surface efg h, &c.; for the planes PFG, efp, already joined at P, have the requisite inclination K, as well as the planes efg, efp. Continuing the comparison, in this way, by successive steps, it

Fig. 284.

will appear that the two surfaces adjust themselves perfectly to each other, and form a single continuous convex surface; which will be that of the regular dodecaedron, since it is composed of twelve equal regular pentagons, and has all its solid angles equal to each other.

Construction of the Icosaedron.

577. Let ABC (fig. 284) be one of its faces. We must first form a solid angle with five planes each equal to ABC, and each equally inclined to its adjacent one. To effect this, on the side B'C', equal to BC, construct the regular pentagon B'C'H'I'D'; at the centre of this pentagon, draw a line at right angles to its plane, and 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', &c., will be the solid angle required. For the oblique lines A'B', A'C', &c. are equal; one of them A'B' is equal to the side B'C'; hence all the triangles B'A'C, C'A'H', &c. are equal to each other and to the given triangle ABC.

It is farther manifest, that the planes B'A'C, C'A'H', &c., are all equally inclined to their adjacent planes; for the solid angles B', C', &c., are all equal to each other, being each formed by two angles of equilateral triangles, and one of a regular pentagon. Let K be the inclination of two planes, forming the equal angles, which inclination may be determined by art. 361; the angle K will at the same time be the inclination of each of the planes composing the solid angle A' to their adjacent planes.

This being granted, if at each of the points A, B, C, a solid angle be formed equal to the angle A', we shall have a convex surface DEFG, &c., composed of ten equilateral triangles, every one of which will be inclined to its adjacent triangle by the quantity K; and the angles D, E, F, &c., of its contour will alternately combine three angles and two angles of equilateral triangles. Conceive a second surface equal to the surface DEFG, &c.; these two surfaces will adapt themselves to each other, if each triple angle of the one is joined to each double angle of the other; and, since the planes of these angles have already the common inclination K, requisite to form a quintuple solid angle equal to the angle A, this junction will require no change in the state of either 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 equal to each other.t

PROBLEM.

578. To find the inclination of two adjacent faces of a regular polyedron.

Solution. This inclination is deduced immediately from the construction we have just given of the five regular polyedrons, taken in connexion with art. 361, by means of which the three plane angles that form a solid angle being given, the angle which two of these plane angles form with each other may be determined.

In the tetraedron. Each solid angle is formed of three angles of equilateral triangles; therefore seek, by the problem referred to, the angle which two of these planes contain between them, and it will be the inclination of two adjacent faces of the tetraedron.

In the hexaedron. The angle contained by two adjacent faces is a right angle.

In the octaedron. Form a solid angle with two angles of equilateral triangles and a right angle; the inclination of the two planes, in which the triangular angles are situated, will be that of two adjacent faces of the octaedron.

In the dodecaedron. Every solid angle is formed by three angles of regular pentagons; the inclination of the planes of two of these angles will be that of two adjacent faces of the dodecaedron.

In the icosaedron. Form a solid angle with two angles of equilateral triangles and one of a regular pentagon; the inclination of the two planes, in which the triangular angles are situated, will be that of two adjacent faces of the icosaedron.

† If the figures 287, 288, 289, 290, 291, be accurately drawn on pasteboard and the fine lines be cut through, and the full lines cut only half through, the edges of the several polygons in each figure may be brought together and glued, the shaded one remaining fixed. Models of the several regular polyedrons may thus be easily obtained.