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 OT = OS = 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 show, 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 show 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 QB 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 pƒ will apply itself to its equal PF, and at the point F will be found three plane angles PFG, pfe, ef g, 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 ef gh, &c.; for the planes PFG, e fp, already joined at P, have the requisite inclination K, as well as the planes eƒg, eƒ p. Continuing the comparison, in this way, by successive steps, it GEOM.

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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.

Fig. 284.

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'ID' ; 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', AH, AI, 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.*

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.

Fig. 292.

PROBLEM.

579. The side of a regular polyedron being given, to find the radius of the inscribed and that of the circumscribed sphere.

Solution. It must first be shown, that every regular polyedron is capable of being inscribed in a sphere, and of being circumscribed about it.

Let AB (fig. 292) be the side common to two adjacent faces; C and E the centres of those faces; CD, ED, the perpendiculars let fall from these centres upon the common side AB, and therefore terminating in D, the middle point of that side. The two perpendiculars CD, DE, make with each other an angle which is known, being the inclination of two adjacent faces, and determinable by the last problem. Now, if in the plane CDE, at right angles to AB, two indefinite lines CO and OE be drawn perpendicular to CD and ED, and meeting each other in O, this point O will be the centre of the inscribed and of the circumscribed sphere, the radius of the first being OC, that of the second OA.

For, since the perpendiculars CD, DE, are equal, and the hypothenuse DO is common, the right-angled triangle CDO must (56) be equal to the right-angled triangle ODE, and the perpendicular OC to OE. But, AB being perpendicular to the plane CDE, the plane ABC (349) is perpendicular to CDE, or CDE to ABC; likewise CO, in the plane CDE is perpendicular to CD, the common intersection of the planes CDE, ABC; hence (351) CO is perpendicular to the plane ABC. For the same reason, EO is perpendicular to the plane ABE; hence the two straight lines CO, EO, drawn perpendicular to the planes of two adjacent faces, through the centres of those faces, will meet in the same point O, and be equal to each other. Now, suppose that ABC and ABE represent any other two adjacent faces; the perpendicular CD will still continue of the same magnitude; and also the angle CDO, the half of CDE; consequently the rightangled triangle CDO, and its side CO will be equal in all the faces of the polyedron; hence, if from the point O as a centre, with the radius OC, a sphere be described, it will touch all the faces of the polyedron at their centres, the planes ABC, ABE, &c., being each perpendicular to a radius at its extremity; therefore the