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CONSTRUCTION OF THE FIVE REGULAR POLYEDRONS OR SOLIDS.

A regular polyedron is one whose faces are all equal regular polygons, and whose solid angles are all equal to each other. (Def. 3. 7.) These conditions can be fulfilled only in five cases: that is,

There can be only five regular polyedrons, viz. the tetraedron, the hexaedron, the octaedron, the dodecaedron and the icosaedron. The tetraedron is bounded by four equal equilateral triangles; the hexaedron is bounded by six equal squares; the octaedron by eight equal equilateral triangles; the dodecaedron by twelve equal pentagons; the icosaedron by twenty equal equilateral triangles.

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 cannot form a solid angle. (20. 6.)

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 form a solid angle.

four right-angles, and cannot

3. If the faces are regular pentagons, their angles likewise may be arranged by threes: the regular dodecaedron will result.

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

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

This problem subdivides itself into five cases.

CASE 1.

Construction of the Tetraedron.

Let ABC be the equilateral triangle which is to form 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: the pyramid SABC will be the tetraedron required.

B

S

For, by reason of the equal distances OA, OB, OC, the oblique lines SA, SB, SC are equally removed from the perpendicular SO, and consequently equal. One of them SA= AB; hence the four faces of the pyramid SABC 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.

CASE II.

Construction of the Hexaedron.

Let ABCD 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, each being formed with three right-angles: hence this prism is D a regular hexaedron or cube.

H

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CASE III.

Construction of the Octaedron.

Let AMB be a given equilateral triangle. On the side AB, describe a 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=OA; then join SA, SB, TA, &c., you will have a solid SABCDT,

M

A

D

C

composed of two quadrangular pyramids SABCD, TABCD, 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; hence those triangles are equal, hence AS=AD. In the same manner we could show, that all the other right-angled triangles AOT, BOS, COT, &c., are equal to the triangle AOD; hence all the sides AB, AS, AT, &c., are equal, 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, comparing the pyramid BASCT with the pyramid SABCD, 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 apply to each other in all points; hence the solid angle S is equal to the solid angle B; therefore 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.

CASE IV.

Construction of the Dodecaedron.

Let ABCDE be a given regular pentagon; let ABP, CBP be two plane angles each equal to the angle ABC. With these plane angles form the solid angle B; and call the mutual

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inclination of two of those planes K. In like manner, at the points C, D, E, A, form solid angles, equal to the solid angle B, and 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, and each inclined to its adjacent plane by the same quantity K. Let pfgh, &c., be a second surface equal to PFGH, &c., we assert 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, 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. 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 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 efgh, &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 appears that the two surfaces will adjust themselves completely 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.

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