2. Of all the spherical polygons, formed with sides all given except one, which may be assumed at pleasure, the greatest is that polygon which may be inscribed in a semicircle, having for its diameter the chord of the undetermined side. The demonstration is deduced from Prop. 26, in the manner exhibited in Prop. 4. of the Appendix just quoted. It is requisite for the existence of a maximum, that the sum of the given sides be less than the semicircumference of a great circle. 3. The greatest of all the spherical polygons, formed with given sides, is that which can be inscribed in a circle of the sphere. Same demonstration as in Prop. 4. App. Book IV. 4. The greatest of all the spherical polygons, having the same perimeter and the same number of sides, is that which has its angles equal and its sides equal. This results from the first and the third of these corollaries. Note. All the propositions about maxima in spherical polygons, are, at the same time, applicable to solid angles, of which those polygons are the measures. APPENDIX TO BOOKS VI. AND VII. THE REGULAR POLYEDRONS. PROPOSITION I. THEOREM. There can only be five regular polyedrons. FOR, regular polyedrons were defined as having equal regular polygons for their faces, and all their solid angles 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 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 (21. V.) cannot form a solid angle. 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. Thirdly. In fine, if the faces are regular pentagons, their angles likewise may be arranged by threes: the regular dodecaedron will re sult. 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. Scholium. In the following Proposition, we shall prove that these five polyedrons actually exist; and that all their dimensions may be determined, when one of their faces is known. PROPOSITION II. PROBLEM. 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, which we shall now solve in succession. 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 per- A pendicular 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. 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 a regular hexaedron or cube. 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, 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 21 H at O, and likewise the triangle AŎD; the sides AO, OS, OD are equal; hence those triangles are equal, hence AS=AD. In the same manner we could shew, 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 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 right; 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; 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. 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 by Prop. 24, V., determine the mutual inclination of two of those planes; which inclination we shall name 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. Construction of the Iosa edron. Το Let ABC be one of its faces. We must first form a solid angle with five planes each equal to ABC, H and each equally inclined to its adjacent one. 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, F H E 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 each equally inclined to their adjacent planes; for the solid angles B', C', &c. 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 named K; which K may be determined by Prop. 24. V.; 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 or rim 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 mutual 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 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 likewise equal. PROPOSITION III. PROBLEM. To find the inclination of two adjacent faces of a regular polyedron. This inclination is deduced immediately from the construction we have just given of the five regular polyedrons; taken in connexion with Prop. 24. V., by means of which, the three plane angles that |