The same demonstration would lead to the same conclusion, Fig. 277. if, taking always the arc CA' CA, the angle BCA' (fig. 277) were made less than BCA; hence ABC is the greatest of all those triangles, which have two sides given, and the third to be assumed at pleasure. Fig. 278. 559. Scholium 1. The triangle ABC (fig. 278), the greatest of all those which have two given sides CA, CB, may be inscribed in a semicircle, the diameter of which is the chord of the third side AB; for O being the middle point of AB, the distances OC, OB, as we have seen, are equal; hence the circumference of a small circle, described from the point O as a pole, with the distance OB, will pass through the three points A, B, C. Moreover, the straight line AB is a diameter to this small circle; for the centre, which must be at once in the plane of the small circle, and (456) in the plane of the arc of the great circle BOA, must of necessity be found in the intersection of those two planes, which is the straight line BA; hence BA will be a diameter. 560. Scholium II. In the triangle ABC, the angle C being equal to the sum of the other two A and B, the sum of all the three angles must be double of the angle C. But (489) that sum is always greater than two right angles; hence C is always greater than one. 561. Scholium III. If the sides CB, CA, are produced till they meet in E, the triangle BAE will be equal to the fourth part of the surface of the sphere. For the angle E=C ABC + CAB; hence the three angles of the triangle BAE are equivalent to the four ABC, ABE, CAB, BAE, whose sum is equal to four right angles; therefore (505) the surface of the triangle BAE 4-2 = 2, which is the fourth part of the surface of the sphere. 562. Scholium IV. There could be no maximum, if the sum of the two given sides CA, CB, were equal to, or greater than, the semicircumference of a great circle. For, since the triangle ABC must be capable of being inscribed in a semicircle of the sphere, the sum of the two sides CA, CB, will be less (460) than the semicircumference BCA, and consequently less than half the circumference of a great circle. The reason why there can be no maximum, when the sum of the two given sides is greater than the semicircumference of a great circle, is that in this case the triangle continues to augment, as the angle contained by its two given sides augments; and at last, when this angle becomes equal to two right angles, the three sides are all in the same plane, and form a whole circumference; the spherical triangle has then increased to a hemisphere, but it has at the same time ceased to be a triangle. THEOREM. 563. Of all the spherical triangles, formed with a given side and a given perimeter, the greatest is that in which the two undetermined sides are equal. Demonstration. Let AB (fig. 279) the given side be common Fig. 279. to the two triangles ACB, ADB, and let AC+ CB=AD+DB ; we are to show that the isosceles triangle ACB, in which ACCB, is greater than ADB, which is not isosceles. Since these triangles have the common part AOB, it will be sufficient to prove that the triangle BOD is less than AOC. Now, the angle CBA, equal to CAB, is greater than OAB; therefore (497) the side AO is greater than OB. Take OI=OB, make OK = OD, and join KI; the triangle OKI (497) will be equal to DOB. Now, if the triangle DOB, or its equal KOI, is not admitted to be less than OAC, it must be either equal or greater; in both which cases, since the point I is between A and O, the point K must be found in the prolongation of OC, otherwise the triangle OKI would be contained in the triangle CAO, and therefore less than CAO. This granted, since the shortest way from C to A is CA, we have CK + KI + IA > CA. But CK=OD-CO, AIAO-OB, KI = BD; hence OD CO+AO OB + BD > CA, or by reduction, AD — CB+BD> CA, or AD + BD > CA + CB. But this inequality is at variance with the supposition of AD+BDCA + CB; hence the point K cannot fall in the prolongation of OC; consequently it falls between O and C, and the triangle KOI or its equal ODB is less than ACO; therefore the isosceles triangle ACB is greater than ADB having the same base and perimeter, which is not isoscele. 564. Scholium. The two last theorems are analogous to those of art. 63 and 69, of the appendix to section. fourth; and from them may be deduced, with regard to spherical polygons, the same consequences as we have obtained respecting plane polygons. The chief are as follows: 565. Among spherical polygons of the same perimeter and the same number of sides, that is the greatest which has its sides equal. The demonstration is the same as that of art. 301. 566. Among spherical polygons formed of known sides and one side taken at pleasure, the greatest is that which can be inscribed in a semicircle the diameter of which is equal to the chord of the undetermined side. The demonstration is deduced from art. 559, in the manner exhibited in art. 303. 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. 567. Among spherical polygons formed of given sides, the greatest is that which can be inscribed in a circle of the sphere. The demonstration is the same as that of art. 303. 568. Among spherical polygons which have the same perimeter and the same number of sides, the greatest is that which has its angles equal, and its sides equal. This results from the first and the third of the above proposi tions. Note. All the propositions relating to the maxima of spherical polygons, are also applicable to solid angles, of which these polygons are the measures. Appendix to Sections Second and Third. OF THE REGULAR POLYEDRONS. THEOREM. 569. There can be only five regular polyedrons. Demonstration. 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. 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 SAAB; 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 tetraedrøn. 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 0, 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 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. |