199. From the expressions for E, F, and S, given in the two preceding Articles, combined with the result 2F+ 25 = 4 + 2E, we obtain + + 2(a+b+c+d+ ...) + 2(a +B+y+8+...)=4+3a+46 +5c+6d+..., 2 (a+b+c+d + ...) + 2(a +B+y+8+...)=4+3a+43+5y+68+..., therefore 2 (a+B+y+8+ ...) – (a + 2b + 3c + 4d + ...) = 4 ... (1), 2(a+b+c+d+ ...)-(a + 2B + 3y + 48+ ...)= 4 ... (2). Therefore, by addition a+a- (c+y) – 2 (d+8) - 3(e+e) - ...... = 8. 8. Thus the number of triangular faces together with the number of solid angles formed with three plane angles cannot be less than eight. Again, from (1) and (2), by eliminating a, we obtain 3a + 26 + -e-2f- – 2B – 4y - ...... = 12, so that 3a + 2b + c cannot be less than 12. From this result various inferences can be drawn; thus for example, a solid cannot be formed which shall have no triangular, quadrilateral, or pentagonal faces. In like manner, we can shew that 3a + 2B + y cannot be less than 12. + 200. Poinsot has shewn that in addition to the five wellknown regular polyhedrons, four other solids exist which are perfectly symmetrical in shape, and which might therefore also be called regular. We may give an idea of the nature of Poinsot's results by referring to the case of a polygon. Suppose five points A, B, C, D, E, placed in succession at equal distances round the circumference of a circle. If we draw a straight line from each point to the next point, we form an ordinary regular pentagon. Suppose however we join the points by straight lines in the following order, A to C, C to E, E to B, B to D, D to A; we thus . form a star-shaped symmetrical figure, which might be considered a regular pentagon. = It appears that, in a like manner, four, and only four, new regular solids can be formed. To such solids, the faces of which intersect and cross, Euler's theorem does not apply. 201. Let us return to Art. 195, and suppose e the number of edges in the bounding contour, and é the number of edges within it; also suppose : the number of corners in the bounding contour, and s' the number within it. Then E = e+e'; S = 8+8'; therefore 1+e+ = 8 + 8 + F. But e=8; therefore 1+ d = 8 + F. We can now demonstrate an extension of Euler's theorem, which has been given by Cauchy. 202. Let a polyhedron be decomposed into any number of polyhedrons at pleasure; let P be the number thus formed, S the number of solid angles, F the number of faces, E the number of edges : then S + F= E + P + 1. For suppose all the polyhedrons united, by starting with one and adding one at a time. Let e, f, 8 be respectively the numbers of edges, faces, and solid angles in the first ; let é', f", be respectively the numbers of edges, faces, and solid angles in the second which are not common to it and the first; let é", f", $" be respectively the numbers of edges, faces, and solid angles in the third which are not common to it and the first or second; and so on. Then we have the following results, namely, the first by Art. 196, and the others by Art. 201; 8+f= e + 2, s'+f'=é' +1, s”+f"=e"+1, + By addition, since 8 + 8' +8" +... = S, f+f' +f" + ... = F, and e + é' + e + ... = E, we obtain S + F= E + P + 1. 203. The following references will be useful to those who study the theory of polyhedrons. Euler, Novi Commentarii Academic.... Petropolitano, Vol. iv. 1758; Legendre, Géométrie ; Poinsot, Journal de l'Ecole Polytechnique, Cahier x; Cauchy, Journal de l'Ecole Polytechnique, Cahier XVI; Poinsot and Bertrand, Comptes Rendus...de l'Académie des Sciences, Vol. XLVI; Catalan, Théorèmes et Problèmes de Géométrie Elémentaire ; Kirkman, Philosophical Transactions for 1856 and subsequent years; Listing, Abhandlungen der Königlichen Gesellschaft...zu Göttingen, Vol. x, MISCELLANEOUS EXAMPLES. 1. Find the locus of the vertices of all right-angled spherical triangles having the same hypotenuse; and from the equation obtained, prove that the locus is a circle when the radius of the sphere is infinite. 2. AB is an arc of a great circle on the surface of a sphere, C its middle point: shew that the locus of the point P, such that the angle APC = the angle BPC, consists of two great circles at right angles to one another. Explain this when the triangle becomes plane. tan-1 + tan On a given arc of a sphere, spherical triangles of equal area are described : shew that the locus of the angular point opposite to the given are is defined by the equation (tan (a + )) 'tan (a -) sin 0 tan A + tan ß, sin (a ++) sin (a -0) where 2a is the length of the given arc, o the arc of the great circle drawn from any point P in the locus perpendicular to the given arc, $ the inclination of the great circle on which is T. S. T. L 1 + tan {sin measured to the great circle bisecting the given arc at right angles, and ß a constant. 4. In any spherical triangle cot A cot a + cot B cot b tan c= cot a cot b-cos A cos B' 5. If 0, 0, y denote the distances from the angles A, B, C respectively of the point of intersection of arcs bisecting the angles of the spherical triangle ABC, shew that cos 6 sin (6 - c) + cos o sin (c – a) + cos y sin (a – 6.) = 0. 6. If A', B', C' be the poles of the sides BC, CA, AB of a spherical triangle ABC, shew that the great circles AA', BB, CC" meet at a point P, such that cos PA cos BC= = cos PB cos CA = cos PC cos AB. 7. If O be the point of intersection of arcs AD, BE, CF drawn from the angles of a triangle perpendicular to the opposite sides and meeting them at D, E, F respectively, shew that 8. If p, q, r be the arcs of great circles drawn from the angles of a triangle perpendicular to the opposite sides, (a, a'), (B, B'), (y, y) the segments into which these arcs are divided, shew that tan a tan a' = tan B tan ' = tan y tan y'; a 9. In a spherical triangle if arcs be drawn from the angles to the middle points of the opposite sides, and if a, á be the two parts of the one which bisects the side a, shew that 10. The arc of a great circle bisecting the sides AB, AC of a spherical triangle cuts BC produced at Q: shew that 11. If ABCD be a spherical quadrilateral, and the opposite sides AB, CD when produced meet at E, and AD, BC meet at F, the ratio of the sines of the arcs drawn from E at right angles to the diagonals of the quadrilateral is the same as the ratio of those from F. 12. If ABCD be a spherical quadrilateral whose sides AB, DC are produced to meet at P, and AD, BC at Q, and whose diagonals AC, BD intersect at R, then sin AB sin CD cos P~sin AD sin BC cos Q=sin AC sin BD cos R. 13. If A' be the angle of the chordal triangle which corresponds to the angle A of a spherical triangle, shew that 14. If the tangent of the radius of the circle described about a spherical triangle is equal to twice the tangent of the radius of the circle inscribed in the triangle, the triangle is equilateral. 15. The arc AP of a circle of the same radius as the sphere is equal to the greater of two sides of a spherical triangle, and the arc AQ taken in the same direction is equal to the less; the EM sine PM of AP is divided at E, so that the natural cosine PM |