if x, y, z are the perpendiculars from this point on the sides a, b, c respectively, sin a sin y sin z sin (S - B) sin (S-C) sin (S-C) sin (S-A) sin (S-A) sin (S-B)* ( 9. Through each angle of a spherical triangle an arc is drawn so as to make the same angle with one side which the perpendicular on the base makes with the other side. Shew that these arcs meet at a point; and that if x, y, z are the perpendiculars from this point on the sides a, b, c respectively, 10. Shew that the points determined in Examples 8 and 9, and the point N of Art. 146 are on a great circle. State the corresponding theorem in Plane Geometry. 11. If one angle of a spherical triangle remains constant while the adjacent sides are increased, shew that the area and the sum of the angles are increased. 12. If the arcs bisecting two angles of a spherical triangle and terminated at the opposite sides are equal, the bisected angles will be equal provided their sum be less than 180°. [Let BOD and COE denote these two arcs which are given equal. If the angles B and C are not equal suppose B the greater. Then CD is greater than BE by Art. 58. And as the angle OBC is greater than the angle OCB, therefore OC is greater than OB; therefore OD is greater than E. Hence the angle ODC is greater than the angle OEB, by Example 11. Then construct a spherical triangle BCF on the other side of BC, equal to CBE. Since the angle ODC is greater than the angle OEB, the angle FDC is greater than the angle DFC; therefore CD is less than CF, so that CD is less than BE. See the corresponding problem in Plane Geometry in the Appendix to Euclid, page 317.] T. S. T. I XIII. POLYHEDRONS. 149. A polyhedron is a solid bounded by any number of plane rectilineal figures which are called its faces. A polyhedron is said to be regular when its faces are similar and equal regular polygons, and its solid angles equal to one another. 150. If s be the number of solid angles in any polyhedron, F the number of its faces, E the number of its edges, then S+F=E+2. Take any point within the polyhedron as centre, and describe a sphere of radius r, and draw straight lines from the centre to each of the angular points of the polyhedron ; let the points at which these straight lines meet the surface of the sphere be joined by arcs of great circles, so that the surface of the sphere is divided into as many polygons as the polyhedron has faces. Let s denote the sum of the angles of any one of these polygons, m the number of its sides ; then the area of the polygon is god {8-(m - 2)a} by Art. 99. The sum of the areas of all the polygons is the surface of the sphere, that is, tope. Hence since the number of the polygons is F, we obtain 47= Is - 12m + 2FT. Х Now Is denotes the sum of all the angles of the polygons, and is therefore equal to 271 the number of solid angles, that is, to 29S; and Em is equal to the number of all the sides of all the polygons, that is, to 2E, since every edge gives rise to an are which is common to two polygons. Therefore 47=27S-26E+2F; therefore S+F= E +2. F= n 151. There can be only five regular polyhedrons. Let me be the number of sides in each face of a regular polyhedron, n the number of plane angles in each solid angle; then the entire number of plane angles is expressed by mF, or by ns, or by 2E; thus mFunS=2E, and S+F=E+2; from these equations we obtain - 4m 2mn 4n S: E= 2 (m + n) - mn 2 (m + n) – mn 2 (m+n) - min These expressions must be positive integers, we must therefore have 2 (m+n) greater than mn; therefore 1 1 1 must be greater than ;; 2 1 1 but n cannot be less than 3, so that cannot be greater than 3' 1 1 and therefore must be greater than and as m must be an 6 6) integer and cannot be less than 3, the only admissible values of m are 3, 4, 5. It will be found on trial that the only values of m and n which satisfy all the necessary conditions are the following: each regular polyhedron derives its name from the number of its plane faces. + m n n m It will be seen that the demonstration establishes something more than the enunciation states ; for it is not assumed that the faces are equilateral and equiangular and all equal. It is in fact demonstrated that, there cannot be more than five solids each of which has all its faces with the same number of sides, and all its solid angles formed with the same number of plane angles. 152. The sum of all the plane angles which form the solid angles of any polyhedron is 2 (S-2). For if m denote the number of sides in any face of the polyhedron, the sum of the interior angles of that face is (m-2) by Euclid 1. 32, Cor. 1. Hence the sum of all the interior angles of all the faces is (m-2), that is Emir - 2Fm, that is 2 (E – F), that is 2 (S – 2). 153. To find the inclination of two adjacent faces of a regular polyhedron. Let AB be the edge common to the two adjacent faces, C and D the centres of the faces; bisect AB at E, and join CE and DE; CE and DE will be perpendicular to AB, and the angle CED is the angle of inclination of the two adjacent faces; we shall denote it by I. In the plane containing CE and DE draw CO and DO at right angles to CE and DE respectively, and meeting at 0; about 0 as centre describe a sphere meeting 04, OC, OE at a, c, e respectively, so that cae forms a spherical triangle. Since AB is perpendicular to CE and DE, it is perpendicular to the plane CED, therefore the plane AOB which contains AB is perpendicular to the plane CED; hence the angle cea of the spherical triangle is a right angle. Let m be the number of sides in each face of the polyhedron, n the number of the plane angles which form each solid T m T 21 angle. Then the angle ace = ACE = and the angle cae 2m is half one of the n equal angles formed on the sphere round a, that is, cae = From the right-angled triangle cae T that is sin m n T n 154. To find the radii of the inscribed and circumscribed spheres of a regular polyhedron. Let the edge AB=a, let OC=r and 0A =R, so that p is the radius of the inscribed sphere, and R is the radius of the circumscribed sphere. Then also T TT r= R cos aOc= R coteca cot eac = Rcot cot ; me 2 T п a I OT tantan n therefore T m T m 155. To find the surface and volume of a regular polyhedron. maa The area of one face of the polyhedron is cot - ,.and 4 m Fas therefore the surface of the polyhedron is cot 4 Also the volume of the pyramid which has one face of the polyhedron for base and for vertex is cot and 3 4 mFra therefore the volume of the polyhedron is cot 12 go ma? |