mon measure which is contained m times in MN and n times in MNPM. Then Apply the measure to the circumference, and through the points of division and the axis AB pass planes; they will divide the whole surface of the sphere into n equal lunes (67, Exercise), of which the given lune ANBMA will contain m. Therefore 2d. We can extend the proof to the case where MN and MNPM are incommensurable by our usual method. (v. VII., Proposition VII.) 71. COROLLARY. The area of a lune is expressed by twice its angle, the angular unit being the degree, and the unit of surface the spherical degree. For, by (69), the area of the surface of the sphere is 720 spherical degrees. We have, then, if S is the area and A the angle of the lune, 72. Scholium. If the angle A contains a whole number of degrees, and each of the parts of the arc MN in the figure above is one degree, each of the small lunes is made up of two spherical degrees, and the lune AMBN obviously contains twice as many spherical degrees as the arc MN contains degrees of arc. PROPOSITION XXII.—THEOREM. 73. The area of a spherical triangle is equal to the excess of the sum of its angles over two right angles. For, let ABC be a spherical triangle. Complete the great circle ABA'B', and produce the arcs AC and BC to meet this circle in A′ and B'. We have, by the figure, ABC + A'BC = lune A, B Α' B and, by Proposition XX., ABC + A'B'C — lune C. The sum of the first members of these equations is equal to twice the triangle ABC, plus the four triangles ABC, A'BC, AB'C, A'B'C, which compose the surface of the hemisphere, whose area is 360 spherical degrees. Therefore, denoting the area of the triangle ABC by T, we have (Proposition XXI., Corollary) 2T+360° 2A + 2B + 20, = 74. Scholium. The excess of the sum of the angles of a spherical triangle over two right angles is sometimes called its spherical excess. EXERCISE. Theorem. The area of a spherical polygon is measured by the sum of its angles minus the product of two right angles multiplied by the number of sides of the polygon less two. A B C D E 75. Scholium. It must not be forgotten that Propositions. XXI. and XXII. merely enable us to express our areas in spherical degrees; that is, in terms of of the surface of the whole sphere. If the area is required in terms of the ordinary unit of surface (IV., 1), the area of the surface of the sphere must first be given in terms of the unit in question. SHORTEST LINE ON THE SURFACE OF A SPHERE BETWEEN TWO POINTS. PROPOSITION XXIII.—THEOREM. 76. The shortest line that can be drawn on the surface of a sphere between two points is the arc of a great circle, not greater than a semi-circumference, joining the two points. Let AB be an arc of a great circle, less than a semi-circumference, joining any two points A and B of the surface of a sphere; and let C be any arbitrary point taken in that arc. Then we say that the shortest line from A to B, on the surface of the sphere, must pass through C. B F G From A and B as poles, with the polar distances AC and BC, describe circumferences on the surface; these circumferences touch at C and lie wholly without each other. For, let M be any point other than C in the circumference whose pole is A, and draw the arcs of great circles AM, BM, forming the spherical triangle AMB. We have, by Proposition XVI., AM + BM > AB, and subtracting from the two members of this inequality the equal arcs AM and AC, we have BM > BC; therefore M lies without the circumference whose pole is B. Now let AFGB be any line from A to B, on the surface of the sphere, which does not pass through the point C, and which therefore cuts the two circumferences in different points, one in F, the other in G. Then a shorter line can be drawn from A to B, passing through C. For, whatever may be the nature of the line AF, an equal line can be drawn from A to C; since, if AC and AF be conceived to be drawn on two equal spheres having M F G B a common diameter passing through A, and therefore having their surfaces in coincidence, and if one of these spheres be turned upon the common diameter as an axis, the point A will be fixed and the point F will come into coincidence with C; the surfaces of the two spheres continuing to coincide, the line AF will then lie on the common surface between A and C. For the same reason, a line can be drawn from B to C, equal to BG. Therefore a line can be drawn from A to B, through C, equal to the sum of AF and BG, and consequently less than AFGB. The shortest line from A to B therefore passes through C; that is, through any, or every, point in AB. Consequently it must be the arc AB itself. EXERCISES ON BOOK VIII. THEOREMS. N M D 1. A SPHERE can be circumscribed about any tetraedron. Suggestion. The locus of the points equally distant from A, B, and C is the perpendicular EM erected at the centre of the circle circumscribed about ABC (VI., Exercise 15.) The locus of the points equally distant from B, C, and D is the perpendicular FN, and both EM and FN lie in the plane perpendicular to BC at its middle point, since that plane contains all the points equally distant from B and C. EM and FN therefore intersect, and O, their point of intersection, is equally distant from the four vertices of the tetraedron. There is only one such point. Therefore only one sphere can be circumscribed about a tetraedron. A E H B 2. The perpendiculars erected at the centres of the four faces of a tetraedron meet in a point. 3. A sphere can be inscribed in any tetraedron. Suggestion. The locus of the points equally distant from two faces of the tetraedron is the plane bisecting the diedral angle between them. 4. The planes bisecting the six diedrai angles of a tetraedron intersect in a point. |