PROPOSITION XVII 642. The shortest line that can be drawn on a sphere between two points is the arc of a great circle, not greater than a semicircle, joining the two points. Let A and B be any two points on a sphere, and AFB the arc of a great circle, not greater than a semicircle, joining them; also let ADCEB be any other path on the sphere from A to B. It is required to prove that the path AFB is shorter than the path ADCEB. Proof. In the path ADCEB take any point C, and draw the arcs of great circles AC and CB. Then ACB is a spherical triangle, and the sum of AC and CB is greater than AB. (Prop. VII.) In the path ADC choose any point D, and draw the arcs of great circles AD and DC. Then AD + DC is greater than AC. Therefore AD + DC + CB is greater than AC + CB, which is again greater than AB. By repeating this process indefinitely the broken path of arcs of great circles increases in length with each repetition, and so differs more and more from the path AFB. Also the broken path of arcs of great circles can eventually be made to differ as little as we please from the given path ADCEB. Therefore the path AFB is shorter than the path ADCEB. PROPOSITION XVIII 643. To find the length of the diameter of a given material sphere. Let PABC be a given material sphere. It is required to find the length of its diameter. Construction. With any point P on the given sphere as pole describe a circle ABC. With the compasses take the lengths of the chords AB, BC, CA, and construct in any plane a triangle A'B'C' whose sides are equal to these three lengths. About A'B'C' describe a circle which will be equal to the circle ABC. Find its centre Q' and the length of its radius B'Q' = BQ. With the compasses take the length of the chord BP and construct a right triangle pbq, having the hypotenuse bp = BP and the side bq = BQ. Draw bp' perpendicular to bp, meeting pq produced at p'. Then pp' is equal in length to a diameter of the given sphere. The proof is left to the pupil. EXERCISES 1. Three great circles of a sphere which do not intersect in a common point, divide the surface of the sphere into four pairs of symmetrical triangles. 2. A tri-rectangular spherical triangle coincides with its polar triangle. SECTION IV AREAS OF SPHERICAL TRIANGLES 644. DEFINITION. The area of a spherical triangle is that portion of the sphere which is enclosed by the triangle. 645. AXIOM 15. Two spherical triangles which are identically equal have equal areas. PROPOSITION XIX 646. Two symmetrical spherical triangles have equal areas. A Let ABC and A'B'C' be two symmetrical spherical triangles lying on the same sphere. It is required to prove that ABC and A'B'C' have equal areas. Proof. Two symmetrical spherical triangles can always be placed so as to be vertically opposite on the sphere. Why? Assume that the given triangles are so placed. Let P be the pole of the small circle through A, B, C, and draw the diameter POP!. Draw also the arcs of great circles PA, PB, PC, P'A', P'B', P'C'. Now also PA P'A', PB = P'B', PC = P'C', (Art. 157.) = = PA PB: = PC. (Prop. II.) As PAB and P'A'B' are symmetrical triangles on the sphere, and they are also isosceles triangles. Therefore As PAB and P'A'B' are identically equal and have equal areas. Similarly As PBC and P'B'C' have equal areas. (Prop. X.) Therefore, adding, As ABC and A'B'C' have equal areas. If the pole P falls outside of ▲ ABC, instead of adding all three isosceles triangles, two should be added and the third subtracted from their sum, in order that the result may give Δ ΑΒΓ. NOTE. The triangles mentioned in Propositions IX, XI, and XII, as being either identically equal or symmetrical, are in every case equal in area. DEFINITIONS 647. Two great circles on a sphere divide the sphere into four parts, each of which is called a lune. A lune then is that portion of a sphere enclosed by two halves of great circles. The spherical angle made by the semicircles is called the angle of the lune. Of the four lunes formed by two great circles the opposite ones have equal angles. Opposite lunes on a sphere are equal, since they can be superposed. 648. THEOREM. Any two lunes, having equal angles, on the same sphere or on equal spheres are identically equal. Of two lunes having unequal angles, on the same sphere or on equal spheres, that which has the greater angle is the greater; and of two lunes having equal angles on unequal spheres, that which lies on the greater sphere is the greater. PROPOSITION XX 649. The ratio of two lunes on the same sphere or on equal spheres is equal to the ratio of their angles. Let PAP'B and PBP'C be two lunes on the same sphere whose angles are APB and BPC. It is required to prove that lune PAP'B: lune PBP'C = ZAPB: Z BPC. Outline of proof. There are two cases to be considered. First, when Ls APB and BPC are commensurable. Take a common measure of the two angles and divide them into equal parts. By planes through the centre of the sphere divide the lunes into parts having equal angles, and hence equal. Next, when Ls APB and BPC are incommensurable. Take a measure of one angle and apply it as often as possible to the other. Then proceed as above, and finally make the unit of measure indefinitely small. As a model take the proof of Prop. X in Chap. III. 650. COROLLARY. The area of a lune is to the area of the whole sphere in the same ratio as the angle of the lune is to four right angles. |