Thus the sum of the squares of the cosines of the arcs which join any point on the surface of the sphere to the solid angles of the regular polyhedron is one third of the number of the solid angles. 178. Since P=Q= R in the preceding Article, it will follow that when the fixed points of Art. 174 are the solid angles of a regular polyhedron, then for any position of the spherical triangle ABC we shall have p = 0, q = 0, and r= 0. For taking any position for the spherical triangle ABC we have Σ=Px2 + Qu2 + Rv*+2pμv + 2qvλ + 2rλμ; then at A we have O and v= = με = 0, so that P is then the value of ; similarly Q and R are the values of Σ at B and C respectively. But by Art. 177 we have the same value for Σ whatever be the position of T; thus therefore P =P (\3 + μ2 + v3) + 2pμv + 2gvλ + 2rλμ ; 0 = 2ρμν + 2ηνλ + 2λμ. This holds then for every position of T. Suppose T is at any point of the great circle of which A is the pole; then λ=0: thus we get puv = 0, and therefore p = 0. Similarly q=0, and r=0. 29 from any ; 179. Let there be any number of fixed points on the surface of a sphere; denote them by H1, Н„ Н„ ........ two points T and U on the surface of the sphere arcs are drawn to the fixed points: it is required to find the sum of the products of the corresponding cosines, that is cos TH1 cos UH ̧ + cos TH ̧ cos UH2+ cos TH ̧ cos UH ̧+ 2 2 3 ... v' Let the notation be the same as in Art. 174; and let X', μ', be the cosines of the arcs which join U with A, B, C respectively. Then by Art. 166, 1 1 cos TH ̧ cos UH1 = (\l ̧ + μm ̧ + vn ̧) (X'7 ̧ + μ'm ̧ + v'n ̧) = XX'l ̧2+μμ'm ̧3+vv'n‚ ̧2 + (\μ'+μλ')? ̧m ̧ + (μv′+vμ')m ̧n ̧+(vλ'+dv′) n ̧? ̧. 2 Similar results hold for cos TH, cos UH,, cos TH, cos UH ̧,... Hence, with the notation of Art. 174, the required sum is XX'P + μμ'Q + vv'R + (μv' + vμ')p + (vλ' + dv') q + (λμ' + μd') r. Now by properly choosing the position of the triangle ABC we have p, q, and reach zero as in Art. 174; and thus the required sum becomes XX'P + μμ'Q + vv'R. 180. The result obtained in Art. 174 may be considered as a particular case of that just given; namely the case in which the points T and U coincide. 181. A sphere is described about a regular polyhedron; from any two points on the surface of the sphere arcs are drawn to the solid angles of the polyhedron: it is required to find the sum of the products of the corresponding cosines. With the notation of Art. 179 we see that the sum is Thus the sum of the products of the cosines is equal to the product of the cosine of TU into a third of the number of the solid angles of the regular polyhedron. 182. The result obtained in Art. 177 may be considered as a particular case of that just given; namely, the case in which the points T and U coincide. 183. If TU is a quadrant then cos TU is zero, and the sum of the products of the cosines in Art. 181 is zero. The results p=0, q= 0, r=0, are easily seen to be all' special examples of this particular case. XV. MISCELLANEOUS PROPOSITIONS. 184. To find the locus of the vertex of a spherical triangle of given base and area. =c Let AB be the given base, c suppose, AC = 0, BAC = 4. $. Since the area is given the spherical excess is known; denote it by E; then by Art. 103, cot Ecotcotc cosec + cot & ; Comparing this with equation (1) of Art. 133, we see that the required locus is a circle. If we call a, ẞ the angular co-ordinates of its pole, we have It may be presumed from symmetry that the pole of this circle is in the great circle which bisects AB at right angles; and this presumption is easily verified. For the equation to that great circle is 0 = cos 0 cos (-)+ sin 8 sin (-) cos (-) and the values 0=a, ẞ satisfy this equation. = 185. To find the angular distance between the poles of the inscribed and circumscribed circles of a triangle. Let P denote the pole of the inscribed circle, and the pole of the circumscribed circle of a triangle ABC; then PABA, by Art. 89, and QAB=S-C, by Art. 92; hence cos PAQ = cos(B − C') ; and cos PQ = cos PA cos QA + sin PA sin QA cos † (B – C). Now, by Art. 62 (see the figure of Art. 89), cos PA = cos PE cos AE= cos r cos (s − a), cos PQ = cos R cos r cos (s − a) + sin R sin r cos (B-C) cosec A. Therefore, by Art. 54 cos PQ = cos R cos r cos (8 − a) + sin R sin r sin † (b + c) cosec 1⁄2 a, = cotr cos (s− a) + tan R sin 1 (b + c) cosec 1⁄2 a. 186. To find the angular distance between the pole of the circumscribed circle and the pole of one of the escribed circles of a triangle. Let denote the pole of the circumscribed circle, and Q, the pole of the escribed circle opposite to the angle A. Then it may be shewn that QBQ, = 1⁄2 π + 1⁄2 (C − A), and cos QQ, П = cos R cos r, cos (s — c) — sin R sin r, sin † (C – A) sec B = cos R cos r ̧ cos (8 — c) — sin R sin r ̧ sin 1⁄2 (c − a) cosec 1⁄2 b. = = cot 2 cos (s—c) - tan R sin (c− a) cosec b; by reducing as in the preceding Article, the right-hand member of the last equation becomes 187. The arc which passes through the middle points of the sides of any triangle upon a given base will meet the base produced at a fixed point, the distance of which from the middle point of the base is a quadrant. Let ABC be any triangle, E the middle point of AC, and F the middle point of AB; let the arc which joins E and F when produced meet BC produced at Q. Then sin BQ sin AQF = sin AQ sin BQF' |