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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
= PX? + Qua + Rv* + 2 pur + 2qvd + 2rdu; then at A we have p= 0 and v=0, so that P is then the value of <; similarly Q and R are the values of Eat B and C respectively. But by Art. 177 we have the same value for whatever be the position of T; thus
P=P(1® + M* + 1) + 2pur + 2q2 + 2rλμ; therefore
0 = 2ρμν + 2φνλ + 2xλμ. 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 d=0: thus we get pur= 0, and therefore p=0. Similarly q = 0, and r=0.
179. Let there be any number of fixed points on the surface of a sphere; denote them by H, I,, H., ...; from any 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 TH, cos UH, +cos TH, cos UH, + cos TH cos UH, + ...
Let the notation be the same as in Art. 174; and let X', re', v' be the cosines of the arcs which join U with A, B, C respectively. Then by Art. 166,
cos TH, cos UH, = (1, + um, + vn,) (^l, + p'm, + v'n,) = λλί,8 + μμ'm, +ννη, + (λμ' + μλ), m, + (μν' +νμ)m,η, +(νλ' +λν)η...
Similar results hold for cos TH cos UH,, cos TH, cos UH,,... Hence, with the notation of Art. 174, the required sum is
λXP+ μμQ + ν'R + (μν' + νμ)p + (νλ' + λν)g + (λμ' + μλ)r.
Now by properly choosing the position of the triangle ABC we have p, q, and r each zero as in Art. 174; and thus the required sum becomes
MP + pepe'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
MP + repe'Q + vv'R.
S And here P=Q R
by Art. 177. 3'
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.
Let AB be the given base, =c suppose, AC = 0, BAC = $. Since the area is given the spherical excess is known; denote it by E; then by Art. 103,
cot 1 E=cot 10 cot 1c cosec $ + cot ; therefore sin ($- 1 E)=cot 10 cot 1 c sin | E;
2 cot } c sin | E cos = sin 0 sin ($- 1 E);
cos O cot 1 c sin } E + sin 0 cos (9–} E + =- cot £ ¢ sin } E.
Comparing this with equation (1) of Art. 133, we see that the required locus is a circle. If we call a, B the angular co-ordinates of its pole, we have
tan c cot }c sin 1 E sin ? E'
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 & coe (1-1) + sin o sin ( - )cos (4- =)
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 Q the pole of the circumscribed circle of a triangle ABC; then PAB = 1 A, by Art. 89, and QAB=S-C, by Art. 92; hence
cos PAQ=cos 1 (B-C);
cos PA = cos PE cos AE = cos r cos (s – a),
cos PQ = cos R cos r cos (8 – a) + sin R sin r cos 1 (B-C) cosec 1 A.
Therefore, by Art. 54
cos PQ = cos R cos r cos (8 - a) + sin R sin r sin 1 (b + c) cosec 1 a,
cos PQ cos R sin r
cot r cos (8 – a) + tan R sin 1 (b+c) cosec ja.
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 Q 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 + 1 (C-A), and cos QQ, = cos R cosr, cos (8 – c) – sin R sin r, sin } (C – A) sec · B
= cos R cosr, cos (8-c) – sin R sin r, sin 1 (c - a) cosec } b. Therefore
cot r, cos (8 – c) – tan R sin 1 (c— a) cosec ] b;
by reducing as in the preceding Article, the right-hand member of the last equation becomes
(sin b + sin c - sin a);
cos QQ, hence
- 1 = (tan R – cotr.), (Art. 94); cos R sin r. therefore
coso QQ, = cos* R sinor, + cos* (R+r,), and
sino QQ, =sin' (R+r) - cos' R sin
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 BFQ sin AQ sin AFQ
sin BQ sin AQF sin AQ sin BQF'