Page images
PDF
EPUB

COS a

cos y=G:

170. We shall now transform the result of Art. 168.
Let

G= J(P + Q*+ R");
and let a, B, y be three arcs determined by the equations

Р
Q

R
cos B
G'

G then

E=G(cos a + p cos B+v cos y). Since cos® a + cosaB+ cosøy=1, it is obvious that there will be some point on the surface of the sphere, such that a, b, y are the arcs which join it to A, B, C respectively; denote this point by U: then, by Art. 166,

cos TU = d cos a tu cos ß + v cos y ; and finally

=G cos TU.

=

[ocr errors]

Thus, whatever may be the position of T, the sum of the cosines of the arcs which join T to the fixed points varies as the cosine of the single arc which joins T to a certain fixed point U.

We might take G either positive or negative; it will be convenient to suppose it positive.

171. A sphere is described about a regular polyhedron; from any point on the surface of the sphere arcs are drawn to the solid angles of the polyhedron : to shew that the sum of the cosines of these arcs is zero.

From the preceding Article we see that if G is not zero. there is one position of T which gives to its greatest positive value, namely, when T coincides with U. But by the symmetry of a regular polyhedron there must always be more than one position of T which gives the same value to £. For instance, if we take a regular tetrahedron, as there are four faces there will at least be three other positions of T symmetrical with any assigned position.

Hence G must be zero; and thus the sum of the cosines of the arcs which join T to the solid angles oy the regular polyhedron is zero for all positions of T.

172. Since G= 0, it follows that P, Q, R must each be zero; these indeed are particular cases of the general result of Art. 171. See Art. 169.

[ocr errors]

1

2

3

173. The result obtained in Art. 171 may be shewn to hold also in some other cases. Suppose, for instance, that a rectangular parallelepiped is inscribed in a sphere; then the sum of the cosines of the arcs drawn from any point on the surface of the sphere to the solid angles of the parallelepiped is zero. For here it is obvious that there must always be at least one other position of T symmetrical with any assigned position. Hence by the argument of Art. 171 we must have G = 0.

174. Let there be any number of fixed points on the surface of a sphere; denote them by H, 1,, H ,, ... Let T be any point on

; H the surface of the sphere. We shall now investigate a remarkable expression for the sum of the squares of the cosines of the arcs which join T with the fixed points. Denote the sum by <; so that

=cos TH, + cos TH,+ cos* TH, + i... Take on the surface of the sphere a fixed spherical triangle ABC, having all its sides quadrants, and therefore all its angles right angles.

Let dg Mo v be the cosines of the arcs which join T with A, B, C respectively; let lī, my, n, be the cosines of the angles which join H, with A, B, C respectively; and let a similar notation be used with respect to H., H,... Then, by Art. 166,

=(1,2 + m,+ n, v) + (.X + mogu + nov) + ... Expand each square, and rearrange the terms; thus

Σ == Ρλ' + Ομ3 + Bν + 2ρμν + 2qνλ + 2xλμ,
where P stands for 1, +1, +7 +
and p stands for m, n, + m,

n, + m ng
with corresponding meanings for Q and q, and for R and r.
T. S. T.

K

2

...)

+

We shall now shew that there is some position of the triangle ABC for which p, q, and r will vanish; so that we shall then have

= PX+ Qu + Rv.

Since is always a finite positive quantity there must be some position, or some positions, of T for which has the largest value which it can receive. Suppose that A has this position, or one of these positions if there are more than one. When T is at A we have

M

and v each zero, and d equal to unity, so that is then equal to P.

[ocr errors]

2

Hence, whatever be the position of T,

P is never less than PXP + Que + Rve + 2pur + 2qvd + 2rdu, that is, by Art. 165, P (1® + pe® + v®) is never less than

PXP + Qu + Rv + 2pur + 2q1 + 2rdu; therefore

(P-Q) p® + (P-R) v is never less than 2pur + 2qvd + 2rdu.

Now suppose v=0; then T is situated on the great circle of which AB is a quadrant, and whatever be the position of Twe have (P-Q) u not less than 2rdwig

2ra and therefore P-Q not less than

д

COS TA
cos TB

λ But now

is equal to ; this is numerically equal to

M tan TB, and so may be made numerically as great as we please, positive or negative, by giving a suitable position to T. Thus

2rà P-Q must in some cases be less than if r have any value dif

р ferent from zero.

[merged small][ocr errors]

In like manner we can shew that q must = 0.

Hence with the specified position for A we arrive at the result that whatever may be the position of T

= P1° + Que + Rv + 2puv. Let us now suppose that the position of B is so taken that when T coincides with B the value of is as large as it can be for

any point in the great circle of which A is the pole. When T is at B we have d and v each zero, and y equal to unity, so that

is then equal to Q. For any point in the great circle of which A is the pole d is zero ; and therefore for any such point

Q is not less than Qu + Rv + 2 puv, that is, by Art. 165,

Q (M +09) is not less than Qu + Rve + 2 puv; therefore Q - R is not less than 2pu.

[ocr errors]

Hence by the same reasoning as before we must have p= 0.

Therefore we see that there must be some position of the triangle ABC, such that for every position of T

= PA' + Qu + Rv,

175. The remarks of Art. 169 are applicable to the result just obtained,

176. In the final result of Art. 174 we may shew that R is the least value which can receive. For, by Art. 165,

I=PX? + Qu? + R (1-1-Mee)

= R + (P R) 1+ (Q-R) j'; and by supposition neither P-R nor Q-R is negative, so that

cannot be less than R.

[ocr errors]

177. A sphere is described about a regular polyhedron; from any point 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 squares of the cosines of these arcs.

With the notation of Art. 174 we have

= PX2 + Qu + Rve.

[ocr errors]

We shall shew that in the present case P, Q, and R must all be equal. For if they are not, one of them must be greater than each of the others, or one of them must be less than each of the others.

If possible let the former be the case; suppose that P is greater than Q, and greater than R.

V =

Now

=P(1 - M - 9) + Qu + Ry*

=P-(P-Q) u (P– R) vo; this shews that is always less than P except when j = 0 and v=0: that is is always less than P except when T is at A, or

a at the point of the surface which is diametrically opposite to A. But by the symmetry of a regular polyhedron there must always be more than two positions of T which give the same value to 2. For instance if we take a regular tetrahedron, as there are four faces there will be at least three other positions of T symmetrical with any assigned position. Hence P cannot be greater than Q and greater than R.

In the same way we can shew that one of the three P, Q, and R, cannot be less than each of the others.

Therefore P=Q=R; and therefore by Art. 165 for every position of T we have = P. Since P=Q= R each of them = }(P+Q+R) =

(

3
1
=>{?,+mo+n,° +1° + m.+n+...}

S
=z, by Art. 165,

[ocr errors]

2

2

where S is the number of the solid angles of the regular polyhedron.

« PreviousContinue »