Therefore PQ2= † ( 1 − a)2 +‡ a2+‡ a2 = † (12 − al+a2) = a2.

(§ 4, p. 586.)

Therefore PQ is equal to a. Similarly every other edge of the figure may be proved equal to a.

Therefore every face of the figure is an equilateral triangle.

And all the points lie on a sphere whose centre is O, since OP2=(a2 + 12).

Each solid pentahedral angle is composed of five equal trihedral angles; for instance the pentahedral angle P(P'r' QʻQR) is composed of the five equal trihedral angles P (P'r'O), P(r'Q'O), P(Q'QO), P(QRO), P (RP'O).

And therefore all the solid angles of the figure are equal and the figure is a regular icosahedron.

Each of these regular solids has a circumscribed sphere passing through all the angular points, an inscribed sphere touching all the faces, and an intrascribed sphere touching all the edges. The squares on the radii of the circumscribed, the inscribed and the intrascribed spheres, which we will call R, r, p respectively are given in the following Table.

In this Table a is equal to the edge of the figure, and 7 satisfies

the equation 12=al+a2.

ANHARMONIC PROPERTY OF FOUR PLANES.

The four points in which four fixed planes which pass through a straight line are cut by any other straight line form an anharmonic range of constant ratio.

Let OAA', OBB', OCC', ODD' be four planes passing through a common straight line OO' and cutting two straight lines ABCD, A'B'C'D'.

b

a

Take any two points O, O' in OO' and draw the planes OABCD, O'A'B'C'D',

and let OA, O'A' meet in a; OB, O'B' in b;

OC, O'C' in c and OD, O'D' in d.

a, b, c, d lie in a straight line, the common section of the planes OABCD, O'A'B'C'D'.

Now O (ABCD) is an anharmonic pencil cut by two transversals ABCD, abcd.

Therefore ABCD, abcd are two like anharmonic ranges. (VI. Add. Prop. 13.) And O'(A'B'C'D') is an anharmonic pencil cut by two transversals A'B'C'D', abcd.

Therefore A'B'C'D', abcd are two like anharmonic ranges, therefore ABCD, A'B'C'D' are two like anharmonic ranges.

EXTENSION OF THE THEOREM OF MENELAUS.

If a plane cut the four sides of a skew quadrilateral, produced if necessary, the ratio compounded of the ratios of the segments of the sides taken in order is equal to unity.

Let a plane PQRS cut the sides AB, BC, CD, DA of a skew quadrilateral ABCD in P, Q, R, S respectively.

Let the plane PQRS cut the two planes ABC, ADC in the straight lines TPQ, TSR; they necessarily intersect in some point T in AC.

(Prop, 3.)

B

P

T

R

Because TPQ is a transversal of the triangle ABC,

the ratio compounded of the ratios AP to PB and BQ to QC is equal to the ratio AT to TC:

(VI. Add. Prop. 1.)

and because TSR is a transversal of the triangle ADC,

the ratio compounded of the ratios CR to RD and DS to SA is equal to the ratio CT to TA.

Therefore the ratio compounded of the ratios AP to PB, BQ to QC, CR to RD, and DS to SA is equal to the ratio compounded of the ratios AT to TC and CT to TA, that is, to unity.

It follows from the result just proved that if AP be to PB as DR to RC,

then AS is to SD as BQ to QC.

(Prop. 17.)

Now if the three lines AD, PR, BC be parallel to the same plane: then AP is to PB as DR to RC, and if AS be to SD as BQ to QC, the three lines AB, QS, DC are parallel to one plane. (Ex. 2 on Prop. 17.)

We may therefore generalize this proposition thus,

Every straight line which can be drawn to intersect three given nonintersecting straight lines which are all parallel to a fixed plane is parallel to a second fixed plane.

SURFACE OF SPHERE.

Let AB be a diameter of a sphere whose centre is 0, and let APB be a great circle. Let P be the middle point of QPR a tangent to the circle APB. Through P, Q and R let planes LST, MCF, NDE be drawn at right angles to AB, cutting AB in L, M and N. Let the tangent plane to the sphere at P cut these planes in ST, CF and DE, and let two planes through AB equally inclined to APB cut the tangent plane at P in CD and FE.

Now the area of the quadrilateral CDEF is equal to QR. ST.

Let LP produced cut the enveloping cylinder of the sphere, whose generating lines are parallel to AB, in p; and the lines MC, ND, NE, MF, MQ, NR, LS, LT cut the tangent plane to the cylinder at p in c, d, e, f, q, r, s, t.

Now the area of the quadrilateral cdef is equal to qr.st.

And

QR: qr=PO: PL=pL: PL=st : ST.

Therefore QR. ST is equal to qr.st;

or, the quadrilaterals CDEF, cdef are equal in area.

Now suppose the surface of the sphere to be divided into any number of lunes of equal angles by planes drawn through AOB, and between the planes forming each of these lunes let a quadrilateral be described corresponding to and equal to the quadrilateral CDEF; and also corresponding to and equal to the rectangle cdef.

If the number of the lunes be increased, the quadrilaterals CDEF... can be made to approach as nearly as we like to a strip of the surface of a cone whose axis is in AOB, which touches the sphere at P, and which is generated by the rotation of the line QR round AOB through four right angles, and at the same time the quadrilaterals cdef... approach as nearly as we like to a strip of the surface of the enveloping cylinder.

Hence the curved surface of the portion of a right circular cone, between two parallel planes at right angles to the axis, is equal to the surface of the portion of a co-axial circular cylinder, intercepted by the same planes; where the radius of the cylinder is equal to the radius of a sphere which touches the cone at points equidistant from the parallel planes.

Let a polygon of an even number of sides be described about the circle APB touching at A and B and let P be the point of contact of one of the sides QPR.

By rotation round AOB a strip of a cone will be generated by each of the other sides of the polygon, except those which touch at A and B.

The surface of all these strips of cones will be equal to the surface of the cylinder.

Now if the number of the sides of the polygon be increased the strips of the cones can be made to approach as near as we like to the surface of the sphere, hence the surface of the sphere is equal to the curved surface of the cylinder.

Similarly it may be proved that if two closed curves be drawn on the sphere and on the cylinder so that the straight line joining corresponding points on the two curves always cuts the axis of the cylinder at right angles, the surfaces of the portions of the sphere and the cylinder bounded by these curves are equal.

T. E.

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