sphere; there will be formed a polyhedron inscribed in the sphere, composed of pyramids, the bases of which are the quadrilaterals already mentioned, and the triangle YRX, and those formed in the same manner in the rest of the sphere, the common vertex of them all being A. Also, the surface of this polyhedron does not meet the less sphere. For, from A draw (XI. 11.) AZ perpendicular to the plane of the quadrilateral KBOS, meeting it in Z, and join BZ, ZK. Then, because AZ is perpendicular to the plane KBOS, it is perpendicular to BZ and ZK: and (I. 47. cor. 5.) because AB is equal to AK, and AZ common to the two right-angled triangles, AZB, AZK; ZB is equal to ZK. In the same manner it may be demonstrated, that the straight lines drawn from Z to O, S, are equal to BZ or ZK. Therefore the circle described from the centre Z, at the distance ZB will pass through the points K, O, S, and KBOS will be a quadrilateral in the circle. Now, because KB is greater than QV, and QV equal to SO, therefore KB is greater than SO: but KB is equal to each of the straight lines BO, KS; wherefore each of the arcs cut off by KB, BO, KS, is greater than that cut off by OS; and these three arcs, together with a fourth equal to one of them, are greater than the same three together with that cut off by OS; that is, than the whole circumference of the circle; therefore the arc subtended by KB is greater than the fourth part of the whole circumference of the circle KBOS, and consequently the angle BZK at the centre is obtuse : and (I. 19.) BK is greater than BZ. But it has been seen already that GU is greater than BK: much more then is GU greater than BZ. Therefore in the two right-angled triangles AĞU, AZB, the hypotenuses AU, AB are (I. def. 30.) equal; but the leg GU is greater than the leg BZ: hence (I. 47. cor. 5.) the remaining leg AZ is greater than the remaining leg AG. But AZ is perpendicular to the plane KBOS, and is therefore the shortest of all the straight lines that can be drawn from A, the centre of the sphere, to that plane. Therefore the plane KBOS does not meet the less sphere. And that the other planes between the quadrants BX, KX, fall without the less sphere, is thus demonstrated. From the point A draw AI perpendicular to the plane of the quadrilateral SOPT, and join 10 ; and, as was demonstrated of the plane KBOS and the point Z, in the same way it may be shown, that the point I is the centre of a circle described about SOPT, and that OS is greater than PT: and PT was shown to be parallel to OS. Therefore, because the two trapeziums KBOS, SOPT inscribed in circles, have their sides KB, OS parallel, as also OS, PT; and their other sides B0, KS, OP, ST all equal to one another, and that BK is greater than Os, and OS than PT; therefore (XII. lem. 2.) ZB is greater than 10. Join AO, which will be equal to AB. Then, in the right-angled triangles AIO, AZB, the hypotenuses AO, AB are equal, but the leg ZB is greater than the leg 10: wherefore (I. 47. cor. 5.) the leg AI is greater than the leg AZ But it has been proved that AZ is greater than AG; much more then is AI greater than AG. Therefore the plane SOPT falls wholly wthout the less sphere. In the same manner, it may be demonstrated, that the plane TPRY falls without the same sphere, as also the triangle YRX, viz., by the corollary of the second lemma. And in a similar way it may be demonstrated, that all the planes which contain the polyhedron, fall without the interior sphere. Therefore, in the greater of the two concentric spheres, a polyhedron is described, the surface of which does not meet the less : which was to be done. Cor. If in the less sphere a polyhedron be described by drawing straight lines between the points in which the radii of the sphere, drawn to all the angles of the polyhedron in the greater sphere, meet the surface of the less; in the same order in which the points are joined in which the same radii meet the surface of the greater sphere ; the polyhedron in the sphere BCDE has to this other polyhedron the triplicate ratio of that which the diameter of the greater sphere has to the diameter of the less. For, if these two solids be divided into the same number of pyramids, and in the same order ; the pyramids will be similar, each to each : because they have the solid angles at their common vertex, the centre of the sphere, the same in each pyramid, and (XI. B.) their other solid angles at the bases equal to one another, each to each, because they are contained by three plane angles, equal each to each ; and the pyramids are contained by the same number of similar planes; and (XI. def. 8.) are therefore similar to one another, each to each. But (XII. 8. cor.) similar pyramids have to one another the triplicate ratio of their homologous sides. Therefore the pyramid, of which the base is the quadrilateral KBOS, and the vertex A, has to the pyramid in the other sphere of the same order, the triplicate ratio of their homologous sides; that is, of that ratio, which the radius of the greater sphere has to the radius of the less. In like manner, each pyramid in the greater sphere has to each of the same order in the less, the triplicate ratio now mentioned. And as one antecedent is to its consequent, so are all the antecedents to all the consequents : wherefore the whole polyhedron in the greater sphere, has to the whole polyhedron in the less, the triplicate ratio of that which AB, the radius of the first, has to the radius of the other; that is, which the diameter BD of the greater has to the diameter HF of the less. PROP. XVIII. THEOR. SPHERES have to one another the triplicate ratio of that which their diameters have. Let ABC, DEF be two spheres, of which the diameters are BC, EF: the sphere ABC has to the sphere DEF the triplicate ratio of that which BC has to EF. For, if not, the sphere ABC will have to a sphere either less or greater than DEF, the triplicate ratio of that which BC has to EF. First, let it have that ratio to a less, viz., to the sphere GHK; and let DEF have the same centre with GHK, and (XII. 17.) in the greater DEF describe a polyhedron, the surface of which does not meet the less : and in ABC describe another similar to that in DEF. Therefore (XII. 17. cor.) the polyhedron in ABC has to the polyhedron in DEF, the triplicate ratio of that which BC has to EF. But ABC has to GHK, the triplicate ratio of that which BC bas to EF: therefore (V. 11.) as ABC to GHK, so is the polyhedron in ABC to the polyhedron in DEF. But the sphere ABC is greater than the polyhedron in it; therefore, also, (V. 14.) the sphere GHK is greater than the polyhedron in the sphere DEF: but it is also less, because it is contained within it; which is impossible. There fore ABC has not to any sphere less than DEF, the triplicate ratio of that which BC has to EF. In the same manner, it may be demonstrated, that DEF has not to any sphere less than ABC, the triplicate ratio of that which EF has to BC. Nor can ABC have to any sphere greater than DEF, the triplicate ratio of that which BC has to EF. For, if it can, let it have that ratio to a greater sphere LMN. Then, by inversion, LMN has to ABC, the triplicate ratio of that which EF has to BC. But as LMN to ABC, so is DEF to some sphere, which (V. 14.) must be less than ABC, because LMN is greater than DEF. Therefore DEF has to a sphere less than ABC the triplicate ratio of that which EF has to BC; which was shown to be impossible. ABC, therefore, has not to any sphere greater than DEF, the triplicate ratio of that which BC has to EF: and it was demonstrated, that it has not that ratio to any sphere less than DEF. Therefore the sphere ABC has to DEF, the triplicate ratio of that which BC has to EF: wherefore spheres, &c. |