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In the greater of two spheres which have the same centre, to inscribe a polyhedron the superficies or faces of which shall not meet the less.

Let there be two spheres having the same centre A. It is required to inscribe in the greater sphere a solid polyhedron, the superficies of which shall not meet the less.

Let the spheres be cut by a plane passing through the centre; the common sections of it with the spheres are circles; for the sphere is described by the revolution of a semicircle about the diameter remaining fixed. Therefore, in whatever position the semicircle is supposed to be, the common section of the plane in which it is, with the superficies of the sphere, is the circumference of a circle; and this section is a great circle of the sphere, because the diameter of the sphere, which is also the diameter of the circle, is greater (III. 15) than any chord in the circle or the sphere. Let the circle made by the section of the plane with the greater sphere be, BCD E, and that with the less, FG H. Draw the two diameters B D and CE at right angles to one another; and in BCDE, the greater of the two circles, inscribe (XII. 16) a polygon of an even number of equal sides not meeting the less circle F G H. Let the sides of this polygon in B E, the fourth part of the circle, be B K, KL, LM, and M E. Join K A, and produce it to N. From A draw (XI. 12) AX at right angles to the plane of the circle BCD E, and let it meet the superficies of the sphere in the point X. Let planes pass through the straight line A X, and each of the straight lines BD and K N; their sections with the superficies of the sphere are great circles. Let BXD and KX N be their semicircles, and BD and K N their diameters. Because XA is at right angles to the circle B C D E, every plane which passes through X A is at right (XI. 18) angles to the circle BCDE. Therefore the semicircles BXD and KX Ñ are at right angles to BCDE. Because the semicircles BED, BXD, and KXN upon the equal diameters BD and K N, are equal, their halves B E, B X, and K X, are equal. Therefore as many sides of the polygon as are in B E, so many are there in B X and K X, equal to the sides BK, K L, L M, and ME. Let these polygons be described, and let their sides be B O, OP, PR, and RX; KS, ST, TY, and Y X. Join O S, PT, and RY; and from the points O and S, draw O V and SQ perpendiculars to AB and A K.

Because the plane BOX D is at right angles to the plane B C D E, and in BOX D, O V is drawn perpendicular to A B, the common section of the planes, therefore O V is perpendicular (XI. Def. 4) to the plane BCDE. For the same reason SQ is perpendicular to the same plane, the plane KSX being at right angles to the plane B CD E. Join V Q. Because in the equal semicircles B X D and K X N, the arcs BO and KS are equal, and OV and SQ are perpendiculars to their diameters. Therefore (I. 26) O V is equal to S Q, and BV to KQ. But the whole BA is equal to the whole K A. Therefore the remainder V A is equal to the remainder QA and BV is to VA as KQ is to QA. Wherefore V Q is parallel (VI. 2) to B K. Because OV and SQ are each of them at right angles to the circle BCDE, O V is parallel (XI. 6) to SQ; and it has been proved that it is also equal to it. Therefore QV and SO are equal (I. 33) and parallel. Because QV is parallel to S O, and also to KB; OS is parallel (xi. 9) to BK. Therefore BO and K S, which join them, are in the same plane, and the quadrilateral figure K BOS is

P

R

T

H

in that plane. If P B and TK be joined and perpendiculars be drawn from the points P and T to the straight lines A B and A K, it may be proved, in the same manner, that TP is parallel to K B. Therefore (XÍ. 9) TP being parallel to SO, the quadrilateral figure SOPT is in one plane. For the same reason, the quadrilateral TPRY is in one plane, and the figure YRX (XI. 2) is also in one plane. Therefore, if from the points O, S, P, T, R, and Y, straight lines be drawn to the point A, a polyhedron will be formed between the arcs B X and K X, composed of pyramids, whose bases are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YR P, and whose common vertex is the point A. If the same construction be made upon each of the sides K L, LM and ME, as upon B K, and the same be done also in the other three quadrants, and in the other hemisphere; a solid polyhedron will be inscribed in the sphere, composed of pyramids, whose bases are the

.

D

F

E

G

B

quadrilaterals K BOS, &c., and triangle Y R X, and those formed in the same manner in the rest of the sphere, and common vertex the point A. Also the superficies on this polyhedron does not meet the less sphere of which the circle FGH section. For, from the point A draw (XI. 11) AZ perpendicular to the piane of the quadrilateral K BOS, meeting it in Z, and join B Z and Z K. Because AZ is perpendicular to the plane K BOS, it makes right angles with every straight line meeting in that plane. Therefore AZ is perpendicular to B Z and KZ; and it may be shown, as in Prop. XIV, Book III, that the straight line B Z is equal to the straight line ZK. In the same manner it may be shown that the straight lines drawn from the point Z to the points Ō and S, are equal to BZ or ZK. Therefore the circle described from the centre Z, and distance Z B, will pass through the points K, O and Z, and K BO'S will be a quadrilateral inscribed in the circle. Because K B is greater than Q V, and QV equal to SO. Therefore K B is greater than SO. But K B is equal to each of the straight lines BO and KS. Therefore each of the arcs cut off by K B, BO and KS, is greater than that cut off by OS; and these three arcs together with a fourth equal to each, 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 K B is greater than the fourth part of the whole circumference of the circle about KBOS, and the angle BZK at the centre is greater than a right angle. From the point G draw G U at right angles to A G, and join A U. If then the circumference B E be bisected, and its half again bisected, and so on, there will at length be left an arc less than the arc subtended by a straight line equal to G Ü, incribed in the circle BCDE; let this be the circumference K B. Therefore the straight line K B is less than G U. Because the angle BZK is obtuse, BK is greater than B Z; in the triangles AGU and A B Z, it may be shown, as in Prop. XV, Book III, that A Z is

greater than A G. But A Z is perpendicular to the plane K BO S, 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 K BOS does not meet the less sphere.

Also the other planes between the quadrants BX and KX do not meet the less sphere. From the point A draw AI perpendicular to the plane of the quadrilateral S O PT, and join IO. In the same manner, as it was proved of the plane K BOS and the point Z, it may be shown that the point I is the centre of a circle described about SOPT; and that OS is greater than PT. But PT was shown to be parallel to OS; and because the two trapeziums KBOS and SOPT inscribed in circles have their sides BK and OS parallel, as also OS and PT; and their other sides BO, KS, OP, and ST all equal to one another, but B K is greater than OS, and OS greater than PT. Therefore the straight line ZB is greater (XII. Lemma 2) than IO. Join A O. In the triangles AB Z and AIO, it may be shown, as above, that A Z is less than A I. But it was proved that A Z is greater than AG; much more then is AI greater than A G. Therefore the plane SOPT does not meet the less sphere. In the same manner it may be proved, that the plane TPRY does not meet the same sphere, as also the triangle YR X (XII. Lem. 2, Cor.) In the same manner it may be demonstrated, that all the planes, or faces the polyhedron, do not meet the less sphere. Therefore in the greater of two spheres, which have the same centre, a polyhedron is inscribed, the superficies of which does not meet the lesser sphere. Q. E. F.

COROLLARY.-If in the less sphere a similar polyhedron be inscribed by joining the points in which the radii of the sphere drawn to the angles of the polyhedron in the greater sphere meet the superficies of the less, the polyhedron in the sphere BCDE shall have to this polyhedron, the triplicate ratio of that which the diameter of the sphere BCDE has to the diameter of the less sphere. For if these two solids be divided into the same number of pyramids, and in the same order, the pyramids shall be similar to one another, each to each: because they have the solid angles at their common vertex, the centre of the sphere, the same in each pyramid, and their other solid angles at the bases, equal to one another, each to each (XI.B), because they are contained by three plane angles, each equal to each; and the pyramids are contained by the same number of similar planes; and are therefore similar (XI. Def. 11) to one another, each to each. But similar pyramids have to one another the triplicate (XII. 8, Cor.) ratio of their homologous sides. Therefore the pyramid of which the base is the quadrilateral K BOS, and vertex A, has to the pyramid in the inner sphere of the same order, the triplicate ratio of their homologous sides, that is, of that ratio which the radius AB, the greater sphere, has to the radius of the less sphere. In like manner, each pyramid in the greater sphere has to each of the same order in the less, the triplicate ratio of that which AB has to the radius of the less sphere. But 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 A B has to the radius of the less; that is, which the diameter BD of the greater has to the diameter of the less sphere.

Corollary 2.-Every face of a polyhedron inscribed in a sphere is inscribable in a circle.

PROP. XVIII. THEOREM.

Spheres have to one another the triplicate ratio of that which their diameters have.

Let ABC and DEF be two spheres, of which the diameters are BC and EF. The sphere A B C has to the sphere DEF the triplicate ratio of that which B C has to EF.

A

D

L

For, if not, the sphere ABC must 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 G H K; and let the sphere DEF have the same centre with GHK. In the greater sphere DEF inscribe (XII. 17) a polyhedron, the superficies of which does not meet the lesser sphere GHK; and in the sphere A B C inscribe another similar to that in the sphere DEF. Therefore

B

CE

FMA

the polyhedron in the sphere ABC has to the polyhedron in the sphere DEF the triplicate ratio (XII. Cor. 17) of that which BC has to E F. But the sphere A B C has to the sphere G HK, the triplicate ratio of that which BC has to EF. Therefore, as the sphere A B C is to the sphere GHK, so is the polyhedron in the sphere A B C to the polyhedron in the sphere DEF. But the sphere ABC is greater than the inscribed polyhedron. Therefore (V. 14), also 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. Therefore, the sphere ABC has not to any sphere less than DEF, the triplicate ratio of that which BC has to E F. In the same manner it may be demonstrated, that the sphere DEF has not to any sphere less than ABC, the triplicate ratio of that which EF has to B C. Nor can the sphere ABC have to any sphere greater than D E F, the triplicate ratio of that which B C has to EF; for, if it can, let it have that ratio to a greater sphere I. MN. Therefore, by inversion, the sphere L M N has to the sphere A B C, the triplicate ratio of that which the diameter EF has to the diameter B C. But the sphere L M N is to A B C as the sphere DEF is to some sphere, which must be less (V. 14) than the sphere A B C, because the sphere LMN is greater than the sphere DEF. Therefore, the sphere DEF has to a sphere less than A B C the triplicate ratio of that which E F has to BC; which was shown to be impossible. Therefore, the sphere A B C has not to any sphere greater than DE F, the triplicate ratio of that which B C has to EF; and it was demonstrated, that neither has it that ratio to any sphere less than DEF. Therefore, the sphere A B C has to the sphere DE F, the triplicate ratio of that which BC has to EF. Wherefore, spheres have to one another, &c.

Every sphere is two-thirds of its circumscribed cylinder.-Archimedes.

FINIS.

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