straight line AB is greater than EF, which is greater than MN. much more is AB greater than MN; therefore the circumference AB is greater than MN: and for the same reason, the circumference DC is greater than PO; therefore the whole circumference ABCD is greater than the whole MNOP: but it is likewise equal to it, which is impossible; therefore KA is not less than LE: nor is it equal to it; therefore, the straight line KA must be greater than LE. COROLLARY. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the center to the circumference of the circle described about the triangle. PROPOSITION XVII. PROBLEM.-In the greater of two given spheres which have the same center (A), to inscribe a solid polyhedron, the superficies of which shall not meet the lesser sphere. SOLUTION. Let the spheres be cut by a plane passing through the center; the common sections of it with the spheres shall be circles, because the sphere is described by the revolution of a semicircle about the diameter remaining immovable; so that in whatever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater than any straight line in the circle or sphere (a). Let then the circle made by the section of the plane with the greater sphere be BCDE, and with the lesser sphere be FGH: and draw the two diameters BD, CE at right angles to one another; and in BCDE, the greater of the two circles, inscribe a polygon of an even number of equal sides not meeting the lesser circle FGH (b); and let its sides in BE, the fourth part of the circle, be BK, KL, LM, ME; join KA, and produce it to N; and from ▲ draw AX at right angles to the plane of the circle BCDE (c), meeting the superficies of the sphere in the point X: and let planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall produce great circles on the superficies of the sphere, aad let BXD, KXN be the semicircles thus made upon the diameters BD, KN: therefore, because XA is at right angles to the plane of the circle BCDE, every plane which passes through XA is at right angles to the plane of the circle BCDE (d); wherefore the semicircles BXD, KXN are at right angles to that plane: and because the semicircles BED, BXD, KXN upon the equal diameters BD, KN are equal to one another, their halves BE, BX, KX are equal to one another; therefore as many sides of the polygon as are in BE, so many are there in BX, KX, equal to the sides BK, KL, LM, ME: let these polygons be described, and their sides be BO, OP, PR, RX; KS, ST, TY, YX; and join OS, PT, RY; and from the points O, S draw OV, SQ perpendiculars to AB, AK: and because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, OV is drawn perpendicular to AB the common section of the planes, therefore OV is perpendicular to the plane BCDE (e): for the same reason, SQ is perpendicular to the same plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ, and because in the equal semicircles BXD, KXN the circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefore OV is equal to SQ (f), and BV equal to KQ: but the whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder QA: therefore as BV is to VA, so is KQ to QA; wherefore VQ is parallel to BK (g): and because OV, SQ are each of them at right angles to the plane of the circle, BCDE, OV is parallel to SQ (h); and it has been proved that it is also equal to it; therefore QV, SO are equal and parallel (2); and because QV is parallel to SO, and also to KB, OS is parallel to BK (k); and therefore BO, KS, which join them, are in the same plane in which these parallels are, and the quadrilateral figure KBOS is in one plane: and if PB, TK be joined, and perpendiculars be drawn from the points P, T, to the straight lines AB, AK, it may be demonstrated, that TP is parallel to KB in the very same way that SO was shown to be parallel to the same KB; wherefore TP is parallel to SO (k), and the quadrilateral figure SOPT is in one plane: for the same reason, the quadrilateral TPRY is in one plane: and the figure YRX is also in one plane (7): therefore, if from the points O, S, P, T, R, Y, there be drawn straight lines to the point A, there will be formed a solid polyhedron between the circumferences BX, KX, composed of pyramids, the bases of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRX, and of which the common vertex is the point A: and if the same_construction be made upon each of the sides KL, LM, ME, as has been done upon BK, and the like be done also in the other three quadrants, and in the other hemisphere, there will be formed a solid polyhedron inscribed in the sphere, composed of pyramids, the bases of which are the aforesaid quadrilateral figures, and the triangle YRX, and those formed in the like manner in the rest of the sphere, the common vertex of them all being the point A. Also the superficies of this solid polyhedron shall not meet the lesser sphere in which is the circle FGH. For, from the point A draw AZ perpendicular to the plane of the quadrilateral KBOS (m), meeting it in Z, and join BZ, ZK: and because AZ is perpendicular to the plane KBOS, it makes right angles with every straight line meeting it in that plane; therefore AZ is perpendicular to BZ and ZK and because AB is equal to AK, and that the squares on AZ, ZB are equal to the square on AB, and the squares on AZ, ZK to the square on AK (n); therefore the squares on AZ, ZB are equal to the squares on AZ, ZK: take from these equals the square on AZ, and the remaining square on BZ is equal to the remaining square on ZK; and therefore the straight line BZ is equal to ZK: in the like manner it may be demonstrated, that the straight lines drawn from the point Z to the points O, S are equal to BZ or ZK; therefore the circle described from the center Z, and distance ZB, will pass through the points K, O, S, and KBOS will be a quadrilateral figure in the circle: and 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 circumferences cut off by KB, BO, KS is greater than that cut off by OS; and these three circumferences, 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 circumference subtended by KB is greater than the fourth part of the whole circumference of the circle KBOS, and consequently the angle BZK at the center is greater than a right angle; and because the angle BZK is obtuse, the square on BK is greater than the squares on BZ, ZK (o); that is, greater than twice the square on BZ. Join KV: and because (in the triangles KBV, OBV) KB, BV are equal to OB, BV, and that they contain equal angles, the angle KVB is equal to the angle OVB (p): and OVB is a right angle; therefore also KBV is a right angle: and because BD is less than twice DV, the rectangle contained by BD, BV is less than twice the rectangle DVB; that is, the square on KB is less than twice the square on KV (q): but the square on KB is greater than twice the square on BZ; therefore the square on KV is greater than the square on BZ: and because BA is equal to AK, and that the squares on BZ, ZA are equal together to the square on BA, and the squares on KV, VA to the square on AK; therefore the squares on BZ, ZA are equal to the squares on KV, VA; and of these the square on KV is greater than the square on BZ; therefore the square on VA is less than the square on ZA, and the straight line AZ is greater than VA: much more then is AZ greater than AG; because, in the preceding proposition, it was shown that KV falls without the circle FGH: and AZ is perpendicular to the plane KBOS, and is therefore the shortest of all the straight lines that can be drawn from A, the center of the sphere, to that plane: therefore the plane KBOS does not meet the lesser sphere. And that the other planes between the quadrants BX, KX fall without the lesser 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 center 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 BK, OS parallel, as also OS, PT; and their other sides, BO, KS, OP, ST, all equal to one another, and that BK is greater than OS, and OS greater than PT, therefore the straight line ZB is greater than IO (r). Join AO, which will be equal to AB; and because AIO, AZB are right angles, the squares on AI, IO are equal to the square on AO or of AB; that is, to the squares on AZ, ZB; and the square on ZB is greater than the square on IO, therefore the square on AZ is less than the square on AI; and the straight line AZ less than the straight line AI: and it was proved that AZ is greater than AG; much more then is AI greater than AG: therefore the plane SOPT falls wholly without the lesser sphere. In the same manner it may be demonstrated, that the plane TPRY falls without the same sphere (s), as also the triangle YRX. And after the same way it may be demonstrated, that all the planes which contain the solid polyhedron, fall without the lesser sphere. Therefore, in the |