1 PROP. XVI. PROB. In the greater of two circles that have the same centre, to Let ABCD, EFGH be two given circles having the same centre K: it is required to inscribe in the greater circle ABCD, a polygon of an even number of equal sides, that shall not meet the lesser circle. * 3. Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser circle draw GA at right angles to BD, and produce it to C; therefore AC touches the circle * Cor. 16. EFGH: then, if the circumference BAD be bisected, and the half of it be again bisected, and so on, there must at length remain a circumference less than AD: *Lemma. let this be LD; and from the point L draw LM perpendicular to BD, and produce it to N; and join LD, DN: therefore LD is equal to DN: and because LN is parallel to AC, and that AC touches the circle EFGH; therefore LN does not meet the circle EFGH; and much less shall the straight lines LD, DN, meet the circle EFGH: so that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be inscribed in the circle a polygon of an even number of equal sides not meeting the lesser circle. Which was to be done. LEMMA II. H L BE KGM D F If two trapeziums ABCD, EFGH be inscribed in the If it be possible, let KA be not greater than LE; 28.3. 2. 6. * two equal circles AD, BC, in the one, are equal to EH, FG in the other, the circumferences AD, BC, are equal to the circumferences EH, FG; but because the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG; therefore the whole circumference ABCD is greater than the whole EFGH: but it is also equal to it, which is impossible: therefore the straight line KA is not equal to LE. But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM which are respectively parallel * to and less than EF, FG, GH, HE: then because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal; therefore the circumference AD is greater than MP: for the same reason the circumference BC is greater than NO: and because the straight line AB is greater than EF, which is K B greater than MN, much more is AB greater than MN: therefore the circumference AB is greater than MN; and for the same reason, the cirference 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. Q. E. D. COR. 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 centre to the circumference of the circle described about the triangle. PROP. XVII. PROB. In the greater of two spheres which have the same centre, See N. to inscribe a solid polyhedron, the superficies of which shall not meet the lesser sphere. Let there be two spheres about the same centre A: it is required to inscribe in the greater a solid polyhedron, the superficies of which shall not meet the lesser sphere. Let the spheres be cut by a plane passing through the centre; 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 unmoveable; 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 15. 3. any straight line in the circle or sphere. 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 * 16. 12. equal sides not meeting the lesser circle FGH; 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 A draw† AX at right angles to the plane of the † 12. 11. circle BCDE, 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, and 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; where- * 18. 11. fore 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 *4 Def. 11. of the planes, therefore OV is perpendicular to the plane BCDE: for the same reason SQ is perpendicu lar to the same plane, because the plane KSX 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, and BV equal to KQ: but the whole BA is equal to the whole KA, therefore the remainder VA is equal to the re 26.1. * 2.6. * 6. 11. * 33. 1. * 9. 11. * 9. 11. *2.11. D * mainder QA: therefore as BV is to VA, so is KQ to QA; wherefore VQ is parallel to BK: and because OV, SQ are each of them at right angles to the plane of the circle BCDE, OV is parallel to SQ; and it has been proved, that it is also equal to it; therefore QV, SO are equal and parallel: and because QV is parallel to SO, and also to KB; OS is parallel to BK; 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 shewn to be parallel to the same KB; wherefore* TP is parallel to SO, 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: 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 *11.11. plane of the quadrilateral KBOS, 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 of AZ, ZB are equal to the square of AB, and the squares of AZ, ZK, to the square of AK*; therefore the squares of AZ, ZB, are * 47. 1. equal to the squares of AZ, ZK: take from these equals the square of AZ, and the remaining square of BZ is equal to the remaining square of 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 centre 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 con |