[ocr errors]





[ocr errors]




[ocr errors]


them be equal; and the cylinders AX, EO being also equal, and (XII. 11.) cones and cylinders of the same altitude being to one another as their bases ; therefore (V. A.) the base BD is equal to the base FH; and as BD is to FH, so is MN to KL. But let the altitudes KL, MN be unequal, and MN the greater ; from MN take MP equal to KL, and through P cut the cylinder EO by the plane YS parallel to the planes of the circles FH, RO; therefore the common section of the plane YS and the cylinder EO is a circle, as is shown in the demonstration of the 13th proposition of this book : and consequently ES is a cylinder, the base of which is the circle EFGH, and altitude MP. Then, (V.7.) because the cylinders AX, EO are equal, as AX is to the cylinder ES, so is EO to the same ES. But (XII. 11.) as AX to ES, so the base BD to the base FH; for AX, ES are of the same altitude; and (XII. 13.) as EO to ES, so is MN to MP, because EO is cut by the plane YS parallel to its bases. Therefore (V. 5.) as BD to the base FH, so is MN to the altitude MP: but MP is equal to KL; wherefore as BD to FH, so is MN to KL; that is, the bases and altitudes of the equal cylinders AX, EO, are reciprocally proportional.

2. But let the bases and altitudes of the cylinders AX, EO, be reciprocally proportional, viz., D to FH, as MN to KL: the cylinders are equal.

First, let BD be equal to FH; then because as BD is to FH, so is MN to KL, MN (V. A.) is equal to KL, and therefore (XII. 11.) AX is equal to Eo.

But let the bases BD, FH be unequal, and let BD be the greater; and because, as BD is to FH, so is MN to KL; therefore (V. A.) MN is greater than KL. Then, the same construction being made as before, because as BD to FH, so is MN to KL; and because KL is equal to MP; therefore (XII. 11.) BD is to FH, as AX to ES; and as MN to MP or KL, so is EO to ES: therefore AX is to ES, as EO is to the same ES: whence the cylinder AX is equal to EO: and the same reasoning holds

Therefore the bases and altitudes, &c.

in cones.

PROP. XVI. PROB. . In the greater of two concentric circles, to describe a polygon of an even number of equal sides, not meeting the less circle.*

Let ABCD, EFGH be two given circles having the same * This proposition is merely a lemma to the 17th ; and it may be omitted by the student, as, in this edition, the 17th is demonstrated in such a manner as not to require it.








centre K: it is required to inscribe in the greater circle a polygon of an even number of equal sides, not meeting the less circle.

Through K draw the straight line BD, and from G, the point in which it meets the circumference of the less circle, draw AGC perpendicular to BD; then (III. 16.) AC touches the circle FH. Now, if the arc BAD be bisected, and the half of it be again bisected, and so on, there must at length remain (XII. lem. 1.) an arc less than AD: let this be LD; and from L draw LMN perpendicular to BD; and join LD, DN. Therefore (1. 4.) LD is equal to DN; and because LN is parallel to AC, and that AC touches the circle FH; therefore LN does not meet the circle FH: and much less will the straight lines LD, DN meet it. If therefore straight lines equal to LĎ be applied in the circle ABCD from the point L around to N, there will be described in the circle a polygon of an even number of equal sides not meeting the less circle : which was to be done.



sides are,

If two trapeziums ABCD, EFGH be inscribed in the circles, the centres of which are K, L; and if the sides AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG, be all equal ; but the side AB greater than EF, and DC greater than HG: the radius KA of the circle in which the greater

is greater than LE the radius of the other. If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less. First, let KA be equal to LE. Therefore, because, in two equal circles, AD, BC in the one are equal to EH, FG in the other, the arcs AD, BC are equal (III. 28.) to the arcs EH, FG; but because the straight lines AB, DC are respectively greater than EF, GH, the arcs AB, DC are greater than EF, HG. Therefore the whole circumfer. ence ABCD is greater than the whole EFGH: but it is also equal to it, which is impossible. Therefore KA is not equal to LE.

But let KA be less than LE, and make LM equal to KA, and from the centre L, at the 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 (VI. 2.) are respectively parallel

to EF, FG, GH, HE, and less than those lines. Then, because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal; therefore the arc AD is greater than the arc MP. For the same reason, the arc BC is greater than the arc NO; and because the straight line AB is greater than EF, which is greater than MN, much more is AB greater than

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

MN. Therefore the arc AB is greater than the arc MN; and, for the same reason, the arc DC is greater than the arc 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; it must therefore be greater. If two trapeziums, therefore, &c.

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 parallel sides; 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 concentric spheres, to describe a polyhedron, the surface of which will not meet the less sphere.

Let there be two spheres about the same centre A; it is required to describe in the greater a polyhedron, the surface of which will not meet the less.

If the spheres be cut by a plane passing through the centre, the common sections of it and the spheres will be circles; because the sphere is described by the revolution of a semicircle about the diameter remaining fixed; so that in whatever position the semicircle is conceived to be, the common section of the plane in which it is, with the surface of the sphere, is the circumference of a circle: and this is a great circle* of the sphere, because (III. 15.) the diameter of the sphere, which is likewise the diameter of the circle, is

See Appendix, V 10.

greater than any chord 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 less be FGH; draw the perpendicular diameters BD, CE: and draw GU perpendicular to AG, and join AU. Then, if the arc BE be bisected in L, its halves in M, K, and so on, an arc will at length be obtained, less than the arc which would have a chord equal to GU. Let BK be this arc : then the chord BK is less than GU. Draw the diameter KAN, and through A draw AX perpendicular to the plane of the circle BCDE, meeting the surface of the sphere in X. Let planes pass through AX and each of the straight lines BD, KN: these, from what has been said, will produce great circles on the surface of

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors]

the sphere ; let BXD, KXN be the semicircles thus made upon the diameters BD, KN. Then (XI. 18.) because XA is at right angles to the plane of the circle BCDE, 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, their halves BE, BX, KX are also equal. Therefore, as many arcs as are in BE, so many there are in BX, KX equal to the arcs BK, KL, LM, ME. Let these arcs be BO, OP, PR, RX; KS, ST, TY, YX; draw their chords, and join OS, PT, RY: draw also 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 their common section, therefore (XI. def. 3.) POV is perpendicular to the plane BCDE. For the same reason, SQ is perpendicular to BCDE, because the plane KSXN is perpendicular to it. Join

VQ; and because in the equal semicircles BXD, KXN, the arcs BO, KS are equal, the angles BAO, KAS are equal ; and (const.) AVO, AQS are right angles ; therefore (I. 26.) OV is equal to SQ, and AV equal to AQ. But the whole BA is equal to the whole KA, therefore the remainders BV, KQ are equal. As, therefore, BV:VA :: KQ: QA; wherefore (VI. 2.) VQ is parallel to BK. Also (XI. 6.) because OV, SQ are each perpendicular to the plane of the circle BCDE, they are parallel ; and they have been proved to be equal; therefore QV, so are (I. 33.) equal and parallel : and because QV is parallel to so, and also to KB; OS is parallel (XI. 9.) to BK; and BO, KS, which join them, are in the same plane in which these parallels

[merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small]

are, and the quadrilateral 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 same way that so was shown to be parallel to the same KB; wherefore (XI. 9.) TP is parallel to SO, and the quadrilateral SOPT is in one plane. For the same reason, the quadrilateral TPRY is in one plane : and (XI. 2.) the triangle YRX is also in one plane. If, therefore, from the points 0, Š, P, T, R, Y, straight lines be drawn to the centre A, there will be formed a polyhedron between the arcs 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 A: and if the same construction be made upon each of the chords of KL, LM, ME, as has been done upon BK, and the like be done also in the other three quadrants, and in the other hemi

« ForrigeFortsett »