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Book XII.

PROP. XVII. PROB.

See N. To describe in the greater of two spheres which

have the same centre, 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 describe 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 • 15. 3. greater a than any straight lige 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 cir16. 12. cles, describeb a polygon of an even number of 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; joiu KA, and produce it to N; and from A draw AX at right angles to the plane of the 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, RN, which, from what has been said, shall produce great circles on the superficies of the sphere, and let BXD, KÉN be the semicircles thus made upon the diameters BD, KN: Therefore, be

cause XA is at right angles to the plane of the circle BCDE, • 18. 11. every plane which passes through XA is at right angles to

the plane of the circle BCDE; wherefore the semicircles BXD, KXN are at right angles to that plane : And because the semicircles BET, 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 there are 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 Book XII. 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 perpendiculara to the plane BCDE: For 4 Def. 11. 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

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BXD, KXN, the circumferences BO, KS are equal, and
OV, SQ are perpendicular to their diameters, therefore d d 26. 1,
OV is equal to SQ, and BV equal to KQ. But the whole
BA is equal to the whole KA, therefore the remainder VA

equal to the remainder QA: As therefore BV is to VA, so is KQ to QA, wherefore VQ is parallel e to BK: And. 2. 6. because OV, SQ are each of them at right angles to the plane of the circle BCDE, OV is parallelf to SQ; and it ' 6. 11. has been proved, that it is also equal to it; therefore QV, SO are equal and parallel 8: And because QV is parallel to * 83. 1. SO, and also to KB; OS is parallel to BK; and there. h 9. 11.

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Book XII. fore BO, KS, which join them are in the same plane in
» which these parallels are, and the quadrilateral figuse

KBOS is in one plane : And if BP, 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 parts * 9. 11. rallel to the same KB; wherefore a 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

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02. 11. the figure YRX is also in one planeb. Therefore, if from

the points O, S, P, T, R, Y, there be drawn straight lines
to the point A, there shall be formed a solid polyhedron
between the circumferences BX, KX, composed of pyra-
mids, the bases of which are the quadrilaterals KBOS,
SOPT, TPRY, and the triangle YRX, and of which the
compion vertex is the point A : And if the same construc.
tion 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
Uhree quadrants, and in the other hemisphere; there shall's

be formed a solid polyhedron described in the sphere, com- Buos XIL posed 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 thein all being the point A: and the superficies of this solid polyhedron does not meet the lesser sphere in which is the eircle FGH: For, from the point A drawa AZ perpendicular to the plane of the quadrilateral * 1111. KBOS, meeting it in 2, 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 b; therefore the squares of AZ, ZB, are 47. 2, equal to the squares of AZ, ZK: Take from these equals the square of AZ, the remaining square of BZ is equal to the reinaising 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 froin the centre 2, and distance ZB, shall pass through the points K, 0, S, and KBOS shall be a quadrilateral figure in the circle: And because KB is greater than QV, and QV equal to SO, therefore KB is greater tlian SO: But KB is equal to each of the straight lines BO, KS; wherefore each of the cireumferences cut off by KB, BO, KS, is greater than that qut 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 centre is greater than a right angle: And because the angle BZK is obtuse, the square of BK is greater than the squares 12.2. of BZ, ZK ; that is, greater than twice the square of BZ. Joio 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: And 4.1. UVB is a right angle ; therefore also KVB is a right angle: And because BD is less than twice DV; the rectangle contained by DB, BV, is less than twice the rectangle DVB; that is, the square of KB is less than twice the square of 8.6 KY: But the square of KB is greater than twice the square

Book XII. of BZ; therefore the square of KV is greater than

the square of BZ: And because BA is equal to AK, and that the squares of BZ, ZA, are equal together to the square of-BA, and the squares of KV, VA, to the square of AK; therefore the squares of BZ, ZA are equal to the squares of KV, VA; and of these the square of KV is greater than the square of BZ; therefore the square of VA is less than the square of ZA, and the straight line AZ 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 centre 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 Al 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 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 * 2 Lem. 12. than PT, therefore the straight line ZB is greater a than IO.

Join AO which will be equal to AB; and because AIO, AZB are right angles, the squares of AI, 10 are equal to the square of A0 or of AB; that is, to the squares of AZ, ZB; and the square of ZB is greater than the square of 10, therefore the square of AZ is less than the square of AI; and the straight line AZ less than the straight line Al: And it was proved, that AZ is greater than AG; much more then is Al 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, as also the triangle YRX, viz. by the Cor. of 2d Lemma. 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 greater of two spheres, which have the same centre, a solid polyhedron is described, the superficies of which does not meet the lesser sphere. Which was to be done.

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