Book 12
17
Now, since KB is greater than WV, while WV is equal to SP, therefore KB is greater than SP.
17
But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP.
17
And, since KBPS is a quadrilateral in a circle, and KB, BP, KS are equal, and PS less, and BX is the radius of the circle, therefore the square on KB is greater than double of the square on BX.
17
Let KZ be drawn from K perpendicular to BV.
17
Then, since BD is less than double of DZ, and, as BD is to DZ, so is the rectangle DB, BZ to the rectangle DZ, ZB, if a square be described upon BZ and the parallelogram on ZD be completed, then the rectangle DB, BZ is also less than double of the rectangle DZ, ZB.
17
And, if KD be joined, the rectangle DB, BZ is equal to the square on BK, and the rectangle DZ, ZB equal to the square on KZ; [III. 31, VI. 8 and Por.] therefore the square on KB is less than double of the square on KZ.
17
But the square on KB is greater than double of the square on BX; therefore the square on KZ is greater than the square on BX.
17
And, since BA is equal to KA, the square on BA is equal to the square on AK.
17
And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [I. 47] therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX; therefore the remainder, the square on ZA, is less than the square on XA.
17
Therefore AX is greater than AZ; therefore AX is much greater than AG.
17
And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface.
17
Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Q. E. F.
