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Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle.

Now, since KB is greater than WV,

while WV is equal to SP,

therefore KB is greater than SP.

But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP. 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.

Let KZ be drawn from K perpendicular to BV.

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.

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.

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. And, since BA is equal to KA,

the square on BA is equal to the square on AK.

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;

[1. 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.

Therefore AX is greater than AZ;

therefore AX is much greater than AG.

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.

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.

PORISM. But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere BCDE, the polyhedral solid in the sphere BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere BCDE has to the diameter of the other sphere.

For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar. But similar pyramids are to one another in the triplicate ratio of their corresponding sides;

[XII. 8, Por.]

therefore the pyramid of which the quadrilateral KBPS is the base, and the point A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius AB of the sphere about A as centre has to the radius of the other sphere.

Similarly also each pyramid of those in the sphere about A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere.

And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents;

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hence the whole polyhedral solid in the sphere about A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere.

Q. E. D.

This proposition is of great length and therefore requires summarising in order to make it easier to grasp. Moreover there are some assumptions in it which require to be proved, and some omissions to be supplied. The figure also is one of some complexity, and, in addition, the text and the figure treat two points Z and V, which are really one and the same, as different.

The first thing needed is to know that all sections of a sphere by planes through the centre are circles and equal to one another (great circles or "greatest circles" as Euclid calls them, more appropriately). Euclid uses his definition of a sphere as the figure described by a semicircle revolving about its diameter. This of course establishes that all planes through the particular diameter make equal circular sections; but it is also assumed that the same sphere is generated by any other semicircle of the same size and with its centre at the same point.

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The construction and argument of the proposition may be shortly given as follows.

A plane through the centre of two concentric spheres cuts them in great circles of which BE, GF are quadrants.

A regular polygon with an even number of sides is inscribed (exactly as in Prop. 16) to the outer circle such that its sides do not touch the inner circle. BK, KL, LM, ME are the sides in the quadrant BE.

AO is drawn at right angles to the plane ABE, and through AO are drawn planes passing through B, K, L, M, E, etc., cutting the sphere in great circles.

OB, OK are quadrants of two of these great circles.

As these quadrants are equal to the quadrant BE, they will be divisible into arcs equal in number and magnitude to the arcs BK, KL, LM, ME.

Dividing the other quadrants of these circles, and also all the quadrants of the other circles through OA, in this way we shall have in all the circles a polygon equal to that in the circle of which BE is a quadrant.

BP, PQ, QR, RO and KS, ST, TU, UO are the sides of these polygons in the quadrants BO, KO.

Joining PS, QT, RU, and making the same construction all round the circles through AO, we have a certain polyhedron inscribed in the outer sphere.

Draw PV perpendicular to AB and therefore (since the planes OAB, BAE are at right angles) perpendicular to the plane BAE; [XI. Def. 4] draw SW perpendicular to AK and therefore (for a like reason) perpendicular to the plane BAE.

Draw KZ perpendicular to BA. (Since BK = BP, and DB. BV = BP2, DB.BZ = BK, it follows that BV = BZ, and Z, V coincide.)

Now, since s PAV, SAW, being angles subtended at the centre by equal arcs of equal circles, are equal,

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But PV, SW are parallel (being both perpendicular to one plane) and equal (by the equal As PAV, SAW),

therefore

VW, PS are equal and parallel.

Therefore BK (being parallel to VW) is parallel to PS.
Consequently (1) BPSK is a quadrilateral in one plane.

Similarly the other quadrilaterals PQTS, QRUT are in one plane; and the triangle ORU is in one plane.

In order now to prove that the plane BPSK does not anywhere touch the inner sphere we have to prove that the shortest distance from A to the plane is greater than AZ, which by the construction in XII. 16 is greater than AG.

Draw AX perpendicular to the plane BPSK.

Then AX2+ XB2 = AX2 + XK2 = AX2 + XS2 = AX2 + XP2 = AB2, whence

XB = XK = XS = XP,

or (2) the quadrilateral BPSK is inscribable in a circle with X as centre and radius XB.

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therefore in the quadrilateral BPSK three sides BK, BP, KS are equal, but PS is less.

Consequently the angles about X are three equal angles and one smaller

angle;

therefore any one of the equal angles is greater than a right angle, i.e. ▲ BXK is obtuse.

Therefore (3)

BK2> 2BX2.

[II. 12]

Next, consider the semicircle BKD with KZ drawn perpendicular to BD. We have

so that

or

BD < 2DZ,

DB. BZ <2DZ. ZB,

BK2 <2KZ2;

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But, by the construction in XII. 16, AZ> AG; therefore, a fortiori, AX> AG.

And, since the perpendicular AX is the shortest distance from A to the plane BPSK,

(6) the plane BPSK does not anywhere meet the inner sphere.

Euclid omits to prove that, a fortiori, the other quadrilaterals PQTS, QRUT, and the triangle ROU, do not anywhere meet the inner sphere.

For this purpose it is only necessary to show that the radii of the circles circumscribing BPSK, PQTS, QRUT and ROU are in descending order of magnitude.

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We have therefore to prove that, if ABCD, A'B'C'D' are two quadrilaterals inscribable in circles, and

AD= BC= A'D' = B'C',

while AB is not greater than AD, A'B' = CD, and AB> CD> C'D',

then the radius OA of the circle circumscribing the first quadrilateral is greater than the radius O'A' of the circle circumscribing the second.

Clavius, and Simson after him, prove this by reductio ad absurdum.

(1) If OA = O'A',

it follows that Ls AOD, BOC, A'O'D', B'O'C' are all equal.

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whence the four angles about O are together greater than the four angles about O', i.e. greater than four right angles;

which is impossible.

H. E. III.

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