Commentary on Proposition XII.17

Back to Proposition XII.17.

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.

O R U Q T P S X B Z V G A D W K L M F E

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 = BK2, it follows that BV = BZ, and Z, V coincide.)

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

and since             $\angle s$ PVA, SWA are right,

while AS = AP, $\Delta s$ PAV, SAW are equal in all respects, [I. 26]

and                     AV = AW.

Consequently         AB : AV = AK : AW ;

and VW, BK are parallel.

But PV, SW are parallel (being both perpendicular to one plane) and equal (by the equal $\Delta s$ 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 :g:`BPSK <0435.g1:< B P S K >>` does not anywhere touch the inner sphere we have to prove that the shortest distance from :g:`A <0435.g1:A>` to the plane is greater than :g:`AZ <0435.g1:/ A Z />`, which by the construction in XII. 16 is greater than :g:`AG <0435.g1:/ A G />`.

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.

Now                         BK $>$ VW

$>$ PS ; 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. $\angle$ BXK is obtuse.

Therefore (3)                                 BK2 $>$ 2 BX2. [II. 12]

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

We have                                 BD $<$ 2 DZ,

so that                               DB . BZ $<$ 2 DZ . ZB,

or                                         BK2 $<$ 2 KZ2 ;

therefore, a fortiori, [by (3) above]

  1.                                         BX2 $<$ KZ2.

Now                                     AK2 = AB2 ;

therefore                     AZ2 + ZK2 = AX2 + XB2.

And                                     BX2 $<$ KZ2 ;

therefore                                 AX2 $>$ AZ2,

or (5)                                       AX $>$ AZ.

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,

  1. 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.

D N C M M' P O L A B D' C' W Z O' X Y A' B'

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 $\angle s$ AOD, BOC, A'O'D', B'O'C' are all equal.

Also                         $\angle$ AOB $>$ $\angle$ A'O'B',

$\angle$ COD $>$ $\angle$ C'O'D', 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.

  1. If O'A' $>$ OA, cut off from O'A', O'B', O'C', O'D' lengths equal to OA, and draw the inner quadrilateral as shown in the figure (XYZW). Then AB $>$ A'B' $>$ XY, CD $>$ C'D' $>$ ZW, AD = A'D' $>$ WX, BC = B'C' $>$ YZ.

Consequently the same absurdity as in (1) follows a fortiori.

Therefore, since OA is neither equal to nor less than O'A', OA $>$ O'A'.

The fact is also sufficiently clear if we draw MO, NO bisecting DA, DC perpendicularly and therefore meeting in O, the centre of the circumscribed circle, and then suppose the side DA with the perpendicular MO to turn inwards about D as centre. Then the intersection of MO and NO, as P, will gradually move towards N.

Simson gives his proof as “Lemma II.” immediately before XII. 17. He adds to the Porism some words explaining how we may construct a similar polyhedron in another sphere and how we may prove that the polyhedra are similar.

The Porism is of course of the essence of the matter because it is the porism which as much as the construction is wanted in the next proposition. It would therefore not have been amiss to include the Porism in the enunciation of XII. 17 so as to call attention to it.