Commentary on Proposition XII.2

Back to Proposition XII.2.

Though this theorem is said to have been proved by Hippocrates, we may with tolerable certainty attribute the proof of it given by Euclid to Eudoxus, to whom XII. 7 Por. and XII. 10 (which Euclid proves in exactly the same manner) are specifically attributed by Archimedes. As regards the lemma used herein (Eucl. X. 1) and the somewhat different lemma by means of which Archimedes says that the theorems of XII. 2, XII. 7 Por. and XII. 18 were proved, see my note on X. 1 above.

The first essential in this proposition is to prove that we can exhaust a circle, in the sense of X. 1, by successively inscribing in it regular polygons, each of which has twice as many sides as the preceding one. We take first an inscribed square, then bisect the arcs subtended by the sides and so form an equilateral polygon of eight sides, then do the same with the latter, forming a polygon of 16 sides, and so on. And we have to prove that what is left over when any one of these polygons is taken away from the circle is more than half exhausted when the next polygon is made and subtracted from the circle.

Euclid proves that the inscribed square is greater than half the circle and that the regular octagon when subtracted takes away more than half of what was left by the square. He then infers that the same thing will happen whenever the number of sides is doubled.

This can be seen generally by taking any arc of a circle cut off by a chord AB. Bisect the arc in C. Draw a tangent to the circle at C, and let AD, BE be drawn perpendicular to the tangent. Join AC, CB. Then DE is parallel to AB, since $\angle$ ECB $= \angle$ CAB, in alternate segment, [III. 32] $= \angle$ CBA. [III. 29, I. 5] Thus ABED is a $\square$; and it is greater than the segment ACB. Therefore its half, the $\triangle$ ACB, is greater than half the segment.

D C E A B

Thus, by X. 1, Euclid’s construction of successive regular polygons in a circle, if continued far enough, will at length leave segments which are together less than any given area.

Now let $X$, $X'$ be the areas of the circles, $d$, $d'$ their diameters, respectively. Then, if

$$X : X' \ne d^2 : {d'}^2,$$

$$d^2 : {d'}^2 = X : S,$$

where $S$ is some area either greater or less than $X'$.

I. Suppose $S < X'$. Continue the construction of polygons in $X'$ until we arrive at one which leaves over segments together less than the excess of $X'$ over $S$, i.e. a polygon such that

$$X' > (\text{polygon in } X') > S.$$

Inscribe in the circle $X$ a polygon similar to that in $X'$. Then

$$(\text{polygon in } X) : (\text{polygon in } X') = d^2 : {d'}^2$$

[XII. 1]

$$= X : S\text{, by hypothesis;}$$

and, alternately,

$$(\text{polygon in } X) : X = (\text{polygon in } X') : S.$$

But

$$(\text{polygon in } X) < X;$$

therefore

$$(\text{polygon in } X') < S.$$

But, by construction,

$$(\text{polygon in } X') > S:$$

which is impossible.

Hence $S$ cannot be less than $X'$ as supposed.

II. Suppose $S > X'$. Since

$$d^2 : {d'}^2 = X : S,$$

we have, inversely,

$${d'}^2 : d^2 = S : X.$$

Suppose that

$$S : X = X' : T,$$

whence, since $S > X'$,

$$X > T.$$

[v. 14] Consequently

$${d'}^2 : d^2 = X' : T,$$

where $T < X$.

This can be proved impossible in exactly the same way as shown in Part I. Hence $S$ cannot be greater than $X'$ as supposed. Since then $S$ is neither greater nor less than $X'$,

$$S = X',$$

and therefore

$$d^2 : {d'}^2 = X : X'.$$

With reference to the assumption that there is some space $S$ such that

$$d^2 : {d'}^2 = X : S,$$

i.e. that there is a fourth proportional to the areas $d^2$, ${d'}^2$, $X$, Simson observes that it is sufficient, in this and the like cases, that a thing made use of in the reasoning can possibly exist, though it cannot be exhibited by a geometrical construction. As regards the assumption see note on v. 18 above.

There is grave reason for suspecting the genuineness of the Lemma at the end of the proposition; though, if it be rejected, it will be necessary to delete the words “as was before proved” in corresponding places in XII. 5, 18.

It will be observed that Euclid proves the impossibility in the second case by reducing it to the first. If it is desired to prove the second case independently, we must circumscribe successive polygons to the circles instead of inscribing them, in the way shown by Archimedes in his first proposition on the Measurement of a circle. Of course we require, as a preliminary, the proposition corresponding to XII. 1, that

Similar polygons circumscribed about circles are to one another as the squares on the diameters.

Let AB, A'B' be corresponding sides of the two similar polygons. Then $\angle\text{s}$ OAB, O'A'B' are equal, since AO, A'O' bisect equal angles.

A C B O A' C' B' O'

Similarly $\angle$ ABO $= \angle$ A'B'O'. Therefore $\triangle\text{s}$ AOB, A'O'B' are similar, so that their areas are in the duplicate ratio of AB to A'B'. The radii OC, O'C' drawn to the points of contact are perpendicular to AB, A'B', and it follows that

$$AB : A'B' = CO : C'O'.$$

Thus the polygons are to one another in the duplicate ratio of the radii, and therefore of the diameters.

Now suppose a square ABCD described about a circle. Make an octagon described about the circle by drawing tangents at the points E etc., where OA etc. meet the circle. Then shall the tangent at E cut off more than half of the area between AK, AH and the arc HEK. For the angle AEG is right, and is therefore $> \angle$ EAG.

A G K D F H E O B C

‘Therefore

$$\begin{aligned} AG &> EG \\ &> GK. \end{aligned}$$

Therefore

$$\triangle\text{">$AGE} > \triangle\text{">$EGK.}$$

Similarly

$$\triangle\text{">$AFE} > \triangle\text{">$EFH.}$$

Hence

$$\triangle\text{">$AFG} > \tfrac{1}{2}\text{ (re-entrant quadrilateral ">$AHEK),}$$

and a fortiori,

$$\triangle\text{">$AFG} > \tfrac{1}{2}\text{ (area between AH, AK and the arc).}$$

Thus the octagon takes from the square more than half the space between the square and the circle. Similarly, if a figure of 16 equal sides be circumscribed by cutting off symmetrically the corners of the octagon, it will take away more than half of the space between the octagon and circle.

Suppose now, with the original notation, that

$$d^2 : {d'}^2 = X : S,$$

where $S$ is greater than $X'$.

Continue the construction of circumscribed polygons about $X'$ until the total area between the polygon and the circle is less than the difference between $S$ and $X'$, i.e. till

$$S > (\text{polygon about } X') > X'.$$

Circumscribe a similar polygon about $X$. Then

$$(\text{polygon about } X) : (\text{polygon about } X') = d^2 : {d'}^2$$

$$= X : S\text{, by hypothesis,}$$

and, alternately,

$$(\text{polygon about } X) : X = (\text{polygon about } X') : S.$$

But

$$(\text{polygon about } X) > X.$$

Therefore

$$(\text{polygon about } X') > S.$$

But

$$S > (\text{polygon about } X'):$$

[above] which is impossible.

Hence $S$ cannot be greater than $X'$.

Legendre proves this proposition by a method equally rigorous but not, I think, possessing any advantages over Euclid’s. It depends on a lemma corresponding to Eucl. XII. 16, but with another part added to it.

Two concentric circles being given, we can always inscribe in the greater a regular polygon such that its sides do not meet the circumference of the lesser, and we can also circumscribe about the lesser a regular polygon such that its sides do not meet the circumference of the greater.

Let CA, CB be the radii of the circles.

I. At A on the inner circle draw the tangent DE meeting the outer circle in D, E. Inscribe in the outer circle any of the regular polygons which we can inscribe, e.g. a square. Bisect the arc subtended by a side, bisect the half, bisect that again, and so on, until we arrive at an arc less than the arc DBE. Let this arc be MN, and suppose it so placed that B is its middle point. Then the chord MN is clearly more distant from the centre C than DE is; and the regular polygon, of which MN is a side, does not anywhere meet the circumference of the inner circle.

B M N D E P A Q C

II. Join CM, CN, meeting DE in P, Q. Then PQ will be the side of a polygon circumscribed about the inner circle and similar to the polygon inscribed in the outer; and the circumscribed polygon of which PQ is a side will not anywhere meet the outer circle.

Legendre now proves XII. 2 after the following manner. For brevity, let us denote the area of the circle with radius CA by $(\text{circ. }$ CA $)$. Then it is required to prove that, if OB be the radius of a second circle, $(\text{circ. }$ CA $) : (\text{circ. }$ OB $) =$ CA2 $:$ OB2.

A C B D O

Suppose, if possible, that this relation is not true. Then CA2 will be to OB2 as $(\text{circ. }$ CA $)$ is to an area greater or less than $(\text{circ. }$ OB)$. I. Suppose, first, that CA2 $:$ OB2 $= (\text{circ. }$ CA $) : (\text{circ. }$ OD $)$, where OD is less than OB.

Inscribe in the circle with radius OB a regular polygon such that its sides do not anywhere meet the circumference of the circle with centre OD; [Lemma] and inscribe a similar polygon in the other circle. The areas of the polygons will then be in the duplicate ratio of CA to OB, or [XII. 1] $(\text{polygon in circ. }$ CA $) : (\text{polygon in circ. }$ OB $)$ $=$ CA2 $:$ OB2 $= (\text{circ. }$ CA $) : (\text{circ. }$ OD $)$, by hypothesis.

But this is impossible, because the polygon in $(\text{circ. }$ CA $)$ is less than $(\text{circ. }$ CA $)$, but the polygon in $(\text{circ. }$ OB $)$ is greater than $(\text{circ. }$ OD)$. ‘Therefore CA2 cannot be to OB2 as $(\text{circ. }$ CA $)$ is to a less circle than $(\text{circ. }$ OB $)$.

II. Suppose, if possible, that CA2 $:$ OB2 $= (\text{circ. }$ CA $) : (\text{some circle} > \text{circ. }$ OB)$. Then inversely OB2 $:$ CA2 $= (\text{circ. }$ OB $) : (\text{some circle} < \text{circ. }$ CA $), and this is proved impossible exactly as in Part I. Therefore CA2 $:$ OB2 $= (\text{circ. }$ CA $) : (\text{circ. }$ OB $)$.