Twoo regular polygons of the same number of sides can be constructed, the one circumscribed about a circle and the other inscribed in it, which shall differ from each other by less than any given surface.

Let ABCE be a circle, O its centre, and Q the side of a square equal to or less than the given surface; then can two similar regular polygons be constructed, the one circumscribed about, and the other inscribed within the given circle, which shall differ from each other by less than the square of Q, and consequently, by less than the given surface. Inscribe a square in the

given circle (P. III.), and by means of it, inscribe, in succes

sion, regular polygons of 8, 16, 32, &c., sides (P. VII., S.), until one is found whose side is less than ; let AB be the side of such a polygon.

Construct a similar circum

scribed polygon abcde: then

D

will these polygons differ from each other by less than the square of Q.

For, from a and b, draw the lines a 0 and 60; they will pass through the points A and B. Draw also OK to the point of contact K; it will bisect AB at I and Prolong 40 to E

be perpendicular to it.

Let P denote the

circumscribed, and p the inscribed

polygon; then, because they are regular and similar, we

shall have (P. IX.),

Pp: OK or OA: OI;

hence, by division (B. II., P. VI.), we have,

But P is less than the square of AE (P. VII., C. 4); hence, P p is less than the square of AB, and consequently, less than the square of Q, or than the given surface; which was to be proved.

Definition. The limit of a variable quantity is a quantity towards which it may be made to approach nearer than any given quantity, and which it reaches under a particular supposition.

Lemma.-Two variable quantities which constantly approach towards equality, and of which the difference becomes less than any finite magnitude, are ultimately equal.

For if they are not ultimately equal, let D be their ultimate difference. Now, by hypothesis, the quantities have approached nearer to equality than any given quantity, as D; hence D denotes their difference and a quantity greater than their difference, at the same time, which is impossible; therefore, the two quantities are ultimately equal.*

Newton's Principia, Book I., Lemma I.

Cor. If we take any two similar regular polygons, the one circumscribed about, and the other inscribed within the circle, and bisect the arcs, and then circumscribe and inscribe two regular polygons having double the number of sides, it is plain that by continuing the operation, two new polygons may be found which shall differ from each other by less than any given surface; hence, by the lemma, the two polygons will become ultimately equal. But this equality cannot take place for any finite number of sides; hence, the number of sides in each will be infinité, and each will coincide with the circle, which is their common limit. Under this hypothesis, the perimeter of each polygon will coincide with the circumference of the circle.

Scholium. The circle may be regarded as a regular polygon having an infinite number of sides. The circumference may be regarded as the perimeter, and the radius as the apothem.

The area of a regular inscribed polygon, and that of a similar circumscribed polygon being given, to find the areas of the regular inscribed and circumscribed polygons having double the number of sides.

Let AB be the side of the given inscribed, and EF that of the given circumscribed polygon. Let C be their common centre, AMB a portion of the circumference of the circle, and M the middle point of the arc AMB. Draw the chord AM, and

at A and B draw the tangents AP and BQ; then will AM be the side of the inscribed polygon, and PQ the side of the circumscribed polygon of double the number of sides (P. VII.). Draw CE, CP, CM, and CF.

E

A

B

Denote the area of the given inscribed polygon by p, the area of the given circumscribed polygon by P, and the areas of the inscribed and circumscribed polygons having double the number of sides, respectively by p' and P'.

2°. Because the triangles CPM and CPE have the

each other as their

CPM: CPE :: PM: PE;

and because CP bisects the angle ACM, we have (B. IV., P. XVII.),

PM : PE :: CM: CE :: CD: ᏟᎪ ;

hence (B. II., P. IV.),

CPM : CPE :: CD : CA or CM.

But, the triangles CAD and CAM have the common altitude AD; they are therefore, to each other as their bases: hence,

CAD : CAM :: CD: CM;

or, because CAD and CAM are to each other as the polygons to which they belong,

Scholium. By means of Equation (1), we can find p', and then, by means of Equation (2), we can find P'.

PROPOSITION XII. PROBLEM.

To find the approximate area of a circle whose radius is 1.

The area of an inscribed square is equal to twice the square described on the radius (P. III., S.), which square is the unit of measure, and is denoted by 1. The area of the circumscribed square is 4. Making p equal to 2, and P equal to 4, we have, from Equations (1) and (2) of Proposition XI.,