ig. 160

PROBLEM.

277. A regular inscribed polygon ABCD &c. (fig. 160) being given, to circumscribe about the same circle a similar polygon.

Solution. At the point T, the middle of the arc AB, draw the tangent GH, which will be parallel to AB (112); do the same with each of the other arcs BC, CD, &c.; these tangents will form, by their intersections, the regular circumscribed polygon GHIK &c., similar to the inscribed polygon.

It will be readily perceived, in the first place, that the three points O, B, H, are in a right line, for the right angled triangles OTH, OHN, have the common hypothenuse OH, and the side OTON; they are consequently equal (126), and the angle TOH= HON, and the line OH passes through the point B, the middle of the arc TN. For the same reason, the point I is in OC produced, &c. But, since GH is parallel to AB, and HI to BC, the angle GHI= ABC (67); in like manner HIK = BCD, &c.; hence the angles of the circumscribed polygon are equal to those of the inscribed polygon. Moreover, on account of these same parallels

and

hence

GH: AB:: OH: OB,

HI: BC :: OH : OB,

GH: AB:: HI : BC.

But AB = BC; consequently GH = HI. For the same reason HI = IK, &c.; consequently the sides of the circumscribed polygon are equal to each other; therefore this polygon is regular and similar to the inscribed polygon.

278. Corollary 1. Reciprocally, if the circumscribed polygon GHIK &c., be given, and it is proposed to construct, by means of it, the inscribed polygon ABCD &c., it is evidently sufficient to draw to the vertices G, H, I, &c., of the given polygon the lines OG, OH, OI, &c., which will meet the circumference at the points A, B, C, &c., and then to join these points by the

etry, or, which amounts to the same thing, by the resolution of equations of the first and second degree. Dut M. Gaus has shown in a work, entitled Disquisitiones Arithmetica, Lipsia, 1801, that we may, by similar methods, inscribe a regular polygon of seventeen sides and in general one of 2′′ + 1 sides, provided that 2n + 1 be a prime number.

chords AB, BC, CD, &c., which will form the inscribed polygon. We might also, in this case, simply join the points of contact, T, N, P, &c., by the chords TN, NP, PQ, &c., which would equally form an inscribed polygon similar to the circumscribed

one.

279. Corollary 11. There may be circumscribed, about a given circle, all the regular polygons which can be inscribed within it; and, reciprocally, there may be inscribed, within a circle, all the polygons that can be circumscribed about it.

THEOREM.

230. The area of a regular polygon is equal to the product of its perimeter by half of the radius of the inscribed circle.

Demonstration. Let there be, for example, the regular polygon GHIK &c. (fig. 160); the triangle GOH, for example, has Fig. 160. for its measure GH× OT, the triangle OHI has for its measure HI× ON. But ON = OT; consequently the two triangles united have for their measure (GH + HI) × ¦OT. By proceeding thus with the other triangles, it is evident that the sum of all the triangles, or the entire polygon, has for its measure the sum of the bases GH, HI, IK, &c., or the perimeter of the polygon, multiplied by OT, half of the radius of the inscribed circle.

281. Scholium. The radius of the inscribed circle is the same as the perpendicular let fall from the centre upon one of the sides.

THEOREM.

282. The perimeters of regular polygons of the same number of sides are as the radii of the circumscribed circles, and also as the radii of the inscribed circles; and their surfaces are as the squares of these same radii.

Demonstration. Let AB (fig. 161) be a side of one of the Fig. 161. polygons in question, O its centre, and OA the radius of the circumscribed circle, and OD, perpendicular to AB, the radius of the inscribed circle; and let ab be the side of another polygon, similar to the former, o its centre, oa and od the radii of the circumscribed and inscribed circles.

The perimeters of the two polygons are to each other as the sides AB, ab (221). Now the angles A and a are equal, being each half of the angle of the polygon; the same may be said of

the angles B and b; therefore the triangles ABO, a bo, are similar, as also the right-angled triangles ADO, ado;

hence

ᎯᏴ : a b:: AO: ao:: DO: do;

consequently the perimeters of the polygons are to each other as the radii AO, a o, of the circumscribed circles, and also as the radii DO, do, of the inscribed circles.

The surfaces of these same polygons are to each other as the squares of the homologous sides AB, ab (221); they are therefore also as the squares of the radii of the circumscribed circles AO, a o, and as the squares of the radii of the inscribed circles DO, do.

LEMMA.

283. Every curved line, or polygon, which encloses, from one Fig. 162. extremity to the other, a convex line AMB (fig. 162), is greater than the enclosed line AMB.

Demonstration. We have already said that, by a convex line, we understand a curved line or polygon, or a line consisting in part of a curve and in part of a polygon, such that a straight line cannot cut it in more than two points (83). If the line AMB had re-entering parts or sinuosities, it would cease to be convex, because, as will be readily perceived, it might be cut by a straight line in more than two points. The arcs of a circle are essentially convex; but the proposition under consideration extends to every line, which fulfils the condition stated.

This being premised, if the line AMB be not smaller than any of those lines which enclose it, there is among these last a line smaller than any of the others, which is less than AMB, or at least equal to AMB. Let ACDEB be this enclosing line; be. tween these two lines draw at pleasure the straight line PQ, which does not meet the line AMB, or at most only touches it; the straight line PQ is less than PCDEQ (3); consequently if, instead of PCDEQ, we substitute the straight line PQ, we shall have the enclosing line APQB, less than APDQB. But, by hypothesis, this must be the shortest of all; this hypothesis then cannot be maintained; therefore each of the enclosing lines is greater than AMB.

284. Scholium. After the same manner, it may be demonstrated, without any restriction, that a line which is convex and

returns into itself, AMB (fig. 163), is less than any line which Fig. 163. encloses it on all sides, whether the enclosing line FHG touches AMB in one or more points, or whether it surrounds it without touching it.

LEMMA.

285. Two concentric circles being given, there may always be inscribed, in the greater, a regular polygon, the sides of which shall not meet the circumference of the smaller; and there may also be circumscribed, about the smaller, a regular polygon, the sides of which shall not meet the circumference of the greater; so that on the whole the sides of the polygon described shall be contained between the two circumferences.

Demonstration. Let CA, CB (fig. 164), be the radii of the two Fig. 164. given circles. At the point A draw the tangent DE terminating, at the greater circumference, in D and E. Inscribe, in the greater circumference, one of the regular polygons, which can be inscribed by the preceding problems, and bisect the arcs subtended by the sides, and draw the chords of these half arcs; and a regular polygon will be described of double the number of sides. Continue to bisect the arcs until one is obtained which is smaller than DBE. Let MBN be this arc, the middle of which is supposed to be in B; it is evident that the chord MN will be further from the centre than DE, and that thus the regular polygon, of which MN is a side, cannot meet the circumference, of which CA is the radius.

The same things being supposed, join CM and CN, which meet the tangent DE in P and Q; PQ will be the side of a polygon circumscribed about the smaller circumference similar to the polygon inscribed in the greater, the side of which is MN. Now it is evident that the circumscribed polygon, which has for its side PQ, cannot meet the greater circumference, since CP is less than CM.

There may, therefore, by the same construction, be a regular polygon inscribed in the greater circumference, and a similar polygon circumscribed about the smaller, which shall have their sides comprehended between the two circumferences.

286. Scholium. If we have two concentric sectors FCG, ICH, we can likewise inscribe, in the greater, a portion of a regular polygon, or circumscribe, about the smaller, a portion of a similar

polygon, so that the perimeters of the two polygons would be comprehended between the two circles. It is only necessary to divide the arc FBG successively into 2, 4, 8, 16, &c., equal parts, until one is obtained smaller than DBE.

By a portion of a regular polygon, as the phrase is here used, is to be understood the figure terminated by a series of equal chords, inscribed in the arc FG, from one extremity to the other. This portion has the principal properties of a regular polygon, it has its angles equal, and its sides equal; it is, at the same time, capable of being inscribed in, and circumscribed about a cirle ; it does not, however, make a part of a regular polygon, properly so called, except when the arc, subtended by one of these sides, is an aliquot part of the circumference.

Fig. 165.

THEOREM.

287. The circumferences of circles are as their radii, and their surfaces are as the squares of their radii.

Demonstration. Denoting, by circ. CA and circ. OB (fig. 165), the circumferences of the circles whose radii are CA and OB, we say that circ. CA: circ. OB:: CA: OB.

For, if this proportion be not true, CA will be to OB as circ. CA is to a fourth term either greater or less than circ. OB. Let us suppose that it is less, and that, if possible,

CA: OB:: circ. CA: circ. OD.

Inscribe, in the circumference of which OB is the radius, a regular polygon EFGKLE, whose sides shall not meet the circumference of the circle, whose radius is OD (285); inscribe a similar polygon MNPSTM in the circle whose radius is CA.

This being done, since the polygons are similar, their perimeters MNPSM, EFGKE, are to each other as the radii CA, OB, of the circumscribed circles (282), and we have

MNPSM: EFGKE:: CA: OB;

but, by hypothesis,

CA: OB: circ. CA: circ. OD;

therefore MNPSM: EFGKE:: circ. CA: circ. OD. Now this proportion is impossible, because the perimeter MNPSM is less than circ. CA (283), while EFGKE is greater than the circ. OD; therefore it is impossible that CA should be to OB as circ. CA is to a circumference less than circ. OB; or, in other words,