passes through all the vertices of the angles of the polygon, and the polygon is inscribed in this circumference.

Furthermore, with respect to this circumference, all the sides AB, BC, CD, &c. are equal chords; they are accordingly equally distant from the centre (109); if therefore from the point O, as a centre, and with the radius OP, a circle be described, the circumference will touch the side BC and all the other sides of the polygon, each at its middle point, and the circle will be inscribed in the polygon, or the polygon will be circumscribed about the circle.

267. Scholium 1. The point O, the common centre of the inscribed and circumscribed circle, may be regarded also as the centre of the polygn; and for this reason we call the angle of the centre the angle AOB formed by the two radii drawn to the extremities of the same side AB.

Since all the chords AB, BC, &c., are equal, it is evident that all the angles at the centre are equal, and that the value of each is found by dividing four right angles by the number of the sides of the polygon.

268. Scholium 11. In order to inscribe a regular polygon of a certain number of sides in a given circle, it is only necessary to divide the circumference into as many equal parts as the polygon has sides; for, the arcs being equal, the chords AB, BC, CD, &c. (fig. 158), will be equal; the triangles ABO, BOC, Fig. 158. COD, &c., will also be equal, for the sides of the one will be respectively equal to those of the other; consequently all the angles ABC, BCD, CDE, &c., will be equal; therefore the figure ABCDE &c. will be a regular polygon.

PROBLEM.

269. To inscribe a square in a given circle.

Solution. Draw the diameters AC, BD (fig. 157), cutting Fig. 157. each other at right angles; join the extremities A, B, C, D, and

the figure ABCD will be the inscribed square.

For, the angles AOB, BOC, &c., being equal, the chords AB, BC, &c. are equal.

270. Scholium. The triangle BOC being right-angled and isosceles, BC: BO:: 2 : 1 (188); therefore, the side of an inscribed square is to radius, as the square root of two is to unity.

Geom.

11

271. To inscribe a regular hexagon and an equilateral triangle in a given circle.

Solution. Let us suppose the problem resolved, and that AB Fig. 153. (fig. 158) is a side of the inscribed hexagon; if we draw the radii AO, OB, the triangle AOB will be equilateral.

For the angle AOB is the sixth part of four right angles; thus, if we consider the right angle as unity, we shall have

=

AOB==.

The two other angles ABO, BAO, of the same triangle, taken together, 2; and, as they are equal to each other, each of them; hence the triangle ABO is equilateral; therefore the side of the inscribed hexagon is equal to radius.

Fig. 159.

It follows from this, that in order to inscribe a regular hexagon in a given circle, the radius is to be applied six times round on the circumference, which will bring it to the point, from which the operation commenced.

The hexagon ABCDEF being inscribed, if the vertices of the alternate angles A, C, E, be joined, an equilateral triangle ACE I will be formed.

272. Scholium. The figure ABCO is a parallelogram, and a rhombus, since AB = BC = CO=AO; therefore, the sum of the squares of the diagonals being equal to the sum of the squares of the sides (195),

-2

--2

AC + BO = 4AB = 4B0;

AC+

therefore the side of an inscribed equilateral triangle is to radius as the square root of three is to unity.

PROBLEM.

273. To inscribe in a given circle a regular decagon, also a pentagon and a regular polygon of fifteen sides.

Solution. Divide the radius AO (fig. 159) in extreme and mean ratio at the point M (240), take the chord AB equal to the

greater segment OM, and AB will be the side of a regular decagon, which is to be applied ten times round on the circumference. For, by joining MB, we have, by construction,

therefore the triangles ABO, AMB, having an angle A common and the sides about this angle proportional, are similar (208). The triangle OAB is isosceles; consequently the triangle AMB is also isosceles, and AB = BM. Besides, AB = OM; hence also BM= OM; therefore the triangle BMO is isosceles.

The angle AMB, the exterior angle of the isosceles triangle BMO, is double of the interior angle O (78). Now the angle AMB = MAB;

consequently the triangle OAB is such that each of the angles at the base OAB, OBA, is double of the angle at the vertex O, and the three angles of the triangle are equal to five times the angle O, and thus the angle O is a fifth part of two right angles, or the tenth part of four right angles; therefore the arc AB is the tenth part of the circumference, and the chord AB is the side of a regular decagon.

274. Corollary 1. If the alternate vertices A, C, E, &c. of the decagon be joined, a regular pentagon ACEGI will be formed. 275. Corollary 11. AB being always the side of a decagon, let AL be the side of a hexagon; then the arc BL will be, with respect to the circumference, ', or '; therefore the chord BL will be the side of a regular polygon of 15 sides. It is manifest, at the same time, that the arc CL is a third of CB.

276. Scholium. A regular polygon being formed, if the arcs subtended by the sides be bisected and chords to these half arcs be drawn, a regular polygon will be formed of double the number of sides. Thus by means of the square we may inscribe successively regular polygons of 8, 16, 32, &c., sides. Likewise by means of the hexagon we may inscribe regular polygons of 12, 24, 48, &c., sides; with the decagon, polygons of 20, 40, 80, &c., sides; with the regular polygon of fifteen sides, polygons of 30, 60, 120, &c., sides.*

* It was supposed, for a long time, that these were the only polygons which could be inscribed by the processes of elementary geom

Fig. 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 OT = ON; 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

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, Lipsiæ, 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.

280. 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 HION. 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 a b 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