PROBLEM. 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. 158. (fig. 158) is a side of the inscribed hexagon; if we draw the radii AO, OB, the triangle AOB will be equilateral. Fig. 159. 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 == 3. 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. It follows from this, that in order to inscribe a regular hexagon in a given cicle, 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), therefore the side of an inscribed equilateral triangle is to radius as the square root of three is to unity. PROBLEM. 1 273. 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 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 2′′ + 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× 1OT, the triangle OHI has for its measure HI × ON. But ON OT; consequently the two triangles united have for their measure (GH + HI) × 1OT. 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 b o, are similar, as also the right-angled triangles ADO, ado; hence AB: ab:: 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; between 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 |