Suppose the problem solved, and that AB is a side of the inscribed hexagon; the radii AO, OB being drawn, the triangle AOB will be equilateral. For, the angle AOB is the sixth part of four right angles; therefore, taking the right angle for unity, we shall have AOB=4= : and the two other angles ABO, BAO, of the same triangle, are together equal to 2=; and being mutually equal, each of them must be equal to ; hence the triangle ABO is equilateral; therefore the side of the inscribed hexagon is equal to the radius. Hence to inscribe a regular hexagon in a given circle, the radius must be applied six times to the circumference; which will bring us round to the point of beginning. And the hexagon ABCDEF being inscribed, the equilateral triangle ACE may be formed by joining the vertices of the alternate angles. Scholium. The figure ABCO is a parallelogram and even a rhombus, since AB-BC-CO-AO; hence the sum of the squares of the diagonals AC+BO2 is equivalent to the sum of the squares of the sides, that is, to 4AB2, or 4BO2 (Book IV. Prop XIV. Cor.): and taking away BO from both, there will remain AC2-3BO2; hence AC2: BO2: : 3: 1, or AC BO :: √3: 1; hence the side of the inscribed equilateral triangle is to the radius as the square root of three is to unity. PROPOSITION V. PROBLEM. In a given circle, to inscribe a regular decagon; then a pentagon, and also a regular polygon of fifteen sides. AM; since the triangles ABO, AMB, have a common angle A, included between proportional sides, they are similar (Book IV. Prop. XX.). Now the triangle OAB being isosceles, AMB must be isosceles also, and AB=BM; but AB=OM; hence also MB=OM; hence the triangle BMO is isosceles. Again, the angle AMB being exterior to the isosceles triangle BMO, is double of the interior angle O (Book I. Prop. XXV. Cor. 6.): but the angle AMB MAB; hence the triangle OAB is such, that each of the angles OAB or OBA, at its base, is double of O, the angle at its vertex; hence the three angles of the triangle are together equal to five times the angle O, which consequently is the fifth part of the two right angles, or the tenth part of four; hence the arc AB is the tenth part of the circumference, and the chord AB is the side of the regular decagon. 2d. By joining the alternate corners of the regular decagon, the pentagon AČEGI will be formed, also regular. 3d. AB being still the side of the decagon, let AL be the side of a hexagon; the arc BL will then, with reference to the whole circumference, be, or; hence the chord BL will be the side of the regular polygon of fifteen sides, or pentedecagon. It is evident also, that the arc CL is the third of CB. Scholium. Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the number of sides: thus it is plain, that the square will enable us to inscribe successively regular polygons of 8, 16, 32, &c. sides. And in like manner, by means of the hexagon, regular polygons of 12, 24, 48, &c. sides may be inscribed; by means of the decagon, polygons of 20, 40, 80, &c. sides; by means of the pentedecagon, polygons of 30, 60, 120, &c. sides. It is further evident, that any of the inscribed polygons will be less than the inscribed polygon of double the number of sides, since a part is less than the whole. H PROPOSITION VI. PROBLEM. A regular inscribed polygon being given, to circumscribe a similar polygon about the same circle. Since T is the middle point of the arc BTA, and N the middle point of the equal arc BNC, it follows, that BT=BN; or that the vertex B of the inscribed polygon, is at the middle point of the are NBT. Draw OH. The line OH will pass through the point B. For, the right angled triangles OTH, OHN, having the common hypothenuse ŎH, and the side OT-ON, must be equal (Book I. Prop. XVII.), and consequently the angle ТOĤ= HON, wherefore the line OH passes through the middle point B of the arc TN. For a like reason, the point I is in the prolongation of OC; and so with the rest. But, since GH is parallel to AB, and HI to BC, the angle GHI ABC (Book I. Prop. XXIV.); in like manner HIKBCD; and so with all the rest: hence the angles of the circumscribed polygon are equal to those of the inscribed one. And further, by reason of these same parallels, we have GH: AB:: OH: OB, and HI: BC :: OH: OB; therefore GH: AB:: HI: BC. But AB=BC, therefore GH-HI. For the same reason, HI=IK, &c.; hence the sides of the circumscribed polygon are all equal; hence this polygon is regular and similar to the inscribed one. Cor. 1. Reciprocally, if the circumscribed polygon GHIK &c. were given, and the inscribed one ABC &c. were required to be deduced from it, it would only be necessary to draw from the angles G, H, I, &c. of the given polygon, straight lines OG, OH, &c. meeting the circumference in the points A, B, C, &c.; then to join those points by the chords AB, BC, &c.; this would form the inscribed polygon. An easier solution of this problem would be simply to join the points of contact T, N, P, &c. by the chords TN, NP, &c. which likewise would form an inscribed polygon similar to the circumscribed one. Cor. 2. Hence we may circumscribe about a circle any regular polygon, which can be inscribed within it, and conversely. Cor. 3. It is plain that NH+HT=HT+TG=HG, one of the equal sides of the polygon. PROPOSITION VII. PROBLEM. A circle and regular circumscribed polygon being given, it is required to circumscribe the circle by another regular polygon having double the number of sides. Let the circle whose centre is P, be circumscribed by the square CDEG: it is required to find a regular circumscribed octagon. Bisect the arcs AH, HB, BF, FA, and through the middle points c, d, a, b, draw tangents to the circle, and produce them till they meet the sides of the square: then will the figure ApHdB &c. be a regular octagon. G Hi iF h E th A C b B P 19 C n H D For, having drawn Pd, Pa, let the quadrilateral PdgB, be applied to the quadrilateral PBfa, so that PB shall fall on PB. Then, since the angle dPB is equal to the angle BPa, each being half a right angle, the line Pd will fall on its equal Pa, and the point d on the point a. But the angles Pdg, Paf, are right angles (Book III. Prop. IX.) ; hence the line dg will take the direction af. The angles PBg, PBf, are also right angles; hence Bg will take the direction Bf; therefore, the two quadrilaterals will coincide, and the point g will fall at f; hence, Bg=Bf, dg=af, and the angle dgB=Bfa. By applying in a similar manner, the quadrilaterals PBfa, PFha, it may be shown, that afah, fB Fh, and the angle Bfa-ahF. But since the two tangents fa, ƒB, are equal (Book III. Prob. XIV. Sch.), it follows that fh, which is twice fa, is equal to fg, which is twice fB. In a similar manner it may be shown that hf-hi, and the angle Fit=Fha, or that any two sides or any two angles of the octagon are equal: hence the octagon is a regular polygon (Def.). The construction which has been made in the case of the and the octagon, is equally applicable to other polygons. square Cor It is evident that the circumscribed square is greater than the circumscribed octagon by the four triangles, Cnp, kDg, hEf, Git; and if a regular polygon of sixteen sides be circumscribed about the circle, we may prove in a similar way, that the figure having the greatest number of sides will be the least; and the same may be shown, whatever be the number of sides of the polygons: hence, in general, any circumscribed regular polygon, will be greater than a circumscribed regular polygon having double the number of sides. PROPOSITION VIII. THEOREM. Two regular polygons, of the same number of sides, can always be formed, the one circumscribed about a circle, the other inscribed in it, which shall differ from each other by less than any assignable surface. Let Q be the side of a square less than the given surface. Bisect AC, a fourth part of the circumference, and then bisect the half of this fourth, and proceed in this manner, always bisecting one of the arcs formed by the last bisection, until an arc is found whose chord AB is less than Q. As this arc will be an exact part of the circumference, if we apply chords AB, a K b E BC, CD, &c. each equal to AB, the last will terminate at A, and there will be formed a regular polygon ABCDE &c. in the circle. Next, describe about the circle a similar polygon abcde &c. (Prop. VI.): the difference of these two polygons will be less than the square of Q. For, from the points a and b, draw the lines aO, b0, to the centre O: they will pass through the points A and B, as was |