Scholium 2. To inscribe a regular polygon of a certain number of sides in a given circle, we have only 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. will also be equal; hence likewise the triangles AOB, BOC, COD, must be equal, because the sides are equal each to each; hence all the angles ABC, BCD, CDE, &c. will be equal; hence the figure ABCDEH, will be a regular polygon. B Η E PROPOSITION III. PROBLEM. To inscribe a square in a given circle. Draw two diameters AC, BD, cutting each other at right angles; join their extremities A, B, C, D: the figure ABCD will be a square. For the angles AOB, BOC, &c. being equal, the A chords AB, BC, &c. are also equal: and the angles ABC, BCD, &c. being in semicircles, are right angles. B PROPOSITION IV. PROBLEM. D C C D Scholium. Since the triangle BCO is right angled and isosceles, we have BC: BO :: √2 : 1 (Book IV. Prop. XI. Cor. 4.); hence the side of the inscribed square is to the radius, as the square root of 2, is to unity. In a given circle, to inscribe a regular hexagon and an equilateral triangle. 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== A and the two other angles ABO, BAO, of the same triangle, are together equal to 2-3 =; 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. E F 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 AC2+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 BO2 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 MBOM; 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 ACEGI 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. K* PROPOSITION VI. PROBLEM. A regular inscribed polygon being given, to circumscribe a sim ilar polygon about the same circle. K Q L 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 arc NBT. Draw OH. The line OH will pass through the point B. For, the right angled triangles OTH, OHN, having the common hypothenuse OH, and the side OT-ON, must be equal (Book I. Prop. XVII.), and consequently the angle TOH= 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 HIK= BCD; 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. G i F h E b 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. 12 k D For, having drawn Pd, Pa, let the quadrilateral PdgB, be applied to the quadrilateral PBfa, so that PB shall fall on PB. с 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 A P H a d B 19 |