PROBLEM I. 462. To divide a given circle into 2, 4, 8, 16, etc., equal parts. Any diameter, as AB, will divide the circle into two equal parts at the points A and B. Drawing through the centre O another diameter perpendicular to this, the circle will be divided into four equal A parts by four radii, at right angles to each other. By joining the ends of these radii a square will be described in the circle. N Bisecting each of the right angles at the centre by radii like ON, the circle will be divided into eight equal parts. By joining the points of division, a regular octagon will be inscribed in the circle. The process of bisection may be continued indefinitely, so as to divide the circle into 2m equal parts, where m may be any positive integer. PROBLEM II. 463. To divide the circle into 3, 6, 12, 24, etc., equal parts. Analysis. 1. Suppose the division into six parts effected, and the points of division to be A, B, C, D, E, F. 2. Draw the radii OA, OB, etc., and join AB, BC, CD, etc., forming a, regular hexagon. D C E B 3. Because each of the three equal angles AOB, BOC, COD, is one third the straight angle AOD, they are each angles of 60°. And because the sides OA, OB, are equal, the angles OAB and OBA are also equal. 4. But the sum of the three angles is 180°. Therefore the three angles are each equal to 60°; that is, the triangle OAB is equiangular and therefore equilateral, and the other five triangles, being identically equal to it, are also equilateral. 5. Therefore each of the sides, AB, BC, CD, DE, EF, FA, is equal to the radius OA of the circle. Hence Construction. 1. Starting from any point A on the circle, cut off the distances AB, BC, CD, etc., each equal to the radius. 2. Six equal measures will extend to the point A, and the circle will be divided into six equal parts at the points A, B, C, etc. 3. The alternate points A, C, E, or B, D, F, divide the circle into three equal parts. 4. By bisecting the arcs AB, BC, etc., the circle will be divided into 12 parts; by bisecting these arcs, the number of parts will be 24, etc. Corollary. The perimeter AB+BC+ CD + DE + EF FA of the hexagon ABCDEF is six times the radius of the circle, and therefore three times its diameter. Because each of the six sides is a straight line, it is less than the corresponding arc; that is, side AB < are AB, etc. Therefore The sum of the six arcs, or the whole circumference of the circle, is more than three times the diameter. PROBLEM III. 464. To divide a circle into 5, 10, 20, etc., equal parts. Analysis. Let O be the centre of the circle, and the arc AB one tenth part of the circumference, or 36°. Join OA, OB, and AB. Then 1. Because the sum of the three angles of the isosceles triangle AOB is 180°, the sum of the angles A and B is 180° 36° 144°, and each of these angles is 72°. Therefore each of the angles OAB and OBA is double the angle at 0. P 2. If we bisect the angle OBA by the line BP, meeting OA in P, the angles OBP and PBA will each be 36°, and we shall have angle PAB - angle PBA = Angle APB 180° (874) 3. Therefore the triangle BAP is isosceles and equiangular, and therefore similar, to the triangle OAB. Also, because angle POB = angle PBO, the triangle OPB is isosceles. Hence PO PB = AB. 4. Because of the similarity of the triangles BAP and OAB, or ᎪᏢ : ᎪᏴ :: AB : A0, AP PO: PO: AO. 5. Therefore the radius OA is divided in extreme and mean ratio at the point P (§ 440). Hence we conclude: 465. If the radius of a circle be divided in extreme and mean ratio, the greater segment will be the chord of one tenth of the circle. Construction. Divide the radius OA of the circle in extreme and mean ratio at the point P (§ 441). Around A as a centre, with a radius equal to the greater segment, OP, describe a circle, and let B and C be the points. in which it intersects the first circle. The arc AB will be one tenth of the circle and BC will be one fifth of the circle. By measuring BC off five times around the circle, the latter will be divided into five equal parts. By successive bisection the circle will be divided into 10, 20, 40, etc., equal parts. Q. E. F. PROBLEM IV. 466. To divide a circle into fifteen equal parts. Construction. 1. From any point A cut off AB equal to one third the circle (§ 463). 2. From the same point A cut off AD, equal to one fifth the circle (§ 465). 3. Arc BD will then be ( – ) of the circle; that is, of it. 4. Therefore, if we bisect the arc BD, we shall have an arc equal to of the circle. By measuring this arc off 15 times, the circle will be divided into 15 equal parts. Q.E.F. Scholium. The foregoing divisions of the circle are all that were known to the ancient geometers. But, about the beginning of the present century, Gauss, the great mathematician of Germany, showed that whenever any power of 2, increased by 1, made a prime number, the circle could be divided into that number of parts by the rule and compass. Thus: 2+1= 3, a prime number. 2+1 17 28 +1=257 Therefore, besides the old solutions, the circle can be divided into 17 or into 257 equal parts. The division into 17 parts by construction is, however, too complicated for the present work, and that into 257 parts is so long that no one has ever attempted to really execute the construction. CHAPTER III. AREAS AND PERIMETERS OF REGULAR POLYGONS AND THE CIRCLE. 467. Def. The apothegm of a regular polygon is the perpendicular from its centre upon any one of its THEOREM VII. 468. The area of a regular polygon is equal to half the rectangle contained by its perimeter and its apothegm. P B C 2. In the same way it may be shown that the area of each of the other triangles into which the polygon is divided is equal to one half of the side into the apothegm. But the apothegms are all equal. Therefore Area ABCDEF={OP.AB÷†OP.BC+†OP. CD + etc. = OP × perimeter. Q.E.D. = 469. Corollary 1. Because the perimeter of each circumscribed regular polygon is less the greater the number of its sides (§ 458), it follows that the area of the circumscribed regular polygon is less the greater the number of its sides. Cor. 2. It is easily shown that the area of the circumscribed square is equal to the square upon the diameter of the circle. Therefore: 470. The area of any circumscribed regular polygon of more than four sides is less than the square upon the diameter of the circle. 471. Scholium. If a regular polygon be inscribed in a circle, and another regular polygon of the same number of sides be circumscribed about it, the area of the outer polygon will be greater than that of the inner one by the surface contained between the perimeters of the two polygons. This surface will be called the included area. When the two polygons are so placed that their respective |