SOLUTION. To bisect the perigon at the point 0, draw any line through 0. This divides the perigon into two straight angles, and all straight angles are equal. 419. COROLLARY. By drawing a second line through O at right angles to the first, we cut the perigon into four equal parts; and as we can bisect any angle, so we can cut the perigon into 8, 16, 32, 64, etc., equal parts. SOLUTION. To trisect the perigon at the point O, to draw any line BO; on BO produced take a sect OC; on OC construct, by 132, an equilateral triangle CDO. X DOB is one-third of a perigon. For X DOC is one-third of a st. X, (174. The three angles of a triangle are equal to a straight angle.) 421, COROLLARY. Since we can bisect any angle, so we may cut the perigon into 6, 12, 24, 48, etc., equal parts. 422. REMARK. To trisect any given angle is a problem beyond the power of strict Elementary Geometry, which allows the use of only the compasses and an unmarked ruler. There is an easy solution of it, which oversteps these limits only by using two marks on the straight-edge. The trisection of the angle, the duplication of the cube, and the quadrature of the circle, are the three famous problems of antiquity. PROBLEM III. 423. To cut a perigon into five equal parts. SOLUTION. By 314, describe an isosceles triangle ABC having ХА * C = 2 * B. Then x A is two-fifths of a st. . (174. The three angles of any triangle are equal to a straight angle.) Therefore, to get a fifth of a perigon at a point O, construct, by 164, X GOH 424. COROLLARY. Since we can bisect any angle, we may cut a perigon into 10, 20, 40, 80, etc., equal parts. PROBLEM IV. 425. To cut a perigon into fifteen equal parts. SOLUTION. At the perigon point 0, by 420, construct the X AOC = one-third of a perigon. By 423, make X AOB = one-fifth of a perigon. Then of such parts, as a perigon contains fifteen, X AOC contains five, and % AOB contains three, therefore x BOC contains two. So bisecting 4 BOC gives one-fifteenth of a perigon. 426. COROLLARY. Hence a perigon may be divided into 30, 60, 120, etc., equal parts. II. Regular Polygons and Circles. PROBLEM V. 427. To inscribe in a circle a regular polygon having a given number of sides. с This problem can be solved if a perigon can be divided into the given number of equal parts. For let the perigon at O, the center of the circle, be divided into a number of equal parts, and extend their arms to meet the circle in A, B, C, D, etc. Draw the chords, AB, BC, CD, etc. Then shall ABCD, etc., be a regular polygon. For if the figure be turned about its center 0, until OA coincides with the trace of OB, therefore, because the angles are all equal, OB will coincide with the trace of OC, and OC with the trace of OD, etc.; then AB will coincide with the trace of BC, and BC with the trace of CD, etc.; AB BC= CD = etc., :. the polygon is equilateral. Moreover, since then ABC will coincide with the trace of BDC, * ABC = X BCD = etc., .. the polygon is equiangular. Therefore ABC.D, etc., is a regular polygon, and it is inscribed in the given circle. .. .. 428. REMARK. From the time of Euclid, about 300 B.C., no advance was made in the inscription of regular polygons until Gauss, in 1796, found that a regular polygon of 17 sides was inscriptible, and in his abstruse Arithmetic, published in 1801, gave the following: In order that the geometric division of the circle into n parts may be possible, n must be 2, or a higher power of 2, or else a prime number of the form 2 + I, or a product of two or more different prime numbers of that form, or else the product of a power of 2 by one or more different prime numbers of that form. In other words, it is necessary that n should contain no odd divisor not of the form 2m + 1, nor contain the same divisor of that form more than once. Below 300, the following 38 are the only possible values of n: 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96, 102, 120, 128, 136, 160, 170, 192, 204, 240, 255, 256, 257, 272. EXERCISES. 81. The square inscribed in a circle is double the square on the radius, and half the square on the diameter. 82. Prove that each diagonal is parallel to a side of the regular pentagon. 83. An inscribed equilateral triangle is equivalent to half a regular hexagon inscribed in the same circle. 84. An equilateral triangle described on a given sect is equivalent to one-sixth of a regular hexagon described on the same sect. 85. If a triangle is equilateral, show that the radius of the circumscribed circle is double that of the inscribed ; and the radius of an escribed, triple. 86. The end points of a sect slide on two lines at right angles : find the locus of its mid-point. |