PROBLEM III. 423. To cut a perigon into five equal parts. H G SOLUTION. BY 314, describe an isosceles triangle ABC having (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, * GOH = 4 A. 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 O, by 420, construct the AOC = one-third of a perigon. By 423, make 4 AOB = one-fifth of a perigon. Then of such parts, as a perigon contains fifteen, five, and AOC contains AOB contains three, therefore & BOC contains two. So bisecting 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. C 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 O, 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., Moreover, since then ABC will coincide with the trace of BDC, .. ABC BCD = etc., .. the polygon is equiangular. Therefore ABCD, 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 2m+ 1, 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 diame ter. 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. PROBLEM VI. 429. To circumscribe about a given circle a regular polygon having a given number of sides. 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 angles, and extend their arms to meet the circle in A, B, C, D, etc. Draw perpendiculars to these arms at A, B, C, D, etc. These will be tangents. Call their points of intersection K, L, M, etc. Then shall KLM, etc., be a regular polygon. For if the figure be turned about its center O until OA coincides with the trace of OB, then, because the angles are all equal, OB will coincide with the trace of OC, and OC with the trace of OD, etc. Therefore the tangents at A, B, C, etc., will coincide with the traces of the tangents at B, C, D, etc. also Hence the polygon will coincide with its trace; :. KL = LM = etc., K = XL = M = etc.; therefore the polygon is regular, and it is circumscribed about the given circle. 430. COROLLARY. Hence we can circumscribe about a circle regular polygons of 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, etc., sides. PROBLEM VII. 431. To circumscribe a circle about a given regular polygon. O. With center O and radius OA describe a circle. This shall be the required circle. PROOF. Join OC. Then ▲ OBCA OBA, (124. Triangles having two sides and the included angle in each equal are con gruent.) .. OCB = OAB = half one of the regular polygon, Similarly prove each .. BCD is bisected. of the polygon bisected, .. OA = OB = OC = OD = OE, (148. In a triangle, sides opposite equal angles are equal.) therefore a circle with radius OA passes through B, C, D, E, and is circumscribed about the given polygon. |