A arc HA Proposition. 14. Problem. 453. To inscribe a regular pentadecagon in a given circle. Given, the ABC. Required, to inscribe in it a regular polygon of 15 sides. AE Cons. Lay off HB equal to a side of a regular inscribed hexagon. (445) Lay off HA equal to a side of a regular inscribed decagon. H B (449) н Join AB. Then AB is a side of the required inscribed pentadecagon. Proof. Arc AB = arc HB = of Oce - it of Oce= 1 of Oce. .. the chord AB is a side of the regular inscribed pentadecagon. (400) ... the figure ABCD ..., formed by applying AB 15 times to the Oce, is the regular inscribed pentadecagon required. Q.E.F. 454. COR. 1. If tangents be drawn at the points at which the circumference is divided, a regular circumscribed pentadecagon is obtained. 455. COR. 2. To inscribe, and circumscribe, regular polygons of 30 sides, bisect the arcs AB, BC, etc., and proceed as before. By repeating this process, regular inscribed and circumscribed polygons of 60, 120, etc., sides may be drawn. 456. SCH. These are the only polygons that could be constructed by the ancient geometers, by the use of the rule and compass. About the beginning of the present century, Gauss, the great German mathematician, proved that whenever 21 +1 is a prime number, and n an integer, a regular polygon of that number of sides could be inscribed in the circle, by the rule and compass. Therefore, it is possible to inscribe regular polygons of 17 sides, and of 257 sides; but the construction of the latter polygon is so lengthy, it is not likely that it has ever been attempted. F B E Proposition 15. Problem. 457. Given the radius and the side of a regular inscribed polygon, to compute the side of a similar circumscribed polygon. Given, AB, a side of the regular in D scribed polygon, and OF = R, the radius of the O ABH. Required, to compute CD, a side of the similar circumscribed polygon. Cons. Join OC, OD. H They will cut the Oce in A and B. (402) The similar As OCF, OAE give If we know the sides of the regular polygons of n, 2n, 4n, .... sides, inscribed in the circle of radius 1, we have, by this formula, the sides, and therefore the perimeters, of the regular circumscribed similar polygons. Proposition 16. Problem. A B E 459. Given the radius and the side of a regular inscribed polygon, to compute the side of the regular inscribed polygon of double the number of sides. Given, AB, a side of the regular inscribed polygon, and OC = R, the radius of the O ABD. Required, to compute AC, a side of the regular inscribed polygon of double the number of sides. Cons. C is the mid. pt. of the arc AB. (403) Produce Co to D, and join 0A. CD is I to AB at its mid. pt. (204) AC is a mean proportional between CD and CE. (325) OE) .. AC = CD X CE =R (PR – 20E). But OF = 1 V 4R– AB. (457) Q.E.F. .:. AC = N R (2R – V4R - ABR). 460. Sch. When R= When R= 1, this gives AC = N 2 -V4 – AB". By repeated applications of this formula, we may compute successively the sides, and therefore the perimeters, of the regular inscribed polygons of 2n, 4n, 8n, 16n, .... sides, n being the number of sides of the first polygon. EXERCISES. 1. If the radius of a circle is 4, find its circumference and area. 2. If the circumference of a circle is 28, find its diameter and area. Proposition 17. Problem. 461. To compute the ratio of the circumference of a circle to its diameter. Given, the Oce C, and the radius R. Required, to find the number . Cons. C= 20R. (436) Let R= 1, then a = {C. That is, the number n = semi o ce of radius 1. Therefore, the semi-perimeter of each polygon inscribed in this Oce is an approximate value of : and as the number of sides of the polygons increases indefinitely, the lengths of the perimeters approach to that of the Oce as the limit. (430) Hence, by the process of (460), we may obtain a succession of nearer and nearer approximations to the length of the semice. It is convenient to begin the computation with the inscribed hexagon. ... making AB = 1, we have, from (460), the following: The last two results show that the first four decimals do not change as the number of sides is increased. Hence the approximate value of n is 3.1415, correct to the fourth decimal place. Q.E.F. In practice we generally take a = 3.14159. 462. Sch. The above is called the method of perimeters. For the method of isoperimeters see Rouché et Comberousse, Edition of 1883, p. 194. which agree Note.- The number 7 is of such fundamental importance in geometry that mathematicians have sought for its value in a great variety of ways, all of the conclusion that cannot be expressed exactly in decimals, but only approximately. Archimedes (born 287 B.c.) was the first to assign an approximate value of . He proved that it is included between the numbers 37 and 311, or, in decimals, between 3.1428 and 3.1408; he therefore assigned its value correctly within a unit of the third decimal place. The number 31, or is often used in rough computations. Adrian Metius, a Dutch geometer of the 16th century, has given us the much more accurate value of 113, which is correct to within a half-millionth, and which is remarkable for the manner in which it is formed by the first three odd numbers 1, 3, 5, each written twice. More recently, the value has been found to a great number of decimals hy the aid of series. Dase and Clausen, German computers, carried the calculation to 200 decimal places, independently of each other, and their results agreed to the last figure. The first 20 figures of their result are as follows : 3.14159 26535 89793 23846. (437) This result is far beyond all the wants of mathematics. Ten decimals are sufficient to give the circumference of the earth to the fraction of an inch, and thirty decimals would give the circumference of the wholė visible universe to a quantity imperceptible with the most powerful microscope.* EXERCISES. 1. The area of the regular inscribed hexagon is equal to twice the area of the inscribed equilateral triangle. 2. The square of a side of the inscribed equilateral triangle is three times the square of a side of the regular inscribed hexagon. 3. The area of a regular inscribed hexagon is half the area of the circumscribed equilateral triangle. 4. If the diameter of a circle be produced to C until the produced part is equal to the radius, the two tangents from C and their chord of contact form an equilateral triangle. * Newcomb's Geom., p. 235. |