10 The construction of ko by dividing the radius in extreme and mean ratio. The first expression for k above shews how the side of a decagon may B The figure shews the construction; bisect the radius OD, make OBOD and join AB; thereupon make AC = AB, then OC will be the side of the decagon, for we have OC AC-AO A line is said to be divided into extreme and mean ratio, when it is divided into two parts, so that the greatest of these is a mean proportional between the least and the whole line; the proportion we got for the determination of k therefore shews that the side of the decagon is the greatest part of the radius thus divided. When the circumference is divided into 10 equal parts, we can, by doubling or continually bisecting the arcs, divide it into 5, 20, 40, 80 5.2" equal parts. ... • 102. If an arc of 36° be subtracted from an arc of 60°, we get an arc of 24°, and thus, therefore, a circle can be divided into 15 and, by continued bisection of the arcs, into 30, 60.... 15.2" equal parts. The expression for k1s will be found below (Ex. 183). 103. When a chord is calculated, its distance from the centre p is found by 95, d; thus if the chord is k, radius r, we get 104. When the radius is given, and the chord of a certain arc has been calculated, we can from that again calculate the chord of half and the chord of double the arc. k is a mean proportional between DE and the diameter CE, therefore = Example. From ker we hereby find k12 r√2-V3; 105. When the side and the least radius of an inscribed regular polygon are calculated, we can calculate the side and the greatest radius of the circumscribed regular polygon with the same number of sides. Let AB be the side of the inscribed polygon, from the centre 0 draw a perpendicular OF to AB; if thereupon the tangent CD be drawn, it becomes the side, and OC the greatest radius of the circumscribed polygon, for we have ABCD, as they both are perpendicular to OF, and therefore the triangles AOB and COD are similar. From this it follows C = A E B F 177. How great are k, and pg, te and R., pg and t ̧? 178. What is the product k12012? 179. What is t, and what is the least radius of an equilateral triangle with the side t? 180. Prove that the line joining BC on the fig. to 101 is the side of the inscribed pentagon. 181. Find the mean proportional between the chord of 30° and the chord of 150°. 182. Find k64, k96, p64, and P96. 183. a, b, and c are sides of an inscribed triangle; find c, when a, b, and r are given. Draw the diameter from the point of intersection of a and b, and lines from the extremities of the diameter to the extremities of a and b. These lines become √4r2 — a2 and V4r2-62, and we then have (98) the upper sign when c is the chord of the sum of the arcs, and the lower sign when c is the chord of the difference of the arcs, of which a and b are the chords (a> b). Hereby we get k1, when a=r, b——(V5—1) (102), r VI. THE LENGTH OF THE CIRCUMFERENCE OF THE CIRCLE. 106. A curved line cannot be measured by a straight line, for the unit must be of the same kind as that which is to be measured; we must therefore more closely explain what is meant when we speak of the length of the circumference. It is evident that an inscribed regular 2n sided figure has greater perimeter than an inscribed n sided figure, whilst the opposite is the case with the circumscribed figures. If we therefore continue to double the number of sides of an inscribed and a circumscribed regular figure, the perimeter will in the first case continually increase and in the second continually decrease; as now the latter always is greater than the former, they must approach nearer and nearer to each other, and we can prove that the difference between them can be made less than any assigned quantity. The perimeter of the inscribed n sided figure with the side k is nk and of the circumscribed figure nt; the difference is therefore, as k pt therefore the difference but here the first factor nt is the perimeter of the circumscribed figure and is accordingly less than the perimeter we had before doubling the number of sides, whilst the second factor can be made as small as we please, by doubling the number of sides till k is sufficiently small. The lengths of the perimeters of the inscribed and circumscribed figures therefore approach nearer and nearer to each other, so that they have a common limit, a value to which they both approach. It is this limit that is meant, when we speak of the length of the circumference of a circle. The limit will be the same, whichever regular polygon we take. Let the circumscribed and inscribed polygons with several sides have the perimeters P and p, whilst G is the limit. By continuing to double the number of sides, we can make P-G and G p less than any ever so small quantity. Let P1, P1, and G1 be the perimeters and the limit, when we take another regular polygon. As p, lies within P and p within P1 we can (Ex. 56 expanded) put 1 where a and ẞ are certain positive quantities; therefore As here P1-P, and P-p by continued doubling become less than any ever so small given quantity, the same must be the case with the positive quantities a and B. P1 must therefore have the same limit as p, or G, must be equal to G. 107. The lengths of the circumferences of two circles are to each other as the radii. For the perimeters of two regular figures with the same number of sides are to each other as the greatest radii, and this proposition continues to hold, how ever often the number of sides of the polygon is doubled; therefore it must also hold good for circles. If we denote the circumferences by P and p, the radii by R and r, and the diameters by D and d, we therefore have |