1 By Case 2, the sine of the given arc is 991/2; and the radius is √2; 154 But, by Case 1, the area of the inscribed polygon is 25 1078 1100 1078 Then, by division and cancellation, we have 175 175 175 175 1100 1078 49 1100 49 Case 1 is of the area of the inscribed polygon of Case 2; but 50 49 the area of the inscribed polygon for Case 1 is of the area of the circumscribed polygon for Case 1; therefore, by (Axiom 1), the area of the inscribed polygon of Case 2 is the same as the area of the circumscribed polygon of Case 1. 1 50 But, by Case 1, the th of the area of the circumscribed polygon was deducted; therefore, by Case 2, the same again. 1 -th has been added on 50 1 and there are 44 sines contained in Again, by Case 1, the sine is 5' the given polygon; and, by Case 2, each of these sines is divided into 14 parts, each equal to tangent No. 2; therefore there are 44 X 14= 616 tangents contained in the polygon for Case 1, each equal to 1 5' of the tangents for Case 1 has become the sum of the sines for Case 2. Again, by Case 1, the sine is and there are 22 sides, each of two complete the polygon; thus + the sines for Case 1 has become the sum of the tangents for Case 2. 7 Again, by Case 1, the cosine is; and, by Case 2, the secant is 99 62 ; therefore the sum of 1 1 98 the cosine; but is of therefore the diameter of the circle has 70 98 70' Consequently the circumference of the circle has been increased its 1 98 Consequently there was th added to the area of the given circle. 19601V2, and there are 616034 sides contained in the given polygon, each of which has two sines. Consequently the sum of the sines for Case 3 is the same as the sum of the sines for Case 2. By Case 3, the sine is 1 1 Then 196012 × 1232062 = 1232064 √2 =sum of the sines for Case 3. 1 By Case 4, the sine is 7683984012, and there are 24149664031 sides contained in the given polygon, each of which has two sines. the sines for Case 4, = 48299328062 V2 sum of 1 Then 768398401√2 × 48299328063 Therefore the sum of the sines for Case 3 is exactly the same as the sum of the sines for Case 4, and the length of the perimeter of the inscribed polygon is not changed. 11808722053187136011 2, and the number of sides contained in the given polygon is 37113126452873856031, each of which has two sines. Then, multiplying the sine by double the number of sides, we have 74226252905747712062 1117949582616008653052412173072062 17785561541618319480379284571601 Thus we perceive that the sum of the sines, that is the perimeter of the inscribed polygon, from Case 1 to Case 5, inclusive, is constantly changing the number of its sides at a rate much more rapid than by doubling, while the length of the perimeter, as well as the radius, and the area of the inscribed polygon constantly remains the same; and, if the number of sides were increased ad infinitum, they would still continue to be the same, although at every successive step the usual number of sides, according to hypothesis, has been deducted. 19602 3842188021/2. 19602 19601V 3842188021 2. Then, by subtraction, we have 1 3842188021/2. Now, by Case 3, there are 616034 sides, each of two sines. Then 62 1 384218802V 2 X 1232062 = 19602; therefore 62 19602 123206 862444 3842188021/2= 2689531614 2 has been deducted from the circumfer ence of the given polygon, and still the perimeter of the inscribed polygon remains the same. the assumed diameter; consequently, by hypothesis, ducted from the circumference. By Case 2, the tangent is ; and, by Case 3, the tangent is 13860 Then, dividing tangent No. 2 by tangent No. 3, we have by cancel gents for Case 2 has been divided into 198 equal parts, each of which is equal to tangent No. 3, or 1 13860' Now, by Case 2, there are 3114 sides contained in the given poly. gon, each of which has two tangents; then 622 × 1981232124 = the number of tangents in the polygon for Case 3. But, by Case 3, the given polygon has only 1232062 tangents; therefore been deducted from the given polygon; and, in Case 4, |