Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
238 SECTION 43. INDIANA BILL 246the diameter produces the side of a square which equals nine whenthe arc of ninety degrees equals eight. By taking the quadrant of thecircle’s circumference for the linear unit, we fulfill the requirementsof both quadrature and rectification of the circle’s circumference.Furthermore, it has revealed the ratio of the chord and arc of ninetydegrees, which is as seven to eight, and also the ratio of the diagonaland one side of a square which is as ten to seven, disclosing the fourthimportant fact, that the ratio of the diameter and circumference isas five-fourths to four; and because of these facts and the further factthat the rule in present use fails to work both ways mathematically, itshould be discarded as wholly wanting and misleading in its practicalapplications.Section 3 In further proof of the value of the author’s proposed contributionto education and offered as a gift to the State of Indiana, isthe fact of his solutions of the trisection of the angle, duplication ofthe cube and quadrature of the circle having been already acceptedas contributions to science by the American Mathematical Monthly,the leading exponent of mathematical thought in this country. Andbe it remembered that these noted problems had been long sincegiven up by scientific bodies as insolvable mysteries and above man’sability to comprehend.Legislative HistoryHouse: Passed, 67-0, 5 Feb. 1897Senate: Postponed indefinitely, 12 Feb. 1897.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
Section 44Estimating πInscribed PolygonsWhile there are many ways of calculating π using calculus (e.g., as the sumof an infinite series), here we outline a procedure based purely on geometriccalculations. It is iterative, in the sense that we need to calculate a sequenceof estimates based on previous estimates, slowly converging to the desiredvalue. We will define π n as the ratio of the perimeter to the diameter ofa succession of inscribed polygons with 2 n sides, starting with n = 2. Theidea is that as more sides are added the inscribed polygon looks more andmore like a circle (rigorously, the limit of the perimeters is the circumferenceof the circle). In general, the perimeter of the n th polygon isP n = 2 n H nwhere H n is the length of the outer edge. Since the diameter of the unitcircle is always 2, we have an estimate for π ofπ n = 1 2 P n = 2 n−1 H nFor n = 2, we inscribe a square in a circle of radius 1, and divide the squareinto four right triangles with hypotenuse H 2 .ThenH 2 = √ 2π 2 = 2 2−1 H 2 = 2 √ 2 ≈ 2.82843For n = 3 we divide each central angle in half. Define the distance a asshown in figure 44.2.239
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Section 44Estimating πInscribed PolygonsWhile there are many ways of calculating π using calculus (e.g., as the sumof an infinite series), here we outline a procedure based purely on geometriccalculations. It is iterative, in the sense that we need to calculate a sequenceof estimates based on previous estimates, slowly converging to the desiredvalue. We will define π n as the ratio of the perimeter to the diameter ofa succession of inscribed polygons with 2 n sides, starting with n = 2. Theidea is that as more sides are added the inscribed polygon looks more andmore like a circle (rigorously, the limit of the perimeters is the circumferenceof the circle). In general, the perimeter of the n th polygon isP n = 2 n H nwhere H n is the length of the outer edge. Since the diameter of the unitcircle is always 2, we have an estimate for π ofπ n = 1 2 P n = 2 n−1 H nFor n = 2, we inscribe a square in a circle of radius 1, and divide the squareinto four right triangles with hypotenuse H 2 .ThenH 2 = √ 2π 2 = 2 2−1 H 2 = 2 √ 2 ≈ 2.82843For n = 3 we divide each central angle in half. Define the distance a asshown in figure 44.2.239