Yes, but no need to know what a sine is. E.g. for the 6-gon, we only need to know that the base of an equilateral triangle is equal to the other two sides. So half of it gives `sin pi/6 = 1/2', but this statement adds nothing to the understanding of a beginner. The essentials are Pythagoras's theorem and the angle bisector theorem. Suppose OAB is a triangle, right-angled at B and angle BOA = 30 degrees. If OB = 1, then AB = 1/2 and the inscribed 6-gon has perimeter 6. If OA = 1, then AB = 2 / root 3, and the circumscribed 6-gon has perimeter 4 root 3 To get the 12-gon, we need to bisect the angle. For the circumscribed 12-gon it's `tan pi/12' we want, and for the inscribed one, `sin pi/12', but the names and notation don't help. Bisect the angle BOA and let the bisector meet AB in C. Then AC : CB = OA : OB = root 3 : 2 and the (`tangent') length we want is 2 - root 3, twenty-four of which give approximately 6.43, while the inscribed (`sine') length is (root(2 - root 3)) / 2, twenty-four of which give approximately 6.21. I don't recall how many stages Archimedes went through, but he used (better and better) rational approximations to root 3, and gave 3 1/7 > pi > 3 10/71 whose mean is 3.14185. No need to know any trigonometry or analysis. R. On Wed, 12 Jul 2006, Jason Holt wrote:
On Wed, 12 Jul 2006, Richard Guy wrote:
Surely best to follow Archimedes and calculate the perimeters of inscribed and circumscribed regular 6-, 12-, 24-, ...-gons? R.
Working that out computationally seems to require sines, so that the problem of calculating pi reduces to the problem of calculating sines, yes?
-J
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