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        riddles >> medium >> Construction of a Regular Polygon
        
(Message started by: ronnodas on Oct 2^nd, 2008, 4:01am)

Title: Construction of a Regular Polygon
Post by ronnodas on Oct 2^nd, 2008, 4:01am

This is not really a puzzle. I have a geometric (compass and straight edge) construction that can closely approximate regular polygons of any number of sides. I want to know why this method works. All replies are welcome.

AB is the diameter of a circle with center O. CAB is an equilateral triangle. D divides AB in the ratio 2/(n-2) where n is the number of sides of the polygon. That is, AD/AB=2/n. CD extended cuts the circle at E. AE is the side of the polygon that is replicated along the circle.

The attached picture is for n=7. As you may see, the last point differs slightly from A. This construction works very well for all n that I have tested.

Title: Re: Construction of a Regular Polygon
Post by Eigenray on Oct 2^nd, 2008, 4:41am

If E were the correct point, and AB = n, then according to my calculations
AD = n sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n) [ cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n) ] / [ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 + sin(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n) ]
= 1.996 for n=7.
AD is exactly 2 for n=2,3,4,6; for n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 5 we have
AD = 2.001, 2.000, 1.996, 1.992, 1.986, 1.980, 1.974, 1.968, ...
But AD actually converges to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/sqrt3 = 1.814 as n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif:
AD = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²/(3n) + O(1/n²).

Title: Re: Construction of a Regular Polygon
Post by ronnodas on Oct 2^nd, 2008, 5:26am

Thanks. It seems I was going in the reverse way. I tried comparing the angle AOE from my construction with 2Pi/n.