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17 gon
Posted:
Jul 16, 1996 1:20 PM


Does anyone know why this works?
a. Construct the circle which is to be divided into 17 equal parts, and in it construct two perpendicular diameters. Find by repeated bisection the point A on one of the diameters (D2) which is onequarter of the radius from O, the centre. b. Using the point A as the center, draw the arc XY as shown. (X is one of the intersections of D1 with the given circle. Y is the intersection of the new circle just drawn with the diameter D2 inside the given circle, i.e. below O on D2.) c. Find by repeated angle bisection the quarter of this arc, as shown, (starting from point Y, onequarter of the way to X) (the following assumes a line drawn from point A to the arc) and hence the point B of intersection with the other diameter(D1) d. Draw a line at 45 degrees to AB to cut the same diameter(D1) at C. (The point C is to be on the opposite side of the diameter from B.) e. Construct a circle, taking CX as diameter to obtain the points of intersection D and E (with diameter D2). (I have left D off the diagram to reduce clutter; it is on the opposite side of O close to A. In any event it is not needed.) f. Construct the circle of radius BE and centre B. g. Construct the two tangents to this small circle parallel to DE. (i.e. parallel to diameter D2.) h. Take the five points of the circumference as shown (the point X together with the four points of intersection of the two tangents from (g) with the given circle) to be the vertices 1(X), 4, 6, 13, and 15. The other 12 vertices of the regular 17gon are now easily found. (Bisecting the angle between vertices 4 and 6 is one approach.)
This is a construction for a regular 17 gon. Coxeter says the proof has to do with x^2+2*x*cot2C1=0 and its roots tanC,cotC. (in his book introduction to Geometry.)
I am looking for a graphic file of the regular 17gon or software to create one. I would like any graphic files in jpg,gif or bmp format.



