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17 gon
Posted:
Jul 16, 1996 1:20 PM
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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
one-quarter 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, one-quarter 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 17-gon 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*cot2C-1=0 and its roots tanC,-cotC. (in his book
introduction to Geometry.)
I am looking for a graphic file of the regular 17-gon or software to
create one. I would like any graphic files in jpg,gif or bmp format.
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