From: ram@tiac.net (robert a moeser)
Subject: Re: A 17-sided polygon the hardcore way?
Date: Tue, 25 May 1999 09:35:17 -0500
Newsgroups: sci.math
Keywords: construction of the heptadecagon (pointers)
In article <7ibk6r$11u@freenet-news.carleton.ca>, hmw@ncf.SPAMNIXca (Hank
Walker) wrote:
> I know it's possible, but is it unfeasibly difficult
> to construct a regular 17-sided polygon with a
> straightedge and compass?
i got this when i asked long ago. it works and is mostly straightforward.
i have another version with some ambiguities clarified, but it has
turnt up missing. i got this to go OK.
i had fun using a CAD system to do the construction in a virtual way,
and can state that "empirically", the thing works.
-- rob
Rob,
The construction below is attributed to Richmond with no further
citation in the Dover publication Mathographics by Robert Dixon
ISBN 0-486-26639-7 copyright 1987.
D2
.|.
. | .
. | .
. | .
. A .
. | .
D1 _________C__|_B__________ X
. O| .
. E .
. | .
. | .
. Y .
.|.
Sorry about the ascii graphics. Since I cannot reliably reproduce
pictures in the book I have labelled the two perpendicular diameters
of the above circle to clarify a couple of descriptions in the text.
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.)
I hope the above is informative enough. Dixon's book is a worthy
addition to any geometer's library; it has the added virtue that
it is cheap and in print.
Happy mathing,
Bruce
bappleby@world.std.com
==============================================================================
From: James Buddenhagen
Subject: Re: A 17-sided polygon the hardcore way?
Date: Mon, 24 May 1999 11:24:58 -0500
Newsgroups: sci.math
The construction of the regular 17-gon can be found several places on
the net. E.g.
http://www.forum.swarthmore.edu/news.archives/geometry.pre-college/article1181.html
http://www.seanet.com/~ksbrown/kmath487.htm
http://www.nevada.edu/~baragar/geom/sevteen.htm
--Jim Buddenhagen