From - Sun Sep  7 18:51:00 1997
From: ikastan@alumni.caltech.edu (Ilias Kastanas)
Newsgroups: sci.math
Subject: Re: Regular Heptagon Problem
Date: 7 Sep 1997 01:47:23 GMT
Organization: Caltech Alumni Association
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Expires: Sep 24, 1997
Message-ID: <5ut13b$es4@gap.cco.caltech.edu>
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In article <dcu128er9e23@forum.swarthmore.edu>,
Diomataris Agiaromelitis <amaskali@agia.romeli.gr> wrote:
>(A_1A_2A_3A_4A_5A_6A_7): Reg. heptagon
>
>(A_5A_3) section (A_1A_2):= M
>(A_4A_7) section (A_1A_2):= N
>(A_4A_7) section (A_5A_3):= P
>
>Prove that (N,M,P,A_6) are homocyclic points,
>that is, the polygon (NMPA_6) can be inscribed in a circle.



	Someone suggested analytic geometry or trigonometry.  That works,
of course, but the point is to see whether there is a geometric proof.


	Let angle PNM =  A_3 A_1 A_2  = a  (= pi/7);  then  NMP = 2a
and  MPN = 4a.   We need to show that  M A_6 N  is also  4a.


	Let  A_6 A_2  meet  A_7 A_3  at T, and  A_6 A_7  meet  N A_1 at U.
A_1 A_2 T A_7  is a rhombus,  and the lengths  A_1 M = A_7 A_3 = A_6 A_2,
so  A_2 M = A_6 T.

	Also, length  N A_7 = A_7 A_2 = A_6 A_1 = A_6 U  (angle A_6 A_1 U =3a).

	So  A_6 A_2 / A_2 M  =  A_6 A_2 / A_6 T  =  A_6 U / A_6 A_7  =
= N A_7 / A_6 A_7.   Since also  angle N A_7 A_6 = 5a = angle A_6 A_2 M,
the triangle   N A_7 A_6  is similar to   A_6 A_2 M.

	Therefore  angle M A_6 N  =  M A_6 A_2 + 2a + A_7 A_6 N =
= 2a + (pi - N A_7 A_6) = 4a.   Hence  M, P, A_6, N  lie on a circle.



						Ilias