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
Lines: 39
Distribution: inet
Expires: Sep 24, 1997
Message-ID: <5ut13b$es4@gap.cco.caltech.edu>
References:
NNTP-Posting-Host: alumni.caltech.edu
In article ,
Diomataris Agiaromelitis 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