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