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- Path: sparky!uunet!gatech!darwin.sura.net!sgiblab!sgigate!odin!fido!graphack.asd.sgi.com!zhu
- From: zhu@graphack.asd.sgi.com (Benjamin Zhu)
- Newsgroups: sci.math
- Subject: Simple convex polygon
- Date: 16 Oct 1992 00:22:11 GMT
- Organization: Silicon Graphics, Inc.
- Lines: 32
- Distribution: world
- Message-ID: <1bl1vjINN97g@fido.asd.sgi.com>
- NNTP-Posting-Host: graphack.asd.sgi.com
-
-
- Hi!
-
- I have a conjecture that I have not been able to either prove or
- disprove so far. I want to utilize the net wisdom here.
-
- Conjecture:
- Given an arbitrary (might be simple, might be complex, i.e., self-
- intersecting) polygon, p[0], p[1], p[2], ..., p[n-1], no two
- vertices coincide, no three vertices are co-linear. Suppose the
- first three vertices p[0], p[1], and p[2] go counter-clockwise,
- i.e., we have a left turn at p[1]. Define the interior angle at
- a vertex p[i] to be the sweep angle from vector p[i]p[(i-1)%n]
- to vector p[i][(i+1)%n] clockwise.
-
- If every interior angle is less than \phi, and the summation of
- all interior angles is (n-2)*\phi, the polygon in question is
- convex and simple (non self-intersecting).
-
- If this is indeed true, please either give me a pointer or a
- reasonably-detailed sketch of proof. If not, can you give me a
- counter-example? Please respond by emails.
-
- Thanks a lot,
-
- Ben
-
- --
- Benjamin Zhu ;;; Priest of ucode and self-appointed bug terminator
- Silicon Graphics, Inc. ;;; ``Just call my name, I will be there.''
- zhu@graphack.asd.sgi.com ;;; ``Projective geometry is all geometry.'' - Cayley
- (415) 390-1187 ;;; ``Learn forever, young forever.'' - Anonymous
-