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Title: Diagonals of a polygon. Post by Grimbal on May 13th, 2007, 3:11pm I took part in a logical and mathematical games contest. One of the problem was (story removed): There is a convex hexagon such that - all sides have a different length, - all 3 diagonals are concurrent - the vertices lie on the vertices of a regular N-sided polygon. What is the minimum possible N? As a starter, the given answer was an odd number, but in my opinion it can not be because by experience (i.e. computer), no 3 digaonals of a regular N-sided polygon are concurrent for N odd. Does anybody know a good argument for that? |
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Title: Re: Diagonals of a polygon. Post by Barukh on May 13th, 2007, 11:05pm on 05/13/07 at 15:11:11, Grimbal wrote:
Shouldn't this be "on the sides of a regular N-gon? |
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Title: Re: Diagonals of a polygon. Post by towr on May 14th, 2007, 12:46am on 05/13/07 at 23:05:50, Barukh wrote:
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Title: Re: Diagonals of a polygon. Post by Grimbal on May 14th, 2007, 2:10am As towr said. The vertices of the hexagon are a subset of those of a regular N-gon. |
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Title: Re: Diagonals of a polygon. Post by Barukh on May 14th, 2007, 8:32am I see. Grimbal, your assumption is true. I don't know if there is a simple argument to prove it, though. Maybe, the following (http://citeseer.ist.psu.edu/cache/papers/cs/9081/ftp:zSzzSzftp.msri.orgzSzpubzSzpublicationszSzpreprintszSz1995zSz1995-060zSz1995-060.pdf/poonen98number.pdf) is of some help. |
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Title: Re: Diagonals of a polygon. Post by Grimbal on May 14th, 2007, 9:56am Thanks. It shows that identifying concurrent diagonals is far from trivial and cannot be solved in the hour I had, not without prior knowledge of the problem. But in the introduction it refers to an earlier paper: Herman Heineken, "Regelmässige Vielecke und ihre Diagonalen", 1962. He proves that an odd-gon has no 3 concurrent diagonals, using complex polynomials. |
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Title: Re: Diagonals of a polygon. Post by balakrishnan on Aug 5th, 2007, 5:37pm I get [hide]8[/hide] as the smallest N. |
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Title: Re: Diagonals of a polygon. Post by Obob on Aug 5th, 2007, 10:55pm You cannot choose six vertices of an octagon in such a way that the consecutive distances between adjacent vertices are all different. |
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Title: Re: Diagonals of a polygon. Post by balakrishnan on Aug 6th, 2007, 5:05am Sorry I overlooked the problem. I get N=[hide]24[/hide]. |
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Title: Re: Diagonals of a polygon. Post by Grimbal on Aug 8th, 2007, 5:39am Yep. That's what I got later with a computer program. |
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