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        riddles >> putnam exam (pure math) >> Pentagon Puzzle (11/26/2002)
        
(Message started by: william wu on Nov 27^th, 2002, 9:25pm)

Title: Pentagon Puzzle (11/26/2002)
Post by william wu on Nov 27^th, 2002, 9:25pm

Joining every vertex of a convex pentagon to every other vertex forms three nested figures:

• The original outer pentagon,
• An inner pentagon, and
• A five-pointed star between them.

http://www.ocf.berkeley.edu/~wwu/images/riddles/pentagonPuzzle.gif

Suppose the perimeters of these three figures are all primes; show that the three primes’ sum is at least 20.

Title: Re: Pentagon Puzzle
Post by towr on Nov 28^th, 2002, 4:31am

I suppose you can't put points over eachother?
i.e. two sides of the pentagram are zero length, which would make your pentagon look like a triangle.. Cause then I could get a minimum of 10 (2+3+5 for a pentagram with sides 0,1,0,a,2-a with 0.5 <= a <=1.5)

Title: Re: Pentagon Puzzle (11/26/2002)
Post by Eigenray on Apr 25^th, 2003, 3:30am

Call the inner pentagon X, outer pentagon Y, and the 5-pointed star Z, and use these letters for the perimeters as well.
By considering the triangles with one edge from Y, and the other two edges each of Z, we get that Z > Y.
By considering the triangles with two edges each from Y, and the third edge made up of 2 from Z and one from X between those, we get that 2Y > X + Z.
Since 2Y > X + Z > X + Y, we get Y > X.
So X < Y < Z < 2Y - X,
and suppose X + Y + Z < 20.
The only primes we need consider are 2,3,5,7,11,13.
It's easy to check that the above inequalities rule out everything except (X,Y,Z) = (2,5,7).
The only way to obtain a sum of 20 subject to the above inequalities is (2,7,11), so judging from the problem statement there must be another condition that rules out (2,5,7), but allows (2,7,11); conceivably Z > X + Y, but I don't see any way of obtaining that at the moment.

Title: Re: Pentagon Puzzle (11/26/2002)
Post by Eigenray on Apr 8^th, 2005, 3:54pm

*bump*

Title: Re: Pentagon Puzzle (11/26/2002)
Post by Barukh on Apr 9^th, 2005, 9:21am

on 04/25/03 at 03:30:43, Eigenray wrote:

...conceivably Z > X + Y, but I don't see any way of obtaining that at the moment.

This condition is not true. In fact, the relation may be arbitrary. So, the proof must use another property of the configuration.

Title: Re: Pentagon Puzzle (11/26/2002)
Post by Deedlit on Apr 9^th, 2005, 4:34pm

on 04/09/05 at 09:21:32, Barukh wrote:

This condition is not true. In fact, the relation may be arbitrary. So, the proof must use another property of the configuration.

Right - for example, using regular pentagons with X = 1, we get Y = phi + 1, Z = 2 phi, where phi is the golden ratio ~ 1.618; so Z < X + Y.

I wonder if William meant the "five-pointed star" to be the nonsimple five sided polygon? Then the perimeter is increased by X, so we have 2Y > Z > X + Y, and the smallest solution is (2,7,11).