Author |
Topic: Pentagon Puzzle (11/26/2002) (Read 2050 times) |
|
william wu
wu::riddles Administrator
Gender:
Posts: 1291
|
|
Pentagon Puzzle (11/26/2002)
« on: Nov 27th, 2002, 9:25pm » |
Quote Modify
|
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. Suppose the perimeters of these three figures are all primes; show that the three primes’ sum is at least 20.
|
« Last Edit: Dec 15th, 2002, 12:16am by william wu » |
IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Pentagon Puzzle
« Reply #1 on: Nov 28th, 2002, 4:31am » |
Quote Modify
|
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)
|
« Last Edit: Nov 28th, 2002, 4:33am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Pentagon Puzzle (11/26/2002)
« Reply #2 on: Apr 25th, 2003, 3:30am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Pentagon Puzzle (11/26/2002)
« Reply #3 on: Apr 8th, 2005, 3:54pm » |
Quote Modify
|
*bump*
|
|
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Pentagon Puzzle (11/26/2002)
« Reply #4 on: Apr 9th, 2005, 9:21am » |
Quote Modify
|
on Apr 25th, 2003, 3:30am, 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.
|
|
IP Logged |
|
|
|
Deedlit
Senior Riddler
Posts: 476
|
|
Re: Pentagon Puzzle (11/26/2002)
« Reply #5 on: Apr 9th, 2005, 4:34pm » |
Quote Modify
|
on Apr 9th, 2005, 9:21am, 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).
|
|
IP Logged |
|
|
|
|