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**zehao1000****Member**- Registered: 2012-11-16
- Posts: 34

Here's a random problem I found:

An integer is assigned to each vertex of a regular polygon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer m from each of the integers at two neighbouring vertices and adding 2m to the opposite vertex, which is not adjacent to either of the first two vertices.( The amount m and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integer, there is exactly one vertex which the game can be won.

(Note: it's not like I know how to do it, it's just for fun, because this would challenge most mathematicians.)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,727

Hi;

I have seen the proof of this but it is tougher for me than the problem itself!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**zehao1000****Member**- Registered: 2012-11-16
- Posts: 34

This is a problem from the extreme USAMO contest, which is 6 problems, 2 days, here's another one if you want another one:

Let ABC be a triangle with angle A=90degrees. Points D and E lie on AC and AB, respectively, such that angle ABD=angle DBC and angle ACE=angle ECB. Segments BD and CE meet I. Determine whether or not it is possible for segments AB, AC, BI, ID, CI, IE to all have interger side lengths.

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