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March 16

IMO problem

1)show that out of any 5 points chosen within a square of side length 2 there are two whose distance apart is at most √2

2)Find the remainder when 2^1990 is divided by 1990.

Solomon7968 (talk) 14:55, 16 March 2013 (UTC)[reply]

1) The configuration of the square will be:

   *-------*
   |       |
   |   *   |
   |       |
   *-------*

You can use some basic trig from there to show the result. StuRat (talk) 16:54, 16 March 2013 (UTC)[reply]

Can you prove that there is no other configuration with wider separation? What about an inscribed pentagon, for example? Looie496 (talk) 17:49, 16 March 2013 (UTC)[reply]

Right. StuRat's claim is not a proof. You're going to need to use the pigeonhole principle for this. —Bkell (talk) 20:07, 16 March 2013 (UTC)[reply]

1)Divide the square to 4 subsquares, each one with side length 1, then there must be two points on the same subsquare therefore the distance between them is at most square root of 2.

2)Factor 1990, and use Chinese remainder theorem and Fermat's little theorem --84.229.98.135 (talk) 20:14, 16 March 2013 (UTC)[reply]

I've added a space to the original question, as "atmost" looked like "almost".86.186.142.172 (talk) 09:32, 17 March 2013 (UTC)[reply]

In my opinion, there's no way either of these could have been an IMO problem. 64.140.122.180 (talk) 09:55, 17 March 2013 (UTC)[reply]

Pollution

what is pollution how it is caused what is the 4 types of pollution — Preceding unsigned comment added by 190.80.8.6 (talk) 21:49, 16 March 2013 (UTC)[reply]

This is the mathematics reference desk. This is probably not the appropriate place to ask your question. You might try the science reference desk. —Bkell (talk) 21:54, 16 March 2013 (UTC)[reply]