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January 25th, 2010, 12:52 PM  #1 
Member Joined: Dec 2009 Posts: 34 Thanks: 0  A few olympiads problems I can't solve
1. Prove that there is no real function such as for every real x,y : $ \frac{f(x)+f(y)}{2}  f( \frac{x+y}{2} ) >= xy $ 2. Find 2 positive numbers x,y with the same number of digits such as the product xy is a 50 digits number at least that all of them are 1 (a number of the form 111111111) ... 3. Calculate the square root of the number: 11111...1222...25 where the digit 1 appears 2008 times and the digit 2 appears 2009 times. Good Luck! 
January 28th, 2010, 12:39 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
3. 333...5

January 28th, 2010, 01:07 PM  #3 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: A few olympiads problems I can't solve
2. 1633466135458167330677291 * 6802168021680216802168021

April 4th, 2010, 08:13 AM  #4 
Newbie Joined: Mar 2010 Posts: 7 Thanks: 0  Re: A few olympiads problems I can't solve
1. let g(x,y) = f(x) + f(y)  f(x+y) let h(x,y) = 2*xy restate the problem as: prove that there does not exist an f, such that for each point in the x,y plane g >= h Note that along the x and y axes, h is unbounded. Note that along the x and y axes, g(x, 0) = f(x) + f(0)  f(x) = f(0) g(0, y) = f(0) + f(y)  f(0) = f(0) If g>=h at every point in the x,y plane, it is greater than every point along the x and y axes as well. this implies f(0) >= 2*x, for every x (or f(0) >= 2*y, for every y) There is no real number f(0) with this property, therefore no f exists. I'm very keen on the explanation to the other two answers. I don't know how you would go about these without a calculator. 

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