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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 : $\frac{f(x)+f(y)}{2} - f( \frac{x+y}{2} ) >= |x-y|$ 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: 20,931 Thanks: 2205 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*|x-y| 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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