Download Mathematical Olympiads 1999-2000: Problems and Solutions by Titu Andreescu PDF

By Titu Andreescu

Contained listed below are ideas to hard difficulties from algebra, geometry, combinatorics and quantity thought featured within the prior e-book, including chosen questions (without strategies) from nationwide and neighborhood Olympiads given through the yr 2000. meant for the intense student/problem solver, those books will help to enhance functionality within the Mathematical Olympiad pageant. even though, for these no longer getting into the contest, there's a lot to problem any mathematician, even people with complicated levels. diverse countries have diverse mathematical cultures, so that you will locate that a number of the questions are super tricky and a few relatively effortless. There are a large choice of difficulties in particular from these nations that experience usually performed good within the IMO. a person drawn to mathematical challenge fixing will come across a few appealing arithmetic within the pages of this ebook. while you are as much as a true problem, take a few of these difficulties on!

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This condition implies that P (u, v) = (u + v)(u3 + v 3 ) = (u + v)2 (u2 − uv + v 2 ) = (u2 + 2uv + v 2 )(1 − uv) = (1 + 2uv)(1 − uv). Under the condition u2 + v 2 = 1 the product uv can attain all values in the interval [− 12 , 12 ] (if u = cos α and v = sin α, then uv = 21 sin 2α). Hence it suffices to find the range of values of the function f (t) = (1 + 2t)(1 − t) on the interval t ∈ [− 12 , 12 ]. From the formula 2 1 9 f (t) = −2t2 + t + 1 = −2 t − + 4 8 it follows that this range of values is the closed interval with endpoints f − 12 = 0 and f 14 = 98 .

16, 2}, one of the numbers is a factor in A and the other is a factor in B. The resulting value V can then be written as a product 29 15 1 28 14 2 ··· 16 2 14 , where each i equals ±1, and where 1 = 1 and 2 = −1 no matter 26 how the parentheses are placed. Since the fractions 27 13 , 12 , . . , 16 2 are greater than 1, the resulting value V (whether an integer or not) has to satisfy the estimate 29 14 27 26 16 · · · · ... · = H, 15 28 13 12 2 where H is number determined in part (a). It follows that H is the only possible integer value of V !

The entry in the second row and second column equals either a, c, or d, yielding the Latin squares     a b c d a b c d  b a d c   b a d c       c d a b ,  c d b a , d c a b d c b a  a  b   c d b c d a c d a b   a b c d  b d a a  ,  b   c a d d c b c  d c  . b  a Thus x = 4, and there are 4! · 3! · 4 = 576 four-by-four Latin squares, and 576 interesting colorings. 6 Czech and Slovak Republics Problem 1 In the fraction 29 ÷ 28 ÷ 27 ÷ · · · ÷ 16 15 ÷ 14 ÷ 13 ÷ · · · ÷ 2 parentheses may be repeatedly placed anywhere in the numerator, granted they are also placed on the identical locations in the denominator.

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