This publication is an absolute should have for these folks who love hard mathematical difficulties. The e-book claims in its preface to be a continuation of Mathematical Contests 1997-1998: Olympiad difficulties and options from world wide, released via the yankee arithmetic Competitions. I made a couple of try and discover a replica of that booklet whilst getting ready this overview and was once unsuccessful. (I have now been expert that the books from 1995-1996, 1996-1997, and 1997-1998 should be ordered from the yankee arithmetic Competitions internet page.) we're lucky that the Mathematical organization of the US has determined to start a brand new sequence of books, the MAA challenge Books sequence, and has started that sequence with this glorious selection of difficulties from world wide.
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Additional resources for Mathematical Olympiads 1997-1998: Problems and Solutions from around the World
Sample text
What is the least possible area of the region where the paper covers itself? Solution: Label the vertices of the triangle which covers itself A, B, C, where AB is the folding axis and ∠BAC is acute. Drop altitudes from A and B and label the points across the strip from them A and B , respectively. Note that ∠BAB folds onto ∠BAC, so these angles have the same measure, which we call x. We consider two cases. First, suppose 0 < x ≤ π/4. Then C is between A and B, ∠ACA = 2x and ∠ABA = x. Then [ABC] = [ABA ] − [ACA ] 1 1 (3)(3 cot x) − (3)(3 cot 2x) = 2 2 9 9 (cot x − cot 2x) = csc 2x.
Then x − y = (2k − 1)(2b − 1)(2a ), z − w = (2k − 1)(2d − 1)(2c ). Also x/y = 2b , z/w = 2d . Now x/y = z/w if and only if b = d if and only if x − y and z − w have the same largest odd divisor. 6 Czech and Slovak Republics 1. Let ABC be a triangle with sides a, b, c and corresponding angles α, β, γ. Prove that the equality α = 3β implies the inequality (a2 − b2 )(a − b) = bc2 , and determine whether the converse also holds. Solution: By the extended law of sines, a = 2R sin α, b = 2R sin β, c = 2R sin γ, where R is the circumradius of ABC.
Prove that no matter how the edges were colored initially, it is possible to make the number of blue segments at each vertex even. Prove also that the resulting coloring is uniquely determined by the initial coloring. Solution: All congruences are taken modulo 2. First, changing the order in which we choose the vertices does not affect the end coloring. Also, choosing a vertex twice has no net effect on the coloring. Then choosing one set of vertices has the same effect as choosing its “complement”: the latter procedure is equivalent to choosing the first set, then choosing all the vertices.
Mathematical Olympiads 1997-1998: Problems and Solutions from around the World
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