# Get Mathematical Olympiads 1997-1998: Problems and Solutions PDF This booklet is an absolute should have for these people who love tough mathematical difficulties. The publication claims in its preface to be a continuation of Mathematical Contests 1997-1998: Olympiad difficulties and ideas from world wide, released through the yank arithmetic Competitions. I made multiple try to discover a replica of that e-book whilst getting ready this overview and was once unsuccessful. (I have now been trained that the books from 1995-1996, 1996-1997, and 1997-1998 could be ordered from the yank arithmetic Competitions net page.) we're lucky that the Mathematical organization of the USA has determined to start a brand new sequence of books, the MAA challenge Books sequence, and has all started that sequence with this excellent choice of difficulties from worldwide.

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Extra info for Mathematical Olympiads 1997-1998: Problems and Solutions from around the World

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We are given 1997 distinct positive integers, any 10 of which have the same least common multiple. Find the maximum possible number of pairwise coprime numbers among them. Solution: The maximum number of pairwise coprime numbers in this set is 9. First, suppose there were 10 pairwise coprime numbers n1 , n2 , . . , n10 . Then the least common multiple of any 10 members of this set is lcm(n1 , n2 , . . , n10 ) = n1 n2 · · · n10 . In particular, for any other N in this set, lcm(N, n2 , · · · , n10 ) = n1 n2 · · · n10 is divisible by n1 ; as n1 is relatively prime to nj for 2 ≤ j ≤ 10, n1 divides N .

Let E and F be the points where the tangents from Q meet the circumcircle of ABCD. Prove that points P, E, F are collinear. Solution: Let X denote the tangent of the circle at a point X on the circle. Now take the polar map through the circumcircle of ABCD. To show P, E, F are collinear, we show their poles are concurrent. E and F map to E and F which meet at Q. Since P = AB ∩ CD, the pole of P is the line through A ∩ B and C ∩ D , so we must show these points are collinear with Q. However, by Pascal’s theorem for the degenerate hexagon AADBBC, the former is collinear with Q and the intersection of AC and BD, and by Pascal’s theorem for the degenerate hexagon ADDBCC, the latter is as well.

Let ABCD be a cyclic quadrilateral. The lines AB and CD meet at P , and the lines AD and BC meet at Q. Let E and F be the points where the tangents from Q meet the circumcircle of ABCD. Prove that points P, E, F are collinear. Solution: Let X denote the tangent of the circle at a point X on the circle. Now take the polar map through the circumcircle of ABCD. To show P, E, F are collinear, we show their poles are concurrent. E and F map to E and F which meet at Q. Since P = AB ∩ CD, the pole of P is the line through A ∩ B and C ∩ D , so we must show these points are collinear with Q.

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### Mathematical Olympiads 1997-1998: Problems and Solutions from around the World

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