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4.0 out of 5 stars
Best IMO compendium but authors' English is not sufficient, August 28, 2008
This review is from: The IMO Compendium: A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2004 (Problem Books in Mathematics) (Hardcover)
This is a great collection of IMO problems, but I am disappointed by the authors' command of English and sloppiness, especially since they have rewritten the problem statements to save space, often leading to unclear or even incorrect statements. For example, problem 4, IMO 1965 is incorrectly stated as: "Find four real numbers x1,x2,x3,x4 such that the sum of any of the numbers and the product of other three is equal to 2." The statement is not as clear as the original one found on the IMO website, the last four words can be replaced with the two words "equals 2" to save space, and the article "the" is missing (a typical native Slavic language speaker grammar error in English). Much more serious is that this formulation has the trivial answer 1,1,1,1 while the original question was to find all solutions, not just a single one. An error in problem statement is the worst mistake possible in a problem book, even worse than a solution error, and it is even worse here as it is for an original IMO problem, not the long or shortlisted one where this could have been understandable due to the large number of problems and the lesser attention given to them over the years. I just got the book yesterday, so have not checked it out thoroughly, but I expect to find many more inaccuracies and mistakes. For me this is not such a problem and I even find it motivating to look for errors, but it can be a serious problem for students using the book to learn about doing IMO problems.
Update, November 17, 2008. I have found more errors in IMO contest statements 1981.4, 1987.4, 1971.3, 1979.1, 1978.3, 1978.5, 1985.4, 1989.1, 1991.2. This includes errors in English (some of which make the statement longer than the original) as well as changing the problem statement for 1987.4 and 1989.1. The case 1989.1 is particularly serious. The original problem asked to partition the numbers 1,...,1989 into 117 sets each with equal sum, while the book asks for 17 sets instead. It seems that the book formulation avoids the subtleties of this problem by considering a less complicated formulation, the general case seems much harder when the number of sets in the partition is greater than the number of elements in each set (in particular I give a solution when this is not the case). Here is an outline of my analysis:
Consider the general problem of partitioning {1,...,n} into d sets of equal size and equal element sum, so n = c*d. Clearly, the problem is the same for the set {0,...,n-1}, and write these numbers as a cxd array whose jth row is 117j + i, i=0,...,d-1. Each permutation of {0,...,d-1} induces a permutation on a row. One then considers permutations x |-> x + a mod d (cyclical shift by a) and x |-> d-1-x (reverse a row). If c is even, then one changes the array by reversing c/2 rows and leaving c/2 rows unchanged. If c is odd and d <= c, then one changes the array by cyclically shifting d rows (row j shifted by j), and then, since c-d is even, one can reverse (c-d)/2 rows and leave (c-d)/2 rows unchanged. It is clear that the columns of the resulting array give a solution. This gives a solution for n = c*d with c even or n = c*d odd and d <=c. In particular, this solves the book version with d = 17, c = 117.
For n = c*d, odd and d > c, I could not find a simple analysis in this case. For example if n = 15, d = 5, I found an ad hoc solution using Sudoku style trial and error, but I don't see a pattern:
{1,10,13}, {2,8,14},{3,9,12},{4,5,15},{6,7,11}.
However, I did find a general solution when d = r*s and r <= s <= c. This solves the IMO problem since r = 9 < s = 13 < c = 17. Once again, consider the cxd array with jth row jd + i, i = 0,...,d-1. By the first part above, since r <= s, one can partition {0,...,d-1} into r sets B1,...,Br each of size s and all with equal sum. There is a permutation sigma of {0,...,d-1} in which each Bk is shifted cyclically by 1, note that sigma has order s. One repeatedly applies this permutation to rows 1,...,s of the array (that is, apply sigma^j to row j, j =0,...,s-1) and on the remaining c-s rows, half will be left the same and the other half reversed. The columns of the resulting array give the required partition of {0,...,n-1}.
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