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Hungarian Problem Book IV
Translated and edited by Robert Barrington Leigh and Andy Liu
Catalog Code: HP4
Print Edition ISBN: 978-0-88385-831-8
Electronic Edition ISBN: 978-0-88385-913-0
115 pp., Paperbound, 2011
List Price: $40.95
Member Price: $33.95
Series: Problem Books
Table of Contents | Excerpt | About the Editors | Buy on Amazon | Buy in MAA Bookstore | More Hungarian Problem Books
This book, which is a continuation of Hungarian Problem Book III, contains the problems from the Kürschák Mathematics Competition in the period 1947-1963. In all, 48 problems are presented.
Hungarian Problem Book IV is intended for beginners, who are encouraged to work the problems in each section and then to compare their results against the solutions presented in the book. They will find ample material in each section to help them improve their problem-solving techniques.
The volume’s problems are classified under 12 subject headings: combinatorics, graph theory, number theory, divisibility, sums and differences, algebra, geometry, tangent lines and circles, geometric inequalities, combinatorial geometry, trigonometry, and solid geometry.
Solutions to the problems are presented along with background material. Furthermore, the last section, titled “Looking Back,” provides additional insights into the problems.
1. Kürschák Mathematics Competition Problems: 1947; 1948; 1949; 1950; 1951; 1952; 1953; 1954; 1955; 1957; 1958; 1959; 1960; 1961; 1962; 1963.
3. Solutions to Problems:
4. Looking Back:
Problem 1950.1 (p. 23):
On a certain day, a number of readers visited a library. Each went only once. Among any three readers, two of them met at the library on that day. Prove that there were two particular instants such that each reader was in the library at one of the two instants.
First Solution. Consider the moment a when the first reader A leaves and the moment b when the last reader B arrives. We may assume that
a < b as otherwise every two readers meet. Suppose a reader C is not present at either moment. If C arrives before a, C must be present at a since C leaves after A. Hence C arrives after a. If C leaves after b, C must be present at b since C arrives before B. Hence, C leaves before b. It follows that among A, B, and C, no two have met. This is a contradiction.
Second Solution. Suppose the librarian makes an announcement twice, trying to catch all readers, Clearly, the first moment a should be when the first reader A leaves. If every reader has heard the first announcement, there is nothing else to prove. Otherwise, the second announcement would be made at a later moment b when the first reader B who has not heard the first announcement leaves. Suppose some reader C has not heard either announcement. Then C must have arrived after b, so among A. B, and C, no two have met. This is a contradiction.
About the Editors
Robert Barrington Leigh (1986-2006), representing Canada, was a Bronze Medalist in two International Mathematical Olympiads and a three-time participant in the William Lowell Putnam Mathematical Competition. This book is dedicated to his memory.
Andy Liu (University of Alberta) was Deputy Leader of the USAMO Team from 1981-84. He edited Math Horizon’s problem section for seven years and won MAA’s Deborah and Franklin Tepper Haimo Teaching Award in 2004.
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