Published on July 10th, 2013 | by Daphane Ng0
AustMS Gazette Puzzle Corner 33
Welcome to the Australian Mathematical Society Gazette’s Puzzle Corner number 33. Each puzzle corner includes a handful of fun, yet intriguing, puzzles for adventurous readers to try. They cover a range of difficulties, come from a variety of topics, and require a minimum of mathematical prerequisites for their solution. Should you happen to be ingenious enough to solve one of them, then you should send your solution to us.
For each puzzle corner, the reader with the best submission will receive a book voucher to the value of $50, not to mention fame, glory and unlimited bragging rights! Entries are judged on the following criteria, in decreasing order of importance: accuracy, elegance, difficulty, and the number of correct solutions submitted. Please note that the judge’s decision — that is, my decision — is absolutely final. Please email solutions to firstname.lastname@example.org or send paper entries to: Gazette of the Australian Mathematical Society, School of Science, Information Technology & Engineering, University of Ballarat, PO Box 663, Ballarat, Vic. 3353, Australia.
The deadline for submission of solutions for Puzzle Corner 33 is 1 September 2013. The solutions to Puzzle Corner 33 will appear in Puzzle Corner 35 in the November 2013 issue of the Gazette.
Notice: If you have heard of, read, or created any interesting mathematical puzzles that you feel are worthy of being included in the Puzzle Corner, I would love to hear from you! They don’t have to be difficult or sophisticated. Your submissions may very well be featured in a future Puzzle Corner, testing the wits of other avid readers.
Let S be a set of 10 distinct positive integers no more than 100. Prove that S contains two disjoint non-empty subsets which have the same sum.
Knights and knaves
In the following problems, knights always tell the truth and knaves always lie.
(i) There is a queue of people, each of whom is either a knight or a knave. It is known that there are more knights than knaves. Apart from the first person, every person points to someone in front of them in the queue and declares the status of that person (being a knight or a knave). Is it possible for a bystander to determine the actual status of everyone in the queue?
(ii) There is a group of people, each of whom is either a knight or a knave. Each person makes the following two statements: ‘All my acquaintances know each other’, and ‘Among my acquaintances, the number of knights is no more than the number of knaves.’ We assume that knowing is mutual. Prove that the number knaves in the group is no more than the number of knights.
In a regular nonagon, prove that the length difference between the longest diagonal and the shortest diagonal is equal to the side length. In other words, prove c−b = a in the diagram below.
Scissors and shapes 2
Edward is playing with scissors again. At each move, he chooses a polygon in front of him, and cuts it into two polygons with a single straight cut. Starting with a single rectangle, determine the minimal number of cuts required to obtain, among other shapes, at least 106 polygons with exactly 22 sides.
There are n distinct non-negative integers written on the board. Jack memorises these numbers before erasing them and replacing them with the (n 2) pairwise sums. Now Jill enters the room and studies the sums on the board. Find all positive integers n for which it is possible for Jill to recover the original n integers uniquely.
About the author
Ivan is a PhD student in the School of Mathematics and Statistics at The University of Sydney. His current research involves a mixture of multi-person game theory and option pricing. Ivan spends much of his spare time playing with puzzles of all flavours, as well as Olympiad Mathematics.
The Puzzle Corner is a regular feature in the AustMS Gazette, puzzles are created by Ivan Guo unless otherwise stated.