Our Dearest Problems

A Colloquium on Recreational Mathematics took place in Lisbon this week. The meeting, RMC-VI (G4GEurope), a great success, was organised by the Ludus Association, with support from several other agencies: MUHNAC, ULisboa, CMAF-IO, CIUHCT, CEMAPRE, and FCT. It was the third meeting integrated in the Gathering for Gardner movement, which celebrates the great populariser of maths, Martin Gardner. For more information about the meeting, see http://ludicum.org/ev/rm/19 .

At the closing session on Tuesday, several participants described their favourite problems. The session, entitled Our Dearest Problems, was chaired by Jorge Buescu (CMAF-IO). The problems are summarised below.

Adam Atkinson

Adam talked about the sub-heaps problem. You are given a heap of n tokens, each valued at, say, €1.00. You may divide the heap into smaller heaps. Then you receive an amount equal to the product of the numbers in each pile (if you stick with a single heap, you win €n.00). Find the optimal division strategy for arbitrary n.

A variation of this problem asks about a strip of material of length L meters. You may accept €L.00, or cut the strip into shorter pieces and receive an amount equal to the product of the lengths of each piece. Find the optimal cutting strategy for arbitrary length L.

Andreas Hinz

Andreas discussed variations on the well-known Tower of Hanoi problem. In the original form, there were three pegs in a triangle. In the linear form, the pegs, perhaps more than three, are in a row and movement of discs is permitted only between neighbouring pegs. The stack must be transferred form one end peg to the other. Can an expression be found for the minimum number of moves?

Guido Ramellini

Guido posed a problem about finding the box of maximum volume to be constructed from a unit square of material by cutting away the corners and folding up the sides. There are many approaches, the best one depending on the mathematical maturity of the solver. The solution forms a box that is one quarter of a cube. Why?

Ricardo Teixeira

Ricardo spoke about the well-known Collatz Conjecture. Given any positive integer n, generate a sequence as follows: for an even number, divide by 2. For an odd number, multiply by 3 and add 1 (a mnemonic for the process is HOTPO: Half-Or-Triple-Plus-One). The conjecture is that the sequence always reaches 1. But this has never been proved: although the problem is easily understood by a young child, its proof has defied all attempts to date. See an earlier post on the Collatz Conjecture at thatsmaths.com.



Thane Plambeck

Thane started with a simply stated geometric puzzle. A triangle (ABE) is cut from a square (ABCD) and placed at one side to form a smaller square. Find a way of restoring the square by a single action different from the exact reversal of this transformation.  (Thane got this problem from David Klarner over 25 years ago who, in turn, got it from N G de Bruijn. The ultimate source is unknown.).

Guessing Red or Black: Thane also posed a card-game problem. The 52 cards of a deck are turned over one by one. Before each turn, you must choose a colour, red or black. For each correct prediction, you win a dollar. For each wrong guess, you neither win or lose anything. What is your expected win? How much would you be willing to pay to play?

Feedback

Comments and proposed solutions to any of the above problems are welcome. Comments will remain open for a reasonable period of time.

2 Responses to “Our Dearest Problems”


  1. 1 thomasdevlin February 8, 2019 at 15:39

    For the square puzzle, it is acceptable to move ADCBE to the right of BCF, to make a square, in a different location?

  2. 2 thatsmaths February 8, 2019 at 16:33

    The elementary solution is to cut the triangle BCF from the shape on the right and to use it to fill the gap AEB.

    There is an additional, completely different, way of cutting the shape on the right into two pieces that can be repositioned to form a square. No tricks are involved.


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