Measuring with unscaled pots - algorithm versus chance Szil´ardAndr´as 1 , ¨ Ors Nagy 2 Babe¸ s-Bolyai University, Cluj Napoca, Romania Motto: I cook every chance in my pot. And only when it hath been quite cooked do I welcome it as my food. Friedrich Nietzsche Abstract. The central focus of this paper is on the following problem: Consider three unscaled pots, with volumes a, b and c > max{a, b} liters, where a, b ∈ N * . Initially the third pot is filled with water and the other ones are empty. Characterize all quantities that can be measured using these pots. In the first part of the paper we solve this problem by using the motion of a billiard ball on a special parallelogram shaped table. In the second part we generalize the initial problem for n + 1 pots (n ∈ N,n ≥ 2) and we give an algorithmic solution to this problem. This solution is also based on the properties of the orbit of a billiard ball. In the last part we present our observations and conclusions based on a problem solving activity related to this problem. The initial problem for 3 pots is mentioned in [2] (The three jug problem on page 89), but the solution is not detailed and the general case (with several pots) is not mentioned. The visualization we use is a key element in developing the proof of our results, so the proof can be viewed as a good example of visual thinking used in arithmetic (see [3], [1]). Keywords: linear diophantine equation, visualization, billiard Mathematical Subject Classification: 97F60, 97E50, 97B20 Introduction The following problem was solved by Sim´ eon Denis Poisson using graphs in the 18 th century ([7]): A man has 12 pt 3 wine and he wants to give to a neighbor 6 pints but he has only a 5 pt and an 8 pt empty pot. How can he measure 6 pt to the 8 pt pot? Poisson’s idea was to represent the possible states of the pots as vertices of a graph while every possible filling corresponds to an oriented edge in this graph. In this way starting from the initial state (12, 0, 0) one can obtain all possible states as follows. For the first filling we have two possibilities, so we obtain two possible states: (4, 8, 0) and (7, 0, 5). From these states we can obtain the states (0, 8, 4), (4, 3, 5), (0, 7, 5), (7, 5, 0) and so on. In figure 1 we illustrated a few vertices and edges of this graph (at each level we put the states that were not already in the graph and for the simplicity we omitted the backward edges between different branches). However this representation also leads to an algorithmic solution (the generation of this graph level by level) we need a different approach in order to solve some general problems: 1 Email address: andraszk@yahoo.com 2 Email address: ors nagy@yahoo.com 3 1 pt (pint) is equivalent to 568.26125 ml 1