In this paper, we collect some results on a problem which has been considered by many authors over the years, including Ed Pegg Jr., R. M. Kurchan, William Rex Marshall, David Neary, Richard Korf and Eric Huang. The results have been rediscovered many times by many people, including the present authors, but very little has been published, except in a little known journal which no longer appears. Some of the cases have been solved by hand while others have been covered by Korf [1, 2], using a computer. We see our main role as that of collectors of information and presenters of results, in the hope that others will be inspired to study what is an exciting and very interesting area of combinatoric geometry; we believe, however, that some of our observations, variations and upper bounds are new.

One of the problems that will vex any President of the Mathematical Association is the topic of the address with which he or she closes his or her year of office. This occupied me, on and off, for more than a year. In my case, in addition to my desire to acknowledge the honour of the invitation made to me, I was deeply conscious of the fact that I would be the 100th individual to serve as President. I dabbled with some pet themes, typically concerning the lack of genuine problem-solving or proof in English school mathematics, before concluding that the most sensible thing would be to talk on the topic about which I know most. My research interests are in comparative mathematics education. I have been fortunate, over the last twenty years or so, to have been able to visit and videotape mathematics classrooms in several European countries. In so doing I have had my understanding of mathematics teaching transformed in ways that led, almost inevitably, to the theme of both this talk and the conference which brought my Presidency to an end: Learning from Others.

The beautiful representation of π by Wallis's product formula has attracted many authors who have given a variety of proofs (see [1, 2, 3, 4, 5, 6, 7]). The purpose of this note is to revisit some elementary derivations with a focus on probabilistic/statistical flavours of this formula. The famous Wallis's product formula for is given by

There are a number proofs available for this formula scattered in the literature. Here we will concentrate on the probabilistic nature of this formula.

Among the many astonishing formulae stated by Ramanujan we find in his Notebooks the following sequence (see [1], p 96, Entry 43):

which hold for any triple of complex numbers (a, b, c) such that a + b + c = 0.

Ramanujan concluded this list by ‘And so on’, which suggests that he had some kind of method or algorithm allowing him to extend this list to any even power of the quadratic form p(a, b, c) = ab + bc + ca, when a + b + c = 0. To explain this, Bruce Brendt details a theorem by S. Bhargava [2], which can be used to produce identities of the kind above, and infers that Ramanujan is likely to have used the same proof to establish his identities (see [1], pp. 97-100)

Suppose that we have a computing machine that can only deal with the operations of addition, subtraction and multiplication, but not division, of integers. Can such a machine be used to find the decimal expansion of a given real number α to an arbitrary length? The answer is ‘yes’, at least in the sense that α is the limit of a sequence of rational numbers, and one can obtain the decimal expansion of a positive rational number a/b without division. For example, for any positive k, we try all non-negative c < b and d < 10ka, and see if 10ka = db + c. There will be success because all we are doing is reducing 10ka modulo b in a particularly moronic way. This guarantees the existence and uniqueness of d, and d/10k is then the desired expansion. Take, for example, = and k = 12; then, after a tedious search, we should find that, for c = 3, there is d = 230 769 230 769 because 3 × 1012 = 230769230769 × 13 + 3, so that = 0.230769230769 + × 10-12. Such a procedure of searching for c, d is hopelessly inefficient, of course, and a more efficient method is given in the next section.

Notation: If A, B are points, then AB denotes the line AB rather than just the segment from A to B, unless stated otherwise. If l, m are lines, then their meet is denoted l m. The circle through points A, B and C (the circumcircle of ΔABC) is denoted ⊙ABC.

Given ΔABC, let the points Xi, Yi, Zi, lie on BC, CA, AB, respectively, for i = 1, 2, 3; that is, three distinct points on each side of the triangle, and suppose

The game ‘toss up’, manufactured by Patch Products (http://www.patchproducts.com/). is played with ten fair, six-sided dice whose sides are coloured green, yellow and red. In particular, each die has 3 green sides, 2 yellow sides and I red side. Ordinary fair six-sided dice may be used instead, of course, provided three sides are identified as ‘green’ (e.g. 1, 2, 3), two sides are identified as ‘yellow’ (e.g. 4,5) and one side is identified as ‘red’ (e.g. 6). Players take turns with each turn consisting of (potentially) several rolls of dice. All the dice are used on the first roll of a turn. Players accumulate points as they play and the first player to reach 100 points wins.

How many times must a die be thrown to get four consecutive sixes? More generally, and more precisely,

Q1. What is the expected number of times we must throw a die to get n consecutive sixes?

The answer, rather suggestively for the numerologically inclined, turns out to be

A1. The expected number of throws is 6 + 62 + 63 + … + 6n.

In some well-known dice games players throw several dice at the same time: in craps a pair (brace) of dice is thrown, while in yahtzee, yam and balut, up to five dice are thrown at a time. However, Q1 is not the same as

What is the most symmetric quadrilateral you can think of? Most people would say ‘a square’, but having given this (or any other) answer, can you create your own definition of a ‘quadrilateral’, and of ‘most symmetric’, so that you can justify your answer as a rigorous piece of mathematics? This is a challenging investigation because the answer may not be the obvious one.

Often the main difficulty in solving a problem lies in finding appropriate definitions; here we must first consider the questions ‘what is a quadrilateral?’, and ‘what is a symmetry of a quadrilateral?’.

Pascal's triangle is well known for its numerous connections to probability theory [1], combinatorics, Euclidean geometry, fractal geometry, and many number sequences including the Fibonacci series [2,3,4]. It also has a deep connection to the base of natural logarithms, e [5]. This link to e can be used as a springboard for generating a family of related triangles that together create a rich combinatoric object.

2. From Pascal to Leibniz

In Brothers [5], the author shows that the growth of Pascal's triangle is related to the limit definition of e.

Specifically, we define the sequence sn ; as follows [6]:

When William Wynne Willson died in the summer of 2010 he left behind an incomplete text covering a number of geometrical topics, including the so-called ‘Simson line’. Geoff Wain, who was completing and editing the text with a view to possible publication, sent me a copy. One chapter, entitled ‘Whose line is it anyway?’ contains the following paragraph.

‘In 1799 the Scottish mathematician William Wallace … published a geometric theorem. Nowadays this result is known as ‘Simson's line’. It is not clear why it should have this name as Robert Simson, another Scottish mathematician, had died aged 80 in 1768, the year in which Wallace was born. Although Simson published a considerable body of mathematical writings … there seems to be no trace of this theorem among them.

The lower polynomials are inextricably linked to the symmetries of polyhedra and Platonic solids [1, 2, 3], and the quartic is no exception; its alter ego is the regular tetrahedron [4]. In this article we present a solution to the problem of aligning the vertices of a tetrahedron with the roots of a particular quartic. After establishing the size of a quartic-equivalent tetrahedron, we derive a triple-angle expression for the alignment rotation, analogous to that for the cubic [5].

In an article entitled ‘A Mathematical Theorem about Golf’ [1] G.H. Hardy introduced a simple model of golfing. He assumed, that, at one hole, a golfer has probability p of gaining a stroke with a single shot, and probability q that his shot costs him a stroke. Such strokes will be described as good (G) or bad (B), respectively, leaving probability 1 − p − q for an ordinary (O) stroke (see also [2]). For example, on a par four hole, successive strokes OGO will result in a birdie (a score which is one stroke less than par) and BBGOO in a bogey (a score which is one stroke more than par). In this paper the probability distribution P(Tk = n) will be derived for the number of strokes T a player may take on a hole of par k.

The forgotten continued fractions

Lord Brouncker's continued fraction for π is

(In this paper we will use the more convenient notation This fraction first appeared in the Arithmetica Infinitorum [1] by John Wallis. In this book, along with topics that lead to Newton's calculus, Wallis derives his famous product

Wallis tells us that he showed this product to Lord Brouncker, and in turn Brouncker converted it not only into (1), but into an infinite sequence of continued fractions.

In 1838, the Belgian mathematician Eugene C. Catalan (1814-1894) discovered that the number Cn of well-fonned sequences, with n pairs of left and right parentheses, is given by where n > 0 [1, 2]. For example, there are exactly five well-formed sequences with three pairs of left and right parentheses: ()()(), ()(()), (())(), (()()), ((())). The case n = 0 yields the null sequence, often denoted by λ. Notice that ()) and ((()()), for example, are not correctly parenthesised.

Some years ago I was invited to take part in a fund-raising competition, [1]. The rules had an interesting feature that led me to analyse the situation mathematically. This resulted in some interesting formulae and one inequality, which I will describe here.

The competition offered a single prize and the entry form looked similar to the partially-completed example illustrated below.

The form listed 40 categories similar to the ones shown. Participants were required to write down an ‘answer’ for each category, e.g. for ‘an animal’, you might choose DOG. No other indications were given, so each answer was a completely free choice.

Given that it took until the 19th century for the case n = 5 of Fermat's last theorem to be settled, it is not surprising that Fermat's claim of having a proof for all exponents greater than 2 is nowadays treated with considerable scepticism. However, perhaps the most important aspect of his claim has been the impetus it has given to the development of mathematical techniques over the three and a half centuries leading up to the proof by Wiles and Taylor [1, 2]. Not least amongst these techniques has been Fermat's own idea of proof by descente infinie [3,4].

You see a Martian walking on the street. What do you do?

(i) seek shelter

(ii) take some pictures

(iii) introduce yourself and offer him/her/it a ride

(iv) go online to buy lots of stocks.

In this article we analyse the long term behaviour of stock investments.The above unusual question will be answered too. In the discussion the notion of ‘almost surely’ will be often used. An event happens almost surely if it happens with probability 1. Note that it is still possible that an almost certain event A does not happen, but the probability of that (i.e. the probability of the complement of A) is zero.