Good morning and thank you for coming to listen to my Presidential Address. You have been listening to Meat Loaf’s Bat out of Hell and for those of you who are used to seeing me in various costumes I apologise for not appearing as Batman on a Batbike, but I know my limits (like .

The illustration on the front cover of the Gazette is a bit of a turkey. It comes from an old mathematics exercise book by Henry Smith who started it in 1843 and appears just before he does some examples on the single rule of three direct in vulgar fractions. I wonder how long it took him to do it, but there has surely been some implicit mathematics done in managing to get the wording arranged fairly equally around a circle! The rest of the exercise book is very neatly presented and illustrated, though without any indication that the work has been marked in a summative or formative way. Hence it would not meet today’s standards.

Consider two related tasks:

A) The diagram below displays a fragment of a pattern known to extend indefinitely and to be generated by an endlessly repeating pattern. What kind of shading will the 137th cell have? What number cell will the 137th white cell be in?

B) What is the next term in the sequence 3, 6, 11, 18, 27, …?

Tasks like the first can provide excellent experience of detecting and expressing generality as one route to algebra, but only if learners are first asked to describe a structural relationship which generates the succeeding items. Tasks like the second feature in intelligence tests, but it is well known that such tasks are mathematically incomplete.

In 1840 C. L. Lehmus sent the following problem to Charles Sturm: ‘If two angle bisectors of a triangle have equal length, is the triangle necessarily isosceles?’ The answer is ‘yes’, and indeed we have the reverse-comparison theorem: Of two unequal angles, the larger has the shorter bisector (see [1, 2]).

Sturm passed the problem on to other mathematicians, in particular to the great Swiss geometer Jakob Steiner, who provided a proof. In this paper we give several proofs and discuss the old query: ‘Is there a direct proof?’ before suggesting that this is no longer the right question to ask.

We go on to discuss all cases when an angle bisector (internal orexternal) of some angle is equal to one of another.

In the July 2011 Gazette Paseau [1] considers a population in which each birth is an independent event with probability g of being a girl and b of being a boy (b + g = 1). He supposes that, due to a preference for girls, all couples stop after producing k girls, where k is an integer greater or equal to 1. Each new birth will, even after some couples have stopped producing, still have the probability g of being a girl and b of being a boy. Paseau correctly states and shows that this will mean that the ratio

Expected number of girl births : Expected number of boy births is g : b.

So it is the same as if all couples stopped randomly. Paseau was showing that this result was a consequence of the ratio g : b always being the same for all families.

However in humans the ratio g : b is not always the same. The ratio g : b can vary significantly with many variables between couples, which influence either the ratio at conception or the subsequent implantation and survival of male or female foetuses to birth. For example, mothers whose menstrual cycles have a short follicular phase (normally days 1-14 of the 28-day cycle) tend to produce boys [2], resulting in a smaller g : b ratio. Also the timing of insemination within the menstrual cycle is associated with the gender outcome such that insemination early or late in the fertile period means that the offspring is more likely to be male; if insemination occurs in the middle it is more likely to be female [3]. It is therefore likely that changes in parental hormone concentrations during the fertile period can affect the sexes of the children [4].

It is an elementary fact that a particle sliding on a rough horizontal table experiences a deceleration of magnitude μg, where μ is the coefficient of friction between the particle and the table. Thus, if the initial speed u of the particle and the braking distance d are both known, then μ can be determined by the formula

The idea of using braking distances to determine the coefficient of friction led the second author to ask the following question:

A particle slides on the interior surface of a rough hemispherical bowl starting from rest. If the starting point of the particle and the point at which it comes to instantaneous rest after its slide are both known, is it possible to determine the coefficient of friction between the particle and the bowl?

In [1] Deshpande and Shiwalkar explore the number of double heads that may appear in sequences of coin tosses. They sought to answer questions such as,

‘If a (fair) coin is tossed r times, how many possible outcomes result in 0 double heads, how many result in 1 double head, …, how many result in (r – 1) double heads?’

Their answer involves Fibonacci expressions. Hirschhorn [2] extended these results using generating functions which on expansion led to some closed form expressions and the statistics of their distribution. Griffiths [3] continues this theme. Another article that touches on the basic idea – a run of particular outcomes – is [4] which is also the source of other references; [5] is also concerned with the same idea.

Every tetrahedron has a set of quadritangent spheres, that is, spheres that touch the four planes containing its faces. In this article I discuss three questions about them.

1. How many quadritangent spheres does a given tetrahedron have?

2. How can we find their sizes and positions with as little calculation as possible?

3. What relations exist between the vertices and points of contact that lie on anyone of the planes?

1. Introduction

In the article [1], a follow-up to my article [2], Christian and Trustrum cite empirical evidence that the probability of a family giving birth to a boy or to a girl may vary from family to family. Letting bi and gi = 1 − bi respectively denote the probability of the ith family giving birth to a boy or to a girl, the suggestion is that bi may be distinct from bj for distinct i and j. As they point out, this has implications for the expected society-wide number of boys and girls. Following my discussion of the ‘stop after k girls’ policy, Christian and Trustrum introduce the related policy ‘stop after k girls or N children’. They argue that the expected number of children under this latter policy is an increasing function of bi (when 0 < k < N). The present article complements their discussion by examining this alternative stopping policy in more detail.

The perimeter of the ellipse x2/a2 + y2/b2 = 1 is 4J (a, b), where J (a, b) is the ‘elliptic integral’

This integral is interesting in its own right, quite apart from its application to the ellipse. It is often considered together with the companion integral

Of course, we may as well assume that a and b are non-negative.

Consider the problem of defining a continuous function f(x) which agrees with factorials at integers. There are many possible ways to do this. In fact, such a function can be constructed by taking any continuous definition of f(x) on [0,1] with f(0) = f(1) = 1 (such as f(x) = 1), and then extending the definition to all x > 1 by the formula f(x + 1) = (x + 1)f(x). This construction was discussed by David Fowler in [1] and [2]. For example, the choice f(x) = ½x(x − 1) + 1 results in a function that is differentiable everywhere, including at integers.

However, this approach had already been overtaken in 1729, when Euler obtained the conclusive solution to the problem by defining what we now call the gamma function. Among all the possible functions that reproduce factorials, this is the ‘right’ one, in the sense that it is the only one satisfying a certain smoothness condition which we will specify below. Admittedly, Euler didn't know this. It is known as the Bohr-Mollerup theorem, and was only proved nearly two centuries later.

First, a remark on notation: the notation Γ (x) for the gamma function, introduced by Legendre, is such that Γ (n) is actually (n − 1)! instead of n!. Though this might seem a little perverse, it does result in some formulae becoming slightly neater. Some writers, including Fowler, write x! for Γ (x + 1), and refer to this as the ‘factorial function’. However, the notation Γ (x) is very firmly entrenched, and I will adhere to it here.

Many students can get a lot of answers in A-level statistics right. But some of the ideas behind the calculations are somewhat subtle. Some of the subtleties are occasionally brought out in examination questions, usually those requiring verbal or discursive answers; unfortunately answers to these questions tend to be poor. Such questions are often dismissed as ‘waffle’ by strong students, while weaker ones generally try to avoid issues of understanding by memorising what they hope will be relevant slogans. Even respected textbooks are open to criticism. This is all very sad as I think it is here that much of the interest of the subject lies; ignoring the subtleties risks reducing the subject to a series of meaningless formulae. There should be time to focus on what is actually happening, as well as on mere technique. This article looks at some common problem areas.

In this note we define and study certain subsets of the solutions of the Diophantine equations of the type

(1)

the coefficients a, b, c and d being integers, with gcd(a, b, c) = 1, and values of the unknowns X, Y and Z sought in rational numbers or integers, with X, Y, Z not all zero. Since (1) is a homogeneous equation it is elementary to pass from rational solutions to integral ones, and also in reverse. We chose to work with rational numbers because of the flexibility in computation they allow. This flexibility will enable us to prove that, under certain conditions, with three distinct solutions we can build infinite subsets of new solutions of (1), the properties of which are the main topic of this note.

As documented by L. E. Dickson [1], formulae said to give all the solutions, either of this general equation (1) or some particular cases, have appeared in the literature at least since the first half of the 17th century. Dickson himself provided the general formulae expressing all the solutions of (1) in rational numbers from an already found particular solution [2, pp. 44-45]. Similar study and formulae are part of a recent article [3, Lemma 2-1] where the solutions of (1) are applied to the study of a particular lattice. In essence these formulae prove that from any single solution of (1) we can compute all the others, which form an infinite set.

In this paper we show how to derive Vieta’s famous product of nested radicals for π [1] from the Archimedean iterative algorithm for π, [2, 3]. Only simple algebraic manipulations are needed.

The Archimedean iterative algorithm for calculating π uses a method involving the two equations:

and

In using this algorithm, we start with a circle of diameter 1. Imagine two regular polygons each with the same number of sides, circumscribed and inscribed to this circle. The larger one has perimeter a0, the smaller has perimeter b0. Since the perimeter of the circle is π we have b0 < π < a0. Now consider regular polygons with twice the number of sides that circumscribe and inscribe the circle and call their perimeters a1 and b1 respectively. These can be calculated from (1) and (2). Continuing in this way we generate the perimeters of inscribed and circumscribed regular polygons, and in each case the number of sides is twice the sides of the previous polygon. Clearly the sequence {an} approaches π from above and {bn} approaches π from below.

In the autumn of 2011 the movie Sherlock Holmes: A Game of Shadows appeared (see [1]). It was a sequel to the Holmes movie from two years earlier. Unlike the first movie, A Game of Shadows Holmes' arch enemy Professor Moriarty featured more prominently than Holmes himself.

One scene, in particular, has Holmes sizing up his opponent by making a visit to Moriarty at his university office. But in the background sits Moriarty's blackboard, filled completely with mathematical equations, symbols, drawings, and formulas. One of these deserves our special attention.

We came across the ‘two girls’ version of the children's gender problem nearly 35 years ago. How we came to it we cannot remember, but Martin Gardner had published a variant of it in the Scientific American in 1959. It re-emerged for us in the summer of 2010, following the publication of an article in Science News [1]. Subsequently Keith Devlin wrote about how this re-emergence impacted on him, and noting that ‘Probability Can Bite“ [2]. The mathematics herein reflects and extends that in Devlin's article.

In case the reader has not encountered the problem before, we first pose four problems.

1. A family has two children. One of them is a girl. What is the probability that they are both girls?

2. A family has two children. The younger is a girl. What is the probability that they are both girls?

3. A family has two children. One of them is a girl, and she was born on a Tuesday. What is the probability that they are both girls?

4. A family has two children. One of them is a girl, and she has green hair. What is the probability that they are both girls?

Recall that for integers n ≥ 2, ζ (n) is defined by

Of course, ζ (1) is not defined, since is divergent. A well-known particular value is ζ (2) = π2/6: numerous alternative proofs of this fact have been presented in the Gazette, e.g. the recent notes [1], [2].

We consider two connected problems:

• • For a given but otherwise arbitrary triangle in the plane, to construct similar triangles which ‘meet’ this triangle.

• • To find the triangle so formed which has least area.

1. Constructing a triangle which meets another

These problems beg the question of what is meant by ‘meet’ and we now aim to make this precise:

When triangle XYZ is actually ‘in’ the triangle ABC, ‘meet’ is synonymous with the traditional ‘inscribe’ (such as in case (1) below). For ‘inscribe’ we understand that some of X, Y, Z may coincide with the vertices of ABC (such as case (2) below).

More generally we use ‘meet’ to extend these possibilities by allowing XYZ to meet triangle ABC with its sides produced externally (such as case (3) below).

In a recent paper [1] Michael Hirschhorn and the present author had to prove that the polynomial

is positive for all integers n ≥ 4. We did so by writing down the identity

Since all the coefficients of the quotient polynomial are positive as well as the remainder, this shows by inspection that f(n) > 0 for n ≥ 5 and computing f(4) = 12025 completes the proof. This example illustrates a useful and efficient method for proving that polynomials have strictly positive values for all real numbers exceeding a given one. We leamed this method from a paper by Chen [2], (see [3] and the references cited there), and we have subsequently used it in our own researches [1,4].

We also conjectured in [1] that the theoretical basis of the method is a true theorem and the present paper is dedicated to proving this conjecture.

Theorem 1: Let f(x) be a polynomial with real coefficients whose leading coefficient is positive and with at least one positive root x = a. Then there exists an x = b ≥ a such that

where f(b), and all the coefficients of the quotient polynomial g(x), are non-negative.

The following three inequalities hold:

The first inequality is trivial. The second one was proved without calculating aids in note [1], and the third along similar lines in note [2]. The author of note [2] also suggested an extension to the relation between

How best to continue the sequence of inequalities is not obvious and we return to that point shortly. Before doing so, we note that an interesting geralisation is to replace π by a variable x, and to determine the precise interval of x in which the regularity of the ‘alternation of inequality signs’ is maintained. We need no longer consider particular properties of π.