The classical river crossing problem of the jealous husbands involves three couples who have to cross a river using a boat that holds just two people. The jealousy of the husbands requires that no wife can be in the presence of another man without her husband being present. This can be accomplished in 11 crossings (i.e. one-way trips). Tartaglia gave a sketchy solution for four couples but Bachet pointed out that this was erroneous and that four couples could not get across the river. In 1879, De Fontenay pointed out that four or more couples could cross the river if there was an island in the river and gave a solution for n couples in 8n – 8 crossings. Dudeney improved the solution for n = 4 and Ball noted that this gives 6n – 7 crossings for n couples.

“Find the modulus and the argument of each root of the equation:

(University of London A level, part question.)

We were investigating iterative methods of solving the equation, using a micro-package (in BASIC) which performs complex arithmetic and displays the results in an Argand diagram. The facility for “doodling” which this provides led to some interesting conjectures and results, of which this note is an account.

Since the introduction in 1978 of the idea of a “common-core” for A-level mathematics, there was been a continuing debate over the structure and content of the subject at this level. Much of this debate remains pertinent in 1989: see, for example, [10], [15] and [28]. However, the setting-up of the Higginson Committee to review A-levels in March 1987 had the effect of putting much of the discussion specific to mathematics on the back-boiler. It is well known that the committee’s report of May 1988 recommended a wider subject-spread, roughly speaking on the model of the International Baccalaureate, and that its immediate rejection by the government created considerable furore in educational circles.

There are some surprising connections between properties of solid convex polyhedra and the “nets” which can be used to make their shapes. These connections may be used as a starting-point for some interesting classroom investigations to “spot the pattern” or “guess and test”.

For a convex polyhedron, such as the pyramid shown below on the left, Euler's relation says that

Vertices + Faces = Edges + 2

In the elementary theory of collisions of two small particles, it is a well-known result that the momentum of the system as a whole is the same after impact as before. This is because the impulses on the two particles are taken to be equal and opposite, thereby implying zero total impulse on the system. It is not obvious that the same will be true after rolling snooker balls collide because, apart from the impulsive forces due to the impact, frictional forces also act on the balls. These are caused by the friction of the table when the balls are in a temporary sliding phase immediately following the impact.

In exploring the probabilities involved in a roll-a-penny game, some properties turned out to be a little more complicated than expected. The main aim here is to find the probability that the player will lose after a certain number of rolls.

Suppose we have a board upon which squares are drawn, and that the side of each square is four times the radius of a penny. (In practice, this means that the squares are of side 4 cm.) The first question is, “What is the probability that the penny lands entirely within a square, assuming that the dividing lines are of negligible thickness?”

The mensuration formulae for curved objects ought to be one of the highlights of everyone’s elementary mathematical education. I get my pupils at 13 and they've already been taught that the area of a circle is πr2. It is very upsetting to find that they usually have seen no sort of justification of this result, so they don’t think of it as surprising when π, which was introduced in the circumference formula, mysteriously pops up again in the area formula.

This year I thought I would put some effort into justifying the surface area formula for the sphere. I don’t claim that what follows is very original, and of course the main idea goes back to Archimedes, but I hope this article will save readers a bit of work at least.

Suppose we are given a sequence of numbers defined by a linear recurrence relation with constant coefficients, for example the Fibonacci numbers which are defined by: