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.

A number of authors have considered the following (and related) problems:

In the March 1981 edition of Mathematics Teacher Lee Markowitz gives the following full set of solutions (though not the considerable calculations involved):

(5, 12, 13) (6, 8, 10) (6, 25, 29) (7, 15, 20) (9, 10, 17).

However, in many formulae involving the triangle, it is the semi-perimeter rather than the perimeter that arises naturally.

You should never judge a book by its title - but I nearly always do. A mathematical colleague recommended a book to me Concrete Mathematics [1]. Just recently, someone else also referred me to the same book. Well a wink’s as good as a nod to a blind man and I now realise what a mistake I’d made; concrete is a conflation of ‘continuous’ and ‘discrete’. Like all good textbooks in mathematics, it distinguishes itself not just by the quality of its contents and presentation, but also by the superb collection of problems that it includes. One such problem - question 58 in Chapter 6 - inspired this article. There was a hint towards a generalised result, but there were no details. It did give a reference, [2, 3], but despite some efforts I found this unobtainable. Hence this article and my attempt to cover the same ground. The question posed in the book is concerned with the linear recurrence relations satisfied by the successive powers of Fibonacci numbers - so we start with the Fibonacci sequence itself.

It is widely, almost universally, believed that the Hebrew Bible gives the value of π as the crude approximation 3, a value much less accurate than those adopted by other ancient civilisations, such as the Babylonian, the Egyptian and the Chinese. In the English of the Authorised Version, the biblical source goes:

‘And he made a molten sea, ten cubits from one brim to the
other: it was round all about, and his [i.e. its] height was five
cubits: and a line of thirty cubits did compass it round about.’

In the geometry of the triangle, the Feuerbach quadrangle comprises the four points of contact between the nine point circle and the four tritangent circles. By omitting each of the four points in turn, we create four Feuerbach triangles. When the basic triangle is isosceles we find that two of the four points of contact coincide and the resultant single Feuerbach triangle is also isosceles. In the spirit of Steiner-Lehmus we address the converse question – if one of the Feuerbach triangles is isosceles, is the basic triangle necessarily isosceles? The answer is negative.

Doublet tasks consist of changing one word to another by altering single letters at each step to make a different word. The two words at the beginning and end of such a chain must, of course, be the same length, and they should be related to each other in some obvious way. They must not have identical letters in the same positions. All words in the chain should be common English words, proper names excluded. A perfect solution has a number of steps equal to the number of places the given words differ. For example: COLD, CURD, CARD, WARD, WARM. If a perfect change is not possible, the best solution is the shortest chain. For playing doublets as a game with two or more players, Lewis Carroll invented a set of scoring rules to determine who wins.