Consider a square array of 25 grid squares, the grid squares being of two types distinguished by the colours black and white. A basic square of four adjacent grid squares can be selected from such a square array in 16 possible ways. If these 16 basic squares are all distinguishably coloured, then the 5x5 square array is said to be pantactic.

A graph consists of a set of vertices some pairs of which are joined by a single edge. A path is a sequence of distinct vertices (x1, x2, …, xm) such that consecutive vertices xi and xi+1 are joined by an edge (xi, xi+1), for i = 1, 2, .., m — 1. One of the unsolved problems of graph theory is to obtain criteria for determining whether any given graph has a Hamilton path, i.e., a path which passes through each vertex exactly once. (For pertinent references and further definitions, see [1].)

A bag contains m > 0 black balls and n > 0 white balls. A sequence of balls from the bag is discarded in the following manner:

• (i) A ball is chosen at random and discarded.

• (ii) Another ball is chosen at random from the remainder. If its colour is different from the last it is replaced in the bag and the process repeated from the beginning (i.e. (i)). If the second ball is the same colour as the first it is discarded and we proceed from (ii). Thus the balls are sampled and discarded until a change in colour occurs, at which point the last ball is replaced and the process starts afresh.

Let (α, β) and (X, Y) be two pairs of values satisfying y = xn, where n is a positive integer and where we may at present regard α and β as constant.

We have

where the last bracket contains exactly n terms.

After the first part of this article had been written, it was suggested to the writer that indeterminate forces and the initial conditions might have something to do with the discrepancy between the two different answers obtained to Example 2. Initially the disc might not just be resting on the pegs; it might have been rammed down.

Applying this method we evaluate (see Fig. 1) the projections of the vector (the height at P2 of the given triangle), and also the co-ordinates of the point in which the perpendicular form

the origin meets the plane of the triangle. In this way, as will be shown, we find immediately the co-ordinates of the perpendicular projection of the given point P2(x2, y2, z2) on the straight line passing through P3 and P1, and also the equation of the plane of the triangle.

According to orthodox teaching texts the tangential component of the reaction between two surfaces in contact is μ times the normal component when slipping is just about to take place. The tangential component cannot exceed this value, so it is asserted, and is equal to this value when slipping is occurring. The coefficient μ is a property of the surfaces in contact and is a constant, independent of the normal component. This may be a basis for some mathematical exercises, but typical texts even go so far as to say that these statements are true within the limits of experimental error (see for instance Quadling and Ramsey, Elementary Mechanics Vol. I, Bell, 1963, p. 58). Yet no experiments on this are normally performed or discussed in schools. One could argue that since the coefficient varies according to the nature of the surfaces it is so particular as to be not worth measuring, but this would equally justify not doing any exercises on the idea.

We consider here three geometrical problems, the solutions of which are not hard, but each involves some rather unobvious point. The reader is invited to attempt them before looking at the solutions. The names of the original proposers are noted in brackets.

Suppose that a1, …, an and c1, …, cn are two sets of n positive numbers. If the arithmetic mean of the ci is at least equal to that of the ai, when can we say non-trivially that the geometric mean of the ci is at least equal to that of the ai and that equality will require equality (in some order) of the ai and ci?

A well-known answer of course occurs when each c is equal to the arithmetic mean of the ai. When the two sets each consist of three numbers I can give an answer including this important case. To obtain this result it is sufficient to confine the ci to the interval from the least to the greatest of the ai.

A problem arose recently concerning the derivation of Rodrigues’ formula for the generalized Laguerre polynomials from the corresponding formula for the ordinary Laguerre polynomials. The derivation involved a generalized form of Leibniz’s theorem which I think may be of interest.

In a Note published in the Mathematics Student, Vol. XXI, (1953), Hansraj Gupta gave a method of determining the centre and axes of a conic without recourse to the usual method of shifting the origin and rotation of the axes of reference. The present author in a Note published in the same journal, Vol. XXIII, (1955), has treated the problem by a still more simple method, and has also derived directly the equations to the asymptotes of a hyperbola. The present paper deals with the determination of the parameter and the location of the parabola by the new method.

The method of finding the maxima and minima of a function of two variables f(x, y) in the unrestricted case is straightforward in theory One solves and then substitutes the roots in Hessian function

This investigation grew out of the need to design a piece of electronic test-equipment. The purpose of the equipment was to study the effect of varying capacity upon the behaviour of a number of electric circuits, and it consisted basically of an automatically-stepping switch which presented at its output terminals a controlled sequence of values of capacity. The nature of the test demanded that the set of capacity values should as nearly as possible cover a given range in a logarithmic scale, and to economise in components the equipment was to generate the set of values from a basic set of condensers, by creating all possible parallel-connected combinations. The problem was, how to choose the basic set? The analysis has a number of points of combinatorial interest, and illustrates fairly well the ad hoc approach of the industrial mathematician.

In a previous article [2] I gave a synthetic geometric proof of theorem which had been proved algebraically by T. G. Room [1] In the second part of my proof I used Pascal’s theorem, but Professor Room has since told me that he would have liked a proof which does not involve the theory of conies. I give such a proof here having first restated the theorem in two parts, as in [2],

This note deals with three differential equations referred to in The Mathematical Gazette, Vol. XLVI, No. 356, May 1962.

1. H. E Tester (Mathematical Notes, No 3010) solves the differential equation

from a geometric aspect, stating that it cannot be solved by a standard procedure.