Geometrically considered the sateen arrangement is a certain regular method of arranging points on squared paper, each point being placed at the centre of a square. In textile work the sateen arrangement is used, (1) as a basis for actual weaves, (2) as a mode of distributing over the surface of a cloth the repeats of an ornamental design.

The extent of the complete domain can be estimated by comparing it with its real part, that is, with the visible universe. For the sake of this comparison, von Staudt speaks as though it were possible to specify the number of real elements in any base, but gives an explicit warning as to the meaning he attaches to the symbol n which he uses for an infinitely great quantity. The number of real points on a line he denotes by n + 1; the one is apparently the ideal point, the n belonging to the actual universe. Now every imaginary pair can be assigned by means of a harmonic cast with one point, A, fixed. If this be (ABA′B′), we have a choice of n(n − 1) pairs for BB′, but no further choice; the number of imaginary points is therefore 2 × n(n−1), that is, n(n−1). Hence if the number of real points on a line is denoted by n+1, the total number of points on the line is n2+1. On an imaginary line of the first kind the points are one real, n imaginary; on an imaginary line of the second kind all are imaginary.

Suppose you are trying to tile an infinite grid of unit squares with 2 x 1 rectangles so that each rectangle covers two squares. You place two of the rectangles without thinking. Can the tiling now be accomplished? As the answer is “yes” I call the rectangle an “idiot-proof tile”. In general a polyomino will be called idiot proof if the infinite grid can be tiled irrespective of the positions of two of the tiles (the idiotic tiles), except that the idiotic tiles shall not be overlapping and shall not completely surround an area that is less than that of the tile (not even an idiot would do that!). The first interesting idiot-proof tile (IPT) is the L-tromino.

The work of writers like Rouse Ball in England, and Cajori and D. E. Smith in America, in revealing the fascinating story of the history of- arithmetic, has shown the subject developing slowly but surely along the corridor of time. This development, which culminates in present-day arithmetic, proceeds not only by increase of subject-matter, but involves suppression and modification both of matter and methods. It is not the haphazard growth of the wild plant, but the steady and systematic development which results from the pruning and shaping of the skilled and sympathetic gardener.

I. There are three ways of trisecting an angle bounded by OA and OB ( = 3αor 3α +2π or 3α +4π), namely:

(i) OX1 and OY1,

(ii) OX2 and OY2,

(iii) OX3 and OY3,

as shown in Fig. 1.

It is well known that if the Euclidean angles of a triangle ABC are trisected as in Fig. 2, then the triangle xyz (commonly quoted as Morley’s Triangle) is equilateral. But angles CAB, ABC, BCA may be trisected each in three ways, so that there are really 27 “Morley” triangles to be considered. The purpose of the present paper is to show that 18 of these 27 triangles are equilateral.

Peter is a lively lad. All systems are at ‘go’. He will usually eat when he’s hungry, like any healthy boy.

There was no trouble about lunch last Saturday. It was roast chicken with bread sauce—one of his favourites. He tackled it with relish, and we were smoothly through to pudding at a brisk rate of knots.

But today things are different. He’s hungry all right, and it’s roast chicken again. But he won’t make a start. Why not? Now that he’s six he knows what he likes, and Mum has forgotten the bread sauce. Would you believe it? But it’s too late to make some now, so we shall all have to be satisfied without it.

In two articles in the Mathematical Gazette, Pedal Triangles and Pedal Circles, July 1919, and Some Properties Relative to a Tetrastigm, Oct. 1920, J. H. Lawlor gave proofs of a number of properties of the pedal circles of the four vertices of a plane quadrangle ABCD with respect to the triangles formed by the remaining three vertices. The articles concluded with the statement of ten results, the first nine of which were fairly easily deducible from those already proved, but No. (10) was given as a conjecture. This was that the common chord of the pedal circles of A and D bisects BH1 and CH2, where H1and H2 are the orthocentres of the triangles BAD and CAD, and hence that BH2, CH1 meet on the common chord of the pedal circles of A and D. It can be seen that the second result would imply the former, for if h is the rectangular hyperbola through ABCD, which is unique unless the points are orthocentric, and if O is the centre of h, then H1 and H2 also lie on h, and it is known that the pedal circles pass through O. The meet of BH2, CH1 is a diagonal point of the quadrangle BH1H2C inscribed in h, and its join to the centre of h is the diameter conjugate to the parallel chords BH1CH2 and therefore bisects them.

What are your mathematical gems? Which results or ideas in mathematics give you the greatest pleasure? Which of them would you choose to take with you to while away the hours on a desert island? Is there a mathematical book above all others that you would wish to have with you? How would you make your choices? And what would be the reason for each specific choice? Similar questions are asked of celebrities (and noncelebrities to boot) in a long-running and very popular radio programme in Britain called Desert Island Discs. The difference is that the radio guests are asked to choose a number of pieces of music, together with one book (other than the Bible or Shakespeare, which are routinely provided), and one luxury.