Diana runs from A to B with speed u and then from B to C with speed v. Eric runs from A to C with speed w. They start together and finish together.

If we now draw three lines from a point O, with OP parallel to AB and of length u, OQ parallel to BC and of length v and OR parallel to AC and of length w, then we find P, Q and R are collinear.

It is not difficult to construct a proof of the theorem so that is left as an exercise for the reader.

It was at the Fifth Southeast Asian Conference on Mathematical Education (June 1990, in Brunei Darussalam) that Brian Bolt posed the question: Can the numbers 1 to 12 be arranged within the hexagram below to give equal sums in the six indicated directions? This is the Magic Hexagram Problem.

MacKinnon introduced the interesting concept of an idiot-proof tile. This is a polyomino-shaped tile P which has the property that wherever two copies are placed (“by an idiot”) on the square grid, then—so long as the copies do not overlap or enclose a region (“even an idiot would not do that”)—it is possible to adjoin further copies of P so as to complete a tiling of the whole plane. MacKinnon proved that the L-triomino is an idiot-proof tile, and indicated how one could prove that the W-pentomino has the same property. The purpose of this note is to extend MacKinnon’s results, and to formulate some open problems.

Some time ago, I came across a book intended to popularise mathematics, whose last chapter dealt with the calculus of one variable. Its final section, evidently intended to climax the whole work, solved the problem of designing the proportions of a tin can so as to obtain the greatest volume out of a given amount of material.

The celebrated Steiner-Lehmus theorem states that if two internal bisectors of a triangle are equal then the triangle is isosceles. It beautifully illustrates the point that the converse of a theorem may be far more difficult to prove than the direct statement. The literature of the Steiner-Lehmus theorem is considerable with many contributions in the Mathematical Gazette.

An interesting variation on this theme may be found by examining the triangle formed by the second intersections of the medians with the circumcircle. Let the median point of a triangle ABC be G and let the medians meet the opposite sides at DEF and the circumcircle again at LMN.

Cyclic pairs

When you transpose the digits of a two digit number and obtain a multiple of the original, the original may be called a “cyclic pair”. I have before me, a computer printout obtained by Mr F. B. Satterthwaite of 439 East 51st Street, New York 10022, exhibiting all the cyclic pairs for which the multiple of the original obtained by transposing the digits is ⩽ 16. This computer printout exhibits patterns of many kinds. The most striking pattern that Satterthwaite identified was that cyclic pairs do not exist when the base is one less than a prime number, but do exist for every other base ⩾ 5. His claim may be substantiated as follows.

Ever since I was a schoolboy, my favourite polynomial has been

It is a wonderful polynomial, with lots of beautiful and useful properties which, alas, do not seem to be known to the present generation of mathematical students. To remedy this, we collect together some of the known properties of this polynomial and present some additional properties which appear to be new.

The array shown in the diagram, where each integer appearing shows the number of routes to that point, assuming there is a North-South wall that cannot be crossed, but otherwise constructed using the same one-step, no back-tracking rules as in the production of Pascal’s triangle, may be used as the starting point of an interesting investigation, provided one’s pupils know a bit about combinations, the binomial theorem and discrete probability distributions.

The Russian Multiplication Algorithm is frequently discussed in history of mathematics courses and in math education courses for elementary and secondary school teachers. Students are intrigued by the algorithm as it allows one to multiply any two whole numbers knowing only the multiplication and division facts for the digit 2.

Unfortunately, most textbooks do not address the natural questions which students ask about the algorithm: “Why are the even numbers crossed out in the left-hand column?” and “Can this be done with numbers other than two?” In a survey of eight textbooks by eight different pubishers, only one textbook answered the first question and none of the textbooks addressed the second question.

A number of deep theorems and unsolved problems ask questions about the relationship of the additive and multiplicative structure of the natural numbers. The famous four squares theorem of Lagrange says that if we form all possible sums of four squares we get the entire sequence of nonnegative integers. Goldbach’s conjecture states that if all possible sums of two odd primes are formed we get every even integer greater than 4.

The depth and difficulty of such problems stand in surprising contrast to the similar question: “Which numbers are not generated when all possible sums of two or more consecutive natural numbers are formed?” The answer is all powers of 2 and the proof is accessible to anyone who can sum arithmetic sequences. This result is an excellent discovery exercise for students, particularly appropriate for teacher training courses.

Soon after the publication of the new edition of the SMP booklet Formulae for advanced mathematics with statistical tables, I received a letter from a colleague objecting to the statement that

“If the origin is taken at a focus of the conic,…, then the equation of the conic has the form

where e is the eccentricity …”

As part of a daily game-show on Channel 4 television, contestants on Countdown have to combine up to six numbers to produce a specified (but randomly chosen) three-digit result. For example, a few weeks ago they were given the numbers 4, 5, 7, 8, 9 and 10. By use of addition, subtraction, multiplication and division (provided that the result is an integer) they had to use each number at most once to obtain the number 526. And they have only 30 seconds in which to do it! One possible solution turned out to be

(7 x 9 + 4) x 8 - 10 = 526.