Pythagoras, however, had a rival, named Pyfudgerous.

He too had a theorem for working out the hypotenuse of

a right-angled triangle from the length of the other two sides.

Figure 1 contains the theorem that you probably used, namely:

Figure 1 also features a second theorem that I suspect is new to you. It too is a theorem for working out the length of the hypotenuse of a right-angled triangle from the lengths of the other two sides.

This pythagorean alternative I have named Pyfudgerous because I do not want you to trust it !

The truth is that the theorem of Pyfudgerous is simply the theorem of Pythagoras in an altered form. This is a sad moment. Something new and intriguing has turned out to be something old in a repackaged form.

in factorised form as

if it is the x-axis crossing points that are of interest or in the completed square form as

when it is the minimum point that is under consideration. Similarly, although the theorem of Pyfudgerous is not a piece of fundamentally new mathematics, it may still be of some significance.

What is the length of the hypotenuse of a right-angled triangle in which the difference between the two shorter sides is 97 and the product of the two shorter sides is 1680 ?

( I hope that you are busily working out a solution at this point, before reading on ! )

The Pyfudgerean solution is effortless, for the question, in effect, states that (B - A) = 97 and AB = 1680. So, double the 1680 and add on a squared 97. Now square root and up pops the answer: H = 113. Try out this problem on a friend. I have had many a happy moment watching two simultaneous equations awkwardly being solved from which it transpires that A = 15 and B = 112 and then, via Pythagoras, being informed that H = 113.

In fact, if it is required to find A and B, there is a further clever

to quickly find (B + A) = 127, which can be very easily solved simultaneously with (B - A) = 97 to give the correct solutions.

The grey square formed upon the hypotenuse of the black triangle contains the same number of tiles (i.e has the same area as) the two grey squares formed upon the other sides of the black triangle.

Figure 3 is a picture from which it is clear that the total area, (B + A)², is made up from four rectangles of dimension A by B plus a square of side (B - A). This is the previously used

Likewise, Figure 4 is a visual representation of the theorem of Pyfudgerous.

The large square of side H is of the same area, H², as the four triangles, ½AB each, plus the small square, (B - A)².

Hence:

One of the interesting features of Figure 4 is the fact that it is a square expressed in terms of a ring of four triangles

Figure 5 shows one such shrinkage and Figure 6 shows an infinity of such stepped shrinkages.

This is a fascinating picture. It spirals inwards and indefinite zooming in upon the centre reveals an unending, repeating structure. I am going to call Figure 6 the

The mathematical principle behind this process is outlined in Box 1.

Notice that it could be applied to any ringed square such as Figure 3, for example. In fact, it can be applied to any ringed

Mathematics Review, November 1993, Volume 4, Number 2.

© Philip Allan Publishers Limited. ISSN 0957-1280

The following Figure was submitted with the article but not published.