Report on 9 December prize question

No-one got the prize this fortnight. Daniel Huang discovered how to solve the problem, but didn’t write up his answer. Often mathematical problems are best done by looking at simple cases first, and then doing the more complicated working necessary for the general case. This is an example where the general case is actually easier than the simple cases.

circle-in-triangle

The problem was to show that if the radius, p, of the circular shape, is 1, then there is only one right-angled triangle exactly fitting around the circle. If p=2 there are two triangles; if p≥3 and p is prime, there are three.

By Pythagoras, (a+b)2=(a+p)2+(b+p)2

Therefore ab=ap+bp+p2

Or (a−p)(b−p)=2p2

2p2 can be factorised only as 1×2p2, 2×p2, or p×2p, and (a−p) and (b−p) must be whole numbers

Therefore the pair ((a−p),(b−p)) is (1,2p2), (2,p2), or (p,2p)

When p=1 all these three options reduce to (1,2)
When p=2 they reduce to two options, (1,8) and (2,4)
For p≥3, 2p2 is bigger than p2, which is bigger than 2p, so the three options are distinct.

The triple (a+p,b+p,a+b) defining the right-angled triangle is

(2p+1, 2p2+2p, 2p2+2p+1) or (2p+2, p2+2p, p2+2p+2), or (3p, 4p, 5p)

There is an odd pattern here. The right-angled triangle fitting exactly round the circle is either one with relatively similar non-hypotenuse sides, (3p,4p,5p), or a very lop-sided one, with the longer non-hypotenuse side only 2 or 1 less than the hypotenuse.

I don’t know why that pattern emerges.