Show that for infinitely many positive integers n, we can find a triangle with integral sides whose semiperimeter divided by its inradius is n.
Solution
Let the sides be a, b, c the semiperimeter s = (a + b + c)/2, the inradius r, and the area A. We have A = sr, A2 = s(s - a)(s - b)(s - c) (Heron's formula) and s = nr (given). So nA = nsr = s2. So s4 = n2s(s - a)(s - b)(s - c). Hence (2s)3 = n2(2s - 2a)(2s - 2b)(2s - 2c). Putting x = -a + b + c, y = a - b + c, z = a + b - c, this becomes (x + y + z)3 = n2xyz (*).
Conversely, if we can find positive integers x, y, z satisfying (*), then 2x, 2y, 2z also satisfy (*) so we may assume x, y, z are even. Then we can find a, b, c such that x = -a + b + c etc by taking c = (x+y)/2, b = (x+z)/2, a = (y+z)/2. So s4 = n2A2 and hence s2 = nA = nsr. Hence s = nr. Thus it is sufficient to show that we can find a positive integer solution (in x, y, z) to (*) for infinitely many n.
We look for solutions with z = k(x + y), n = 3(k + 1). We require (x + y)3(k + 1)3 = 9(k + 1)2xyk(x + y), or (x + y)2(k + 1) = 9kxy. Put w = x/y. Then (1 + w)2(k + 1) = 9kw or w2(k + 1) - (7k - 2)w + (k + 1) = 0. We require w to be rational, so it is sufficient if (7k - 2)2 - 4(k + 1)2 = 9k(5k - 4) is a square. So it is sufficient if k = u2 and 5u2 - 4 = v2.
This equation has infinitely many solutions, for (u, v) = (1, 1) is a solution and if (u, v) is a solution, then u' = (3u + v)/2, v' = (5u + 3v)/2 is also a solution. To establish this we show that (1) 5u'2 - 4 = v'2, (2) u > u' and v > v', (3) u' and v' are both integers. For (1), we have 5(3u + v)2 - (5u + 3v)2 = 45u2 + 30uv + 5v2 - 25u2 - 30uv - 9v2 = 20u2 - 4v2 = 4(5u2 - v2) = 16, as required. For (2), we note that u' and v' are certainly positive, so u' > 3u/2 > u, and v' > 3v/2 > v. For (3) we claim that u and v are integral and have the same parity. Hence 3u + v and 5u + 3v are both even and so u' and v' are integral. Also u' + v' = (8u + 4v)/2 = 4u + 2v, which is even, so u' and v' have the same parity.
Thus we can find an infinite sequence of integral pairs (u, v) with u and v strictly increasing. That gives an infinite strictly increasing sequence of values of k which yield rational w. Then if w = r/s, we may take x = r, y = s, z = k(x + y) as a solution for n = 3(k + 1). So we get solutions for infinitely many n.
For example, the second member of the sequence (u, v) is (2, 4). That gives k = 4, and w satisfies 5w2 - 26w + 5 = 0, so w = (26 ± 24)/10 = 5 or 1/5. So n = 15 and we may take x = 1, y = 5, z = 24. Then (x + y + z)3 = 303, and n2xyz = 1525.3.8 = 303. So x = 2, y = 10, z = 48 is also a solution. So a = 29, b = 25, c = 6. Hence s = 30. So A2 = s(s - a)(s - b)(s - c) = 30.1.5.24 = 602. So r = A/s = 2 and s = nr, as required.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002