Find an infinite set of incongruent triangles each of which has integral area and sides which are relatively prime integers, but none of whose altitudes are integral.
Solution
One possible answer is: r2 + 4, r4 + 3r2 + 1, r4 + 4r2 + 3 with r any integer not ±2 mod 5.
We use Heron's formula. Take the sides as a + b, n, n + a - b. This involves no loss of generality and makes Heron's expression more convenient. Let the area be A. Heron gives A2 = (n+a)(n-b)ab. Suppose we take n - b = ar2, n + a = bs2. Then A = abrs. We require a(r2 + 1) = b(s2 - 1). The simplest idea is not to take a = s2 - 1, b = r2 + 1, but then the area is rs(s2 - 1)(r2 + 1), n = s2r2 + 1 and n + a - b = (s2 - 1)(r2 + 1) which divides the area, so one of the altitudes is integral.
The next simplest idea is to take b = 1, s - 1 = r2 + 1, a = s + 1 = r2 + 3. This gives sides r2 + 4, r4 + 3r2 + 1, r4 + 4r2 + 3 and area A = r(r2 + 2)(r2 + 3). We need to check that each pair of sides is relatively prime and that each altitude is non-integral.
Any common factor of r2 + 4 and r4 + 3r2 + 1 must also divide r2(r2 + 4) - (r4 + 3r2 + 1) = r2 - 1 and hence also (r2 + 4) - (r2 - 1) = 5. But 5 only divides r2 + 4 if r = ±2 mod 5, so for other values of r, these two sides are relatively prime.
Any common factor of r2 + 4 and r4 + 4r2 + 3 must also divide r4 + 4r2 + 3 - r2(r2 + 4) = 3. But any square is 0 or 1 mod 3, so r2 + 4 cannot be divisible by 3. Hence these two sides are relatively prime. Finally, any common factor of (r4 + 3r2 + 1) and (r4 + 4r2 + 3) must also divide their difference r2 + 2. Hence also (r4 + 3r2 + 1) - r2(r2 + 2) = r2 + 1 and hence 1. So these two sides are relatively prime.
r2 + 4 is relatively prime to r2 + 3, has at most a factor 2 in common with r2 + 2 and at most a factor 4 in common with r. So r2 + 4 can only divide A if r = 2. If (r4 + 3r2 + 1) divides A then it also divides (r2 + 2)(r4 + 3r2 + 1) - rA= r2 + 2. But (r4 + 3r2 + 1) > r2 + 2. So (r4 + 3r2 + 1) does not divide A. Finally, if (r4 + 4r2 + 3) = (r2 + 3)(r2 + 1) divides A, then r2 + 1 divides r(r2 + 2), but that is impossible since r2 + 1 and r2 + 2 are relatively prime and r2 + 1 > r.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002