The parameterisation given by Buchholz (spelling!) --- which he ascribes to Carmichael, incidentally --- does not guarantee a primitive triangle with GCD(edge-lengths) = 1, only some triangle similar to it. In fact, this is a major problem when generating lots of these things: there is no usably small bound on the generated scale-up GCD. For example, the smallest GCD for (any perm of) case [a, b, c] = [26, 73, 97] is 2275, using [m, n, k] = [175, 7, 30] . Fred Lunnon On 11/17/11, Warren Smith <warren.wds@gmail.com> wrote:
Re Fred Lunnon's Heronian triangle problem.
Lunnon says the Heronian triangles = the triangles with all 3 sides a,b,c, and the area, being integers.
Wikipedia http://en.wikipedia.org/wiki/Heronian_triangle says: Theorem: Given a Heronian triangle, one can split it into two right triangles, whose sidelengths form Pythagorean triples with rational entries.
This proves any Heronian triangle can be made to have rational vertex coordinates. Therefore after scale-up by LCM(denominators of coordinates), it can be made to have all coordinates integer.
Lunnon: Bucholtz' parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), semiperim=s = (m + n)m n, area=d = k m n(m + n)(m n - k^2); constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m n^2/(m + 2 n) ) enforce one (nonprimitive) triangle in each Heronian similarity class. The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a).
The wikipedia Pythag proof shows the altitude of the triangle is given by altitude = 2*area / base and the two legs of the right triangles which both lie on the base are ( hypot1^2 - hypot2^2 + base^2 ) / (2*base) and ( hypot2^2 - hypot1^2 + base^2 ) / (2*base).
Taking base=c=leg1+leg2 and hypot1=a and hypot2=b we find
altitude = 2*k*m*n leg1 = m*(n^2-k^2) leg2 = n*(m^2-k^2)
all of which are integers, proving Lunnon's "integer poseability" conjecture.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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