Re: [math-fun] Heronian triangles can be made to have integer vertex coordinates
Lunnon: 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. --so let me understand your question. What you want is a proof that every PRIMITIVIZED heronian triangle is integer-posable? I.e. you conjecture any Bucholz triangle AFTER dividing by GCD(a,b,c) is integer-posable? Call that "revised Lunnon conjecture." To repeat the formulas so far Bucholz: a := (n*n + k*k)*m; b := (m*m + k*k)*n; c := (m + n)*(m*n - k*k); semiperim=s = (m + n)*m*n, area=d = k*m*n*(m + n)*(m*n - k*k) = k*m*n*c; constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m*n*n/(m + 2*n) ) vertex coordinates proposed by WDS: A=(+n*(m-k)*(m+k), 0) B=(-m*(n-k)*(n+k), 0) C=(0, 2*k*m*n) a^2 = |B-C|^2 = |B|^2 + |C|^2, b^2 = |A-C|^2 = |A|^2 + |C|^2, c^2 = |B-A|^2. Well clearly G = GCD(a,b,c) = sqrt(GCD(a^2, b^2, c^2, area)) divides a, and b, and c, and 2*s=2*(m+n)*m*n; and G^2 divides area. If G divides 2*k*m*n=|C|, then G^2 divides |A|^2 and |B|^2 and |C|^2 and a^2 and b^2 and c^2 and area forcing G to divide a, b, c, A, B, and C. So "revised Lunnon" <== "G must divide 2*k*m*n." But we know G divides GCD(2*s, c, a+b). Hence G divides (m+n)*GCD(2*m*n, m*n-k*k, m*n+k*k). Hence G divides (m+n)*GCD(m*n, 2*k*k). But Lunnon-constraint implies that GCD(m*n,2*k*k)=GCD(m*n,2*k). Hence G divides 2*k*m*n. Therefore, the revised Lunnon conjecture, is also proven. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Sorry; I blew it at the end of my "proof." Kill the final two paragraphs please. Proof of "revised Lunnon" retracted. Can replace the killed part with all this: But we know G divides GCD(2*s, c, a+b). Hence G divides (m+n)*GCD(2*m*n, m*n-k*k, m*n+k*k). Hence G divides (m+n)*2*GCD(m*n, k*k). But Lunnon-constraint implies that GCD(m*n,k*k)=GCD(m*n,k)... I think. Hence G divides 2*(m+n)*GCD(k, m*n). We know G divides both a-b = (n-m)*(m*n-k*k) and c = (n+m)*(m*n-k*k) hence G divides 2*(m*n-k*k)*GCD(m,n). Conclude: G must be a multiple of GCD(m+n, m*n-k^2) and must divide 2*GCD( (m+n)*GCD(m*n,k), (m*n-k*k)*GCD(m,n) ). Hence G must divide 2*(m*n-k*k)*m*n and 2*(m+n)*m*n*k. Hence G must divide 2*m*n*GCD( m*n-k*k, (m+n)*k). G must divide a*n-b*m = m*n*(n-m)*(n+m). Hence G must divide m*n*(n+m)*GCD(n-m, 2*k). G must divide a*n+b*m = m*n*(n*n+2*k*k+m*m). consider this and c we find G must divide (m+n)*GCD( m*n*(n-m), m*n-k*k ). Well, I'm clearly going insane here. The question is, can we conclude from some combination of all that crap, that G must divide 2*m*n*k? If so, the revised Lunnon conjecture, is proven.
See the example I gave earlier, [a, b, c] = [5, 29, 30], s, d = 32, 72. Smallest Carmichael GCD for any perm is 60, with [m,n,k] = [12, 8, 9]. The altitude 2 d / a is not an integer for any choice of base a. WFL On 11/17/11, Warren Smith <warren.wds@gmail.com> wrote:
Lunnon: 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.
--so let me understand your question. What you want is a proof that every PRIMITIVIZED heronian triangle is integer-posable? I.e. you conjecture any Bucholz triangle AFTER dividing by GCD(a,b,c) is integer-posable? Call that "revised Lunnon conjecture."
To repeat the formulas so far Bucholz: a := (n*n + k*k)*m; b := (m*m + k*k)*n; c := (m + n)*(m*n - k*k); semiperim=s = (m + n)*m*n, area=d = k*m*n*(m + n)*(m*n - k*k) = k*m*n*c; constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m*n*n/(m + 2*n) )
vertex coordinates proposed by WDS: A=(+n*(m-k)*(m+k), 0) B=(-m*(n-k)*(n+k), 0) C=(0, 2*k*m*n)
a^2 = |B-C|^2 = |B|^2 + |C|^2, b^2 = |A-C|^2 = |A|^2 + |C|^2, c^2 = |B-A|^2.
Well clearly G = GCD(a,b,c) = sqrt(GCD(a^2, b^2, c^2, area)) divides a, and b, and c, and 2*s=2*(m+n)*m*n; and G^2 divides area. If G divides 2*k*m*n=|C|, then G^2 divides |A|^2 and |B|^2 and |C|^2 and a^2 and b^2 and c^2 and area forcing G to divide a, b, c, A, B, and C.
So "revised Lunnon" <== "G must divide 2*k*m*n."
But we know G divides GCD(2*s, c, a+b). Hence G divides (m+n)*GCD(2*m*n, m*n-k*k, m*n+k*k). Hence G divides (m+n)*GCD(m*n, 2*k*k). But Lunnon-constraint implies that GCD(m*n,2*k*k)=GCD(m*n,2*k).
Hence G divides 2*k*m*n. Therefore, the revised Lunnon conjecture, is also proven.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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