[math-fun] Heronian primitive triangles -- single formula describing all
I also point out that (assuming Reid proof valid) by combining my Buchholz-->nonprimitive coordinates formula, and the Gaussian GCD, we can actually write just a single formula in terms of n,m,k, rational operations, and Gaussian GCD, giving coordinates for every PRIMITIVE Heronian triangle's vertices A,B,C. That is quite cool. Here are the nonprimitive coordinate formulas: xC = m*(n-k)*(n+k); yC = 2*k*m*n; xA = n*(m-k)*(m+k) + xC; yA = 0; xB = 0; yB = 0; and now the primitivization as Gaussian integers is 0, xA/G, and (xC+yC*i)/G where G=GaussianGCD(xA, xC+yC*i).
@warren: that is pretty cool if you have n,m,k, but is there a way to make it constructive if you are just given the sides a,b,c? For example, suppose [a,b,c] = [471, 518, 611] which are the sides of a primitive Heron triangle. How can I find appropriate n,m,k and then G, so as to actually compute the vertices A,B,C of this triangle? On Fri, Nov 18, 2011 at 10:03 AM, Warren Smith <warren.wds@gmail.com> wrote:
I also point out that (assuming Reid proof valid) by combining my Buchholz-->nonprimitive coordinates formula, and the Gaussian GCD, we can actually write just a single formula in terms of n,m,k, rational operations, and Gaussian GCD, giving coordinates for every PRIMITIVE Heronian triangle's vertices A,B,C. That is quite cool.
Here are the nonprimitive coordinate formulas: xC = m*(n-k)*(n+k); yC = 2*k*m*n; xA = n*(m-k)*(m+k) + xC; yA = 0; xB = 0; yB = 0; and now the primitivization as Gaussian integers is 0, xA/G, and (xC+yC*i)/G where G=GaussianGCD(xA, xC+yC*i).
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I remarked earlier (Oct 29th) --- Buchholz' (Euler/ Carmichael / Brahmagupta) parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), s = (m + n)m n, 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 proper Heronian similarity class. The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a); the inverse never yields primitive parameters nontrivially, since s^2 and d^2 = s(s-a)(s-b)(s-c) share common factor s. WFL On 11/18/11, James Buddenhagen <jbuddenh@gmail.com> wrote:
@warren: that is pretty cool if you have n,m,k, but is there a way to make it constructive if you are just given the sides a,b,c? For example, suppose [a,b,c] = [471, 518, 611] which are the sides of a primitive Heron triangle. How can I find appropriate n,m,k and then G, so as to actually compute the vertices A,B,C of this triangle?
On Fri, Nov 18, 2011 at 10:03 AM, Warren Smith <warren.wds@gmail.com> wrote:
I also point out that (assuming Reid proof valid) by combining my Buchholz-->nonprimitive coordinates formula, and the Gaussian GCD, we can actually write just a single formula in terms of n,m,k, rational operations, and Gaussian GCD, giving coordinates for every PRIMITIVE Heronian triangle's vertices A,B,C. That is quite cool.
Here are the nonprimitive coordinate formulas: xC = m*(n-k)*(n+k); yC = 2*k*m*n; xA = n*(m-k)*(m+k) + xC; yA = 0; xB = 0; yB = 0; and now the primitivization as Gaussian integers is 0, xA/G, and (xC+yC*i)/G where G=GaussianGCD(xA, xC+yC*i).
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Thanks Fred. I probably wasn't paying attention earlier. On Fri, Nov 18, 2011 at 11:27 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
I remarked earlier (Oct 29th) ---
Buchholz' (Euler/ Carmichael / Brahmagupta) parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), s = (m + n)m n, 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 proper Heronian similarity class.
The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a); the inverse never yields primitive parameters nontrivially, since s^2 and d^2 = s(s-a)(s-b)(s-c) share common factor s.
WFL
On 11/18/11, James Buddenhagen <jbuddenh@gmail.com> wrote:
@warren: that is pretty cool if you have n,m,k, but is there a way to make it constructive if you are just given the sides a,b,c? For example, suppose [a,b,c] = [471, 518, 611] which are the sides of a primitive Heron triangle. How can I find appropriate n,m,k and then G, so as to actually compute the vertices A,B,C of this triangle?
On Fri, Nov 18, 2011 at 10:03 AM, Warren Smith <warren.wds@gmail.com> wrote:
I also point out that (assuming Reid proof valid) by combining my Buchholz-->nonprimitive coordinates formula, and the Gaussian GCD, we can actually write just a single formula in terms of n,m,k, rational operations, and Gaussian GCD, giving coordinates for every PRIMITIVE Heronian triangle's vertices A,B,C. That is quite cool.
Here are the nonprimitive coordinate formulas: xC = m*(n-k)*(n+k); yC = 2*k*m*n; xA = n*(m-k)*(m+k) + xC; yA = 0; xB = 0; yB = 0; and now the primitivization as Gaussian integers is 0, xA/G, and (xC+yC*i)/G where G=GaussianGCD(xA, xC+yC*i).
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participants (3)
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Fred lunnon -
James Buddenhagen -
Warren Smith