From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: 120-degree angles (was: I have a system of equations which urgently require solution) Date: Tue, 14 Dec 1999 20:21:59 GMT Newsgroups: sci.math Keywords: three 120 degree angles in a triangle, all lengths integral In article <8353rh$6g8$1@gannett.math.niu.edu> rusin@vesuvius.math.niu.edu (Dave Rusin) writes: >In article , KK wrote: >>a^2 + b^2 + ab = C^2 >>b^2 + c^2 + bc = A^2 >>c^2 + a^2 + ca = B^2 > >Any positive integer solutions (a,b,c,A,B,C)? > >(Looks like the rational box without a "main-diagonal" constraint.) The geometrical interpretation is three lines segments (lengths a, b, c) with a common endpoint, meeting at 120 degree angles. Draw a triangle connecting the other ends of the three segments. When are all distances integral? One integral solution, found by the Maple program below, is (a, b, c, A, B, C) = (264, 195, 325, 399, 511, 455). I haven't searched whether there are any integral solutions with smaller MAX(a, b, c). QUESTION: Can we have GCD(A, B, C) = 1? ----- Maple program # Choose b, c such that b^2 + b*c + c^2 is a perfect square. # Example: b = 3, c = 5. We want y1 and y2 to be squares. y1 := a^2 + b*a + b^2; y2 := a^2 + c*a + c^2; # Look for rational values of a such that y1*y2 is a square. # An obvious solution is a = 0. # We can derive another by setting y1 = y2, which gives a = -b - c. # Find a quadratic polynomial y in a such that y^2 - y1*y2 # has three known rational roots. Hope the fourth root is useful. # For any k, if we let y := b*c + a*(b + c)/2 + k*a^2; # then y^2 - y1*y2 has a double root at a = 0. Choosing k := (3*b^2 + 2*b*c + 3*c^2)/2/(b + c)^2; # gives another root at a = -b - c. The new value of a is eqn := factor(y1 * y2 - y^2)/a^2 / (a + b + c); a := factor(solve(numer(eqn), a)); # By chance (design?) the new y1 and y2 are both squares. factor(y1); factor(y2); # b = 3 and c = 5 gives a = 264/65. Scale by 65 to get integers. subs({b = 3, c = 5}, a); a := 264; b := 195; c := 325; sqrt(a^2 + a*b + b^2); sqrt(a^2 + a*c + c^2); sqrt(b^2 + b*c + c^2); ;quit; -- The claimed founder of the Internet (after whom Algorithms are named) is now Turing the USA, seeking large cache donations. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI ============================================================================== Date: Thu, 06 Jun 2002 10:08:40 +0100 From: "Allan MacLeod" To: Subject: 120degree angles inside a triangle Dear Prof. Rusin, I am a great admirer of your Mathematical Atlas web-site, and often meander through the pages. A couple of days ago, I was in the section 14J on Surfaces and Higher-Dimensional Varieties and came across the page entitled "three 120 degree angles inside a triangle, all six lengths integral". Having considered similar problems before, I (willingly) stopped the boring administration task I was involved in and looked at the equations a^2 + a b + b^2 = C^2 a^2 + a c + c^2 = B^2 b^2 + b c + c^2 = A^2 I found a parametric solution, and also solutions with GCD(A,B,C)=1 as asked in the text from Peter Montgomery. The ideas are not new but standard that I first learnt from the papers of Andrew Bremner. The equation n^2 = m^2 + m k + k^2 has parametric solution n = p^2 - p q + q^2 m = q^2 - 2 p q k = p^2 - q^2 so we can write a = alpha ( q^2 - 2 p q ) b = alpha ( p^2 - q^2 ) a = beta ( s^2 - 2 r s ) c = beta ( r^2 - s^2 ) b = gamma ( v^2 - 2 u v ) c = gamma ( u^2 - v^2 ) The alpha, beta, gamma can be removed using the identity (a/b) * (b/c) * (c/a) = 1 giving (p*p-q*q)*(u*u-v*v)*(s*s-2*r*s)-(q*q-2*p*q)*(v*v-2*u*v)*(r*r-s*s) = 0 Defining the variables f=p/q , g= r/s , and h=u/v we get f^2*(h^2-1)*(2*g-1)+2*f*(g^2-1)*(2*h-1)+g^2*(1-2*h)+2*g*(1-h^2)+h^2+2*h-2=0 This quadratic in f will have rational solutions if the discriminant is a rational square, so that d^2 = g^4*(2*h-1)^2+2*g^3*(h^2-1)*(2*h-1)+g^2*(4*h^4-2*h^3-15*h^2+10*h+1) +2*g*(1-h^2)*(2*h^2+2*h-3)+h^4+2*h^3+h^2-6*h+3 This quartic has a rational solution with g=1 and d=(h^2-1), so the quartic is birationally equivalent to an elliptic curve. After some standard but tedious algebra we find the elliptic curve j^2 = i^3 -i^2*(8*h^8+8*h^7-31*h^6-2*h^5+31*h^4-20*h^3-h^2+14*h-7) +4*i*(4*h^16+8*h^15-27*h^14-33*h^13+84*h^12+39*h^11-151*h^10+22*h^9 +162*h^8-98*h^7-97*h^6+99*h^5+24*h^4-45*h^3+3*h^2+8*h-2) Investigation of this curve shows at least 8 torsion points and rank at least one. I found a parameterisation of a point of infinite order but it gives solutions with at least one of (a,b,c) negative. Doubling this point gives a parametric solution to the original problem which does give (a,b,c) > 0. The formulae are a = (u^2-6*u*v+2*v^2)*(9*u^4-28*u^3*v+32*u^2*v^2-8*u*v^3+4*v^4)* (3*u^6+10*u^5*v-10*u^4*v^2-32*u^3*v^3+28*u^2*v^4+8*u*v^5-8*v^6) b = -8*u*v*(42*u^10-345*u^9*v+1346*u^8*v^2-2992*u^7*v^3+ 3872*u^6*v^4-3032*u^5*v^5+1744*u^4*v^6-1088*u^3*v^7+ 608*u^2*v^8-208*u*v^9+32*v^10) c = 168*u^12-1296*u^11*v+4568*u^10*v^2-8304*u^9*v^3+5952*u^8*v^4+ 2816*u^7*v^5-7104*u^6*v^6+4224*u^5*v^7-2432*u^4*v^8+2304*u^3*v^9- 1152*u^2*v^10+256*u*v^11 Not all values of u and v give a,b,c positive (after removal of common factors), but we find u/v < -1 does. Putting u=-2, v=1 we find a = 2709, b = 110960, c = 66576 which gives C = 112339, B = 67971, A = 155344. u=-3, v=1 gives a solution with GCD(a,b,c)=1. I hope this information is useful. Allan MacLeod Legal disclaimer -------------------------- The information transmitted is the property of the University of Paisley and is intended only for the person or entity to which it is addressed and may contain confidential and/or privileged material. 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