The rational equation that MK discusses can step solutions in at least three
different ways, along with some simple modifications.
A^2 + B^2 + C^2 + 2ABC = 1
With starting solution 1/2, 1/2, 1/2.
The simplest mods are to permute A,B,C; and to negate any pair.
The linear step is to fix A and B; then C is the root of a quadratic whose
two roots sum to -2AB, whence A,B,C' is a solution, C' = -2AB-C.
Applying this twice just regenerates C, but you can step away by
permuting A,B,C.
The quadratic stepper is the fix A, then consider the equation
B^2 + 2A BC + C^2 = 1 - A^2.
For our starting solution of 1/2,1/2,/1/2, the new equation is
B^2 + BC + C^2 = 3/4.
We can generate more rational solutions from this by finding
rational solutions of
X^2 + XY + Y^2 = 1
or integer solutions of S^2 + ST + T^2 = N^2. These come from taking
N = U^2 + UV + V^2 = norm(U - wV), and squaring. Take U=2, V=1, N=7,
giving S = U^2 - V^2 = 3, T = 2UV + V^2 = 5, and then X = 3/7, Y = 5/7.
Multiply (B-wC)*(X-wY) = B'-wC', giving B' = -1/7, C' = 13/14.
Together with the unchanged A, the new solution A,B',C' = (1/2,-1/7,13/14).
A final stepper can be derived from treating the equation as a cubic
surface. Take any two distinct solutions and draw the line connecting them.
As long as the delta of the two solutions has no 0 components, the line
intersects the cubic surface in one more point. That point can be computed
easily: Assume the two solutions are A,B,C and A+a,B+b,C+c. The connecting
line is A+ta,B+tb,C+tc. Write the surface equation,
2 abc t^3 + (2abC+2aBc+2Abc+a^2+b^2+c^2) t^2 + xxx t + yyy = 1
The sum of the three roots for t is - coeff(t^2) / 2abc, and two of the
roots are t=0 and t=1. This is the standard elliptic curve trick, and it
works for cubic surfaces (and cubics with even more variables).
This is already too long, so I won't go through the grommy details.
I've ignored the unphysicalness of solutions with negative numbers.
I'm guessing that half of all our stepped-to solutions will have an even
number of negative components, and we can complement the negative pair
to get a positive solution.
Puzzle: Can all the other solutions be got by starting from 1/2,1/2,1/2 and
using the various stepping methods? Given any solution, is there a path back to
1/2,1/2,1/2 with decreasing denominators?
Rich
--------- copy of MKs message ----------
On 4/16/07, Michael Kleber <michael.kleber(a)gmail.com> wrote:
> A friend recently mentioned to me the following ARML
> problem from 1989: "A convex hexagon is inscribed in
> a circle. If its successive sides are 2, 2, 7, 7, 11, 11,
> compute the diameter of the circumscribed circle."
>
> (Anyone who wants to solve this on their own may go do
> so now, and come back later for the rest of my question.)
>
> Since we can reorder sides of this cyclic hexagon, the
> question is the same as asking for the diameter D of a
> circle in which you can inscribe a quadrilateral with side
> lengths 2,7,11,D. So maybe we should generalze from
> Pythagorean triangle, and say that a convex n-gon is
> Pythagorean if it has integral "hypotenuse" D and edges
> a1, a2,...,a_{n-1}, and can be inscribed in a semi-circle
> with diameter D. The obvious question is, what's the
> generalization of the Pythagorean theorem?
>
> In the quadrilateral case, the relation you get -- necessraily
> symmetric on the edges a_i -- turns out to be
> D * (a1^2 + a2^2 + a3^2) + 2 a1 a2 a3 = D^3
> It's maybe more elegant-looking if you say that you're searching
> for rational solutions to the D=1 version,
> a^2 + b^2 + c^2 + 2abc = 1
> but maybe not, since the LHS isn't homogeneous.
>
> Solutions include all Pythagorean triangles, by setting a3=0,
> and lots of solutions like 2-2-7-8, whose trigonometric meaning
> I'll leave you to ponder. Hmm, I suppose the regular hexagon's
> sol'n 1-1-1-2 also falls into this category.
>
> But the interesting solutions start with the 2-7-11-D from the
> ARML, then 2-9-12-16, 6-11-14-21, 1-12-22-26, 3-14-25-30,
> and so on. Do these come up anywhere else?
>
> And what's the relation between the edges and the hypotenuse
> for the n-gon case?
>
> Likely this is all well-known to those who know it, but not to me...
>
> --Michael
