[math-fun] semicyclic polygons
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@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
Rich had some nice stepping methods to produce "semicyclic quadrilaterals" -- that is (if you weren't paying attention), three edge lengths A,B,C such that a quadrilateral with side lengths A,B,C,1 inscribes in a circle of diameter 1. He said:
I've ignored the unphysicalness of solutions with negative numbers.
Not at all! These are precisely the nonconvex solutions, where you read the points in the wrong order as you go around the circle. For instance,
Together with the unchanged A, the new solution A,B',C' = (1/2,-1/7,13/14).
That is to say, there exists a self-intersecting "bow-tie" quadrilateral with side-lengths 2,7,13,14 whose vertices all lie on a circle of diameter 14, and the length-2 edge is oriented oppositely from the 7 and 13. Coords of its vertices, for example, can be (-7,0), (7,0), (7/2, 7 sqrt(3)/2), (71/14, 39 sqrt(3)/2), with edge lengths 14, 7, "-2", 13, in that cyclic order. Of course, as we've mentioned, that's only one of the three noncongruent possibilities, since you get to permute the edges, eg to choose which one falls opposite the diameter. I'll have to think some to try to understand whether Rich's steppers manifest physically. Shouldn't the linear stepper, at least, be representable by some geometric construction? --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Rich wrote:
A^2 + B^2 + C^2 + 2ABC = 1 [...] 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.
And yes indeed, this is a simple geometric evolution. Suppose the vertices of your original semicyclic quadrilateral are P,Q,R,S in that order, and the line PS is the diameter of the circle, which Q and R also lie on. Now replace Q by its reflection through the line PS to get Q', also on the circle. The question is whether the length of Q'R is still rational, and indeed it is, and we can calculate it, using Ptolemy's Theorem. In PQRS we had |PQ||RS| + |QR||SP| = |PR||QS|; here |PR| and |QS| are the lengths of the two diagonals, which are not necessraily rational, but the product is. Replacing Q by Q' reshuffles things so that the old diagonals are now outside edges and the new diagonals are Q'R and the diameter PS, but |PQ|=|PQ'| and |SQ|=|SQ'| since we reflected in PS. Ptolemy now says that |PR||QS| + |PQ||RS| = |PS||Q'R|, and comparing this to the preceding, we get |Q'R| = 2 |PQ||RS| / |PS| + |QR|, exactly Rich's linear solution stepping (except that here the sign manifests geometrically in the bow-tie result). --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
participants (2)
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Michael Kleber -
Richard Schroeppel