From: "Chip Eastham" <eastham@bellsouth.net>
Subject: Re: Rational Distances to the Vertices of a Square
Date: Tue, 24 Oct 2000 21:44:37 -0400
Newsgroups: sci.math
Summary: [missing]

Executive Summary:

I reconstruct for phunt the particulars of Hugo's
approach, make what seems to be a correction, and
show that there are no solutions along the midline
of the square.

"phunt" <phunt@interpac.net> wrote in message
news:8t48e1$72l$1@nnrp1.deja.com...
>
>
> Thanks for trying to help us with proofing things.  Here
> are some questions we need answered.
>
>
> > 1) The most obvious place to look would be on one of
> >    the orthogonal symmetry lines.
>
> Here we should define what is meant by lines of orthogonal
> symmetry, including examples for the case of a square (and a
> rectangle).
>
> Also we should explain where the next most obvious place to
> look might be if we chose not to look at lines of symmetry.
>

phunt,

I believe that Hugo is suggesting we first look for solutions
along a line that perpendicularly bisects one pair of edges
of the square.  Since these points are equidistant from both
pairs of endpoints of those edges, we might kill four birds
with two stones.

> > 2) If I've calculated correctly, solutions exist on such
> >    a line if there exist angles theta and phi
>
> Can theta and phi be referred to the coordinate axes? ...
> Or alternativety to the lines of symmetry defined above?
> Or can they be referred to the figure (the unit square)?
>
> Then we should also actually define the pairs that give
> the vertices of the unit square in the plane, so that we
> are all talking about the very same thing.
>
>
> > 3) such that tan theta + tan phi = 2, and cos theta and
> >    sin phi are both rational
>
> Now this is very elegant condition for discovering rational
> points in a plane.  Can you show how it was derived?  Can
> you explain why sin theta and cos phi need not be rational,
> and are there any other restrictions on their ranges?
>
>

Let's suppose point P is on the midline of the square as we
proposed above.  Theta and phi presumably correspond to
the acute angles made by lines drawn from P to one set of
(generally nonequidistant) corners on the same side of the
midline and the line at P perpendicular to the midline.

The sum of the tangents of these two angles will be 2, and
the distances to the corners will be twice the respective
secants (but the secant is rational iff the cosine is).  So
here I part company a bit with Hugo, as I think we must
require both cos(theta) and cos(phi) rational.

> > 4) (in which case the point (1/2, 1/2 tan theta) is an
>      example of a point that solves the problem).
>
> This will make more sense when we have your definition of
> the points that describes the square.  Also you may wish
> to replace tan theta with a real number thus identifying
> one such point that answers the original question.
>

{(0,0),(1,0),(0,1),(1,1)}.  Apparently theta is the angle that
Hugo associates with the line to (0,0).

>
> 5) Further calculation suggests that this requires rational
>    solutions to the equation 4(r^2-1)(s^2-1) = (6-r^2-s^2)^2,
>    but I'm not sure where to take it from there.
>
> We can talk about this another time, but it looks awfully
> sinister for some undefinable reason.
>
> Thanks,
>
> /ph
>

Sinisterly suitable for the Halloween season, yes?  However
I believe that we can reconstruct the equation as follows (at
least modulo my choice of rational cosines for both angles):

tan(theta) + tan(phi) = 2

Let r = sec(theta), s = sec(phi).

sqrt(r^2 - 1) + sqrt(s^2 - 1) = 2

(r^2 - 1) + 2sqrt([r^2 - 1][s^2 - 1]) + (s^2 - 1) = 2

2sqrt([r^2 - 1][s^2 - 1]) = (4 - r^2 - s^2)

4(r^2 - 1)(s^2 - 1) = (4 - r^2 - s^2)^2

r^4 - 2(r^2)(s^2) + s^4 - 4r^2 - 4s^2 + 12 = 0

(r^2 - s^2)^2 - 4(r^2 + s^2) + 12 = 0

for which we want to locate any rational points.

Now we have tackled a simplified version of a hard
problem, which leads to the natural conjecture that
either this equation has no solutions or else it has
them but they'll be terrifically hard to find.  But
there is yet another possibility.  Because of the
squarings involved in our derivation, its possible
we introduced some "artifact" solutions, so we may
find solutions of this that yet don't satisfy the
original requirements.

Note that if we replace r^2 by x and s^2 by y,
then x,y should still be rational and we want to
locate (as partial solution) rational points on the
quadratic curve:

(x - y)^2 - 4(x + y) + 12 = 0

This much is trivial, since any rational value for
x - y will determine a rational value for x + y:

(x + y) = 3 + (x - y)^2/4

Something of the sort can be attempted directly
with r,s by the substitutions u = r + s, v = r - s:

r^2 - s^2 = uv

2(r^2 + s^2) = u^2 + v^2

Thus:

(r^2 - s^2)^2 - 4(r^2 + s^2) + 12 = 0

becomes:

(uv)^2 - 2(u^2 + v^2) + 12 = 0

(u^2 - 2)(v^2 - 2) + 8 = 0

The expression is symmetric in u,v but requires
for solution that u^2 and v^2 lie on opposite sides
of 2.  If we suppose v^2 < 2, say:

u^2 = 2 + 8/(2 - v^2) = 2(1 + 4/(2 - v^2))

If we express v as usual as a ratio m/n of relatively
prime integers:

u^2 = 2(1 + 4n^2/(2n^2 - m^2))

u^2 = 2(6n^2 - m^2)/(2n^2 - m^2)

If m is odd then easily there is no rational solution u,
so assume m = 2k and n odd (since coprime to m):

u^2 = 4(3n^2 - 2k^2)/(2n^2 - 4k^2)

u^2 = 2(3n^2 - 2k^2)/(n^2 - 2k^2)

Ooops!  Since n is odd, there is again no rational u
that will solve this equation.  This is perhaps the
best of outcomes that might reasonably be expected.
We looked in the place where the light is good, and
consistent with a degree of competence of others who
must have looked there before, the problem remains
unsolved.

Regards,
Chip