From: jbuddenh@artsci.wustl.edu (Jim Buddenhagen)
Newsgroups: sci.math
Subject: Re: Rational Triangles of fixed area
Date: 1 Oct 1996 15:22:23 GMT
Kevin Whyte (kwhyte@hans.math.upenn.edu) wrote:
: This may be obvious, but are there infinitely many triangles
: with rational side lengths all with the same area? For the
: more symbolically minded, this boils down to: are there infinitely
: many rational solutions to 2p^2q^2+2p^2r^2+2q^2r^2-p^4-q^4-r^4=C
: for any fixed C>0?
There are infinitly many square Heron triangles (i.e. integer sides
and square integer area). By scaling, these are equiv to rational
triangles of area 1. For any rational k>1/sqrt(2) the triangle with
sides a,b,c (below) has area 1. There are infinitely many others
as well. Essentially, this is a problem of finding rational points
on certain families of elliptic curves.
4 2
20 k + 4 k + 1
a = ---------------------------
2 2
2* (2 k + 1) k ( 2 k - 1 )
4 2
k (4 k + 4 k + 5)
b = -----------------------
2 2
(2 k + 1) ( 2 k - 1 )
2
2 k - 1
c = ----------
2 * k
--Jim Buddenhagen (jbuddenh@artsci.wustl.edu)
==============================================================================
From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Rational Triangles of fixed area
Date: 1 Oct 1996 22:22:07 GMT
Kevin Whyte (kwhyte@hans.math.upenn.edu) wrote:
> This may be obvious, but are there infinitely many triangles
> with rational side lengths all with the same area? For the
> more symbolically minded, this boils down to: are there infinitely
> many rational solutions to 2p^2q^2+2p^2r^2+2q^2r^2-p^4-q^4-r^4=C
> for any fixed C>0?
Jim Buddenhagen wrote:
> There are infinitly many square Heron triangles (i.e. integer sides
> and square integer area). By scaling, these are equiv to rational
> triangles of area 1. For any rational k>1/sqrt(2) the triangle with
> sides a,b,c (below) has area 1. There are infinitely many others
> as well. Essentially, this is a problem of finding rational points
> on certain families of elliptic curves.
[Interesting parameterization deleted]
I think the scaling argument is inappropriate here; for example, one
cannot in this way obtain rational triangles with area sqrt(3)/4,
although this does occur with an equilateral triangle of side length 1.
Whyte seeks infinitely many rational points on a quartic surface. I
don't know for which positive rational numbers C the surface has rational
points; possibly all C, I suppose. But if there is a rational point,
there are almost surely infinitely many. I haven't carried out all
the calculations in full, but the following seems to work.
Suppose given a rational triangle with sides of lengths p, q, and r.
The area A may be determined by Heron's formula:
(4A)^2 = (p+q+r)(p+q-r)(p-q+r)(-p+q+r).
Any other rational triangle with this area gives another rational
point on this surface. The converse isn't true of course but is true
if p, q, and r are positive and satisfy the triangle inequality.
Note that we may just as well set x = p+q, y=p-q, and C = -(4A)^2<0,
and look for more rational solutions to
C = (x^2-r^2)(y^2-r^2)
given that at least one exists. Observe that only the one given rational
point (x0, y0, r0) need be given; C may then be computed, or omitted
altogether: we seek solutions to
(x^2 - r^2) (y^2 - r^2) = (x0^2 - r0^2) (y0^2 - r0^2)
other than (x0, y0, r0).
(Note that with C < 0 we must have r between x and y in magnitude.
Changing signs as necessary to have x, y, and r positive, we then have
the triangle inequality indeed met. Details left to the reader.)
Well, I'm almost convinced that given any single rational point
(x0, y0, r0) there are infinitely many more on its surface. There are some
particular combinations of (x0, y0, r0) for which the method below
does not appear to work, but I will address these points in the
conclusion.
The basic idea is to fix r and look for other points (x,y,r)
on the same quartic surface as (x0, y0, r). For a fixed r, these points
form an elliptic curve. Indeed, setting
X=x
Y=(y/r)(x^2-r^2)-x^2
i.e.
x=X
y=r(Y+X^2)/(X^2-r^2)
will yield a cubic in X and Y with rational coefficients.
Thus we can use the methods of elliptic curves to find many points given
just one (using the standard chord and tangent process).
I will spare the reader the unenlightening calculations. The curve has
not only the rational point (X0, Y0) but also (-X0, Y0), and
two more points (+-X0, -Y0 + 2 X0^2). In addition there are two points
at infinity; the line joining them is actually tangent to one of them.
Thus the procedure outlined in Cassel's little book on Elliptic Curves
can be used to render the curve into the canonical form Z^2 = a cubic.
If Dr. Maplev and I have done this calculation correctly, this cubic
may be expressed in terms of the previous variable Y; it's
4( -2 (X0^2 - r^2) Y + (r^4 + Y0^2 + 2 Y0 X0^2) ) *
( (X0^2 - r^2) Y^2 + r^2( Y0^2 + 2 Y0 X0^2 + r^2 X0^2) ).
(If you want to carry out the calculations yourself, use the cubic in
X and Y, substitute X = 1/W - Y and solve for W; there is a quadratic
equation to be solved, and the above cubic in Y is its discriminant. This
transformation carries one of the points at infinity to the point of
order 2 on the Y-axis; the other point at infinity is now the identity
element of the group. The remaining four initial points are negatives of
each other in pairs.)
With the curve in standard form it's rather easy to carry out the
calculations explicitly. For example, we can double the initial point
(X0, Y0) on the curve and get a new point with Y coordinate
2 2 2 2 2 6 2 4 4 2
(X0 + Y0) (r Y0 + 2 r Y0 X0 + X0 - r X0 + r X0 )
---------------------------------------------------------
3
X0
Now, the critical bit of theory to answer Whyte's query is that an
elliptic curve is almost sure to have an infinite number of points. By a
theorem of Mazur, the torsion subgroup of an elliptic curve over Q
can only be cyclic of order at most 12 (and not equal to 11) or
isomorphic to (Z/2) X (Z/2N) for N=1, 2, 3, or 4. Thus one needs only
demonstrate that, for example, the curve has more than 16 rational points,
and then conclude that it has infinitely many.
And indeed, the sums and doubles of known rational points on the curve
will surely give us more points, whose coordinates we can even calculate.
The difficulty is that in general these points are not guaranteed to
be distinct. For example, the newly-construct point indicated above
might be equal to (X0, Y0) or its negative, causing (X0, Y0) to
have order 3. On the other hand, it's a simple matter to test whether
this new coordinate does equal Y0. My sources indicate the necessary
condition is that
8 6 2 6 4 4 4 2 2 2 2 4 2 2 3
X0 + X0 Y0 - r X0 + r X0 + X0 r Y0 + 3 r Y0 X0 + r Y0 X0 + r Y0
3
- Y0 X0 = 0
which of course will not hold unless your initial choice of triangle is
extremely unlucky.
In a similar way one can compute other multiple of the original point
and ask if these multiples equal some other, previously computed point.
In each case, the answer is "no" unless a polynomial in X0, Y0, and r
is satisfied.
Although there are more efficient ways to test this, one need only, at
worst, check to see whether nP = P for n = 2, 3, ..., 13; if none is
true, then P is a point of infinite order on the curve.
So we conclude that virtually every initial choice of X0, Y0, and r
will give a curve with infinitely many rational points, the only exceptions
being those on the union of a few algebraic varieties. Working
backwards we can likewise state that:
For almost all choices of (p, q, r) there are infinitely many
triangles of the same area and having rational sides (p', q', r).
Of course, there is no reason to hold r fixed rather than one of the
other variables. Thus there is reason to hope that for _every_ initial
choice of (p,q,r) one can find other rational triangles of the same
area (even looking only among those triangles sharing at least one
side length.)
Since it's the end of the day here let me leave the matter in this
partially unfinished state. Two computational questions remain:
(1) Find explicitly the variety of (p,q,r) for which the elliptic
curve described above really is finite.
(2) Demonstrate that the three varieties obtained from the previous one
by permuting p q and r have empty intersection
as well as a geometrical challenge:
(3) Describe the construction of a second triangle having the same area
as the original one, which results from the tangent-process
(doubling) on the elliptic curve. Can a succint construction with
straightedge and compass be described?
(You may view the side of length r as lying on the horizontal
axis; then the problem is to move the top point horizontal so that
both remaining sides have rational length.)
dave