There is no Heronian triangle (or simply consider triangles on an integer lattice
which also have integer side lengths) for which one side is half the length of
another side. What other "side-side ratios" are impossible?


Here are the impossible ratios I've come across in researching this:
 1/2, 2/5, 2/3, 62/63, 6/7


There seem to be infinitely more from the generation method given by Fine
in his "On Rational Triangles" paper. But, I would like a general check for any
given ratio. For example, I've used Matlab to evaluate this ratio for millions of
triangles and never seen the ratios 1/3, 1/4, or 1/5; so, does anybody know
if these are truly impossible?


I have a hunch that these impossible ratios have a neat pattern...so please
help! Thanks!


ADDED 1/27/2012:
From the answers of Alan and Jamie, I've learned that most ratios can quickly be
checked with Sage (using descents in elliptic curves):
 <https://picasaweb.google.com/107800252627134603876/Math#5702303639089321090>
From the plotted summary, you can see that all the ratios I mentioned truly are
impossible. I'm guessing that the few uncheckable ratios in the plot could be
checked with some new creative descent method if that were one's lifetime goal...
it even sounds like Magma already has the capability to clarify some of these
points.



Impossible Heronian Triangles (Ratio of 2 Sides)
Suppose the ratio is called $k$, and let the sides be $g, kg, h$, with the angle between sides $g$ and $kg$ called $A$, with all these variables rational.


Then $\Delta = \frac{1}{2}kg^2 \sin A$, so if the triangle is a Heron triangle with rational area, then $\sin A$ must be rational. Also, by the cosine rule $\cos A$ must be rational.


We can set $\cos A=(1-t^2)/(1+t^2)$ and $\sin A=2t/(1+t^2)$, and $t$ must be rational.


For $h$ to be rational, we must have $k^2-2k\cos A+1$ a rational square, and substituting the rational form for $\cos A$ means that the quartic


$(k+1)^2t^4+2(k^2+1)t^2+(k-1)^2$


must be a rational square. For fixed $k$, the value $t=0$ gives a solution, so this quartic is birationally equivalent to an elliptic curve. Setting $k=m/n$, with $m,n \in \mathbb{Z}$ and doing some standard algebra, leads to the elliptic curve


$W^2=Z(Z-m^2)(Z-n^2)$


with $t=W/(Z(m+n))$.


The curve has (usually) only $3$ finite torsion points at $(0,0)$, $(m^2,0)$ and $(n^2,0)$, none of which give a non-trivial value of $t$. Thus, to get solutions, we need the rank of the elliptic curve to be greater than $0$.


Testing gives the rank equal to $0$ for $k=1/2,1/3,1/4,1/5$ so no solutions will exist. For $k=1/6$, however, the rank is $1$, with generator $(81,540)$. This gives $t=20/21$, and, after clearing denominators a triangle with sides $174, 29, 175$ which has area $2520$.


