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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.

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Reference for Fine's paper? –  Gerry Myerson Jan 20 '12 at 10:57
    
And definition of Heronian triangle (I am assuming that it is: all sides and the area are integral) –  Igor Rivin Jan 20 '12 at 15:02
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Actually, this paper: wm.uni-bayreuth.de/fileadmin/Sascha/Publikationen/… has a definition AND a parametrization. –  Igor Rivin Jan 20 '12 at 15:04
    
Never mind, I'll do it myself: MR0414484 (54 #2585) Fine, N. J. On rational triangles. Amer. Math. Monthly 83 (1976), no. 7, 517–521. –  Gerry Myerson Jan 22 '12 at 0:00
    
The Heron triangles that I can think of are either right triangles or are formed from joining two right triangles. Could this help in characterizing the possible ratios? Gerhard "Has Limited Heronian Triangle Vision" Paseman, 2012.01.27 –  Gerhard Paseman Jan 27 '12 at 15:01

5 Answers 5

up vote 10 down vote accepted

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 $2520$.

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Thanks! But, can you elaborate on "Testing gives the rank equal to 0"? My impression is that there is no general way to determine the rank but sometimes "infinite descent" luckily shows it's zero. So, if you need to be specific, can you show why 1/3 gives zero, and what the rank of a more random ratio like 6/17 is? –  bobuhito Jan 21 '12 at 18:48
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This works; still, it's easier to get from "Heronian triangle with side-lengths $a,ra,b$" to "rational point on $y^2 = x(x-1)(x-r^2)$" using Heron's formula than using the formulas for $\sin A$ and $\cos A$ (granted that these formulas are one approach to Heron's formula too). –  Noam D. Elkies Jan 21 '12 at 23:47

Bob, you are correct, there is no known algorithm guaranteed to compute the rank of an elliptic curve. There are however very effective algorithms for obtaining an upper bound on the rank by computing a Selmer group.

This can be done in either Sage or Magma. Since Sage is free, I will put some sample code here:

sage: k = 6/17
sage: E = EllipticCurve([0,-1 - k^2, 0, k^2, 0]).minimal_model()
sage: E.selmer_rank() - E.two_torsion_rank() # Upper bound from 2-Selmer Group
0    
sage: E.rank_bound() # Computes a sometimes better bound
0

In these cases the 2-Selmer group is relatively easy to compute, because every point of order 2 on the elliptic curve is rational. It's behavior is determined, in a somewhat complicated way, by the prime factorization of $n*m*(n+m)*(n-m)$ and whether those primes are squares modulo one another.

The going conjecture is that if you pick $k$ "at random", the the rank of the elliptic curve Allan wrote down has about a 50% chance of being 0 and a 50% percent chance of being 1.

Unfortunately, some results of Gang Yu, and certain elliptic curve analogues of Malle's corrections to the Cohen-Lenstra heurstics, suggest that if you pick $k$ at random and compute the upper bound you will quite often get an upper bound of 2, or 4, or 6, or so on, even when the rank is 0. If you want to do something more exhaustive, you might need to do so-called higher descents. These are, at present, best done with Magma.

EDIT STARTS HERE:###

For simplicity, I'm changing coordinates a bit starting from $W^2 = Z(Z-m^2)(Z-n^2)$ let $x = Z/m^2$ and $y = W/m^3$ then we see from Allan's answer above that a Heron triangle with side lengths $a, ka, b$ will give rise to a rational point on the elliptic curve $$E_k : y^2 = x(x-1)(x-k^2)$$ other than $(0,0)$, $(1,0)$, or $(k.0)$.

The Mordell-Weil theorem (since we are working over $\Bbb Q$ I should probably call it Mordell's theorem) tells us that the group of rational points $E_k(\Bbb Q)$ is a finitely generated abelian group, thus isomorphic to $T_k \times \Bbb Z^{r_k}$ for some finite abelian group $T_k$ and some non-negative integer $r_k$.

Given an elliptic curve, the torsion subgroup is easy to compute in Sage:

sage: k = 1/3
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()
sage: E.torsion_subgroup().invariants()
(2, 2)

but let me tell you what $T_k$ is for all $k$. Since we know that are $T_k$ contains the 3 points $(0,0)$, $(1,0)$, and $(k,0)$ of order 2, a deep theorem of Mazur assures us that $T_k$ is isomorphic to $\Bbb Z / 2 \Bbb Z \times \Bbb Z /2j\Bbb Z$ where $j = 1, 2, 3,$ or $4$.

I'm going to use the fact that Noam mentioned in the comments that $E_k$ precisely the quadratic twists by $-1$ of those elliptic curves with torsion subgroup $\Bbb Z/2 \Bbb Z \times \Bbb Z/4 \Bbb Z$. In particular, this means that there is a rational cyclic 8-isogeny among those elliptic curves isogenous to $E_k$ over $\Bbb Q$. If we had $j = 3$ then each of those isogenous curves would also possess a rational cyclic 3-isogeny. So there would be a rational cyclic 24-isogeny among these curves. This does not happen over $\Bbb Q$.

No let's consider the possibility of $j = 2$ or $j = 4$. If $E_k$ possesses a point of order 4 over $\Bbb Q$ then $E_k$ would have to possess 3 points of order 4 over $\Bbb Q(i)$. This will happen if and only if $k^2 - 1$ is a square up to sign. See for example:

sage: k = 3/5
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()    
sage: k^2 - 1
-16/25
sage: E.torsion_subgroup().invariants()
(2, 4)
sage: k = 5/4
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()
sage: k^2 - 1
9/16
sage: E.torsion_subgroup().invariants()
(2, 4)

The condition that $E_k(\Bbb Q(i))$ contains 3 points of order 4 rules out the possibility of a point of order 8. Since I've already used Mazur's theorem classifying torsion subgroups over elliptic curves over $\Bbb Q$. I'll go ahead say that Kamienny's theorem classifying torsion subgroups of elliptic curves over quadratic fields and say that $E_k(\Bbb Q(i))$ cannot have 3 points of order 4 and a point of order 8. You rule out $j = 4$ in a simpler way that comes down to Fermat's $a^4 + b^4 = c^2$, but I won't do that now.

So here is the best answer I can give to your question at present:

Let $k$ be a rational number. If there are infinitely many non-congruent Heron triangles with sides of the form $a, ka, b$ if then the Mordell-Weil rank of the elliptic curve $$E_k : y^2 = x(x-1)(x-k^2)$$ is positive. If the Mordell-Weil rank of $E_k$ is zero, then there can, and will be finitely many congruence classes of Heron triangles with side lengths of the form $a, ak, b$ if and only if $k^2 - 1$ is a square up to sign. In particular one can take the right triangle with sides $1, k, \sqrt{|k^2 -1|}$.

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(Edited only to prefix each line of code with four spaces so it gets formatted correctly) –  Noam D. Elkies Jan 21 '12 at 23:39
    
(...and to correct spelling. While I'm at it: These curves are the quadratic twists by ${\bf Q}(i)$ of the family of curves $y^2 = x(x+1)(x+r^2)$ with torsion group $2 \times 4$, so there's even more 2-isogeny descents to try than usual for $y^2 = x (x-A) (x-B)$, which might make the mwrank bound sharp more often.) –  Noam D. Elkies Jan 21 '12 at 23:57
    
Wow! I am new to elliptic curves, so many of these terms send me to Wikipedia...it will probably take me a few weeks to grasp the definition of the Selmer group, then longer to see how prime factors exactly limit it... In the meantime, could someone give the underlying logic (details of Jamie's "somewhat compicated way", proved via infinite descent?) of why one specific case, like 6/17, is impossible? Also, I'm curious, is there a "simple" ratio for which Sage finds an upper bound greater than zero but Magma, or further research, finds it to be zero? –  bobuhito Jan 22 '12 at 2:46
    
@Bob: A lot of the ideas which Jamie refers to can be found in Silverman's book "The arithmetic of elliptic curves". Its a good read and well worth it. –  Daniel Loughran Jan 22 '12 at 10:52
    
@Jamie: I've been playing with Sage (online version), but there seems to be a bug in the "rank_bound" function. Your simple code with k=3/5 (or 4/5) gives a bound of zero (contradicting the 3-4-5 triangle). I get a reasonable answer for other checks so I don't think it's user error, but I am a rookie, which is why I'm hoping you could check also and officially report this if you agree. By the way, I am a bit shocked that Matlab doesn't support this stuff...I'd rather not pay for Magma too. –  bobuhito Jan 24 '12 at 10:07

I did some calculations this morning on the elliptic curve

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

I have several codes in Ubasic or Pari which use the Birch & Swinnerton-Dyer conjecture to predict ranks.

For $1 \le m \le n \le 99$ with $\gcd(m,n)=1$, the results suggest 645 curves have rank 0, 730 curves have rank 1 and 126 curves have rank greater than 1.

The codes also use the BSD conjecture to predict the height of the generator of the rank 1 curves. For this range, the highest height comes from m=2 n=81, with height 34.9/69.8 (depending on which height normalisation you like). m=1 n=83 gives a higher BSD value but I found a point with height a quarter of this value.

For (2,81), the triangle produced has integer sides with over 30 digits.

Allan MacLeod

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Thanks again! So, my Matlab method will take trillions of years to find the 2/81 ratio... If I understand you and Jamie correctly, your 645 ratios truly are impossible, and some of your remaining 856 ratios might be impossible. If you could confidently bin all ratios, it might be neat to somehow "scatter plot" them (maybe as m,n points in the plane without including multiples?)...to me this highlights triangles that elude coordinates but could exist in some higher level of math/physics. –  bobuhito Jan 23 '12 at 13:02

Starting with $Z(Z - m^2) = (Z - n^2) b^2$, to which Allan MacLeod's elliptic curve can be reduced by taking $W = (Z - n^2)b$, one can find a general parametrization of m, n as follows.

Letting $a, x = Z/b, m/b$ gives $a^2 - a b x^2 = a b - n^2$, in which then letting $n = a y$ gives $b (x^2 + 1) = a (y^2 + 1)$. This implies $x^2 + 1, y^2 + 1 = a z, b z$ for some rational $z$, and multiplying these gives after composition $(\frac{x y + 1}{z})^2 + (\frac{x - y}{z})^2 = a b$.

Letting $a, b = k A, k B$, where $A, B$ are coprime integers, the factor $k^2$ in the preceding equation can be absorbed into each square, and we can conclude that $A, B$ are each a sum of two squares, say $p^2 + q^2, r^2 + s^2$ resp, so that $B (x^2 + 1) = A (y^2 + 1)$ becomes $(p x + q)^2 + (p - q x)^2 = (r y + s)^2 + (r - s y)^2$

The latter has general solution as follows, for rational $u, v$ with $u^2 + v^2 = 1$ :

$p x + q, p - q x = u (r y + s) + v (r - s y), v (r y + s) - u (r - s y) w$

So that:

$ x = \frac{u (r y + s) + v (r - s y) - q}{p} = \frac{p - v (r y + s) - u (r - s y)}{q}$

which expresses $y$ and then $x$ rationally in terms of $u, v$ and $p, q, r, s$ (and the latter appear homogenously, so one of them is disposable i.e. can be assumed equal to 1).

edit: I should clarify that this isn't a rational parametrization of one elliptic curve, which of course is impossible if $m^2, n^2$ are non-zero and distinct. What it does is start with a supposed "symbolic" solution and express the roots parametrically in a way consistent with that solution. In other words it constructs a multi-dimensional pencil of all elliptic curves having the required form.

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With reference to Noam's result, equivalent to $a(a - x^2) = (a - 1) y^2$, it isn't hard to show that this is rational, and $a, x, y$ can be expressed in terms of two parameters.

From this one can churn out possible values of $x$ by the bushel. Deciding whether any given value of $x$ is possible is harder, and may be no more practical using this result than those already discussed.

Anyway, I'll sketch how the solution is obtained.

Firstly, the N&S condition for the above, considered as a quadratic in $a$, to have rational $a$ when $x, y$ are rational is $(x^2 + y^2)^2 - 4 y^2 = z^2$ for some rational $z$.

This implies some rational $p$ for which:

$$\frac{y}{x^2 + y^2} = \frac{p}{p^2 + 1},\qquad \frac{z}{x^2 + y^2} = \frac{p^2 - 1}{p^2 + 1}$$

Reciprocating the first and completing squares expresses it in the form:

$$x^2 + (y - \frac{p^2 + 1}{2 p})^2 = (\frac{p^2 + 1}{2 p})^2$$

so there must be some rational $q$ for which:

$$y = \frac{p^2 + 1}{2 p} + \frac{q^2 - 1}{2 q} x, \qquad \frac{p^2 + 1}{2 p} = \frac{q^2 + 1}{2 q} x$$

From the last one can express $x = x(p, q)$, and plugging this into the preceding equation gives $y = y(p, q)$. Finally, we can obtain $z$ and then $a$ also both in terms of $p, q$.

NOTE: As the PC I am typing this on doesn't format equations, I am "flying blind". So if the result looks hideous due to any missing dollars and suchlike, perhaps someone could edit it.

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John, after getting your vision back, please merge your $\ge2$ MO accounts. –  Wadim Zudilin Jan 24 '12 at 23:58

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