Is it possible to get arbitrarily near any acute angle with Pythagorean triangles?

Solution 1:

If you mean a pythagorean triangle having one of the angles arbitrary close to a given acute angle then the answer is yes.

Notice that $\displaystyle (m^2-n^2, 2mn, m^2+n^2)$ is always a pythagorean triangle for any positive integers $\displaystyle m \gt n$.

We have that $\displaystyle \tan \theta = \frac{2mn}{m^2-n^2} = \frac{2t}{1-t^2}$ where $\displaystyle t = \frac{n}{m}$.

By continuity (and surjectivity) of $\displaystyle \frac{2t}{1-t^2}$ we can find an interval of $\displaystyle t$ (which is infact a subinterval of $\displaystyle (0,1)$) for which $\displaystyle \frac{2t}{1-t^2}$ is arbitrarily close to any positive number we want.

Since the rationals are dense, we can find $\displaystyle m,n$ which make the tangent of one the angles arbitrarily close to the tangent of a given acute angle, and hence the angles can be made arbitrarily close.

Solution 2:

In fact, more is true -- consider the triples $(3,4,5), (-7,24,25), (-117,44,125), (-527,-336,625), \ldots$, where the $n$th triple is

$$ (a_n, b_n, c_n) := (Re((3+4i)^n), Im((3+4i)^n), 5^n). $$

Since $\tan^{-1} 4/3$ is irrational, the sequence of pairs $(a_n/c_n,b_n/c_n)$

$$ (3/5, 4/5), (-7/25, 24/25), (-117/125, 44/125), (-527/625, -336/625), \ldots $$

is equidistributed on the unit circle. Therefore the sequence $(|a_1/c_1|, |b_1/c_1|), (|a_2/c_2|, |b_2/c_2|)$ obtained by taking absolute values is equidistributed on the arc of the unit circle in the first quadrant.

I'm not sure if we can say that the Pythagorean triples (as opposed to some subsequence of them) are equidistributed in the first quadrant in this sense. Anyone?

Solution 3:

The other answers here explain that it is possible to find arbitrarily good approximations to any angle using Pythagorean triples. Here's a footnote on how to turn this into a constructive procedure.

Moron above says to pick a rational number $n/m$ near $t$ where $t$ satisfies $\tan \theta = 2t/(1 - t^2)$. You could pick $n$ and $m$ using a Farey algorithm to find the best rational approximation to $t = (1 - \cos \theta)/\sin\theta$ with a given denominator.

Start with $a = 0$, $b = c = d = 1$. The number $t$ is between $a/b$ and $c/d$. Next compute the "mediant" $(a+c)/(b+d)$. If $t$ is bigger than the mediant, update $a$ and $b$ with $a+c$ and $b+d$. Otherwise update $c$ with $a+c$ and $d$ with $b+d$. Each iteration of this process produces the best rational approximation to $t$ for a given denominator size.