How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$? How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$?
I've been reading the Plimpton 322 news, and this fits in the gap in the Babylonian table between 0.6996 ($a=1679,b=2400$) and 0.75 ($a=3,b=4$).  The Babylonians apparently tabulated a lot of triples and then looked for the best one, but is there a more direct way?

Pythagorean triplets are characterized by being of the form
$$a= m^2-n^2 \\ b= 2mn \\ c=m^2+n^2$$
So you could look for two integers $m>n$ such that
$$\frac{m^2-n^2}{2mn} \approx \frac{5}{7}$$
or, equivalently,
$$\frac{m}{n}- \frac{n}{m} \approx 2 \cdot \frac{5}{7}$$
Now, let $x$ be unique positive solution of
$$x-x^{-1} = 2 \cdot \frac{5}{7}$$
one can easily compute that $x= \frac{5 + \sqrt{74}}{7} \approx 1.94$. The idea is to appximate $x= m/n$with a rational number. The continued fraction of $x$ is periodic and it is $[1; \overline{1, 16, 1,1,1}]$. This gives us the first convergents of $x$: $$\frac{1}{1} , \frac{2}{1} , \frac{33}{17} , \frac{35}{18} , \frac{68}{35} , \frac{103}{53} , \frac{171}{88},\dots$$ which are the best rational approximations of $x$. Taking $m= 33$ and $n= 17$ you get the triplet
$$a= 800 ; \ \ b= 1122 ; \ \ c= 1378$$
which is not a primitive triplet. Dividing it by $2$ you get the primitive triplet
$$a= 400 ; \ \ b= 561 ; \ \ c= 689$$
with
$$\left| \frac{a}{b} - \frac{5}{7} \right|= \left| \frac{400}{561} - \frac{5}{7} \right|= \frac{5}{3927} \approx 0.0013$$
Going on like this ($m= 35 , n=18$) you get the next solution
$$a= 901 ; \ \ b= 1260 ; \ \ c= 1549$$
with 
$$\left| \frac{a}{b} - \frac{5}{7} \right|= \left| \frac{901}{1260} - \frac{5}{7} \right|= \frac{1}{1260} \approx 0.0008$$ and so on.
