# Right angle triangles with bases $c/2^{2^n}$

Let's have a right angle triangle where $a=5$, $b=4$, $c=3$. Is it possible to create an infinity of right angle triangles with rational sides from the above triplet, with bases equal to $3/2^{2^n}$ when $n$ takes values from zero to infinity?

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Yes; just scale the entire triangle. If you have a triangle with sides $a$, $b$, and $c$, then there is a similar triangle with sides $\lambda a$, $\lambda b$, and $\lambda c$ for any real number $\lambda\gt 0$. –  Arturo Magidin Dec 2 '11 at 5:12
Divide all the other sides by $2^{2^n}$? –  Ｊ. Ｍ. Dec 2 '11 at 5:13
Is it possible to create an infinity of m.se questions, and only accept answers to 25 per cent of them? –  Gerry Myerson Dec 2 '11 at 5:18
I'm just guessing, but Vassili's first comment to Zev's answer may mean that the OP is seeking a sequence of pairwise distinct (in the sense of similarity of triangles) right triangles with rational side lengths with one side $3/ 2^{2^n}$. The question then boils down to one about Pythagorean triples with one of the two "legs" divisible by 3. –  Willie Wong Dec 2 '11 at 15:54

For any $n$, by scaling the $3$-$4$-$5$ right triangle by $1/2^{2^n}$, we get a $\dfrac{3}{2^{2^n}}$ - $\dfrac{4}{2^{2^n}}$ - $\dfrac{5}{2^{2^n}}$ right triangle.
There are infinitely many $n$, hence infinitely many such triangles. Is that what you mean?
$a_2=6562/216$ $b_2=1640/54$ $c_2=3/4$ –  Vassilis Parassidis Dec 2 '11 at 5:33