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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
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Divide all the other sides by $2^{2^n}$? –  J. M. Dec 2 '11 at 5:13
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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

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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?

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All triangles differ from each other. They are not the same and they are not similar. –  Vassilis Parassidis Dec 2 '11 at 5:18
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That reminds me of a Wittgenstein quote: "To say of two things that they are identical is nonsense, and to say of one thing that it is identical with itself is to say nothing." So, it is true that all triangles differ from each other, in the utterly useless sense that if two triangles did not "differ" from each other (in whatever meaning you choose), they would be "the same" (in whatever meaning you choose). Also, your second sentence is obviously wrong - for example, all equilateral triangles are similar. –  Zev Chonoles Dec 2 '11 at 5:24
    
$a_2=6562/216$ $b_2=1640/54$ $c_2=3/4$ –  Vassilis Parassidis Dec 2 '11 at 5:33
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Of course, it couldn't possibly be that you expressed your question unclearly - no, the only explanation is that we are all dumb, and you are a genius. –  Zev Chonoles Dec 2 '11 at 6:52
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"Those are some nice rational number's you've written down." Brilliant. –  mixedmath Dec 12 '11 at 0:30

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