# An Interesting Question about Pythagorean Triples

Consider such a right-angled trapezium formed by 3 right-angled triangle. Determine does it exist integral solutions for lengths of sides $AB, BC, CD, DE, EA, AC$ and $CE$.

Here's my ideas.

I know that Pythagorean Triple can be generated by substituting integer into $x^2-y^2, 2xy, x^2+y^2.$ So let $AB=m^2-n^2$, $BC=2mn$, $CD=2pq$, $DE=p^2-q^2$, $AC=m^2+n^2=u^2-v^2$, $CE=p^2+q^2=2uv$ and $AE=u^2+v^2.$ To answer the question, I have to show that if $m^2+n^2=u^2-v^2$ and $p^2+q^2=2uv$ have integral solution(*). But I don't know how to show this.

Can anyone tell me if I'm right about (*)? If I'm right, how to show it? If I'm wrong, how to solve the question?

Thank you.

Sorry, I'm a poor question-tagger.

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What if there's no similar triangles? –  ᴊ ᴀ s ᴏ ɴ Aug 14 '12 at 15:26

Note that triangles ABC and CDE must be similar if BCD is to be a straight line, so both must be built from integer multiples of the same primitive pythagorean triple T= {a,b,c} with hypotenuse c.

Suppose ABC is $pT$ and CDE is $qT$ then $AC=pc$ and $CE=qc$ and $AE^2=p^2c^2+q^2c^2$ so that $AE$ is divisible by $c$ and equals $rc$.

You can build your trapezium from any pair of pythagorean triples (not necessarily primitive)

T={$a,b,c$} and U= {$p,q,r$}

with ABC = {$pa,pb,pc$}, CDE = {$qb,qa,qc$}, ACE = {$pc,qc,rc$}.

And this is essentially the only way of doing it.

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Thanks. I get it. –  ᴊ ᴀ s ᴏ ɴ Aug 18 '12 at 5:17

You have a lots of solutions, just take any triples (for example 3,4,5).

Then multiply by the right amount based on another triple (for example 3,4,5).

BC=9, AB=12, AC=15 DE=12, CD=16, CD=20

Then AE = 25

So for any pythagorean triples (a,b,c) (d,e,f) (g,h,i), a solution is (agf,bgf,cgf) (dhc,ehc,fhc) and then (icf) for the last length.

Of course, you can divide the obtained solution by the gcd of all lengths...

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