Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have recently thought about a interesting question about Pythagorean Triples.

enter image description here

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. enter image description here

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.

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

2 Answers 2

up vote 8 down vote accepted

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.

share|improve this answer
    
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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.