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now the question which follows up is : If all the sides and area of a triangle were rational numbers then show that the triangle is got by `pasting' two right-angled triangles having the same property

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same property means?
both triangles should have rational sides.
|dw:1345129143913:dw| if ABC has rational sides, we have to show AMB and MBC also has rational sides.

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Other answers:

AC.MB is rational because area is rational and AC is rational so MB is rational u just proveAM and MC are rational
lol...not that easy
I think it will be enough to show just one side Am or MC rational
beause if one is rational than the other must be rational
if we say tanA = BC/AC = rational also tanA = MB/AM is it good enough to say here that AM will be rational ?
sorry,,tanA=BC/Ab ,,but still rational..
ABC is right triangle !!? work on \[\sqrt{AB^2-MB^2}+\sqrt{BC^2-MB^2}=AM+MC\]
i thought its not right
hmm,,well,,how will sum of sides help ?
sorry that will not help..
AM^2 and MC^2 will be rational
yep..thats where even i was wondering
this ques is a subpart of my previous ques,,do both have any relation ?
how ?
i mean ABC is a right triangle?
Dont think so
aha...thats what i had assumed!! so its not a right triangle!! hmm..
sauravshakya mentiond a good point \[(AM+MC)^2=AC^2=AM^2+2AM.MC+MC^2\]so \(AM.MC\) is rational
now our problem is like this if \(a+b\) and \(ab\) are rational prove that both \(a\) and \(b\) are rational
well why were AM^2 and MC^2 rational ?
Take right triangle AMB and triangle BMC
ohh leave it got it..
if ab and and a+b are rational,,its not necessary both are rational,, eg. -sqrt(2) +1 and sqrt(2) + 1
has to be an other way around..
yeah i was thinkin of that
guys i gotta go...but i love to know what is the answer
|dw:1345131550151:dw| here,,we have AC/sin(a+b) = BC/sinx now sin(a+b) = sina cosb + cosa sinb = AM/AB . BM/BC + BM/AB . MC/BC = (AM+BM)/AB + (BM +MC)/BC also, sin x =BM/AB on substituing, AC . BM/AB = BC . ( (AM+BM)/AB + (BM +MC)/BC ) AC.BM = (AM+BM)BC + (MB +MC)AB AC BM = AM BC + BM BC + MB AB + MC AB I DONT KNOW IF IT HELPS OR NOT!! lol.. :P
hasn't this been solved yet?
its a diff ques..
i lost track of it ... it's fuzzy ... well what's the Q?
|dw:1345132117021:dw|if ABC has rational sides, we have to show AMB and MBC also has rational sides.
@experimentX thats the question
|dw:1345138702333:dw| a^2 + b^2 = c^2 b^2 = rational AM + MC = rational AM^2 + MC^2 = rational. AM*MC = rational.
Let me finish it man!! AM + MC = rational <-- either both are irrational or rational) AM*MC = rational <--- either both are irrational or rational Let's assume both to be rational, AM ( rational - AM) = rational AM (rational) - AM^2 = rational irrational - rational = rational (contradiction) hence both must be rational
typo:- Let's assume both to be irrational,
here is my proof|dw:1345140385316:dw|
\(x+y=c\) is rational Area=\(ch/2\)=rational so \(h\) is rational since \(c\) is rational we have\[x^2+h^2=a^2\]\[y^2+h^2=b^2\]\[(x-y)(x+y)=(x-y).c=a^2-b^2\]so we obtain\[x-y=\frac{a^2-b^2}{c}\]\[x+y=c\]it gives\[x=\frac{a^2+c^2-b^2}{2c}\]\[y=\frac{b^2+c^2-a^2}{2c}\]\(x\) and \(y\) both are rational
lol ... your method is more obvious!!
aha..i see and understand both methods! thanks again @mukushla and @experimentX

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