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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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

Mathematics
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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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AC.MB is rational because area is rational and AC is rational so MB is rational ...now 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
yeah..
beause if one is rational than the other must be rational
hmm..
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
right?
right
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
YEP
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
hmm..
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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