shubhamsrg
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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sauravshakya
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same property means?
shubhamsrg
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both triangles should have rational sides.
shubhamsrg
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|dw:1345129143913:dw|
if ABC has rational sides,
we have to show AMB and MBC also has rational sides.
mukushla
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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
mukushla
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lol...not that easy
sauravshakya
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I think it will be enough to show just one side Am or MC rational
mukushla
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yeah..
sauravshakya
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beause if one is rational than the other must be rational
shubhamsrg
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hmm..
shubhamsrg
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if we say tanA = BC/AC = rational
also tanA = MB/AM
is it good enough to say here that AM will be rational ?
shubhamsrg
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sorry,,tanA=BC/Ab ,,but still rational..
mukushla
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ABC is right triangle !!?
work on
\[\sqrt{AB^2-MB^2}+\sqrt{BC^2-MB^2}=AM+MC\]
mukushla
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i thought its not right
shubhamsrg
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hmm,,well,,how will sum of sides help ?
mukushla
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sorry that will not help..
sauravshakya
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AM^2 and MC^2 will be rational
sauravshakya
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right?
mukushla
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right
shubhamsrg
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yep..thats where even i was wondering
shubhamsrg
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this ques is a subpart of my previous ques,,do both have any relation ?
shubhamsrg
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how ?
mukushla
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i mean ABC is a right triangle?
sauravshakya
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Dont think so
shubhamsrg
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aha...thats what i had assumed!! so its not a right triangle!! hmm..
mukushla
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sauravshakya mentiond a good point
\[(AM+MC)^2=AC^2=AM^2+2AM.MC+MC^2\]so \(AM.MC\) is rational
mukushla
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now our problem is like this
if \(a+b\) and \(ab\) are rational prove that both \(a\) and \(b\) are rational
sauravshakya
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YEP
shubhamsrg
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well why were AM^2 and MC^2 rational ?
sauravshakya
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Take right triangle AMB and triangle BMC
shubhamsrg
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ohh leave it
got it..
shubhamsrg
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if ab and and a+b are rational,,its not necessary both are rational,,
eg. -sqrt(2) +1 and sqrt(2) + 1
shubhamsrg
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has to be an other way around..
mukushla
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yeah i was thinkin of that
shubhamsrg
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hmm..
mukushla
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guys i gotta go...but i love to know what is the answer
shubhamsrg
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|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
experimentX
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hasn't this been solved yet?
shubhamsrg
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its a diff ques..
experimentX
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i lost track of it ... it's fuzzy ... well what's the Q?
sauravshakya
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|dw:1345132117021:dw|if ABC has rational sides,
we have to show AMB and MBC also has rational sides.
sauravshakya
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@experimentX thats the question
experimentX
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|dw:1345138702333:dw|
a^2 + b^2 = c^2
b^2 = rational
AM + MC = rational
AM^2 + MC^2 = rational.
AM*MC = rational.
experimentX
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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
experimentX
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typo:- Let's assume both to be irrational,
mukushla
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here is my proof|dw:1345140385316:dw|
mukushla
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\(x+y=c\) 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
experimentX
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lol ... your method is more obvious!!
shubhamsrg
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aha..i see and understand both methods!
thanks again @mukushla and @experimentX