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

- shubhamsrg

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- anonymous

same property means?

- shubhamsrg

both triangles should have rational sides.

- shubhamsrg

|dw:1345129143913:dw|
if ABC has rational sides,
we have to show AMB and MBC also has rational sides.

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## More answers

- anonymous

AC.MB is rational because area is rational and AC is rational so MB is rational
...now u just proveAM and MC are rational

- anonymous

lol...not that easy

- anonymous

I think it will be enough to show just one side Am or MC rational

- anonymous

yeah..

- anonymous

beause if one is rational than the other must be rational

- shubhamsrg

hmm..

- shubhamsrg

if we say tanA = BC/AC = rational
also tanA = MB/AM
is it good enough to say here that AM will be rational ?

- shubhamsrg

sorry,,tanA=BC/Ab ,,but still rational..

- anonymous

ABC is right triangle !!?
work on
\[\sqrt{AB^2-MB^2}+\sqrt{BC^2-MB^2}=AM+MC\]

- anonymous

i thought its not right

- shubhamsrg

hmm,,well,,how will sum of sides help ?

- anonymous

sorry that will not help..

- anonymous

AM^2 and MC^2 will be rational

- anonymous

right?

- anonymous

right

- shubhamsrg

yep..thats where even i was wondering

- shubhamsrg

this ques is a subpart of my previous ques,,do both have any relation ?

- shubhamsrg

how ?

- anonymous

i mean ABC is a right triangle?

- anonymous

Dont think so

- shubhamsrg

aha...thats what i had assumed!! so its not a right triangle!! hmm..

- anonymous

sauravshakya mentiond a good point
\[(AM+MC)^2=AC^2=AM^2+2AM.MC+MC^2\]so \(AM.MC\) is rational

- anonymous

now our problem is like this
if \(a+b\) and \(ab\) are rational prove that both \(a\) and \(b\) are rational

- anonymous

YEP

- shubhamsrg

well why were AM^2 and MC^2 rational ?

- anonymous

Take right triangle AMB and triangle BMC

- shubhamsrg

ohh leave it
got it..

- shubhamsrg

if ab and and a+b are rational,,its not necessary both are rational,,
eg. -sqrt(2) +1 and sqrt(2) + 1

- shubhamsrg

has to be an other way around..

- anonymous

yeah i was thinkin of that

- shubhamsrg

hmm..

- anonymous

guys i gotta go...but i love to know what is the answer

- shubhamsrg

|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

hasn't this been solved yet?

- shubhamsrg

its a diff ques..

- experimentX

i lost track of it ... it's fuzzy ... well what's the Q?

- anonymous

|dw:1345132117021:dw|if ABC has rational sides,
we have to show AMB and MBC also has rational sides.

- anonymous

@experimentX thats the question

- experimentX

|dw:1345138702333:dw|
a^2 + b^2 = c^2
b^2 = rational
AM + MC = rational
AM^2 + MC^2 = rational.
AM*MC = rational.

- experimentX

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

typo:- Let's assume both to be irrational,

- anonymous

here is my proof|dw:1345140385316:dw|

- anonymous

\(x+y=c\) 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

lol ... your method is more obvious!!

- shubhamsrg

aha..i see and understand both methods!
thanks again @mukushla and @experimentX

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