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Algebra

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25 Mar 2008 19:40:35 IST

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$\text {Solve:} \\ \\ \frac {4x^2} {4x^2+1} = y \\ \\ \frac {4y^2}{4y^2+1} = z \\ \\ \frac {4z^2}{4z^2+1} = x$

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sandeep ramesh

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25 Mar 2008 20:30:45 IST

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okay okay no feel da :D

to know how bad i am in everything see here

sreeraman nagasubramaniyan

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25 Mar 2008 20:35:50 IST

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$\mbox{Substituting} \\ x= \frac{1}{2} \tan(a) , y=\frac{1}{2} \tan(b) , z=\frac{1}{2} \tan(c) \\ \\ \Longrightarrow \sin^2a = \frac{1}{2} \tan(b) \\ \\ \sin^2b = \frac{1}{2} \tan(c) \\ \\\sin^2c = \frac{1}{2} \tan(a) \\ \\ \mbox{This cyclic relation forces the angle to lie within} \ 0 \ \mbox{and} \ \frac{\pi}{4} \\ \\ \mbox{Reason:} \ \sin(\theta)<=1 \\ \\ \mbox{ans the only two angles where} \ \sin^2(\theta)= \frac{1}{2}\tan(\theta) \ (0,\frac{\pi}{4}) \\ \\ \mbox{is} \ 0 \ \mbox{and} \ \frac{\pi}{4} \\ \\ \mbox{Thus the values are} \ 0,\frac{1}{2}$

gokul subramanian

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25 Mar 2008 20:36:44 IST

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rohit thts a bit 2 much u no..

pardesi .svk

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25 Mar 2008 20:37:03 IST

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claim is x=y=z is the only solution provided all are non zero if one of them is 0 say x=0 then y=0 hence z=0 so x=y=z
if say some two are equal say x=y then x=y=z=1/2
so
say x>y>z
then the function can be written as
f(u)=1/(1+(1/4u^{2}))
which is increasing in u
so f(x) >f(y)>f(z) which gives y>z>x a contradiction

shridhar ramachandran

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25 Mar 2008 20:37:25 IST

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edit: soln posted on top

sandeep ramesh

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25 Mar 2008 20:38:22 IST

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shrids, i told ya that was a bit long :D but to each person his own :)

gokul, i was speaking the truth :D

pardesi .svk

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25 Mar 2008 20:44:14 IST

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@sboosy
how do u proove ur claim that only angles for which (tan \theta)/2 and sin^{2} \theta are equal is 0 and pi/4 in that domain
(this may be simple but it surely constitutes the most important part of ur proof)

sreeraman nagasubramaniyan

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25 Mar 2008 21:15:22 IST

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well okay for everyone's satisfaction ...here goes
2sin^2(a) = tan(a)
=> sin(a) [ sin(a) - 1/2cos(a)] = 0
=> sin(a) = 0 (or) sin(2a) = 1
=> a=0 or pi/4
i just took it for for granted as this part is easy

pardesi .svk

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25 Mar 2008 21:16:50 IST

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it wasn't a question of easy or difficult but completeness because ur entire proof laid on this fact only ...that's why i wrote in my prev post that the proof is easy but imp.

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