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Cool goIITian

Joined: 18 Jun 2007

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25 Mar 2008 19:43:07 IST

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one could be x=y=z

Dinesh Babu

Blazing goIITian

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25 Mar 2008 19:43:58 IST

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x=y=z=0 is one soln

Hari Shankar

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Joined: 28 Feb 2007

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25 Mar 2008 19:46:30 IST

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I am looking for all solutions. x=y=z=0 is one. Are there any other. Your working should be able to exhaust all possibilities.

Cool goIITian

Joined: 18 Jun 2007

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25 Mar 2008 19:49:59 IST

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x=y=z=1/2

Cool goIITian

Joined: 18 Jun 2007

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25 Mar 2008 19:51:07 IST

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did u get it
i think it is the last possible solution

Hari Shankar

Forum Expert

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25 Mar 2008 19:51:45 IST

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did who get it?

Akhil

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Joined: 17 Jan 2008

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25 Mar 2008 19:52:23 IST

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i am not sure if we can do this, but by symmetry
x=y=z=a

so

4a²/(4a²+1)=a

so if a not equal to 0

4a=4a²+1

(2a-1)²=0
2a=1
or a=1/2

substituting a=0
works

so values of
a=x=y=z={0,1/2}

Cool goIITian

Joined: 18 Jun 2007

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25 Mar 2008 19:52:47 IST

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solved kya

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

25 Mar 2008 20:11:48 IST

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WLOG assume 1>x>=y>=z>=0

Clearly from eqn1 and eqn3 making eqn3>=eqn1 we get z>=x which is a contradiction So x=y=z

so solving the quadratic eqns, x=y=z ={0,1/2}

Cool goIITian

Joined: 18 Jun 2007

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25 Mar 2008 20:12:59 IST

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thats wht i said

Blazing goIITian

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25 Mar 2008 20:14:35 IST

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sorry, but where did you say that? im sorry i didnt see it :)

Cool goIITian

Joined: 18 Jun 2007

Posts: 79

25 Mar 2008 20:16:16 IST

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i think it was not posted on web
i told after the ans i gave

Blazing goIITian

Joined: 28 Jan 2008

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25 Mar 2008 20:16:45 IST

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{1/x + 1/y}= (1 + 1/2x)^2 1
{1/x + 1/z}= (1 + 1/2z)^2 2
{1/z+ 1/y}= (1 + 1/2y)^2 3
assume 1/x>1/y>1/z
tcompareing RHS hen v shud have 3>2
which will mean 1/z + 1/y>1/x + 1/z but this is false
if v assume 1/x>1/z>1/y
then comp RHs 1>2 => 1/x+1/y>1/x+1/z but this is false..
assume nething else(inequality) v get 1 such relation false
so only possibility is x=y=z

Blazing goIITian

Joined: 28 Jan 2008

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25 Mar 2008 20:18:49 IST

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well i neednt have complicated so much...its obvious

but 1 can say i think differently

shridhar ramachandran

Scorching goIITian

Joined: 30 Jan 2008

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25 Mar 2008 20:19:06 IST

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let $2x = \tan a , 2y = \tan b, 2z = \tan c$
so, the eqns are:

$\frac {\tan ^2 a}{\sec ^2 a} =\frac{\tan b}{2}$

$=> \sin ^2 a = \frac{\tan b}{2}$
similarly

$=> \sin ^2 b = \frac{\tan c}{2}$

$=> \sin ^2 c = \frac{\tan a}{2}$

multiplying all 3, and replacing $\tan a$ as $\frac {\sin a}{\cos a}$ we get
$\sin a. \sin b. \sin c.(8\sin a. \sin b. \sin c. \cos a. \cos b. \cos c - 1) = 0$

$\sin a. \sin b. \sin c.(\sin 2a. \sin 2b. \sin 2c. -1) = 0$

so $\sin a. \sin b. \sin c = 0$ which gives you
x=y=z=0.

OR

$\sin 2a. \sin 2b. \sin 2c. = 1$
which gives you a = b = c = $\frac {\pi}{4}$
$=> x = y = z = \frac 1 2$
Smile

phew!

Blazing goIITian

Joined: 13 Mar 2008

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25 Mar 2008 20:19:54 IST

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gokul, how is the soln any diff?

@sandy1990: Im sorry, but i dont have saintly powers to conceive what people have in their minds or what people mumbled to themselves

Blazing goIITian

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Posts: 1183

25 Mar 2008 20:20:45 IST

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shrids, i thought of the same thing at first but realised that it was tooooooooo tedious man :D

Blazing goIITian

Joined: 28 Jan 2008

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25 Mar 2008 20:28:45 IST

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no i wasn sayin soln is different..
i did huge process to prove x=y=z...
so my polite way of sayin im different is directly related to the rude way in which u wud say im stupid!