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It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1

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nick1816
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It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   Updated on: 22 Apr 2019, 00:38
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It is given that \(x^2+y^2=z+1\), \(y^2+z^2=x+1\) and \(z^2+x^2=y+1\). Find the value of x*y*z?

A. 1
B. -1/8
C. 1 or 1/8
D. 1 or -(1/8)
E. 1/3
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D

Originally posted by nick1816 on 21 Apr 2019, 22:52.
Last edited by Bunuel on 22 Apr 2019, 00:38, edited 1 time in total.
Renamed the topic.
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nick1816
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Re: It is given that x^2 + y^2 = z + 1, y^2 +z^2 = x + 1 [#permalink] New post   22 Apr 2019, 04:52
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\(x^2 + y^2\) = z + 1 and \(y^2 + z^2\) = x + 1
subtracting these equations, we get \(x^2 - z^2\)=z-x
or (x+z )(x-z) + (x-z) =0
or (x-z) (x+z+1 ) =0
Hence either x=z or x+z=-1
Similarly you can prove either y=z or y+z=-1
Either way we will get x=y=z

Now we have \(x^2 + y^\)2 = z + 1 or \(x^2 + x^2\) = x + 1
2\(x^2 - x\) - 1=0
x= 1 or -(1/2)

Whenever you see these symmetrical equations in a particular question, mostly you have to either subtract them or add them. Then factorize the resulting equation.
neha283 wrote:
This seems like a formula based question. Could someone post a solution on how to approach it?
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   22 Apr 2019, 04:02
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This seems like a formula based question. Could someone post a solution on how to approach it?
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   23 Apr 2019, 08:13
Can someone please help with this?
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   30 Jun 2019, 20:40
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Hi nick1816, why did we completely ignore the equation x+z=-1 in the solution? Doesn't that have any significance?
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   30 Jun 2019, 20:51
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x+y=-1, y+z=-1 and z+x=-1
Solve them you'll get
x=y=z= -1/2
I didn't ignore it, just didn't mention it explicitly.

pankaj2k19 wrote:
Hi nick1816, why did we completely ignore the equation x+z=-1 in the solution? Doesn't that have any significance?


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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   30 Jun 2019, 21:02
Ohh correct. But one general doubt that this question gave me is: if I get say 3 sets of eq
x=y OR x+y = -1
y=z OR y-z = -1
x=z OR x-z = -1

So while solving are there 2*2*2 number of possible ways of solving this sets of equation? Where one possible way could be -
x=y AND y-z = -1 AND x=z??

Thanks for your reply

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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   30 Jun 2019, 21:09
Ohh correct. But one general doubt that this question gave me is: if I get say 3 sets of eq
x=y OR x+y = -1
y=z OR y-z = -1
x=z OR x-z = -1

So while solving are there 2*2*2 number of possible ways of solving this sets of equation? Where one possible way could be -
x=y AND y-z = -1 AND x=z??

Thanks for your reply

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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink] New post   19 Jul 2021, 08:44
Expert Reply
1st: \(x^2 + y^2 = z + 1\)

2nd: \(y^2 + z^2 = x + 1\)

3rd: \(z^2 + x^2 = y + 1\)

If x = 1, y = 1 and z = 1, Then all conditions get satisfied.

Therefore, x * y * z = 1

Also, when x = \(\frac{1}{2}\) and y = \(\frac{1}{2}\) then z = \(\frac{-1}{2}\) satisfy first condition.

It also satisfies 2nd and 3rd when two variables = \(\frac{1}{2}\) each and third is \(\frac{-1}{2}\).

Therefore, x * y * z = \(\frac{1}{2}\) * \(\frac{1}{2}\) * (\(\frac{-1}{2}\)) = \(\frac{-1}{8}\)

Answer D
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1 [#permalink]
19 Jul 2021, 08:44
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