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

Question

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

nick1816 · Accepted Answer

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 · Answer

This seems like a formula based question. Could someone post a solution on how to approach it?

arorni · Answer

Can someone please help with this?

pankaj2k19 · Answer

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

nick1816 · Answer

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.

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pankaj2k19 · Answer

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

Posted from my mobile device

pankaj2k19 · Answer

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

Posted from my mobile device

MathRevolution · Answer

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

Kinshook · Answer

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?

x^2 - z^2 = z - x
(x-z)(x+z) + (x-z) = 0
(x-z)(x+z+1) = 0
x = z or x+z=-1

Similarly it can be derived that
y = z or y+z=-1 & 
x = y or x+y=-1

x = y = z; 
x^2 + x^2 = x + 1
2x^2 - x - 1 = 0
x = 1 or x = -1/2

x*y*z = 1 or -1/8

IMO D

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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My Rewards