A geometry problem by A Former Brilliant Member

Geometry Level 4

Let x , y x, y and z z be real numbers satisfying x + y + z = x y z x+y+z = xyz .

Which of the following algebraic expressions must be equal to

x ( 1 − y 2 ) ( 1 − z 2 ) + y ( 1 − z 2 ) ( 1 − x 2 ) + z ( 1 − x 2 ) ( 1 − y 2 ) ? x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z (1-x^2) ( 1-y^2)?

4 x y z 4xyz 12 ( x 2 + y 2 + z 2 ) x + y + z \frac{12 ( x^2 + y^2 + z^2 ) } { x+y+z} 12 3 12 \sqrt{3} 4 3 ( x 3 + y 3 + z 3 ) \frac{4}{3} ( x^3 + y^3 + z^3)

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Kushal Bose
Nov 18, 2016

For the first part:

x ( 1 − y 2 ) ( 1 − z 2 ) = x − y 2 x − z 2 x + y 2 z 2 x = x − y 2 x − z 2 x + y z . ( x y z ) = x − y 2 x − z 2 x + y z . ( x + y + z ) x(1-y^2)(1-z^2)=x-y^2x-z^2x+y^2z^2x \\ =x-y^2x-z^2x+yz.(xyz) \\ =x-y^2x-z^2x+yz.(x+y+z)

Repeat this for other two parts and add them you will get answer as 4 x y z 4xyz

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