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 Symmetric expressions (Posted on 2023-09-13)
Let a, b, c, x, y and z be numbers such that a=(b+c)/(x-2), b=(c+a)/(y-2), c=(a+b)/(z-2). If xy+yz+zx=1000 and x+y+z=2016, find the value of xyz.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Question | Comment 1 of 2
Is this problem correctly set out?

Simply ignoring most of what's given, and substituting for z:
xy+x(2016-x-y)+y(2016-x-y)=1000.

Then (x+y+z)^2-(x^2+y^2+z^2)=2(xy+xz+yz)
Substituting (2016)^2-(x^2+y^2+z^2)=2(1000)
so x^2+y^2+z^2=4062256.
Substituting x^2+y^2+(2016-x-y)^2=4062256

xy+x(2016-x-y)+y(2016-x-y)=1000, x^2+y^2+(2016-x-y)^2=4062256 can be solved for x and y, and so z:
x=(672-4p), y=(672+2p), z=y, p=sqrt(338438/3)
x=(672+4p), y=(672-2p), z=y, p=sqrt(338438/3)

(Switching is possible between these values, but only the product xyz is needed.)

Then xyz=(672-4p)(672+2p)^2=-(5415008 sqrt(338438/3))/3 + 606256896), around -1,212,513,916,
or xyz=(672+4p)(672-2p)^2= (5415008 sqrt(338438/3))/3 - 606256896) around 124.023196

Edited on September 14, 2023, 9:26 am
 Posted by broll on 2023-09-14 01:30:24

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