All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Symmetric expressions (Posted on 2023-09-13) Difficulty: 2 of 5
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

Comments: ( Back to comment list | You must be logged in to post comments.)
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

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information