March 15th, 2011, 08:22 AM  #1 
Newbie Joined: Apr 2010 Posts: 2 Thanks: 0  Geometry help needed :)
Hey guys! Can someone tell me how does this surface looks like > S: abs(xy+z)+abs(yz+x) +abs(zx+y)=1 ?? Maybe some ingenious substitution needed? Thanks in advance. 
March 15th, 2011, 08:53 AM  #2 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Geometry help needed :)
Looking at the crosssection z=0, you get 2xy+x+y = 1. The substitution u=xy, v=x+y (which involves a tilting of the axes by 45°) gives v = 1  2u, and this is a rhombus. By symmetry, the crosssections through x=0 and y=0 are also rhombi, which suggests that the shape is maybe a rhombohedron, a kind of parallelepiped.

March 15th, 2011, 09:54 AM  #3 
Newbie Joined: Apr 2010 Posts: 2 Thanks: 0  Re: Geometry help needed :)
Thanks a lot indeed! 
March 15th, 2011, 08:19 PM  #4 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Geometry help needed :)
On second thoughts, this is a linear transformation of x+y+z=1, which suggests that it is an octahedron. Oops!


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