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Turkey NMO 2010 1st Round - P15 (Algebra)

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July 27, 2012

Problem Statement

If the real numbers x,y,zx,y,z satisfies the equations xyzx+y=−1\frac{xyz}{x+y}=-1, xyzy+z=1\frac{xyz}{y+z}=1, and xyzz+x=2\frac{xyz}{z+x}=2, what can xyzxyz be?
<spanclass=′latex−bold′>(A)</span> −815<spanclass=′latex−bold′>(B)</span> 85<spanclass=′latex−bold′>(C)</span> −835<spanclass=′latex−bold′>(D)</span> 715<spanclass=′latex−bold′>(E)</span> None <span class='latex-bold'>(A)</span>\ -\frac{8}{\sqrt {15}} \qquad<span class='latex-bold'>(B)</span>\ \frac{8}{\sqrt 5} \qquad<span class='latex-bold'>(C)</span>\ -8\sqrt{\frac{3}{5}} \qquad<span class='latex-bold'>(D)</span>\ \frac{7}{\sqrt{15}} \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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