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Find the value of $S$ if $$S = {x\over y} + {y\over z} + {z\over x} = {y\over x} + {z\over y} + {x\over z}$$ and $x + y + z = 0$.

This is a difficult question in my opinion and I was wondering if I could get some help on it. I tried to multiply by $xyz$ but that didn't help. Any help would be appreciated.

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  • $\begingroup$ Please name the source of the problem -- which contest is it from, if any? Even "I heard it make its rounds at IMO training in New Zealand" is better than nothing. $\endgroup$ – darij grinberg Apr 2 at 15:05
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Note that the sum of any two is the negative of third.

Now if we add two times S then we have $$2S = {x\over y} + {y\over z} + {z\over x} + {y\over x} + {z\over y} + {x\over z}$$ $$ ={x+y\over z} +{y+z\over x} +{x+z\over y} = $$ $$ = {-z\over z}+{-x\over x} +{-y\over y} = -3$$

So $$ S = -{3\over 2}$$

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    $\begingroup$ Really really elegant :) $\endgroup$ – Eureka Mar 27 at 20:04
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$$0=\sum_{cyc}\left(\frac{x}{y}-\frac{y}{x}\right)=\frac{\sum\limits_{cyc}(x^2z-x^2y)}{xyz}=\frac{(x-y)(y-z)(z-x)}{xyz}.$$ Since our conditions does not depend on any cyclic permutation of the variables,

we can assume that $x=y$.

Thus, $z=-2y$ and $$S=1-\frac{1}{2}-2=-\frac{3}{2}.$$

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Denote: $y=ax, z=bx$. Then: $$x+y+z=0 \iff x+ax+bx=0 \iff a=-1-b;\\ S={x\over y} + {y\over z} + {z\over x} = {y\over x} + {z\over y} + {x\over z} \iff \\ S={1\over a} + {a\over b} + b = a + {b\over a} + {1\over b} \iff \\ S=-{1\over 1+b} - {1+b\over b} + b = -1-b - {b\over 1+b} + {1\over b} \iff \\ \frac{(b+2)(2b+1)(b-1)}{b(1+b)}=0 \iff \\ b_{1,2,3}=\{\color{red}{-2},\color{green}{-\frac12},\color{blue}1\}; a_{1,2,3}=\{\color{red}{1},\color{green}{-\frac12},\color{blue}{-2}\}\\ S_1=\frac1a+\frac ab+b=\frac1{\color{red}{1}}+\frac{\color{red}1}{\color{red}{-2}}+(\color{red}{-2})=-\frac32.\\ S_2=\frac1a+\frac ab+b=\frac1{\color{green}{-\frac12}}+\frac{\color{green}{-\frac12}}{\color{green}{-\frac12}}+(\color{green}{-\frac12})=-\frac32.\\ S_3=\frac1a+\frac ab+b=\frac1{\color{blue}{-2}}+\frac{\color{blue}{-2}}{\color{blue}{1}}+\color{blue}{1}=-\frac32.\\ $$

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  • $\begingroup$ there is one upvote and one downvote so far. wondering why downvote - long? $\endgroup$ – farruhota Apr 5 at 5:53

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