hello, for $x$ i have found the equation $(x-1)(x^3-7x^2+7x+7)=0$ hence the solution is $x=y=z=1$ and for the sum we get $x+y+z=3$. Sonnhard.

Let $a:=x+y+z$. Since $zx+y=2z$ and two similar equations hold, summing all of them yields $yz+zx+xy=a$ too. Now, write $y=z(2-x)$. Multiply all such three equations to get $1=(2-x)(2-y)(2-z)$, or $xyz=7-2a$. Finally, because $zx=2z-y$, we have $7-2a=xyz=2yz-y^2$. Adding the three of such equations, we have \[3(7-2a)= 4(yz+zx+xy)-(x+y+z)^2=4a-a^2\,.\] That is, $a^2-10a+21=0$. Hence, $a=3$ or $a=7$. If $a=3$, then $x=y=z=1$. If $a=7$, then $xyz=-7$, so not all of $x$, $y$, and $z$ may be positive reals. Therefore, the only possible solution is $x=y=z=1$.

Also, how'd you get these problems so early? Not that there's anything wrong.. it's just that the competition only finished about a couple hours ago.

My solution: We just need to check two cases (the other are analogous): Case 1. $x\geq y \geq z$ $2=x+\frac{y}{z}\geq x+\frac{z}{z}=x+1$ which implies $1 \geq x \geq y \geq z$. Futhermore, we have $1+\frac{z}{x}\geq y+\frac{z}{x}=2$ which implies $z\geq x$. Then we have $x=y=z$. Case 2. $x\geq z \geq y$ $2=y+\frac{z}{x}\leq y+\frac{x}{x}=y+1$ which implies $1 \leq y \leq z \leq x$. Futhermore, we have $1+\frac{x}{y}\leq z+\frac{x}{y}=2$ which implies $x\leq y$. Then we have $x=y=z$. We can conclude like above.

From the given equations we get \begin{eqnarray*}xz+y&=&2z\\ yx+z&=&2x\\ zy+x&=&2y\end{eqnarray*} Adding them yields $xy+xz+yz = x+y+z$ Suppose wlog $ x \ge y \ge z$. Hence $xy \ge xz \ge yz$. If $xy < z => xy+xz+yz < 3z \le x+y+z$. Hence $xy \ge z => x\ge \frac {z}{y}$. Therefore we have $2 = x+ \frac {y}{z} \ge \frac{z}{y} + \frac{y}{z} \ge 2 => y=z$ Finally $x+1 =2 => x=1$ and from $y+z=2$ and $z+\frac{1}{y} = 2$ we get $y=z=1$. Therefore $x+y+z = 3$

This problem has a lot of nice solutions.. here's another one Due to the Cauchy-Schwarz Inequality and the AM-GM inequality we have \[{2(x+yz)=(x+\dfrac{y}{z}})(x+yz)\ge (x+y)^2\ge 4xy\] Furthermore, we know that from the condition we have $x+yz=2y$ so that \[2(2y)\ge 4xy \implies 1\ge x\] applying similar inequalities for $y$ and $z$ we obtain $x,y,z\le 1$. This implies that \[1+\dfrac{y}{z}\ge x+\dfrac{y}{z}=2\implies y\ge z\] again, applying similar inequalities for the other variables we get $y\ge z$, $z\ge x$ and $x\ge y$ which can only hold simultaneously if $x=y=z$, plugging this into the condition gives us that $x=y=z=1$ and we're done! Hmmm.. maybe this problem was too easy for a Centro #5... Also, I found day two to be much easier than day one.

Compared to Centro #5 2010, which is also based on algebraic manipulations yes, this was easier. Problem 4 and 5 are easier than problem 1, at least in problem 1 you had to be careful. I haven't tried problem 3 but 6 is a hard problem.

There is another wonderful solution using AM-GM. You will get $z\ge xy$, $x\ge yz$ and $y\ge xz$ but $x+y+z=xy+xz+yz$ leading to equality in each, forcing $x=y=z$ It's a very short, sweet solution.

tonypr wrote: Also, how'd you get these problems so early? Not that there's anything wrong.. it's just that the competition only finished about a couple hours ago. I have some good friends among the tutors who attended the olympiad, and they kindly sent me the papers right after the tests were over. Anyway, glad you guys liked the problem

it's trivial that $0<x,y,z<2$ by substituting $x=\frac{z}{2-y},z=\frac{2y(y-2)}{y^2-2y-1}$ we get $(y-1)(y^3-7y^2+7y+7)=0$ so $y=1$ or $y=5.493...;2.109...;-0.603...$ since $0<y<2$,so $y=1$ yielding $x=y=z=1$ QED

Adding the three equations yields $$ x + y + z + \frac{y}{z} + \frac{z}{x} + \frac{x}{y} = 6. $$Dividing the first equation by $y$, the second one by $z$ and the third one by $x$ and adding the results yields $$ \frac{y}{z} + \frac{z}{x} + \frac{x}{y} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2 \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right). $$Hence, $$ \frac{y}{z} + \frac{z}{x} + \frac{x}{y} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}. $$Combined with the first obtained equation, we get $$ x + y + z + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 6. $$As $a + \dfrac{1}{a} \geq 2$ for all positive $a$, we can only get equality if $x = y = z = 1$. Hence, $x + y + z = 3$.

This problem is only algebraic manipulation...

Other solution: Since $x + \frac{y}{z} = 2 \rightarrow xz + y = 2z \rightarrow y = z(2-x)$ Doing the same thing, we have $x = (2-z)y \rightarrow y = \frac{x}{2-z} \rightarrow z(2-x)(2-z) = x \rightarrow z(2-z)(2- z(2-z)(2- z(2-z)\cdots) = x \rightarrow z(2-z)x = x \rightarrow z^2 - 2z + 1 = 0 \rightarrow z = 1$. Than, we have that $x = y \text { and } y = 2-x \rightarrow x = 2-x \rightarrow x = 1$ and $y = 1$, so $x + y + z = 3$

assume that wlog $x$ is the largest one among them.Then,$ x \ge y \implies y/z \le z/x \implies xy \le z^2$ implies $y$ is the smallest one among them.again, $y+z/x=z+x/y \implies z/x \ge x/y \implies yz \le x^2 \implies x=y=z \implies x=y=z=1$

$xy+yz+zx= x+y+z$, $x+\frac{y}{z}\geq 2 (\frac{xy}{z})^{\frac{1}{2}}$ $2\geq 2 (\frac{xy}{z})^{\frac{1}{2}}$ $x\geq yz$ equality holds at $AM-GM$ equality which is in $x=yz, y=zx, z=xy --> \frac{x}{y}=z,\frac{y}{z}=x, \frac{z}{x}=y$ summing all equalities $(x+y+z) + \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=6$ $ (\frac{x}{y}+\frac{y}{z}+\frac{z}{x})=3$ $x=y=z=1$ $x+y+z=3$

Adding the equations $xz+y=2z$, $xy+z=2x$, and $yz+x=2y$, we have $x+y+z=xy+yz+zx$. Let this common value be denoted $t$. Adding the equations $xyz+y^2=2yz$, $xyz+z^2=2xz$, and $xyz+x^2=2xy$, we have $3xyz+x^2+y^2+z^2=2(xy+yz+zx)$, or $xyz=\frac{4t-t^2}3$. Notice that this implies $\frac{4t-t^2}3\ne0$ since $xyz\ne0$, or $t\notin\{0,4\}$. Multiplying the equations $x+\frac{y}{z}=y+\frac{z}{x}=z+\frac{x}{y}=2$, we have: $$8=x^2+y^2+z^2+xyz+\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}+1=(x+y+z)^2-2(xy+yz+zx)+xyz+\frac{(xy+yz+zx)^2-2xyz(x+y+z)}{xyz} =t^2-2t+\frac{4t-t^2}3+\frac{t^2-2t\cdot\frac{4t-t^2}3}{\frac{4t-t^2}3}+1=\frac{2t^3-16t^2+23t}{3t-12}+1$$so $t^3-8t^2+t+42=0$, which factors as $(t+2)(t-3)(t-7)=0$. If $t=-2$, then $xyz=-4$ which is impossible for positive reals. If $t=7$, then $xyz=-7$ which is impossible for positive reals. But $t=\boxed3$ is achievable with $x=y=z=1$.