chaotic_iak wrote: Find all triples of reals $(x,y,z)$ satisfying: $$\begin{cases} \frac{1}{3} \min \{x,y\} + \frac{2}{3} \max \{x,y\} = 2017 \\ \frac{1}{3} \min \{y,z\} + \frac{2}{3} \max \{y,z\} = 2018 \\ \frac{1}{3} \min \{z,x\} + \frac{2}{3} \max \{z,x\} = 2019 \\ \end{cases}$$ Ordering $x,y,z$ as $a\le b\le c$, we get $a+2b=3u$ $a+2c=3v$ (and so $c-b=3\frac{v-u}2$ and so $u\le v$) $b+2c=3w$ (and so $b-a=3(w-v)$ and so $v\le w$) and so $(u,v,w)=(2017,2018,2019)$ and so $(x,y,z)=(b,a,c)$ System easily gives $(a,b,c)=\left(u+2v-2w,u-v+w,\frac{-u+v+2w}2\right)$ And so $(a,b,c)=\left(2015,2018,\frac{4039}2\right)$ And so $\boxed{(x,y,z)=\left(2018,2015,\frac{4039}2\right)}$

Why am I getting a contradiction? Please find out my mistake Without loss of generality, assume that $x\le y\le z$. $\frac13 x +\frac23 y =2017 .....(1)$ $\frac13 y+\frac23 z = 2018 .......(2)$ $\frac13 x +\frac23 z = 2019........(3)$ Adding these equations we get, $\frac23 x +y+ \frac43 z =6054$. Subtracting equation (3) from the above equation twice we get,$y=2016$ Plug the value of $y$ in equation (1) to get $x=2019$ and in equation (2) to get $z=2019$. But then $y<x$ which contradicts our assumption. Hence no solution. I can't figure out my mistake, please help me. @Oops! Got it!!

You can't assume $x\leq y\leq z$. What pco did was considering the actual ordering, and naming the variables that way