Let real numbers
![x_1, x_2, \cdots , x_n](/media/m/c/5/c/c5ceeea4536362d946dfa4482a4b88d4.png)
satisfy
![0 < x_1 < x_2 < \cdots< x_n < 1](/media/m/c/3/6/c3690503257082fb4685c3c696a5a18f.png)
and set
![x_0 = 0, x_{n+1} = 1](/media/m/d/8/6/d8621ff4da04ab8570fc0a192d5a671c.png)
. Suppose that these numbers satisfy the following system of equations:
![\sum_{j=0, j \neq i}^{n+1} \frac{1}{x_i-x_j}=0 \quad \text{where } i = 1, 2, . . ., n.](/media/m/3/d/9/3d97fb9d6d77d6f14a09371f3137c6dc.png)
Prove that
![x_{n+1-i} = 1- x_i](/media/m/d/0/3/d0314e96d3fbeed027a18a306d27a270.png)
for
%V0
Let real numbers $x_1, x_2, \cdots , x_n$ satisfy $0 < x_1 < x_2 < \cdots< x_n < 1$ and set $x_0 = 0, x_{n+1} = 1$. Suppose that these numbers satisfy the following system of equations:
$$\sum_{j=0, j \neq i}^{n+1} \frac{1}{x_i-x_j}=0 \quad \text{where } i = 1, 2, . . ., n.$$
Prove that $x_{n+1-i} = 1- x_i$ for $i = 1, 2, . . . , n.$