Consider the following system of 10 equations in 10 real variables $v_{1}, \ldots, v_{10}$ :

$$
v_{i}=1+\frac{6 v_{i}^{2}}{v_{1}^{2}+v_{2}^{2}+\cdots+v_{10}^{2}} \quad(i=1, \ldots, 10) .
$$

Find all 10-tuples $\left(v_{1}, v_{2}, \ldots, v_{10}\right)$ that are solutions of this system.

## Sun

Life Financial
For a particular solution $\left(v_{1}, v_{2}, \ldots, v_{10}\right)$, let $s=v_{1}^{2}+v_{2}^{2}+\cdots+v_{10}^{2}$. Then

$$
v_{i}=1+\frac{6 v_{i}^{2}}{s} \Rightarrow 6 v_{i}^{2}-s v_{i}+s=0
$$

Let $a$ and $b$ be the roots of the quadratic $6 x^{2}-s x+s=0$, so for each $i, v_{i}=a$ or $v_{i}=b$. We also have $a b=s / 6$ (by Vieta's formula, for example).

If all the $v_{i}$ are equal, then

$$
v_{i}=1+\frac{6}{10}=\frac{8}{5}
$$

for all $i$. Otherwise, let $5+k$ of the $v_{i}$ be $a$, and let $5-k$ of the $v_{i}$ be $b$, where $0<k \leq 4$. Then by the AM-GM inequality,

$$
6 a b=s=(5+k) a^{2}+(5-k) b^{2} \geq 2 a b \sqrt{25-k^{2}}
$$

From the given equations, $v_{i} \geq 1$ for all $i$, so $a$ and $b$ are positive. Then $\sqrt{25-k^{2}} \leq 3 \Rightarrow 25-k^{2} \leq 9 \Rightarrow k^{2} \geq 16 \Rightarrow k=4$. Hence, $6 a b=9 a^{2}+b^{2} \Rightarrow(b-3 a)^{2}=0 \Rightarrow b=3 a$.

Adding all given ten equations, we get

$$
v_{1}+v_{2}+\cdots+v_{10}=16
$$

But $v_{1}+v_{2}+\cdots+v_{10}=9 a+b=12 a$, so $a=16 / 12=4 / 3$ and $b=4$. Therefore, the solutions are $(8 / 5,8 / 5, \ldots, 8 / 5)$ and all ten permutations of $(4 / 3,4 / 3, \ldots, 4 / 3,4)$.