Let $n$ be a positive integer, and let $a>0$ be a real number. Consider the equation: \[ \sum_{i=1}^{n}(x_i^2+(a-x_i)^2)= na^2 \] How many solutions ($x_1, x_2 \cdots , x_n$) does this equation have, such that: \[ 0 \leq x_i \leq a, i \in N^+ \]
Problem
Source: Pan African 2001
Tags:
deej21
04.10.2005 05:42
Expanding LHS, we get
$\sum_{i=1}^n (2x_i^2-2ax_i + a^2) = na^2$
$2\sum x_i^2 - 2a\sum x_i + na^2 = na^2$
$\sum_{i=1}^n x_i^2 = a\sum_{i=1}^n x_i$
Dang... can't get any further. I think I'm missing something easy...
socrates
04.10.2005 14:13
Let's continue.... $\sum_{i=1}^n x_i^2 =a\sum_{i=1}^n x_i$ or $\sum_{i=1}^n x_i^2 -\sum_{i=1}^n ax_i=\sum_{i=1}^nx_i(x_i-a)\leq0$ because $0\leq x_i \leq a$ so $x_i=a$ $\forall i=1,2,3, \cdots$
tµtµ
04.10.2005 15:44
An other solution : $f(x) = x^2 + (x-a)^2 < a^2$ for $0 \leq x \leq a$ except when $x = 0$ or $x = a$ So there is equality iff $x_i \in \{ 0,a \}$ and there are $2^n$ solutions