$a_1,a_2,a_3,\ldots,a_n,a_{n+1}$ are fixed real numbers in $(-1,\infty)$.
$x_1$ and $x_2$ are fixed real numbers in $(0,1)$.
Is it possible to prove that there exists or doesn't exists a real number $x_3$ in $(0,1)$ such that
$$\prod_{i=1}^n (1 + x_3a_i) = \frac{1}{2}\prod_{i=1}^n (1 + x_1a_i) + \frac{1}{2}\prod_{i=1}^n (1 + x_2a_i)$$
and
$$\prod_{i=1}^{n+1} (1 + x_3a_i) = \frac{1}{2}\prod_{i=1}^{n+1} (1 + x_1a_i) + \frac{1}{2}\prod_{i=1}^{n+1} (1 + x_2a_i)$$
or is it possible to prove that this problem is unsolvable to begin with?
If there is a simultaneous solution, then either $a_{n+1}=0$ or it must be $$ x_3=\frac{x_1\prod_{i=1}^n (1 + x_1a_i) + x_2\prod_{i=1}^n (1 + x_2a_i)}{\prod_{i=1}^n (1 + x_1a_i) + \prod_{i=1}^n (1 + x_2a_i)}. $$ That this is actually also a solution of the first $n$-th degree polynomial is highly unlikely.