Saudi Arabian Mathematical Competitions

Denote $P(x) = (x - a_1)(x - a_2) \cdots (x - a_{100})$ with $a_1, a_2, \ldots, a_{100}$ are real roots of polynomial $P(x)$. So
$$P\left(2x^{2}-4x\right) = 0 \Leftrightarrow 2x^{2}-4x-a_i = 0 \text{ for } 1 \leq i \leq 100.$$
This equation cannot have $1$ root since $\Delta = 16 + 8a_i \neq 0$. So each equation can have $0$ or $2$ roots. Note that $P\left(2x^{2}-4x\right)$ has $130$ roots so there are $\frac{130}{2} = 65$ equations have $2$ roots, which mean there are $65$ numbers $a_i > -2$ and $35$ numbers $a_i < -2$.

By the same way, we have
$$P\left(4x-2x^{2}\right) = 0 \Leftrightarrow 2x^{2}-4x+a_i = 0$$
and $\Delta = 16 - 8a_i$ for any $1 \leq i \leq 100$. And there are $65$ numbers $a_i < 2$ and $35$ numbers $a_i > 2$.

By applying the principle of inclusion and exclusion, there are
$$65 + 65 - 100 = 30 \text{ numbers } a_i \in (-2 ; 2).$$

Suppose that $a_1 < a_2 < \ldots < a_{35} < -2 < a_{36} < \ldots < a_{64} < 2 < a_{65} < \ldots < a_{100}$; and denote $M, N, K$ as the subsets of these numbers with the indices $1 \rightarrow 35, 36 \rightarrow 64, 65 \rightarrow 100$.

Take $A_1(x) = \prod_{m \in M}(x - m)$, $B(x) = \prod_{n \in N}(x - n)$, $A_2(x) = \prod_{k \in K}(x - k)$ then we will prove that
$$|A_1(x_0) \cdot A_2(x_0)| > |B(x_0)|$$
for any number $x_0 \in (-1 ; 1)$.

Note that $\forall x_0 \in (-1 ; 1)$ then $|x_0 - m| > 1, \forall m < -2$ and $|x_0 - k| > 1, \forall k > 2$. We can suppose that $x_0 > 0$ and for any $n \in N$, we have two cases:

1. If $n > 0$ then $n \in (0 ; 2)$ and $|x_0 - n| < |x_0 - 2| < |x_0 - k|$ with any $k > 2$.
2. If $n < 0$ then $n \in (-2 ; 0)$ and $|x_0 - n| < |x_0 - (-2)| < |x_0 - m|$ with any $m < -2$.

So in all cases of $n$ respect to the factor $x_0 - n$ in $B(x_0)$, we can choose factor from $A_1(x_0)$ or $A_2(x_0)$ with absolute value greater than it, and since $|M|, |K| > |N|$, we always can do that. Hence $|A_1(x_0) \cdot A_2(x_0)| > |B(x_0)|, \forall x_0 \in (-1 ; 1)$ which mean this equation has no solution.

Therefore, we can choose $A(x) = A_1(x)A_2(x)$ and $B(x)$ to satisfy the given condition. $\square$