Exercise 13. A polynomial $P$ is said to be fantastic if there exist real numbers $\mathrm{a}_{0}, \ldots, \mathrm{a}_{2022}$ such that

$$
P(X)=X^{2023}+a_{2022} X^{2022}+\cdots+a_{1} X+a_{0}
$$

if it has 2023 roots $\mathrm{r}_{1}, \ldots, \mathrm{r}_{2023}$ (not necessarily distinct) in $[0,1]$, and if $\mathrm{P}(0)+\mathrm{P}(1)=0$. Determine the maximum value that $r_{1} \cdot r_{2} \cdots r_{2023}$ can take for a fantastic polynomial.
## Solution to Exercise 13

The answer is $2^{-2023}$. We will show this in two steps: first, we will show that we cannot find a polynomial with a larger value, and second, we will exhibit a polynomial that satisfies the conditions of the problem and whose product of roots is indeed equal to $2^{-2023}$.

By hypothesis, we can rewrite $P(X)=\prod_{k=1}^{2023}\left(X-r_{k}\right)$. In particular, $P(0)=-\prod_{k=1}^{2023} r_{k}$ and $P(1)=\prod_{k=1}^{2023}\left(1-r_{k}\right)$. The hypothesis $P(0)+P(1)=0$ becomes:

$$
\prod_{k=1}^{2023}\left(1-r_{k}\right)=\prod_{k=1}^{2023} r_{k}=r_{1} \cdot r_{2} \cdots r_{2023}
$$

Thus, since for all $k$, $r_{k} \geqslant 0$ and $1-r_{k} \geqslant 0$, by the arithmetic-geometric mean inequality,

$$
\left(r_{1} \cdot r_{2} \cdots r_{2023}\right)^{2}=\prod_{k=1}^{2023}\left(1-r_{k}\right) r_{k} \leqslant \prod_{k=1}^{2023}\left(\frac{\left(1-r_{k}\right)+r_{k}}{2}\right)^{2}=\frac{1}{4^{2023}}
$$

Thus, $r_{1} \cdot r_{2} \cdots r_{2023} \leqslant 2^{-2023}$.

Furthermore, if $P(X)=\left(X-\frac{1}{2}\right)^{2023}$, $P$ is a monic polynomial of degree 2023, has 2023 roots in $[0,1]$, and satisfies $P(0)+P(1)=\left(-\frac{1}{2}\right)^{2023}+\left(\frac{1}{2}\right)^{2023}=0$, so $P$ is fantastic. The product of its roots is $2^{-2023}$. Therefore, the maximum value that $r_{1} \cdot r_{2} \cdots r_{2023}$ can take for a fantastic polynomial is $2^{-2023}$.

Grader's Comment: The exercise required finding the maximum value that the product of roots could take for a particular class of polynomials. Almost all students found the case that realizes the maximum (and thus its value), and a satisfactory proportion of these students completed the proof, generally by applying the AM-GM inequality. Vieta's formulas seem to be well mastered but were absolutely not necessary for solving the problem. A key takeaway: the AM-GM inequality is often a judicious inequality to show that a product of positive reals is maximal when all its terms are equal.