Year: 2018
Topic: Calculus Trigonometry

Problem:
Determine the greatest possible value of $\sum_{i=1}^{10} \cos(3x_i)$ for real numbers $x_1,x_2,\dots,x_{10}$ satisfying $\sum_{i=1}^{10} \cos(x_i) = 0$.

Solution:
Since $\cos(3x_i) = 4 \cos(x_i)^3 - 3 \cos(x_i)$, it is equivalent to maximize $4 \sum_{i=1}^{10} y_i^3$ for $y_1,\dots,y_{10} \in [-1,1]$ with $\sum_{i=1}^{10} y_i = 0$;  note that this domain is compact, so the maximum value is guaranteed to exist. For convenience, we establish something slightly stronger: we maximize $4 \sum_{i=1}^{n} y_i^3$ for $y_1,\dots,y_{n} \in [-1,1]$ with $\sum_{i=1}^{n} y_i = 0$, where $n$ may be any even nonnegative integer up to $10$, and show that the maximum is achieved when $n=10$.  We first study the effect of varying $y_i$ and $y_j$ while fixing their sum. If that sum is $s$, then the function $y \mapsto y^3 + (s-y)^3$ has constant second derivative $6s$, so it is either everywhere convex or everywhere concave. Consequently, if $(y_1,\dots,y_{n})$ achieves the maximum, then for any two indices $i<j$, at least one of the following must be true: \begin{itemize} \item one of $y_i$, $y_j$ is extremal (i.e., equal to $1$ or $-1$); \item $y_i = y_j < 0$ (in which case $s<0$ and the local maximum is achieved above); \item $y_i = -y_j$ (in which case $s=0$ above). \end{itemize} In the third case, we may discard $y_i$ and $y_j$ and achieve a case with smaller $n$; we may thus assume that this does not occur. In this case, all of the non-extremal values are equal to some common value $y < 0$, and moreover we cannot have both 1 and -1. We cannot omit 1, as otherwise the condition $\sum_{i=1}^{n} y_i = 0$ cannot be achieved; we must thus have only the terms 1 and $y$, occurring with some positive multiplicities $a$ and $b$ adding up to $n$.  Since $a+b=n$ and $a+by = 0$, we can solve for $y$ to obtain $y = -a/b$; we then have \[ 4\sum_{i=1}^n y_i^3 = a + by^3 = 4a \left( 1 - \frac{a^2}{b^2} \right). \] Since $y > -1$, we must have $a < b$. For fixed $a$, the target function increases as $b$ increases, so the optimal case must occur when $a+b=10$. The possible pairs $(a,b)$ at this point are \[ (1,9), (2,8), (3,7), (4,6); \] computing the target function for these values yields respectively \[ \frac{32}{9}, \frac{15}{2}, \frac{480}{49}, \frac{80}{9}, \] yielding $480/49$ as the maximum value. The maximum value is $\boxed{480/49}$.