, 2022

Solution:
Without loss of generality let the vertices of the pentagon lie on the unit circle. Then, if $f(\theta)=\cos (\theta)^{k}$ and $g(\theta)=\sum_{j=0}^{4} f(\theta+2 j \pi / 5)$, the condition becomes $g(\theta)=g(\pi / 2-\theta)$, or $g(\theta)=g(\theta+\pi / 2)$, since $g$ is an odd function.

Write $f \asymp g$ if $f=c g$ for some nonzero constant $c$ that we don't care about. Since $\cos \theta \asymp e^{i \theta}+e^{-i \theta}$, we find that
$$f(\theta) \asymp \sum_{\ell \in \mathbb{Z}}\binom{k}{\frac{k+\ell}{2}} e^{i \ell \theta}$$
where $\binom{a}{b}$ is defined to be zero if $b$ is not an integer in the interval $[0, a]$. It is also true that
$$\sum_{j=0}^{4} e^{i \ell(\theta+2 j \pi / 5)}= \begin{cases}5 e^{i \theta} & 5 \mid \ell \\ 0 & \text{ else }\end{cases}$$
so
$$g(\theta) \asymp \sum_{\ell \in 5 \mathbb{Z}}\binom{k}{\frac{k+\ell}{2}} e^{i \ell \theta}$$
This is periodic with period $\pi / 2$ if and only if all terms with $\ell$ not a multiple of 4 are equal to 0. However, we know that the nonzero terms are exactly the $\ell$ that (1) are multiples of $5$, (2) are of the same parity as $k$, and (3) satisfy $|\ell| \leq k$. Hence, if $k$ is even, the condition is satisfied if and only if $k<10$ (else the $\ell=10$ term is nonzero), and if $k$ is odd, the condition is satisfied if and only if $k<5$ (else the $\ell=5$ term is nonzero). Our final answer is $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 8=1152$.