\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{USEMO 2021/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 90}
\maketitle

\section*{Problem}
Given a polynomial $p(x)$ with real coefficients,
we denote by $S(p)$ the sum of the squares of its coefficients.
For example, $S(20x+21) = 20^2+21^2 = 841$.

Prove that if $f(x)$, $g(x)$, and $h(x)$ are polynomials with real coefficients
satisfying the identity $f(x) \cdot g(x) = h(x)^2$, then
\[ S(f) \cdot S(g) \ge S(h)^2. \]

\section*{Video}
\href{https://youtu.be/PhNIee2CzdY}{\texttt{https://youtu.be/PhNIee2CzdY}}

\section*{External Link}
\url{https://aops.com/community/p23524112}

\newpage

\section*{Solution}
The following write-up is due to Ankan Bhattacharya,
and is the same as the solution proposed by the author.

\begin{claim*}
  Let $p$ be a polynomial with real coefficients, and $n > \deg p$ an integer.
  Then \[ S(p) = \frac1n \sum_{k=0}^{n-1} |p(e^{2\pi ik/n})|^2. \]
\end{claim*}
\begin{proof}
  Note that
  \[ |p(e^{2\pi ik/n})|^2 = p(e^{2\pi ik/n}) \cdot p(e^{-2\pi ik/n}) \]
  so if we define $q(x) = p(x) p(1/x)$,
  the right-hand side is the sum of $q$ across the $n$th roots of unity.

  Applying a roots of unity filter,
  the right-hand side is the constant coefficient of $q(x)$.
  But that constant coefficient is exactly equal to $S(p)$.
\end{proof}

To solve the problem,
choose $n > \max\{\deg f, \deg g, \deg h\}$,
set $\omega = e^{2\pi i/n}$,
and apply the key claim to all three to get that
the desired inequality is equivalent to
\begin{align*}
  \left[\frac1n \sum |f(\omega^k)|^2 \right] \cdot
  \left[\frac1n \sum |g(\omega^k)|^2 \right] & \ge
  \left[\frac1n \sum |h(\omega^k)|^2 \right]^2\\ \iff
  \left[\sum |f(\omega^k)|^2 \right] \cdot
  \left[\sum |g(\omega^k)|^2 \right] & \ge
  \left[\sum |f(\omega^k)| \cdot |g(\omega^k)| \right]^2.
\end{align*}
This is just Cauchy-Schwarz, so we are done.
\begin{remark*}[Continuous version of above solution]
  To avoid the arbitrary choice of parameter $n$,
  one can make the same argument to show that for any $p \in \RR[x]$,
  \[ S(p) = \frac1{2\pi} \int_0^{2\pi} |p(e^{ix})|^2 \, dx. \]
  Using Cauchy's inequality for integrals,
  we obtain a continuous version of the above solution.
  However, this is technically out of scope for high-school olympiad,
  despite the fact it is really just the limit as $n\to\infty$
  of the above solution.
\end{remark*}

\end{document}
