\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{INMO 2023/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 112}
\maketitle
\section*{Problem}
Suppose $a_0$, \dots, $a_{100}$ are positive reals.
Consider the following polynomial for each $k$ in $\{0,1,\ldots, 100\}$:
\[ a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}
+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k, \]
where indices are taken modulo $101$.
Show that it is impossible that each of these $101$ polynomials
has all its roots real.
\section*{Video}
\href{https://www.youtube.com/watch?v=9WdIHzvXF64&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/9WdIHzvXF64}}
\section*{External Link}
\url{https://aops.com/community/p26888629}
\newpage
\section*{Solution}
Assume for contradiction.
Take the reciprocal polynomial
\[ Q_k(x) = a_k X^{100} + a_{k+1} X^{99} + a_{k+2} X^{98} + \dots \]
which has also $101$ real roots for each $k$.
Differentiate it $98$ times:
\begin{align*}
Q_k^{(98)}(x) &= \frac{100!}{2} a_k X^{2} + 99! a_{k+1} X + 98! a_{k+2} \\
&= 98!( 4950 a_k X^{2} + 99 a_{k+1} X + a_{k+2}).
\end{align*}
By Rolle's theorem this should still have $2$ real roots.
Ergo, the discriminant is nonnegative:
\[ (99 a_{k+1})^2 \ge 4 \cdot 4950a_k a_{k+2}
\implies a_{k+1}^2 > a_k a_{k+2}
\]
since $4 \cdot 4950 > 99^2$.
But multiplying all of these together gives a contradiction
\end{document}