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

\begin{document}
\title{Ukraine 2023/9.4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 122}
\maketitle

\section*{Problem}
Find the smallest real number $C$,
such that for any positive integers $x \neq y$ holds the following:
\[ \min\left( \left\{ \sqrt{x^2 + 2y} \right\},
  \left\{ \sqrt{y^2 + 2x} \right\} \right) < C \]
where $\{ t \}$ denotes the fractional part of the real number $t$.

\section*{Video}
\href{https://www.youtube.com/watch?v=y8k9qrrG-Qc&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/y8k9qrrG-Qc}}

\section*{External Link}
\url{https://aops.com/community/p27437949}

\newpage

\section*{Solution}
The answer is $C = \frac{\sqrt5-1}{2}$.

First, we prove this works.
WLOG, $x < y$.
Then observe two cases:
\begin{itemize}
  \ii Whenever $x < Cy$, we have
  \[ y < \sqrt{y^2+2x} < y + C. \]
  \ii Also, whenever $Cy < x < y$ we have
  \[ x+1 < \sqrt{x^2+2y} < x + (C+1). \]
\end{itemize}
Thus, in each case, one of the fractional parts is less than $C$.

Conversely, for the construction, we really just need $x/y \approx C$.
Consider any increasing sequences $x_n$, $y_n$ of positive integers
with $\lim_{n \to \infty} \frac{x_n}{y_n} = C$.
(For example, one could take $y_n = 10^n$
and $x_n = \left\lfloor 10^n C \right\rfloor$,
or take $y_n$ and $x_n$ to be consecutive Lucas/Fibonacci numbers.)
For brevity, abbreviate $x_n$ to $x$ and $y_n$ to $y$.
Then
\begin{align*}
  \sqrt{y^2+2x} - y
  &= \frac{2x}{\sqrt{y^2+2x}+y}
  = \frac{2}{\sqrt{\left( \frac yx \right)^2 + \frac 2x} + \frac yx} \\
  &\xrightarrow{n \to \infty}
  \frac{2}{\sqrt{(1/C)^2}+(1/C)} = C.
\end{align*}
and similarly
\begin{align*}
  \sqrt{x^2+2y} - (x+1)
  &= \frac{2y-2x-1}{\sqrt{x^2+2y}+x+1}
  = \frac{2 - \frac{1}{y-x}}
  {\sqrt{\left( \frac x{y-x} \right)^2 + \frac2{y-x}} + \frac x{y-x} + \frac{1}{y-x}} \\
  &\xrightarrow{n \to \infty}
  \frac{2}{\sqrt{\left( \frac{C}{1-C} \right)^2} + \frac{C}{1-C}}
  = C.
\end{align*}

\end{document}
