\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{ToT Fall 2011 S-A7}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 72}
\maketitle
\section*{Problem}
Suppose $100$ red points divide a blue circle into $100$ arcs
such that their lengths are all positive integers from $1$ to $100$
in an arbitrary order.
Prove that there exist two perpendicular chords with red endpoints.
\section*{Video}
\href{https://www.youtube.com/watch?v=DgsH-Nn2fOo&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/DgsH-Nn2fOo}}
\section*{External Link}
\url{https://aops.com/community/p14404365}
\newpage
\section*{Solution}
We will assume no two red points are diametrically opposite,
otherwise, the problem is trivial.
\begin{claim*}
If there exist two intervals which have sum of lengths $2525$,
and it is not the case that
\begin{itemize}
\ii the intervals share an endpoint; and
\ii one interval contains the other
\end{itemize}
then the problem is solved.
\end{claim*}
\begin{proof}
If two of the endpoints coincide, but the intervals are disjoint,
then we have diametrically opposite red points, a contradiction.
So assume we have four distinct red points in play.
If the intervals are disjoint, then we get two chords which
are perpendicular inside the circle;
if they are nested, we get two chords perpendicular outside the circle.
\end{proof}
For every red point $R$,
we define its two \emph{attached arcs} to be the
two maximal-by-inclusion arcs with endpoint $R$ and total
length at most $2525$.
See figure below with two attached arcs drawn.
\begin{center}
\begin{asy}
dotfactor *= 2;
size(6cm);
draw(unitcircle);
dot("$R$", dir(180), dir(45), red);
label("$-R$", dir(0), dir(0));
dot(dir(0), grey);
draw(arc(origin, dir(20)*1.1, dir(180)*1.1), deepgreen+1.4);
draw(arc(origin, dir(180)*0.9, dir(330)*0.9), deepcyan+1.4);
dot(dir(20), red);
dot(dir(330), red);
\end{asy}
\end{center}
Then, of these two, we pick the one with longest length
and call it the \emph{main arc} of $R$.
A main arc $\Gamma$
always has length $2525-x$ for some $0 < x \le 50$.
We refer to the given arc of length $x$
as the \emph{killer} of $\Gamma$.
Now, the problem is solved unless we are in a situation
where for every main arc $\Gamma$,
its killer is contained in $\Gamma$ and shares an endpoint with it.
\begin{claim*}
If there exists any main arc with $x=50$, we are done.
\end{claim*}
\begin{proof}
It means the two attached arcs both have length exactly $2475$,
and that the segment of length $100$ is split exactly
by the antipode of $R$.
So the segment of length $50$ can't be killers of both.
\end{proof}
\begin{claim*}
If any arc of length $x < 50$ is the killer of two different
main arcs, we are done.
\end{claim*}
\begin{proof}
If it was the killer of two attached arcs,
then we have the following picture.
\begin{center}
\begin{asy}
size(6cm);
draw(unitcircle);
draw(arc(origin, dir(80), dir(100)), red+1.8);
draw(arc(origin, dir(80)*1.1, dir(240)*1.1), deepgreen+1.4);
draw(arc(origin, dir(-60)*0.9, dir(100)*0.9), deepcyan+1.4);
draw(arc(origin, dir(240), dir(300)), grey+1.8);
label("$x$", dir(90)*0.9, dir(-90), red);
label("$2525-x$", dir(160)*1.1, dir(160), deepgreen);
label("$2525-x$", dir(20)*0.9, -dir(20), deepgreen);
label("$3x$", dir(270)*1.1, grey);
dotfactor *= 2;
dot(dir(80), red);
dot(dir(100), red);
dot(dir(240), red);
dot(dir(300), red);
label("$B$", dir(80), dir(45), red);
label("$A$", dir(100), dir(225), red);
label("$C$", dir(240), -dir(240), red);
label("$D$", dir(300), dir(310), red);
\end{asy}
\end{center}
Since $2x < 100$, we can consider the
arc of length $2x$.
Both arcs $AC$ and $BD$ have length $2525-2x$.
The arc of length $2x$ can't share an endpoint with both at once.
\end{proof}
Finally, note that there are at least $50$ main arcs
(each choice of $R$ generates one,
and each main arc appears at most twice this way),
but only $49$ possible killers.
So this solves the problem.
\end{document}