\documentclass[11pt]{scrartcl}
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\begin{document}
\title{Clock}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 110}
\maketitle

\section*{Problem}
There is a analog clock in front of you.
How many positions of the hands are there,
by which it is impossible to determine the time,
if you do not know which hand is hour and which is minute?

\section*{Video}
\href{https://www.youtube.com/watch?v=16UNuZTrNTc&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/16UNuZTrNTc}}



\newpage

\section*{Solution}
We regard each time as an ordered pair $(H,M)$
where $0 \le H < 12$ is an integer, $0 \le M < 60$ is real.
The position of the hands measured in degrees from 12 clockwise is
\[ (H,M) \mapsto \left\{ 30\left( H+\frac{M}{60} \right), 6M \right\}
  = \left\{ 30H + \half M, 6M \right\}. \]
So we are looking for pairs $(H,M)$ and $(H',M')$ with
\begin{align*}
  60H + M &= 12M' \\
  60H' + M' &= 12M
\end{align*}
Solving,
\begin{align*}
  M &= \frac{60H + 60 \cdot 12 H'}{143} \\
  M' &= \frac{60H' + 60 \cdot 12 H}{143} \\
\end{align*}
For each $(H, H') \in \{0,1,\dots,11\}^2$ with $H \neq H'$
this gives two numbers both less than $60$.
Hence the number of positions of hands is just $\binom{12}{2} = 66$.
(Or $12 \cdot 11 = 132$ if you consider the hands
to be distinguishable when counting,
even when indistinguishable to the viewer.)

\end{document}
