\documentclass[a4paper,10pt]{article}
\title{Clock analysis}
\author{Viktor Engelmann}
\begin{document}
\maketitle
\begin{abstract}
In tenth grade I wrote a program to determine the times of the day at wich the short hand and long hand of the clock
have a given angle $\alpha$ to each other. Lacking math skills, I made the program try out all 1440 minutes of a day and compare
the angles at the given times. Today I have the neccessary math knowledge, to solve this Problem properly.
\end{abstract}
\section{Calculating the time to a given angle}
Observe that every minute the long hand of the clock turns by $\frac{360^\circ}{60}=6^\circ$. The short hand turns just by
$\frac{360^\circ}{12\cdot60}=0.5^\circ$ per minute and moves $\frac{360^\circ}{12}=30^\circ$ per hour. So if it is $h$ o'clock
and $m$ minutes, the hands have the angles
\begin{eqnarray*}
s(h,m) & = & 30h + 0.5m \\
l(h,m) & = & 6m
\end{eqnarray*}
The angle between the hands is
\begin{eqnarray*}
\alpha & = & s(h,m)-l(h,m) \\
& = & 30h + 0.5m - 6m \\
\Leftrightarrow \alpha - 30h & = & (0.5-6)m \\
\Leftrightarrow \frac{\alpha-30h}{-5.5} & = & \frac{30h-\alpha}{5.5} = m
\end{eqnarray*}
This means that to determine the times of the day we simply have to replace $h$ by the availiable hours $\{0,1,2,...,11\}$
and calculate the minute $m$ for this particular hour.\\
Example: the hands have angle $\alpha=90^\circ$ at
\begin{eqnarray*}
m_0 =\frac{30\cdot0-90}{5.5} & = & -16.\overline{36} \Rightarrow 11:43:38.\overline{18} \\
m_1 =\frac{30\cdot1-90}{5.5} & = & -10.\overline{90} \Rightarrow 00:49:05.\overline{45} \\
m_2 =\frac{30\cdot2-90}{5.5} & = & -5.\overline{45} \Rightarrow 01:54:32.\overline{72} \\
m_3 =\frac{30\cdot3-90}{5.5} & = & 0 \Rightarrow 03:00:00 \\
& ...
\end{eqnarray*}
Consider that for example 2 o'clock $-5.\overline{45}$ minutes is
\begin{eqnarray*}
& 1 : (60-5.\overline{45}) \\
= & 1 : 54.\overline{54} \\
= & 1 : 54+0.\overline{54} \\
= & 1 : 54 : 60\cdot 0.\overline{54} \\
= & 1:54:32.\overline{72}
\end{eqnarray*}
\section{Calculating the repeating time of an angle}
Now I was intrigued and wanted to know after what timespan the hands would have the same angle again. At hour $(h+1)$ the
associated minute would have changed by $\Delta$ minutes:
\begin{eqnarray*}
\frac{30(h+1)-\alpha}{5.5} &=& \Delta+\frac{30h-\alpha}{5.5} \\
\Leftrightarrow \frac{30(h+1)-\alpha}{5.5}-\frac{30h-\alpha}{5.5}&=&\Delta \\
\Leftrightarrow \frac{30h+30 - 30h -\alpha +\alpha}{5.5}&=&\Delta \\
\Leftrightarrow \frac{30}{5.5} & = & 5.\overline{45}=\Delta
\end{eqnarray*}
$5.\overline{45}$ minutes equal $5$ minutes and $27.\overline{27}$ seconds, so every $1:05:27.\overline{27}$ hours the
hands have the same angle again - independent of the angle!\\[1ex]
Note: $1:05:27.\overline{27}$ hours are one eleventh of 12 hours.
\section{Angle-alteration by $1^\circ$}
Looking for more applications of the above results I calculated in which timespan $c$ the angle between the hands changes
by exactly $1^\circ$.\\
Let $m=\frac{30h-\alpha}{5.5}$.
\begin{eqnarray*}
\frac{30h-\alpha+1}{5.5} & = & m+c \\
\Leftrightarrow \frac{30h-\alpha+1}{5.5} & = & \frac{30h-\alpha}{5.5}+c \\
\Leftrightarrow 30h-\alpha+1 & = & 30h-\alpha+5.5c \\
\Leftrightarrow 1 & = & 5.5c \\
\Leftrightarrow c & = & \frac{1}{5.5} min = 10.\overline{90} \mbox{sec.}
\end{eqnarray*}
\section{Smiling times}
Years after finding these results I learned from a TV quiz-show, that in advertisements for clocks, the clocks always show
the time $1:50$, because the clock "smiles", which increases the sales figures.
Obviously at $1:50$ the angles of the hands aren't symmetrical, because the long hand has $(360-60)^\circ$, but the short hand only
has $(60-10*0.5)^\circ=55^\circ$, which caused me to think about the question, at which times the angles of the short hand and the
long hand are symmetrical (so $s(h,m)=360-l(h,m)$).
This is the case, iff
\begin{eqnarray*}
6m & = & 360 - (30h+0.5m) \\
\Leftrightarrow 12m & = & 720-60h-m \\
\Leftrightarrow 13m & = & 720-60h \\
\Leftrightarrow m & = & \frac{720-60h}{13}
\end{eqnarray*}
using this result, we can now explicitly calculate the times, at which the hands are symmetrical:
\begin{center}
\begin{tabular}{c|c|c}
h & m & default representation \\ \hline
0 & 55.38 & 0:55:23.0769 \\
1 & 50.76 & 1:50:46.1538 \\
2 & 46.15 & 2:46:09.2307 \\
3 & 41.54 & 3:41:32.3076 \\
4 & 36.92 & 4:36:55.3846 \\
5 & 32.3 & 5:32:18.4615 \\
6 & 27.69 & 6:27:41.5384 \\
7 & 23.08 & 7:23:04.6153 \\
8 & 18.46 & 8:18:27.6923 \\
9 & 13.85 & 9:13:50.7692 \\
10 & 9.23 & 10:09:13.8461 \\
11 & 4.62 & 11:04:36.9230 \\
12 & 0 & 12:00:00
\end{tabular}
\end{center}
Note: the difference between one of these times, and the next, is always $55:23.0769$ minutes, which is a thirteenth of 12 hours.
\end{document}