\documentclass[12pt]{article}
\usepackage{fullpage,amsmath,amsfonts,graphicx}

\def\v{\boldsymbol{v}}
\def\w{\boldsymbol{w}}
\def\b{\boldsymbol{b}}
\def\a{\boldsymbol{a}}
\def\p{\boldsymbol{p}}
\def\q{\boldsymbol{q}}

\begin{document}

\section*{Solution to Ivanov, Chapter 3, Problem 9}

Let $S\v$ denote the rotation by $90$ degrees anticlockwise of the
vector $\v$.  Further, define
\begin{equation}
  S_i \v = \sum_{j=1}^i S^{j-1} \v 
\end{equation}
where $S_0\v=\boldsymbol{0}$.  Then for any vector $\v$, the set
$\{S_0\v, S_1 \v, S_2 \v, S_3 \v \}$ are the four vertices of a
square.  Vice versa, any square can be represented of a translation of
this set for some vector $\v$; see the figure below.
\begin{center}
  \includegraphics[width=6cm]{square-vectors.eps}
\end{center}
It is easy to check that $S_i(\v+\w)=S_i\v + S_i\w$ for any two
vectors $\v$ and $\w$.  (In fact, $S_i$ is a linear map and can be
though of as a matrix.)

The solution to Problem 9 is now an easy application of vector
algebra.  Let the vertices of the first square have coordinates
\begin{equation}
  \p_i=\a + S_i\v \quad \text{for } i=0,\dots,3 \,,
\end{equation}
and the vertices of the second square have coordinates
\begin{equation}
  \q_i=\b + S_i\w \quad \text{for } i=0,\dots,3 \,.
\end{equation}
Then the midpoint of the line from the $i$th vertex of the first
square to the $i$th vertex of the second square (the above formulas
make no assumption on where we start counting, but once a first vertex
is specified, follow anti-clockwise orientation) has coordinates
\begin{equation}
  \frac{\p_i+\q_i}2 
  = \frac{\a + S_i\v + \b + S_i\w}2
  = \frac{\a+\b}2 + S_i \biggl( \frac{\v+\w}2 \biggr) \,.
\end{equation}
This is again the representation for the vertices of a square. 
\hfill $\Box$



\end{document}
