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@@ -339,4 +339,174 @@ which we have seen is $168$.  So we are left with $60\,480 \, / \, 168
 %
 %
 %
+
+\exercise
+
+In this exercise, we consider the affine plane $\mathbb{A}^2$ with
+coordinates $(x,y)$ as a subset of the projective plane $\mathbb{P}^2$
+with coordinates $(T{:}X{:}Y)$ by $(x,y) \mapsto (1{:}X{:}Y)$.  We
+work over the field $\mathbb{R}$ of real numbers, but we will also
+consider some complex points (i.e., $\mathbb{C}$-points).
+
+\textit{Definitions:} A \textbf{translation} of $\mathbb{A}^2$ is a
+map $(x,y) \mapsto (x,y) + (a,b)$ for certain (fixed) $(a,b) \in
+\mathbb{R}^2$.  A \textbf{vector homothety} is a map $(x,y) \mapsto (c
+x, cy)$ for certain (fixed) $c \in \mathbb{R}^\times :=
+\mathbb{R}\setminus\{0\}$.  A \textbf{vector rotation} is a map $(x,y)
+\mapsto (ux + vy,\, -vx + uy)$ for certain (fixed) $u,v \in
+\mathbb{R}^2$ satisfying $u^2+v^2 = 1$.  An \textbf{affine similitude}
+is an element of the group generated by translations, vector
+homotheties and vector rotations (this is a subgroup of the group all
+affine transformations).
+
+\textbf{(1)} Describe the matrices, in $\mathit{PGL}_3(\mathbb{R})$ of
+the extensions to $\mathbb{P}^2$ of the three kinds of transformations
+we just defined (translations, vector homotheties, and vector
+rotations).
+
+\begin{answer}
+The matrices in question are, with the same notations as in the
+definition:
+\[
+\begin{pmatrix}1&0&0\\a&1&0\\b&0&1\\\end{pmatrix}\quad,\quad
+\begin{pmatrix}1&0&0\\0&c&0\\0&0&c\\\end{pmatrix}\quad,\quad
+\begin{pmatrix}1&0&0\\0&u&v\\0&-v&u\\\end{pmatrix}
+\]
+(each one being defined, of course, only up to multiplication by a
+constant).
+\end{answer}
+
+\medskip
+
+We define the complex points $I := (0 : 1 : \sqrt{-1})$ and $J := (0 :
+1 : -\sqrt{-1})$ in $\mathbb{P}^2(\mathbb{C})$, also known as the
+\textbf{cyclic points}.
+
+\textbf{(2)} Show that $I$ and $J$ are fixed under every affine
+similitude (extended to $\mathbb{P}^2$).
+
+\begin{answer}
+We compute the product of the matrices found in (1) with the column
+vectors giving the coordinates of $I$ and $J$.  Translations and
+homotheties fix $I$ and $J$ simply because they are on the line at
+infinity ($T=0$).  For a rotation acting on $I$ we find, with the same
+notations as previously, $(0 : u + v \sqrt{-1} : -v + u \sqrt{-1})$,
+and we observe that these are the coordinates of $I$ multiplied by $u
++ v \sqrt{-1}$, so it is the same point; the same argument works
+for $J$.
+\end{answer}
+
+\textbf{(3)} Conversely, show that every real projective
+transformation of $\mathbb{P}^2$ (i.e., every element of
+$\mathit{PGL}_3(\mathbb{R})$) which fixes $I$ and $J$ is an affine
+similitude.  (\emph{Hint:} You may want to first observe that it
+stabilizes the line $\ell_\infty := IJ$ and conclude that it is an
+affine transformation.)
+
+\begin{answer}
+The line $IJ$ is the line at infinity $\{T=0\}$.  A real projective
+transformation fixing $I$ and $J$ must stabilize the line through them
+(because projective transformations preserve alignment).  This means
+that it is, in fact, an affine transformation, or equivalently, given
+by a matrix of the form:
+\[
+\begin{pmatrix}1&0&0\\a&m&n\\b&p&q\\\end{pmatrix}
+\]
+(with $m,n,p,q\in\mathbb{R}$).  Now fixing $I$ imposes the condition
+that $(0 : m + n \sqrt{-1} : p + q \sqrt{-1})$ is $(0 : 1 :
+\sqrt{-1})$, which means that $p + q \sqrt{-1} = \sqrt{-1}\,(m + n
+\sqrt{-1}) = -n + m \sqrt{-1}$, and by identifying real and imaginary
+parts we conclude $p = -n$ and $q = m$.  So our matrix is now of the
+form:
+\[
+\begin{pmatrix}1&0&0\\a&m&n\\b&-n&m\\\end{pmatrix}
+\]
+and if we let $c = \sqrt{m^2+n^2}$ (a real number) and $u := m/c$ and
+$v = n/c$ (which satisfy $u^2 + v^2 = 1$), the above matrix is the
+product (i.e., the composition)
+\[
+\begin{pmatrix}1&0&0\\a&1&0\\b&0&1\\\end{pmatrix}\,
+\begin{pmatrix}1&0&0\\0&c&0\\0&0&c\\\end{pmatrix}\,
+\begin{pmatrix}1&0&0\\0&u&v\\0&-v&u\\\end{pmatrix}
+\]
+as required.
+\end{answer}
+
+\textbf{(4)} In this question, let $A := (0,0)$ and $B := (1,0)$ in
+$\mathbb{A}^2$.  Compute the equation of the line $(AI\wedge BJ) \vee
+(AJ\wedge BI)$.  (Here, $\ell\wedge m$ denotes the intersection point
+of the lines $\ell,m$, and $P\vee Q$ or just $PQ$ denotes the line
+through $P,Q$.)
+
+\begin{answer}
+Let us write as usual $[\lambda{:}\mu{:}\nu]$ for the line $\{\lambda
+x+\mu y+\nu z = 0\}$.  We find $AI = [0 : 1 : \sqrt{-1}]$ and $BJ = [1
+  : -1 : \sqrt{-1}]$, so $AI\wedge BJ = (2 : 1 : \sqrt{-1})$.  The
+formulas for $AJ$, $BI$ and $AJ\wedge BI$ are obtained by exchanging
+$\sqrt{-1}$ with $-\sqrt{-1}$ so it is not necesary to recompute them.
+Finally, the sought-after line $(AI\wedge BJ) \vee (AJ\wedge BI)$ is
+obtained by joining $(2 : 1 : \sqrt{-1})$ with $(2 : 1 : -\sqrt{-1})$,
+and this gives $[1 : -2 : 0]$.  So it is the line $\{2X=T\}$, or (the
+projective extension of) $\{x = \frac{1}{2}\}$.
+\end{answer}
+
+\textbf{(5)} Show that for any any two distinct points $A,B$ in
+$\mathbb{A}^2(\mathbb{R})$ there is an affine similitude taking
+$(0,0)$ to $A$ and $(1,0)$ to $B$.  (\emph{Hint:} You can use simple
+arguments of standard elementary Euclidean plane geometry for this
+question, independently of all previous questions.  Alternatively, you
+can use the previous questions and a fact from projective geometry
+seen in the course.)
+
+\begin{answer}
+By Euclidean geometry arguments: using a translation we can place any
+point of $\mathbb{A}^2(\mathbb{R})$ in any given place, so we can
+assume without loss of generality that $A = (0,0)$; using a homothety
+we multiply distances by a constant $c\neq 0$, so we can assume
+without loss of generality that the distance $AB$ is $1$, and $B$ is
+now a point on the unit circle; finally, we can rotate around the
+origin to get $B$ in $(1,0)$.  This provides the required affine
+similitude.
+
+By a projective geometry argument: for any two distinct $A,B$ in
+$\mathbb{A}^2(\mathbb{R})$, the points $A,B,I,J$ form a projective
+basis of $\mathbb{P}^2(\mathbb{C})$ (in detail: the points $A,B$ are
+not aligned with $I,J$ because they are not on the line $\ell_\infty$
+at infinity, and the points $I,J$ cannot be on the line $AB$ because
+$AB$ is a \emph{real} line and therefore also $AB\wedge \ell_\infty$
+is a real point).  So if we let $A_0 := (0,0)$ and $B_0 := (1,0)$,
+there is a unique complex projective transformation taking
+$A_0,B_0,I,J$ to $A,B,I,J$; now that complex transformation is real
+because its complex conjugate takes $A_0,B_0,J,I$ to $A,B,J,I$, so it
+is the same.  But we have seen in question (3) that a real projective
+transformation of $\mathbb{P}^2$ which fixes $I$ and $J$ is an affine
+similitude, so we have answered the question.
+
+Alternatively, it is also possible to answer the question by a direct
+computation of the coefficients of the matrix.
+\end{answer}
+
+\textbf{(6)} Conclude that, for any two distinct points $A,B$ in
+$\mathbb{A}^2(\mathbb{R})$, the perpendicular bisector\footnote{In
+French: “la médiatrice”.  The perpendicular bisector of $[AB]$ is the
+line of points at equal distance from $A$ and $B$ in Euclidean
+geometry.  In this context, it is also the perpendicular line to $AB$
+through the midpoint of $[AB]$.} of $[AB]$ can be constructed as the
+line $(AI\wedge BJ) \vee (AJ\wedge BI)$.
+
+\begin{answer}
+We have seen in question (4) that the construction $(AI\wedge BJ) \vee
+(AJ\wedge BI)$ gives the perpendicular bisector $x = \frac{1}{2}$ of
+the two points $A = (0,0)$ and $B = (1,0)$.  Since we have seen in
+question (5) that any two distinct points in
+$\mathbb{A}^2(\mathbb{R})$ can be brought to this position by an
+affine similitude, and since affine similitudes preserve perpendicular
+bisectors (because each one of translations, vector homotheties and
+vector rotations preserve angles and midpoints), the construction
+works for any two distincts points $A,B$.
+\end{answer}
+
+%
+%
+%
 \end{document}