From d130ace9e4f2c025d5f9a54ec3f964f92bb02165 Mon Sep 17 00:00:00 2001 From: "David A. Madore" Date: Sat, 11 Apr 2026 12:57:12 +0200 Subject: An exercise on constructing bisectors using the cyclic points. --- controle-20260415.tex | 170 ++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 170 insertions(+) diff --git a/controle-20260415.tex b/controle-20260415.tex index 99d9223..d1dd792 100644 --- a/controle-20260415.tex +++ b/controle-20260415.tex @@ -336,6 +336,176 @@ which we have seen is $168$. So we are left with $60\,480 \, / \, 168 = 360$ (unlabeled) Fano configurations in $\mathbb{P}^2(\mathbb{F}_4)$. \end{answer} +% +% +% + +\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} + % % % -- cgit v1.2.3