From 011ff6e7e5f080d1c815618e85bc63f0b63ac66f Mon Sep 17 00:00:00 2001
From: Bertrand Benjamin <benjamin.bertrand@opytex.org>
Date: Mon, 23 Dec 2019 18:13:49 +0100
Subject: [PATCH] Feat: add old geometry template

---
 snippets/Geometrie/fig/parcours.pdf         | Bin 0 -> 6652 bytes
 snippets/Geometrie/tpl_Pythagore_thales.tex | 121 ++++++++++++++++++++
 2 files changed, 121 insertions(+)
 create mode 100644 snippets/Geometrie/fig/parcours.pdf
 create mode 100644 snippets/Geometrie/tpl_Pythagore_thales.tex

diff --git a/snippets/Geometrie/fig/parcours.pdf b/snippets/Geometrie/fig/parcours.pdf
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diff --git a/snippets/Geometrie/tpl_Pythagore_thales.tex b/snippets/Geometrie/tpl_Pythagore_thales.tex
new file mode 100644
index 0000000..d48ffe1
--- /dev/null
+++ b/snippets/Geometrie/tpl_Pythagore_thales.tex
@@ -0,0 +1,121 @@
+% vim:ft=tex:
+%
+\documentclass[12pt]{article}
+\usepackage[utf8x]{inputenc}
+\usepackage[french]{babel}
+\usepackage[T1]{fontenc}
+\usepackage{amssymb}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{graphicx}
+
+
+\title{%
+    Snippets pour Opytex \\
+    Pythagore et Thalès
+}
+\author{%
+    Benjamin Bertrand
+}
+
+\begin{document}
+\maketitle
+\section{Pythagore}
+
+
+\section{Thalès}
+
+\section{Mélange des 2}
+\subsection{Longueur du parcours}
+% exo de geometrie comme au brevet blanc.
+
+%- set AD, AC, DC = random_pythagore()
+%- set tourACDA = AC+AD+DC
+%- set AE, AF = round(tourACDA/2*random(), 1), round(tourACDA/2*random(), 1)
+%- set EF = round(tourACDA - AE - AF - randint(20,40)*0.2, 1)
+%- set tourAEFA = round(AE+EF+AF, 1)
+%- set rapport = randint(2,5)
+%- set AE1, AF1, EF1 = round(AE/rapport,2) , round(AF/rapport,2), round(EF/rapport,2)
+%- set objectif = randint(floor(tourAEFA), tourACDA)
+%- if objectif > 100
+    %- set unit = "m"
+%- else
+    %- set unit = "km"
+%- endif
+
+
+Une commune souhaite aménager des parcours de santé sur son territoire. On fait deux propositions au conseil municipale, schématisés ci-dessous:
+\begin{itemize}
+    \item Le parcours ACDA
+    \item Le parcours AEFA
+\end{itemize}
+Ils souhaitent faire un parcours dont la longueur s'approche le plus possible de \Var{objectif}\Var{unit}.
+
+Peux-tu les aider à choisir le parcours? Justifie
+
+\textbf{Attention: La figure proposée au conseil municipale n'est pas à l'échelle, mais les codages et les dimension données sont correctes.}
+
+\begin{minipage}{0.6\textwidth}
+    \includegraphics[scale = 0.4]{./fig/parcours} 
+\end{minipage}
+\begin{minipage}{0.4\textwidth}
+    \begin{itemize}
+        \item $AC = \Var{AC}\Var{unit}$
+        \item $CD = \Var{DC}\Var{unit}$
+        \item $AE' = \Var{AE1}\Var{unit}$
+        \item $AE = \Var{AE}\Var{unit}$
+        \item $AF = \Var{AF}\Var{unit}$
+        \item $E'F' = \Var{EF1}\Var{unit}$
+        \item $(E'F') // (EF)$
+        \item L'angle $\widehat{EAF}$ vaut $30^o$
+    \end{itemize}
+\end{minipage}
+\begin{solution}
+    \begin{itemize}
+        \item Parcours ACDA:
+
+            D'après la figure, on voit que le triangle $ACD$ est rectangle en $C$ donc d'après le théorème de Pythagore, on a
+            \begin{align*}
+                AD^2 &= AC^2 + DC^2 \\
+                AD^2 &= \Var{AC}^2 + \Var{DC}^2 \\
+                AD^2 &= \Var{AC**2} + \Var{DC**2} \\
+                AD^2 &= \Var{AC**2 + DC**2} \\
+                AD &= \sqrt{\Var{AC**2 + DC**2}} = \Var{AD}\Var{unit}
+            \end{align*}
+            Donc le parcours ACDA mesure
+            \begin{align*}
+                AD + AC + CD = \Var{AD} + \Var{AC} + \Var{DC} = \Var{tourACDA}\Var{unit}
+            \end{align*}
+
+        \item Parcours AEFA:
+
+            D'après les données, on sait que $(EF) // (E'F')$. On voit aussi que $A$, $E'$ et $E$ sont alignés. Il en est de même pour les points $A$, $F'$ et $F$. Donc d'après le théorème de Thalès
+
+            \begin{tabular}{|c|c|c|c|}
+                \hline
+                Triangle AEF & AE = \Var{AE} & AF = \Var{AF}  & EF \\
+                \hline
+                Triangle AE'F' & AE' = \Var{AE1} & AF' & E'F' = \Var{EF1} \\
+                \hline
+            \end{tabular}
+            est un tableau de proportionnalité. Donc on peut faire un produit en croix pour calcul $EF$.
+            \begin{align*}
+                EF = \frac{E'F' \times AE}{AE'} = \frac{\Var{EF1} \times \Var{AE}}{\Var{AE1}} = \Var{EF}  \Var{unit}
+            \end{align*}
+
+            Donc le parcours AEFA mesure
+            \begin{align*}
+                AF + AE + EF = \Var{AF} + \Var{AE} + \Var{EF} = \Var{tourAEFA}\Var{unit}
+            \end{align*}
+
+        \item Choix du parcours:
+
+        %- if abs(tourACDA - objectif) < abs(tourAEFA - objectif)
+            Il faudra choisir le tour $ACDA$ car sa longueur est plus proche de \Var{objectif}\Var{unit}.
+        %- else
+            Il faudra choisir le tour $AFEA$ car sa longueur est plus proche de \Var{objectif}\Var{unit}.
+        %- endif
+    \end{itemize}
+\end{solution}
+
+\end{document}