Feat: import snippets and create first tests
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snippets/Geometrie/fig/parcours.pdf
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snippets/Geometrie/fig/parcours.pdf
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121
snippets/Geometrie/tpl_Pythagore_thales.tex
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121
snippets/Geometrie/tpl_Pythagore_thales.tex
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% vim:ft=tex:
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%
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\documentclass[12pt]{article}
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\usepackage[utf8x]{inputenc}
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\usepackage[french]{babel}
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\usepackage[T1]{fontenc}
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\usepackage{amssymb}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{graphicx}
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\title{%
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Snippets pour Opytex \\
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Pythagore et Thalès
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}
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\author{%
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Benjamin Bertrand
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}
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\begin{document}
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\maketitle
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\section{Pythagore}
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\section{Thalès}
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\section{Mélange des 2}
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\subsection{Longueur du parcours}
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% exo de geometrie comme au brevet blanc.
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%- set AD, AC, DC = random_pythagore()
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%- set tourACDA = AC+AD+DC
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%- set AE, AF = round(tourACDA/2*random(), 1), round(tourACDA/2*random(), 1)
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%- set EF = round(tourACDA - AE - AF - randint(20,40)*0.2, 1)
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%- set tourAEFA = round(AE+EF+AF, 1)
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%- set rapport = randint(2,5)
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%- set AE1, AF1, EF1 = round(AE/rapport,2) , round(AF/rapport,2), round(EF/rapport,2)
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%- set objectif = randint(floor(tourAEFA), tourACDA)
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%- if objectif > 100
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%- set unit = "m"
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%- else
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%- set unit = "km"
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%- endif
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Une commune souhaite aménager des parcours de santé sur son territoire. On fait deux propositions au conseil municipale, schématisés ci-dessous:
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\begin{itemize}
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\item Le parcours ACDA
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\item Le parcours AEFA
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\end{itemize}
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Ils souhaitent faire un parcours dont la longueur s'approche le plus possible de \Var{objectif}\Var{unit}.
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Peux-tu les aider à choisir le parcours? Justifie
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\textbf{Attention: La figure proposée au conseil municipale n'est pas à l'échelle, mais les codages et les dimension données sont correctes.}
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\begin{minipage}{0.6\textwidth}
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\includegraphics[scale = 0.4]{./fig/parcours}
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\end{minipage}
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\begin{minipage}{0.4\textwidth}
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\begin{itemize}
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\item $AC = \Var{AC}\Var{unit}$
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\item $CD = \Var{DC}\Var{unit}$
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\item $AE' = \Var{AE1}\Var{unit}$
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\item $AE = \Var{AE}\Var{unit}$
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\item $AF = \Var{AF}\Var{unit}$
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\item $E'F' = \Var{EF1}\Var{unit}$
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\item $(E'F') // (EF)$
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\item L'angle $\widehat{EAF}$ vaut $30^o$
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\end{itemize}
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\end{minipage}
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\begin{solution}
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\begin{itemize}
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\item Parcours ACDA:
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D'après la figure, on voit que le triangle $ACD$ est rectangle en $C$ donc d'après le théorème de Pythagore, on a
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\begin{align*}
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AD^2 &= AC^2 + DC^2 \\
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AD^2 &= \Var{AC}^2 + \Var{DC}^2 \\
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AD^2 &= \Var{AC**2} + \Var{DC**2} \\
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AD^2 &= \Var{AC**2 + DC**2} \\
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AD &= \sqrt{\Var{AC**2 + DC**2}} = \Var{AD}\Var{unit}
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\end{align*}
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Donc le parcours ACDA mesure
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\begin{align*}
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AD + AC + CD = \Var{AD} + \Var{AC} + \Var{DC} = \Var{tourACDA}\Var{unit}
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\end{align*}
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\item Parcours AEFA:
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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
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\begin{tabular}{|c|c|c|c|}
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\hline
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Triangle AEF & AE = \Var{AE} & AF = \Var{AF} & EF \\
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\hline
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Triangle AE'F' & AE' = \Var{AE1} & AF' & E'F' = \Var{EF1} \\
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\hline
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\end{tabular}
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est un tableau de proportionnalité. Donc on peut faire un produit en croix pour calcul $EF$.
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\begin{align*}
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EF = \frac{E'F' \times AE}{AE'} = \frac{\Var{EF1} \times \Var{AE}}{\Var{AE1}} = \Var{EF} \Var{unit}
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\end{align*}
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Donc le parcours AEFA mesure
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\begin{align*}
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AF + AE + EF = \Var{AF} + \Var{AE} + \Var{EF} = \Var{tourAEFA}\Var{unit}
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\end{align*}
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\item Choix du parcours:
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%- if abs(tourACDA - objectif) < abs(tourAEFA - objectif)
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Il faudra choisir le tour $ACDA$ car sa longueur est plus proche de \Var{objectif}\Var{unit}.
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%- else
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Il faudra choisir le tour $AFEA$ car sa longueur est plus proche de \Var{objectif}\Var{unit}.
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%- endif
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\end{itemize}
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\end{solution}
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\end{document}
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96
snippets/tpl_fonctions.tex
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96
snippets/tpl_fonctions.tex
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% vim:ft=tex:
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%
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\documentclass[12pt]{article}
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\usepackage[utf8x]{inputenc}
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\usepackage[francais]{babel}
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\usepackage[T1]{fontenc}
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\usepackage{amssymb}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\title{
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Snippets pour Opytex \\
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Fonctions
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}
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\author{
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Benjamin Bertrand
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}
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\begin{document}
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\maketitle
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\section{Calculer des images}
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\begin{enumerate}
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%-set f = Expression.random("{a}*x^2 + {b}*x + {c}")
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\item $\forall x \in \mathbb{R} \qquad f(x) = \Var{f}$
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Solution:
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\begin{align*}
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f(0) &= \Var{f(0).explain() | join('=')} \\
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f(1) &= \Var{f(1).explain() | join('=')} \\
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f(2) &= \Var{f(2).explain() | join('=')} \\
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f({10}) &= \Var{f(10).explain() | join('=')} \\
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f({100}) &= \Var{f(100).explain() | join('=')}
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\end{align*}
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\end{enumerate}
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\section{Résolution d'équation du 2nd degré}
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%- macro solveEquation(P)
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On commence par calculer le discriminant de $P(x) = \Var{P}$.
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\begin{eqnarray*}
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\Delta & = & b^2-4ac \\
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\Var{P.delta.explain()|calculus(name="\\Delta")}
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\end{eqnarray*}
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\Block{if P.delta > 0}
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comme $\Delta = \Var{P.delta} > 0$ donc $P$ a deux racines
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\begin{eqnarray*}
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x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{\Var{-P.b} - \sqrt{\Var{P.delta}}}{2 \times \Var{P.a}} = \Var{P.roots[0] } \\
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x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{\Var{-P.b} + \sqrt{\Var{P.delta}}}{2 \times \Var{P.a}} = \Var{P.roots[1] }
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\end{eqnarray*}
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Les solutions de l'équation $\Var{P} = 0$ sont donc $\mathcal{S} = \left\{ \Var{P.roots[0]}; \Var{P.roots[1]} \right\}$
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\Block{elif P.delta == 0}
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Comme $\Delta = 0$ donc $P$ a une racine
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\begin{eqnarray*}
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x_1 = \frac{-b}{2a} = \frac{-\Var{P.b}}{2\times \Var{P.a}} = \Var{P.roots[0]} \\
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\end{eqnarray*}
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La solution de $\Var{P} = 0$ est donc $\mathcal{S} = \left\{ \Var{P.roots[0]}\right\}$
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\Block{else}
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Alors $\Delta = \Var{P.delta} < 0$ donc $P$ n'a pas de racine donc l'équation $\Var{P} = 0$ n'a pas de solution.
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\Block{endif}
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%- endmacro
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\begin{enumerate}
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%-set P = Expression.random("{a}*x^2 + {b}*x + {c}", ["b**2-4*a*c>0"])
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\item Étude du polynôme $P$, $\forall x \in \mathbb{R} \quad P(x) = \Var{P}$
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Solution:
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\Var{solveEquation(P)}
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%-set P = Expression.random("{a}*x^2 + {b}*x + {c}", ["b**2-4*a*c==0"])
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\item Étude du polynôme $P$, $\forall x \in \mathbb{R} \quad P(x) = \Var{P}$
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Solution:
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\Var{solveEquation(P)}
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%-set P = Expression.random("{a}*x^2 + {b}*x + {c}", ["b**2-4*a*c<0"])
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\item Étude du polynôme $P$, $\forall x \in \mathbb{R} \quad P(x) = \Var{P}$
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Solution:
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\Var{solveEquation(P)}
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\end{enumerate}
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\end{document}
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snippets/tpl_fraction.tex
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snippets/tpl_fraction.tex
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% vim:ft=tex:
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%
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\documentclass[12pt]{article}
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\usepackage[utf8x]{inputenc}
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\usepackage[francais]{babel}
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\usepackage[T1]{fontenc}
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\usepackage{amssymb}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\title{
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Snippets pour Opytex \\
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Fractions
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}
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\author{
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Benjamin Bertrand
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}
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\begin{document}
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\maketitle
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\section{Simplifications de fractions}
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\begin{itemize}
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\item Trouver le numérateur quand le dénominateur augmente
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\Block{set a,b,ans,c = random_str("{a},{b},{a*c},{b*c}", conditions = ["{a} != {b}"], val_min = 2, val_max = 10).split(',')}%
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\begin{align*}
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\dfrac{\Var{a}}{\Var{b}} = \dfrac{\ldots}{\Var{c}}
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\end{align*}
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Solution
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\begin{align*}
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\dfrac{\Var{a}}{\Var{b}} = \dfrac{\Var{ans}}{\Var{c}}
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\end{align*}
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\item Trouver le numérateur quand le dénominateur diminue
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\Block{set a,b,ans,c = random_str("{a*c},{b*c},{a},{b}", conditions = ["{a} != {b}"], val_min = 2, val_max = 10).split(',')}%
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\begin{align*}
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\dfrac{\Var{a}}{\Var{b}} = \dfrac{\cdots}{\Var{c}}
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\end{align*}
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Solution
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\begin{align*}
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\dfrac{\Var{a}}{\Var{b}} = \dfrac{\Var{ans}}{\Var{c}}
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\end{align*}
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Explications
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\Block{set f = Expression(a + "/" +b)}
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\begin{align*}
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\Var{f.simplify().explain()|join('=')}
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\end{align*}
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\end{itemize}
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\section{Ajouts de fractions}
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\begin{itemize}
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\item Fraction avec le même dénominateur
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\Block{set e = Expression.random("{a} / {b} + {c} / {b}", ["{b} > 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\item Fraction avec un denominateur multiple de l'autre
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\Block{set e = Expression.random("{a} / {b} + {c} / {b*d}", ["{b} > 1","{d} > 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\item Fraction avec des dénominateurs premiers entre eux
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\Block{set e = Expression.random("{a} / {b} + {c} / {d}", ["{b} > 1","{d} > 1", "gcd({b},{d}) == 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\item Une fraction et un entier
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\Block{set e = Expression.random("{a} / {b} + {c}", ["{b} > 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\item Une fraction et un entier
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\Block{set e = Expression.random("{c} + {a} / {b}", ["{b} > 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\end{itemize}
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\section{Multiplications de fractions}
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\begin{itemize}
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\item Une fraction et un entier
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\Block{set e = Expression.random("{c} * {a} / {b}", ["{b} > 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\item Fraction avec des dénominateurs quelconques
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\Block{set e = Expression.random("{a} / {b} * {c} / {d}", ["{b} > 1","{d} > 1"], val_min = 1)}
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\begin{align*}
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A = \Var{e}
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\end{align*}
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Solution
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\begin{align*}
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\Var{e.simplify().explain() | join('=')}
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\end{align*}
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\end{itemize}
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\end{document}
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87
snippets/tpl_suite.tex
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snippets/tpl_suite.tex
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% vim:ft=tex:
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%
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\documentclass[12pt]{article}
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\usepackage[utf8x]{inputenc}
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\usepackage[francais]{babel}
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\usepackage[T1]{fontenc}
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\usepackage{amssymb}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\title{
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Snippets pour Opytex \\
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Suites
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}
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\author{
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Benjamin Bertrand
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}
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\begin{document}
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\maketitle
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\section{Calculs de termes}
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\begin{enumerate}
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\item Calculer les termes $u_0$, $u_1$, $u_2$, $u_{10}$ et $u_{100}$ pour les suites suivantes
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\begin{enumerate}
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%-set u = Expression.random("{a}*n+{b}")
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\item $\forall n \in \mathbb{N} \qquad u_n = \Var{u}$
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Solution:
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\begin{align*}
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u_0 &= \Var{u(0).explain() | join('=')} \\
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u_1 &= \Var{u(1).explain() | join('=')} \\
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u_2 &= \Var{u(2).explain() | join('=')} \\
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u_{10} &= \Var{u(10).explain() | join('=')} \\
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u_{100} &= \Var{u(100).explain() | join('=')}
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\end{align*}
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%-set v = Expression.random("({a}*n+{b})/{c}", ["c>1"])
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\item $\forall n \in \mathbb{N} \qquad v_n = \Var{v|replace("frac","dfrac")}$
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Solution:
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\begin{align*}
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v_0 &= \Var{v(0).explain() | join('=')} \\
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v_1 &= \Var{v(1).explain() | join('=')} \\
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v_2 &= \Var{v(2).explain() | join('=')} \\
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v_{10} &= \Var{v(10).explain() | join('=')} \\
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v_{100} &= \Var{v(100).explain() | join('=')}
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\end{align*}
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%-set v = Expression.random("({a}*n+{b})/{c}", ["c>1"])
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\item $\forall n \in \mathbb{N} \qquad v_n = \Var{v}$
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Solution:
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\begin{align*}
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%- for j in [0, 1, 2, 10, 100]
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v_{\Var{j}} &= \Var{v(j).explain() | join('=')} \\
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%- endfor
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\end{align*}
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%-set f = Expression.random("{a}*x")
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%-set v0 = randint(0, 10)
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\item $\forall n \in \mathbb{N} \qquad v_{n+1} = \Var{f("v_n")} \mbox{ et } v_0 = \Var{v0}$
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Solution:
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\begin{align*}
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v_0 &= \Var{v0} \\
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%-set v = f(v0)
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v_1 &= \Var{v.explain() | join('=')} \\
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%-set v = f(v)
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||||
v_2 &= \Var{v.explain() | join('=')} \\
|
||||
\end{align*}
|
||||
Pour le terme 10, il faut calculer tous les autres avant!
|
||||
\begin{align*}
|
||||
%#- Trick to move around scoping rules
|
||||
%#- https://stackoverflow.com/a/49699589
|
||||
%- set v = namespace(val = v)
|
||||
%- for i in range(8)
|
||||
%- set v.val = f(v.val)
|
||||
v_{\Var{i+3}} &= \Var{v.val.explain() | join('=')} \\
|
||||
%- endfor
|
||||
\end{align*}
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
16
test_no_compile.py
Normal file
16
test_no_compile.py
Normal file
@ -0,0 +1,16 @@
|
||||
from pathlib import Path
|
||||
import os
|
||||
import shutil
|
||||
|
||||
SNIPPETS_DIR = "./snippets"
|
||||
OUTPUT = "./output"
|
||||
|
||||
if __name__ == "__main__":
|
||||
output = Path(OUTPUT)
|
||||
shutil.rmtree(output)
|
||||
output.mkdir()
|
||||
snippets_dir = Path(SNIPPETS_DIR)
|
||||
for snippet in snippets_dir.rglob("tpl_*.tex"):
|
||||
print(snippet)
|
||||
assert os.system(f"bopytex -n {snippet}") == 0
|
||||
|
||||
Loading…
Reference in New Issue
Block a user