{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "![En tête general](img/En_tete_general.png)\n", "\n", "\n", "*(C) Copyright Franck CHEVRIER 2019-2021 http://www.python-lycee.com/*\n", "\n", " Pour exécuter une saisie Python, sélectionner la cellule et valider avec SHIFT+Entrée." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Interprétations géométriques de nombres complexes (corrigé)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Sommaire\n", "\n", "I. Interprétations géométriques de $z_B-z_A$ et $\\displaystyle \\frac{z_C-z_A}{z_B-z_A}$.
\n", "II. Applications directes

\n", "III. Études de configurations\n", "
    \n", "
  1. Deux triangles rectangles isocèles rectangles en un même point
    2. \n", "
  2. Trois triangles équilatéraux ayant un sommet commun
    3. \n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## I. Interprétations géométriques de $z_B-z_A$ et $\\displaystyle \\frac{z_C-z_A}{z_B-z_A}$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Dans le plan complexe muni du repère orthonormé $\\left( O \\;; \\overrightarrow{u} ; \\overrightarrow{v} \\right)$, on considère trois points distincts $A$ ; $B$ et $C$ d'affixes respectives $z_A$ ; $z_B$ et $z_C$.\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "1. En utilisant le point $M$ défini par $\\overrightarrow{OM}=\\overrightarrow{AB}$, démontrer le résultat suivant :\n", "\n", "
\n", " À RETENIR
\n", " Interprétation géométrique de $z_B-z_A$ :\n", "\n", "
\n", "
\n", "
\n", "\n", "$\\overrightarrow{AB}(z_B-z_A)=\\overrightarrow{OM}(z_M)$ donc :\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\n", "\t\n", "\t\t
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\n", "\t\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Sélectionner cette zone puis SHIFT+ENTREE pour obtenir la figure dynamique\n", "from IPython.display import display, HTML ; display(HTML('fig_dyn_GeoGebra/intgeomcomplexeprop1.html'))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "2. À l'aide de la propriété précédente, démontrer que :\n", "\n", "
\n", " À RETENIR
\n", " Interprétation géométrique de $\\displaystyle \\frac{z_C-z_A}{z_B-z_A}$ :\n", "\n", "
\n", "
\n", "
\n", "\n", " \n", "" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\n", "\t\n", "\t\t
\n", "\t\t\t\t\t\t\n", "\t\t\t
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\n", "\t\t\t\t\t\t\t\t\n", "\t\t\t\t\t\t\t\t\n", "\t\t\t\t
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\n", "\t\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Sélectionner cette zone puis SHIFT+ENTREE pour obtenir la figure dynamique\n", "from IPython.display import display, HTML ; display(HTML('fig_dyn_GeoGebra/intgeomcomplexeprop2.html'))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## II. Applications directes" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Exercice 1 :
\n", "On considère les points $A$ ; $B$ et $C$ d’affixes respectives $z_A=1+i$ ; $z_B=2+3i$ et $z_C=-1+2i$.\n", "
    \n", "
  1. Réaliser une figure. Quelle semble être la nature du triangle $ABC$ ?
    2. \n", "
  2. Calculer $\\displaystyle m= \\frac{z_C-z_A}{z_B-z_A}$.
    3. \n", "
  3. En interprétant géométriquement le module et un argument de $m$, en déduire la nature du triangle $ABC$.
    4. \n", "
  4. On souhaite vérifier le résultat du calcul précédent à l'aide de saisies en Python :
    \n", "
    \n", " Le module sympy permet :\n", "
      \n", "
    • d'effectuer des calculs sous forme exacte avec racines carrées et $\\pi$ avec les syntaxes sqrt et pi ;
      • \n", "
    • de créer un nombre complexe avec la syntaxe majuscule I pour le complexe $i$ ;
      • \n", "
    • de simplifier une expression avec la fonction simplify ;
      • \n", "
    • de calculer le module et l'argument d'un complexe respectivement à l'aide des fonctions abs et arg.
      • \n", "
    \n", "
    \n", " Exécuter les cellules suivantes pour vérifier les calculs précédents.\n", "
    5. \n", "
\n", "
\n", "
\n", "\n", "
    \n", "
  1. Le triangle $ABC$ semble isocèle rectangle en $A$.
    2. \n", "
  2. $\\displaystyle m=\\frac{z_C-z_A}{z_B-z_A}=\\frac{-1+2i-(1+i)}{(2+3i)-(1+i)}=\\frac{-2+i}{1+2i}=\\frac{(-2+i) \\overline{(1+2i)}}{(1+2i) \\overline{(1+2i)}}=\\frac{(-2+i)(1-2i)}{1^2+2^2}=\\frac{5i}{5}=i=e^{i\\frac{\\pi}{2}}$
    3. \n", "
  3. $\\displaystyle \\frac{ AC }{ AB }= \\left| \\frac{z_C-z_A}{z_B-z_A} \\right| = \\left| e^{i\\frac{\\pi}{2}} \\right| = 1$ donc $AB=AC$.
    \n", " $\\displaystyle \\left( \\overrightarrow{AB} \\;; \\overrightarrow{AC} \\right) = arg \\left( \\frac{z_C-z_A}{z_B-z_A} \\right) \\;[2\\pi]= arg \\left( e^{i\\frac{\\pi}{2}} \\right) \\;[2\\pi]= \\frac{\\pi}{2} \\;[2\\pi]$
    \n", " Le triangle $ABC$ est donc isocèle rectangle en A.
    4. \n", "
\n", "" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{\\left(-2 + i\\right) \\left(1 - 2 i\\right)}{5}$" ], "text/plain": [ "(-2 + I)*(1 - 2*I)/5" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Import du module sympy\n", "from sympy import *\n", "\n", "# Mise en mémoire des valeurs de z_A; z_B et z_C\n", "z_A = 1+I ; z_B = 2+3*I ; z_C = -1+2*I\n", "\n", "# Calcul de m\n", "m = (z_C-z_A)/(z_B-z_A)\n", "\n", "m" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle i$" ], "text/plain": [ "I" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Évaluation de la valeur de m\n", "simplify(m)" ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 1$" ], "text/plain": [ "1" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Détermination du module de m\n", "abs(m)" ] }, { "cell_type": "code", "execution_count": 6, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{\\pi}{2}$" ], "text/plain": [ "pi/2" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Détermination de l'argument de m\n", "arg(m)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Exercice 2 :
\n", "
    \n", "
  1. De la même manière que dans l’exercice 1, déterminer par le calcul la nature des triangles suivants :\n", "
      \n", "
    • $DEF$ tel que $z_D=-2+i$ ; $z_E=-1+4i$ et $z_F=4-i$.
      • \n", "
    • $IJK$ tel que $\\displaystyle z_I=\\sqrt{3}+i$ ; $ z_J=4i$ et $\\displaystyle z_K=-\\sqrt{3}+i$.
      • \n", "
    \n", "
  2. Vérifier les calculs précédents à l'aide de saisies Python.
    3. \n", "
\n", "
\n", "
\n", "\n", "
    \n", "
  1. On détermine la nature des triangles par le calcul :\n", "
      \n", "
    • $\\displaystyle \\frac{z_F-z_D}{z_E-z_D}=\\frac{4-i-(-2+i)}{(-1+4i)-(-2+i)}=\\frac{6-2i}{1+3i}=\\frac{(6-2i) \\overline{(1+3i)}}{(1+3i) \\overline{(1+3i)}}=\\frac{(6-2i)(1-3i)}{1^2+3^2}=\\frac{-20i}{10}=-2i=2e^{-i\\frac{\\pi}{2}}$
      \n", " $\\displaystyle \\frac{ DF }{ DE }=\\displaystyle \\left| \\frac{z_F-z_D}{z_E-z_D} \\right| = \\left| 2e^{-i\\frac{\\pi}{2}} \\right| = 2$ donc $DF=2DE$.
      \n", " $\\displaystyle \\left( \\overrightarrow{DE} \\;; \\overrightarrow{DF} \\right)=arg \\left( \\frac{z_F-z_D}{z_E-z_D} \\right) \\;[2\\pi]= arg \\left( 2e^{-i\\frac{\\pi}{2}} \\right) \\;[2\\pi]= -\\frac{\\pi}{2} \\;[2\\pi]$
      \n", " Le triangle $DEF$ est donc rectangle en $D$ (et $DF=2DE$).
      • \n", "
    • $\\displaystyle \\frac{z_J-z_K}{z_I-z_K}=\\frac{4i-(-\\sqrt{3}+i)}{\\sqrt{3}+i-(-\\sqrt{3}+i)}=\\frac{\\sqrt{3}+3i}{2\\sqrt{3}}=\\frac{1}{2}+i\\frac{\\sqrt{3}}{2}=e^{i\\frac{\\pi}{3}}$
      \n", " $\\displaystyle \\frac{ KJ }{ KI }=\\displaystyle \\left| \\frac{z_J-z_K}{z_I-z_K} \\right| = \\left| e^{i\\frac{\\pi}{3}} \\right| = 1$ donc $KJ=KI$.
      \n", " $\\displaystyle \\left( \\overrightarrow{KI} \\;; \\overrightarrow{KJ} \\right)=arg \\left( \\frac{z_J-z_K}{z_I-z_K} \\right) \\;[2\\pi]= arg \\left( e^{i\\frac{\\pi}{3}} \\right) \\;[2\\pi]= -\\frac{\\pi}{3} \\;[2\\pi]$
      \n", " Le triangle $IJK$ est donc équilatéral.
      • \n", "
    \n", "
\n", "" ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{\\left(1 - 3 i\\right) \\left(6 - 2 i\\right)}{10}$" ], "text/plain": [ "(1 - 3*I)*(6 - 2*I)/10" ] }, "execution_count": 7, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Utiliser ces zones de saisie pour vérifier les calculs relatifs au triangle DEF\n", "z_D = -2+I ; z_E = -1+4*I ; z_F = 4-I\n", "\n", "n = (z_F-z_D)/(z_E-z_D)\n", "n" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - 2 i$" ], "text/plain": [ "-2*I" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" } ], "source": [ "\n", "simplify(n)" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 2$" ], "text/plain": [ "2" ] }, "execution_count": 9, "metadata": {}, "output_type": "execute_result" } ], "source": [ "abs(n)\n" ] }, { "cell_type": "code", "execution_count": 10, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - \\frac{\\pi}{2}$" ], "text/plain": [ "-pi/2" ] }, "execution_count": 10, "metadata": {}, "output_type": "execute_result" } ], "source": [ "arg(n)" ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - \\frac{\\sqrt{3} \\left(- \\sqrt{3} + 3 i\\right)}{6}$" ], "text/plain": [ "-sqrt(3)*(-sqrt(3) + 3*I)/6" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Utiliser ces zones de saisie pour vérifier les calculs relatifs au triangle IJK\n", "z_I = sqrt(3)+I ; z_J = 4*I ; z_K = -sqrt(3)+I\n", "\n", "p = (z_J-z_I)/(z_K-z_I)\n", "p" ] }, { "cell_type": "code", "execution_count": 12, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{1}{2} - \\frac{\\sqrt{3} i}{2}$" ], "text/plain": [ "1/2 - sqrt(3)*I/2" ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" } ], "source": [ "simplify(p)" ] }, { "cell_type": "code", "execution_count": 13, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 1$" ], "text/plain": [ "1" ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "abs(p)" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - \\frac{\\pi}{3}$" ], "text/plain": [ "-pi/3" ] }, "execution_count": 14, "metadata": {}, "output_type": "execute_result" } ], "source": [ "arg(p)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## III. Études de configurations" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Dans cette partie, on dira qu'un triangle $ABC$ est :\n", "\n", "(on différenciera donc les triangles $ABC$ et $ACB$, qui sont l'un direct et l'autre indirect)\n" ] }, { "cell_type": "code", "execution_count": 15, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\n", "\t\n", "\t\t
\n", "\t\t\t\t\t\t\n", "\t\t\t
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\n", "\t\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Sélectionner cette zone puis SHIFT+ENTREE pour obtenir la figure dynamique\n", "from IPython.display import display, HTML ; display(HTML('fig_dyn_GeoGebra/intgeomcomplexedirindir.html'))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "On admettra également la généralisation des résultats vus précédemment, pour 4 points $A$ ; $B$ ; $C$ et $D$ d'affixes respectives $z_A$ ; $z_B$ ; $z_C$ et $z_D$.\n", "\n", "
\n", " Interprétation géométrique de $\\displaystyle \\frac{z_D-z_C}{z_B-z_A}$ :\n", "
    \n", "
  • $ \\displaystyle \\left| \\frac{z_D-z_C}{z_B-z_A} \\right| = \\frac{CD}{AB}$
    • \n", "
  • $ \\displaystyle arg \\left( \\frac{z_D-z_C}{z_B-z_A} \\right) = \\left( \\overrightarrow{AB} \\;;\\overrightarrow{CD} \\right) \\;[2\\pi]$
    • \n", "
\n", "
\n", "
\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### III.1. Deux triangles rectangles isocèles rectangles en un même point " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "Dans le plan complexe $\\left( O \\;; \\overrightarrow{u} ; \\overrightarrow{v} \\right)$, on considère deux triangles $OAB$ et $OCD$ qui sont rectangles directs.
\n", "On note $M$ le milieu de $[BC]$.

\n", "1. Dans cette question, les points $A$ et $C$ ont pour affixe respectives $z_A=4-i$ et $z_C=-3+5i$.
\n", "a. Activer la cellule ci-dessous pour obtenir la figure.
\n", "$\\;\\;\\;$Quelle conjecture peut-on émettre concernant les droites $(OM)$ et $(AD)$ ?\n", "" ] }, { "cell_type": "code", "execution_count": 16, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\n", "\t\n", "\t\t
\n", "\t\t\t\t\t\t\n", "\t\t\t
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\n", "\t\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Sélectionner cette zone puis SHIFT+ENTREE pour obtenir la figure dynamique\n", "from IPython.display import display, HTML ; display(HTML('fig_dyn_GeoGebra/intgeomcomplexeconfig1.html'))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "$\\quad\\;\\;$b. Déterminer les affixes respectives $z_B$ ; $z_D$ et $z_M$ des points $B$ ; $D$ et $M$.
\n", "$\\quad\\;\\;$c. À l'aide de saisies Python, déterminer la valeur de $\\displaystyle \\frac{z_M}{z_D-z_A}$.
\n", "$\\quad\\quad\\;$Vérifier ensuite par le calcul le résultat obtenu à la question c.
\n", "" ] }, { "cell_type": "code", "execution_count": 17, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{- \\frac{3}{2} + \\frac{i \\left(4 - i\\right)}{2} + \\frac{5 i}{2}}{-4 + i \\left(-3 + 5 i\\right) + i}$" ], "text/plain": [ "(-3/2 + I*(4 - I)/2 + 5*I/2)/(-4 + I*(-3 + 5*I) + I)" ] }, "execution_count": 17, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Utiliser ces zones de saisie pour les calculs\n", "z_A = 4-I ; z_C = -3+5*I ; z_B = z_A*I ; z_D = z_C*I\n", "z_M = (z_B+z_C)/2\n", "\n", "k = (z_M)/(z_D-z_A)\n", "k" ] }, { "cell_type": "code", "execution_count": 18, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - \\frac{i}{2}$" ], "text/plain": [ "-I/2" ] }, "execution_count": 18, "metadata": {}, "output_type": "execute_result" } ], "source": [ "simplify(k)" ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{1}{2}$" ], "text/plain": [ "1/2" ] }, "execution_count": 19, "metadata": {}, "output_type": "execute_result" } ], "source": [ "abs(k)" ] }, { "cell_type": "code", "execution_count": 20, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - \\frac{\\pi}{2}$" ], "text/plain": [ "-pi/2" ] }, "execution_count": 20, "metadata": {}, "output_type": "execute_result" } ], "source": [ "arg(k)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "$\\quad\\;\\;$d. Que peut-on conclure concernant les droites $(OM)$ et $(AD)$ ? et concernant les longueurs des segments $[OM]$ et $[AD]$ ?
\n", " " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "2. Dans cette question, on revient au cas général où les affixes $z_A$ et $z_C$ des points $A$ et $C$ sont quelconques.
\n", "$\\;\\;\\;$a. Déplacer les points $A$ et $C$ sur la figure fournie. Quelle conjecture peut-on émettre ?
\n", "$\\;\\;\\;$b. Démontrer cette conjecture.\n", "\n", "

" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "
    \n", "
  1. Cas où $z_A=4-i$ et $z_C=-3+5i$.
    \n", " $\\;$a. Il semble que les droites $(OM)$ et $(AD)$ sont perpendiculaires.
    \n", " $\\;$b. $OAB$ est isocèle rectangle direct donc $\\displaystyle \\frac{z_B}{z_A}=e^{i\\frac{\\pi}{2}}=i$ donc $z_B=iz_A=1+4i$.
    \n", " $\\quad$De même en considérant $OCD$ on a $z_D=iz_C=-5-3i$
    \n", " $\\quad$ $\\displaystyle z_M=\\frac{z_B+z_C}{2}=\\frac{1+4i-3+5i}{2}=\\frac{-2+9i}{2}=-1+\\frac{9}{2}i$
    \n", " $\\;$c. $\\displaystyle \\frac{z_M}{z_D-z_A}=\\frac{-1+\\frac{9}{2}i}{-5-3i-(4-i)}=\\frac{1}{2}\\times\\frac{-2+9i}{-9-2i}=\\frac{1}{2}\\times \\frac{-i(-9-2i)}{-9-2i}=-\\frac{1}{2}i=\\frac{1}{2}e^{-i\\frac{\\pi}{2}}$
    \n", " $\\;$d. $\\displaystyle \\frac{ OM }{ AD }=\\displaystyle \\left| \\frac{z_M}{z_D-z_A} \\right| = \\left| \\frac{1}{2}e^{-i\\frac{\\pi}{2}} \\right| = \\frac{1}{2}$ donc $AD=2OM$.
    \n", " $\\quad$ $\\displaystyle \\left( \\overrightarrow{AD} \\;; \\overrightarrow{OM} \\right)=arg \\left( \\frac{z_M}{z_D-z_A} \\right) \\;[2\\pi]= arg \\left( \\frac{1}{2}e^{-i\\frac{\\pi}{2}} \\right) \\;[2\\pi]= -\\frac{\\pi}{2} \\;[2\\pi]$ donc $(AD)$ et $(OM)$ sont perpendiculaires. \n", "
    2. \n", "
  2. Cas général.
    \n", " $\\;$a. Il semble que quelles que soient les positions de $A$ et $C$, le résultat vu en 1.d reste valable.
    \n", " $\\;$b. Comme dans la partie 1, on peut établir que :
    \n", " $\\quad$ $z_B=iz_A$ ; $z_D=iz_C$ ; $\\displaystyle z_M=\\frac{z_B+z_C}{2}$
    \n", " $\\quad$ On en déduit :
    \n", " $\\quad$ $\\displaystyle \\frac{z_M}{z_D-z_A}=\\frac{1}{2}\\times \\frac{z_B+z_C}{z_D-z_A} = \\frac{1}{2}\\times \\frac{iz_A+z_C}{iz_C-z_A} = \\frac{1}{2}\\times \\frac{-i(-z_A+iz_C)}{iz_C-z_A} = -\\frac{1}{2}i = \\frac{1}{2}e^{-i\\frac{\\pi}{2}}$
    \n", " $\\quad$ce qui permet de conclure de la même façon que dans la question 1.d.\n", "
    3. \n", "
\n", " \n", " " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### III.2. Trois triangles équilatéraux ayant un sommet commun " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "On se place dans le plan complexe $\\left( O \\;; \\overrightarrow{u} ; \\overrightarrow{v} \\right)$ et on pose $\\displaystyle j=e^{i\\frac{2\\pi}{3}}=-\\frac{1}{2}+i\\frac{\\sqrt{3}}{2}$.
\n", "Dans cette étude, l'affixe d'un point $M$ sera systématiquement désignée par la notation $z_M$.\n", "

\n", "\n", "1. Condition nécessaire et suffisante pour qu'un triangle soit équilatéral

\n", "$\\;\\;\\;$a. Exécuter les cellules Python suivantes (la fonction conjugate permet de calculer le conjugué d'un complexe).
\n", "$\\quad\\;\\;$Quelles relations vérifiées par $j$ obtient-on ?
\n", "$\\quad\\;\\;$Vérifier ces résultats par le calcul, et justifier en particulier que $1+j=e^{i\\frac{\\pi}{3}}$\n", "\n", "
\n", "
\n", "\n", "a.\n", "
    \n", "
  • $\\displaystyle j^3=\\left( e^{i\\frac{2\\pi}{3}} \\right) ^3 = e^{2i\\pi} =1$
    • \n", "
  • $\\displaystyle j^2-\\overline{j}= \\left( e^{i\\frac{2\\pi}{3}} \\right) ^2 -\\overline{e^{i\\frac{2\\pi}{3}}} = e^{i\\frac{4\\pi}{3}} - e^{-i\\frac{2\\pi}{3}}=0$ donc $\\displaystyle j^2=\\overline{j}$
    • \n", "
  • $\\displaystyle 1+j+j^2=1+j+\\overline{j}=1+2Re(j)=1+2\\times \\left( -\\frac{1}{2} \\right) = 0$ donc $\\displaystyle 1+j+j^2=0$
    • \n", "
  • $\\displaystyle 1+j=-j^2=-\\overline{j}=- \\left( -\\frac{1}{2}-i\\frac{\\sqrt{3}}{2} \\right) = \\frac{1}{2}+i\\frac{\\sqrt{3}}{2} = e^{i\\frac{\\pi}{3}}$
    • \n", "
\n", " " ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle - \\frac{1}{2} + \\frac{\\sqrt{3} i}{2}$" ], "text/plain": [ "-1/2 + sqrt(3)*I/2" ] }, "execution_count": 21, "metadata": {}, "output_type": "execute_result" } ], "source": [ "j = (-1+I*sqrt(3))/2\n", "j" ] }, { "cell_type": "code", "execution_count": 22, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 1$" ], "text/plain": [ "1" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "simplify(j**3)" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 0$" ], "text/plain": [ "0" ] }, "execution_count": 23, "metadata": {}, "output_type": "execute_result" } ], "source": [ "simplify(j**2-conjugate(j))" ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 0$" ], "text/plain": [ "0" ] }, "execution_count": 24, "metadata": {}, "output_type": "execute_result" } ], "source": [ "simplify(1+j+j**2)" ] }, { "cell_type": "code", "execution_count": 25, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\frac{1}{2} + \\frac{\\sqrt{3} i}{2}$" ], "text/plain": [ "1/2 + sqrt(3)*I/2" ] }, "execution_count": 25, "metadata": {}, "output_type": "execute_result" } ], "source": [ "simplify(1+j)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "$\\;\\;\\;$b. On considère trois points distincts $M$ ; $N$ et $P$.
\n", "$\\quad\\;\\;$ Démontrer que le triangle $MNP$ est équilatéral direct si et seulement si $z_M+jz_N+j^2z_P=0$.
\n", " \n", "
\n", "
\n", "\n", "b.\n", "$MNP$ est équilatéral direct
\n", "$\\quad \\displaystyle \\Leftrightarrow \\frac{z_M-z_N}{z_P-z_N}=e^{i\\frac{\\pi}{3}}$
\n", "$\\quad \\displaystyle \\Leftrightarrow \\frac{z_M-z_N}{z_P-z_N}=1+j$
\n", "$\\quad \\displaystyle \\Leftrightarrow z_M-z_N=(1+j)(z_P-z_N)$
\n", "$\\quad \\displaystyle \\Leftrightarrow z_M+(-1+(1+j))z_N-(1+j)z_P=0$
\n", "$\\quad \\displaystyle \\Leftrightarrow z_M+jz_N-(-j^2)z_P=0$
\n", "$\\quad \\displaystyle \\Leftrightarrow z_M+jz_N+j^2z_P=0$
\n", " " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "2. Application
\n", "
\n", "$\\;\\;\\;$On considère 6 points $A$ ; $B$ ; $C$ ; $D$ ; $E$ et $F$ tels que $OAB$ ; $OCD$ et $OEF$ sont des triangles équilatéraux directs.
\n", "$\\;\\;\\;$$M$; $N$ et $P$ sont les milieux respectifs de $[BC]$ ; $[DE]$ et $[FA]$.
\n", "
\n", "$\\;\\;\\;$a. Activer la figure dynamique fournie ci-dessous. Quelle conjecture peut-on faire concernant le triangle $MNP$ ? \n", " " ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\t\n", "\t\n", "\t\n", "\t\t
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\n", "\t\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Sélectionner cette zone puis SHIFT+ENTREE pour obtenir la figure dynamique\n", "from IPython.display import display, HTML ; display(HTML('fig_dyn_GeoGebra/intgeomcomplexeconfig2.html'))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "$\\quad\\quad$b. Exprimer $z_A$ ; $z_C$ et $z_E$ respectivement en fonction de $z_B$ ; $z_D$ et $z_F$.
\n", "$\\quad\\quad$c. Exprimer $z_M$ ; $z_N$ et $z_P$ en fonction de $z_A$ ; $z_B$ ; $z_C$ ; $z_D$ ; $z_E$ et $z_F$.
\n", "$\\quad\\quad$d. Calculer $z_M+jz_N+j^2z_P$. Conclure. \n", "\n", "
\n", "
\n", "\n", "b. $OAB$ est équilatéral direct donc $ABO$ aussi et d'après la caractérisation vue en 1. on a : $z_A+jz_B+j^2\\times0=0$ c'est-à-dire $z_A=-jz_B$.
\n", "$\\;\\;\\;$De la même façon, on a $z_C=-jz_D$ et $z_E=-jz_F$.
\n", "c. $\\displaystyle z_M=\\frac{1}{2}(z_B+z_C)$ ; $\\displaystyle z_N=\\frac{1}{2}(z_D+z_E)$ et $\\displaystyle z_P=\\frac{1}{2}(z_F+z_A)$.
\n", "d. $\\displaystyle z_M+jz_N+j^2z_P=\\frac{1}{2} \\left( z_B+z_C+j(z_D+z_E)+j^2(z_F+z_A) \\right)$
\n", " $\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\;\\; \\displaystyle =\\frac{1}{2} \\left( z_B+z_C+jz_D+jz_E+j^2z_F+j^2z_A \\right)$
\n", " $\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\;\\; \\displaystyle =\\frac{1}{2} \\left( z_B-jz_D+jz_D-j^2z_F+j^2z_F-j^3z_B \\right)$ et comme $j^3=1$ :
\n", " $\\displaystyle z_M+jz_N+j^2z_P=0$ donc $MNP$ est un triangle équilatéral direct
\n", "" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*(C) Copyright Franck CHEVRIER 2019-2021 http://www.python-lycee.com/*\n" ] } ], "metadata": { "celltoolbar": "Raw Cell Format", "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.7.10" } }, "nbformat": 4, "nbformat_minor": 2 }