{"version":3,"file":"5G41-9FaYoFGW.js","sources":["../../src/exercices/5e/5G41.js"],"sourcesContent":["import { codageAngleDroit } from '../../lib/2d/angles.js'\nimport { cercle, cercleCentrePoint, traceCompas } from '../../lib/2d/cercle.js'\nimport { cibleCarree, dansLaCibleCarree } from '../../lib/2d/cibles.js'\nimport { afficheLongueurSegment, afficheMesureAngle, codageSegments } from '../../lib/2d/codages.js'\nimport { droite } from '../../lib/2d/droites.js'\nimport {\n  milieu,\n  point,\n  pointAdistance,\n  pointIntersectionCC,\n  pointIntersectionDD,\n  tracePoint\n} from '../../lib/2d/points.js'\nimport { polygoneAvecNom } from '../../lib/2d/polygones.js'\nimport { demiDroite, longueur, segment } from '../../lib/2d/segmentsVecteurs.js'\nimport { labelPoint, texteParPosition } from '../../lib/2d/textes.js'\nimport { rotation, similitude } from '../../lib/2d/transformations.js'\nimport { choice } from '../../lib/outils/arrayOutils'\nimport { miseEnEvidence, texteEnCouleurEtGras } from '../../lib/outils/embellissements'\nimport { choisitLettresDifferentes } from '../../lib/outils/aleatoires'\nimport { arrondi } from '../../lib/outils/nombres'\nimport { lettreDepuisChiffre } from '../../lib/outils/outilString.js'\nimport { texNombre } from '../../lib/outils/texNombre'\nimport Exercice from '../deprecatedExercice.js'\nimport { mathalea2d } from '../../modules/2dGeneralites.js'\nimport { listeQuestionsToContenu, randint, calculANePlusJamaisUtiliser } from '../../modules/outils.js'\n\nexport const titre = 'Construct specific quadrilaterals'\nexport const dateDeModifImportante = '02/09/2023'\nexport const dateDePublication = '03/02/2020'\n/**\n * Construction de quadrilatères avec dispositif d'auto-correction aléatoire\n * @author Jean-Claude Lhote\n */\nexport const uuid = '37e37'\nexport const ref = '5G41'\nexport default function ConstructionsParallelogrammesParticuliers () {\n  Exercice.call(this) // Héritage de la classe Exercice()\n  this.nbQuestions = 1\n  this.nbQuestionsModifiable = false\n  this.nbCols = 1\n  this.nbColsCorr = 1\n  this.sup = 1\n  this.correctionDetaillee = false\n  this.correctionDetailleeDisponible = true\n  this.nouvelleVersion = function () {\n    this.listeQuestions = [] // Liste de questions\n    this.listeCorrections = [] // Liste de questions corrigées\n    this.autoCorrection = []\n    let texte = ''; let texteCorr = ''\n    const celluleAlea = function (rang) {\n      const lettre = lettreDepuisChiffre(randint(1, rang))\n      const chiffre = Number(randint(1, rang)).toString()\n      return lettre + chiffre\n    }\n    // We prepare the figure...\n    const noms = choisitLettresDifferentes(5, 'QOX', true) // on choisit 5 lettres, les 4 premières sont les sommets, la 5e est le centre\n    const nom = noms[0] + noms[1] + noms[2] + noms[3]\n    let A, B, C, D, O, d1, d2, c1, c2, c3, c4, alpha, tri, t1, t2, t3, t4, t5, dd1, dd2\n    const objetsEnonce = []; const objetsCorrection = []\n    let typesDeQuestionsDisponibles\n    let xm, ym, xM, yM\n    if (this.sup === 1) typesDeQuestionsDisponibles = [1, 2, 3]\n    else if (this.sup === 2) typesDeQuestionsDisponibles = [4, 5, 6, 7]\n    else typesDeQuestionsDisponibles = [1, 2, 3, 4, 5, 6, 7]\n    typesDeQuestionsDisponibles = [7, 7]\n    const typeDeQuestion = choice(typesDeQuestionsDisponibles)\n    switch (typeDeQuestion) {\n      case 1:\n        A = point(0, 0, noms[0])\n        c1 = randint(20, 25) * 2 // AB\n        c4 = calculANePlusJamaisUtiliser(randint(20, 30, c1 / 2) / 5) // AD\n        c1 = calculANePlusJamaisUtiliser(c1 / 10)\n        d1 = 5 * (Math.abs(c4 - c1) + 2)\n        d2 = 5 * (c1 + c4 - 3)\n        d1 = calculANePlusJamaisUtiliser(randint(Math.min(d1, d2), Math.max(d1, d2)) / 5) // BD\n        B = pointAdistance(A, c1, randint(-30, 30), noms[1])\n        D = pointIntersectionCC(cercle(A, c4), cercle(B, d1), noms[3])\n        O = milieu(B, D, noms[4])\n        C = rotation(A, O, 180, noms[2])\n        texte = `$${nom}$ is a parallelogram such that`\n        texte += `$${noms[0] + noms[1]}=${texNombre(c1)}$ cm, $${noms[0] + noms[3]}=${texNombre(c4)}$ cm, $${noms[1] + noms[3]}=${texNombre(d1)}$ cm.<br>`\n        objetsEnonce.push(tracePoint(A, B), labelPoint(A, B))\n        if (this.correctionDetaillee) {\n          texteCorr += `Since $${nom}$ is a parallelogram, its diagonals intersect in the middle.<br>`\n          texteCorr += `Let's name $${noms[4]}$, the midpoint of $[${noms[1] + noms[3]}]$. $${noms[2]}$ is the symmetric of $${noms[0]}$ with respect to $${noms[4]}$.`\n          texteCorr += `Let's first construct the triangle $${noms[0] + noms[1] + noms[3]}$.<br>Then let's place $${noms[4]}$, the midpoint of $[${noms[1] + noms[3]}]$ and finally the point $${noms[2]}$.<br>`\n        }\n        if (longueur(B, D) !== longueur(A, C)) {\n          texteCorr += `Like $${noms[0] + noms[3]}\\ne ${noms[0] + noms[1]}$ and $${noms[0] + noms[2]}\\ne ${noms[3] + noms[1]}$, the parallelogram $${nom}$ is neither a rhombus nor a rectangle.<br>`\n          texteCorr += `$${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is not a particular parallelogram')}.<br>`\n        } else {\n          texteCorr += `Since $$${noms[0] + noms[2]} = ${noms[3] + noms[1]}$ and $${noms[0] + noms[3]}\\ne ${noms[0] + noms[1]}$, the parallelogram $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is a rectangle')}.<br>`\n        }\n        objetsCorrection.push(afficheLongueurSegment(A, B, 'black', -0.5), afficheLongueurSegment(A, D, 'black', 0.5))\n        t1 = traceCompas(A, D, 15)\n        t2 = traceCompas(B, D, 15)\n        t3 = traceCompas(O, C, 20)\n        tri = polygoneAvecNom(A, B, D)\n        xm = Math.min(A.x, B.x, D.x) - 0.8\n        ym = Math.min(A.y, B.y, D.y) - 0.8\n        xM = Math.max(A.x, B.x, D.x) + 0.8\n        yM = Math.max(A.y, B.y, D.y) + 0.8\n        break\n      case 2:\n        O = point(0, 0, noms[4])\n        c1 = randint(25, 35) * 2 // AC\n        c4 = calculANePlusJamaisUtiliser(randint((c1 + 4) / 2, 45) / 5) // BD\n        c1 = calculANePlusJamaisUtiliser(c1 / 10)\n        alpha = randint(100, 130)\n\n        A = pointAdistance(O, c1 / 2, randint(-30, 30), noms[0])\n        B = similitude(A, O, alpha, c4 / c1, noms[1])\n        D = rotation(B, O, 180, noms[3])\n        C = rotation(A, O, 180, noms[2])\n        texte = `$${nom}$ is a parallelogram with center $${noms[4]}$ such that`\n        texte += `$${noms[0] + noms[2]}=${texNombre(c1)}$ cm, $${noms[1] + noms[3]}=${texNombre(c4)}$ cm and $\\\\widehat{${noms[0] + noms[4] + noms[1]}}=${alpha}\\\\degree$ counterclockwise.<br>`\n        objetsEnonce.push(tracePoint(A, O), labelPoint(A, O))\n        if (this.correctionDetaillee) {\n          texteCorr += `Since $${nom}$ is a parallelogram, its diagonals intersect $${noms[4]}$ in the middle.<br>`\n          texteCorr += `$${noms[2]}$ is the symmetric of $${noms[0]}$ with respect to $${noms[4]}$. The distance $${noms[4] + noms[1]}$ is equal to half of $${noms[1] + noms[3]}$.<br>`\n          texteCorr += `Let us first construct the point $${noms[2]}$ symmetrical to $${noms[0]}$ with respect to $${noms[4]}$.<br>`\n          texteCorr += `Then let's construct an angle $\\\\widehat{${noms[0] + noms[4] + 'x'}}$ measuring $${alpha}\\\\degree$ counterclockwise.<br>`\n          texteCorr += `Finally, let us construct the point $${noms[1]}$ on $[${noms[4]}x)$ and its symmetric $${noms[3]}$ with respect to $${noms[4]}$ both located at $${texNombre(arrondi(c4 / 2))}$ cm from $${noms[4]}$.<br>`\n        }\n        texteCorr += `$${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is not a particular parallelogram')}.<br>`\n        xm = Math.min(A.x, B.x, C.x) - 0.8\n        ym = Math.min(A.y, B.y, C.y) - 0.8\n        xM = Math.max(A.x, B.x, C.x) + 0.8\n        yM = Math.max(A.y, B.y, C.y) + 0.8\n        break\n      case 3:\n        A = point(0, 0, noms[0])\n        c1 = randint(26, 40) * 2 // AB\n        c4 = calculANePlusJamaisUtiliser(randint(15, 25) / 5) // AD\n        c1 = calculANePlusJamaisUtiliser(c1 / 10)\n\n        B = pointAdistance(A, c1, randint(-30, 30), noms[1])\n        D = similitude(B, A, 90, c4 / c1, noms[3])\n        O = milieu(B, D, noms[4])\n        C = rotation(A, O, 180, noms[2])\n        texte = `$${nom}$ is a parallelogram such that`\n        texte += `$${noms[0] + noms[1]}=${texNombre(c1)}$ cm, $${noms[0] + noms[3]}=${texNombre(c4)}$ cm, $${noms[1] + noms[3]}=${noms[0] + noms[2]}$.<br>`\n        objetsEnonce.push(tracePoint(A, B), labelPoint(A, B))\n\n        if (this.correctionDetaillee) {\n          texteCorr += `Let $${noms[4]}$ be the midpoint of $[${noms[1] + noms[3]}]$. $${noms[2]}$ is the symmetric of $${noms[0]}$ with respect to $${noms[4]}$.<br>`\n          texteCorr += `Let's first construct the triangle $${noms[0] + noms[1] + noms[3]}$ then $${noms[4]}$ in the middle of $[${noms[1] + noms[3]}]$.<br>`\n          texteCorr += `The four vertices of $${nom}$ are on the circle with center $${noms[4]}$ passing through $${noms[0]}$. $[${noms[0]}${noms[2]}]$ and $[${noms[1]}${noms[3]}]$ are diameters of this circle.<br>`\n        }\n        texteCorr += `Since $${nom}$ is a parallelogram whose diagonals have the same length, $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is therefore a rectangle')}.<br>`\n        objetsCorrection.push(afficheLongueurSegment(A, B, 'black', -0.5), afficheLongueurSegment(A, D, 'black', 0.5))\n        t1 = cercleCentrePoint(O, A, 'gray')\n        t1.opacite = 0.5\n        t3 = traceCompas(O, C, 20)\n        tri = polygoneAvecNom(A, B, D)\n        xm = Math.min(A.x, B.x, D.x) - 0.8\n        ym = Math.min(A.y, B.y, D.y) - 0.8\n        xM = Math.max(A.x, B.x, D.x) + 0.8\n        yM = Math.max(A.y, B.y, D.y) + 0.8\n\n        break\n      case 4:\n        A = point(0, 0, noms[0])\n        c1 = randint(15, 30) // AB\n        c4 = calculANePlusJamaisUtiliser(randint(15, 20, c1) / 5) // BD\n        c1 = calculANePlusJamaisUtiliser(c1 / 5)\n\n        B = pointAdistance(A, c1, randint(-30, 30), noms[1])\n        D = pointIntersectionCC(cercle(A, c1), cercle(B, c4), noms[3])\n        O = milieu(B, D, noms[4])\n        C = rotation(A, O, 180, noms[2])\n\n        texte = `$${nom}$ is a parallelogram such that`\n        texte += `$${noms[0] + noms[1]}=${texNombre(c1)}$ cm, $${noms[1] + noms[3]}=${texNombre(c4)}$ cm, $[${noms[0] + noms[2]}]\\\\perp [${noms[1] + noms[3]}]$.<br>`\n        objetsEnonce.push(tracePoint(A, B), labelPoint(A, B))\n\n        if (this.correctionDetaillee) {\n          texteCorr += `Since $${nom}$ is a parallelogram whose diagonals are perpendicular, it is a rhombus and then`\n          texteCorr += `the triangle $${noms[0] + noms[1] + noms[3]}$ is isosceles in $${noms[0]}$.<br>`\n          texteCorr += `Let's first construct the triangle $${noms[0] + noms[1] + noms[3]}$ then $${noms[4]}$, the midpoint of $[${noms[1] + noms[3]}]$ and finally the point $${noms[2]}$.<br>`\n        }\n        texteCorr += `Since $${nom}$ is a parallelogram whose diagonals are perpendicular, $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is therefore a rhombus')}.<br>`\n        objetsCorrection.push(afficheLongueurSegment(A, B, 'black', -0.5), afficheLongueurSegment(A, D, 'black', 0.5))\n        t1 = traceCompas(A, D, 15)\n        t2 = traceCompas(B, D, 15)\n        t3 = traceCompas(O, C, 20)\n        tri = polygoneAvecNom(A, B, D)\n        xm = Math.min(A.x, B.x, D.x) - 0.8\n        ym = Math.min(A.y, B.y, D.y) - 0.8\n        xM = Math.max(A.x, B.x, D.x) + 0.8\n        yM = Math.max(A.y, B.y, D.y) + 0.8\n        break\n      case 5:\n        A = point(0, 0, noms[0])\n        c1 = randint(20, 35) * 2 // AC\n        c4 = calculANePlusJamaisUtiliser(randint((c1 - 4) / 2, 35) / 5) // AD\n        c1 = calculANePlusJamaisUtiliser(c1 / 10)\n        alpha = randint(95, 120)\n        B = pointAdistance(A, c1, randint(-30, 30), noms[1])\n        D = similitude(B, A, alpha, c4 / c1, noms[3])\n        O = milieu(B, D, noms[4])\n        C = rotation(A, O, 180, noms[2])\n        texte = `$${nom}$ is a parallelogram with center $${noms[4]}$ such that`\n        texte += `$${noms[0] + noms[1]}=${texNombre(c1)}$ cm, $${noms[0] + noms[3]}=${texNombre(c4)}$ cm and $\\\\widehat{${noms[1] + noms[2] + noms[3]}}=${alpha}\\\\degree$ counterclockwise.<br>`\n        objetsEnonce.push(tracePoint(A, B), labelPoint(A, B))\n        if (this.correctionDetaillee) {\n          texteCorr += `Since $${nom}$ is a parallelogram, its opposite angles have the same measure, so $\\\\widehat{${noms[3] + noms[0] + noms[1]}}=${alpha}\\\\degree$.<br>`\n          texteCorr += `Let's first construct the triangle $${noms[0] + noms[1] + noms[3]}$.<br>`\n          texteCorr += `Let $${noms[4]}$ be the midpoint of $[${noms[1] + noms[3]}]$.<br>`\n          texteCorr += `Let us then construct the point $${noms[2]}$ symmetrical to $${noms[0]}$ with respect to $${noms[4]}$, midpoint of $[${noms[1] + noms[3]}]$.<br>`\n        }\n        texteCorr += `Since $${nom}$ is a parallelogram which does not have a right angle and its consecutive sides are of different lengths,`\n        texteCorr += `$${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is not a particular parallelogram')}.<br>`\n        t1 = traceCompas(A, D, 15)\n        t2 = traceCompas(A, B, 15)\n        t3 = traceCompas(O, C, 20)\n        tri = polygoneAvecNom(A, B, D)\n        xm = Math.min(A.x, B.x, D.x) - 0.8\n        ym = Math.min(A.y, B.y, D.y) - 0.8\n        xM = Math.max(A.x, B.x, D.x) + 0.8\n        yM = Math.max(A.y, B.y, D.y) + 0.8\n        break\n      case 6:\n        A = point(0, 0, noms[0])\n        c1 = randint(20, 35) * 2 // AC\n        c2 = randint(15, 20) * 2 // AO\n        c3 = calculANePlusJamaisUtiliser(c1 + randint(5, 10) * 2) - c2 // BO\n        c1 = calculANePlusJamaisUtiliser(c1 / 10)\n        c2 = calculANePlusJamaisUtiliser(c2 / 10)\n        c3 = calculANePlusJamaisUtiliser(c3 / 10)\n\n        B = pointAdistance(A, c1, randint(-30, 30), noms[1])\n        O = pointIntersectionCC(cercle(A, c2), cercle(B, c3), noms[4])\n        C = rotation(A, O, 180, noms[2])\n        D = rotation(B, O, 180, noms[3])\n        texte = `$${nom}$ is a parallelogram with center $${noms[4]}$ such that`\n        texte += `$${noms[0] + noms[1]}=${texNombre(c1)}$ cm, $${noms[4] + noms[2]}=${texNombre(c2)}$ cm and $${noms[4] + noms[3]}=${texNombre(c3)}$ cm.<br>`\n        objetsEnonce.push(tracePoint(A, B), labelPoint(A, B))\n        if (this.correctionDetaillee) {\n          texteCorr += `Since $${nom}$ is a parallelogram, its diagonals intersect $${noms[4]}$ in the middle.<br>`\n          texteCorr += `We deduce that $${noms[0] + noms[4]}=${noms[4] + noms[2]}=${texNombre(c2)}$ cm and that $${noms[1] + noms[4]}=${noms[4] + noms[3]}=${texNombre(c3)}$ cm.<br>`\n          texteCorr += `Let's first construct the triangle $${noms[0] + noms[1] + noms[4]}$`\n          texteCorr += `then the respective symmetric points $${noms[2]}$ and $${noms[3]}$ of $${noms[0]}$ and $${noms[1]}$ with respect to $${noms[4]}$.<br>`\n        }\n        if (c1 * c1 !== (c2 * c2 + c3 * c3)) {\n          texteCorr += `The triangle $${noms[0] + noms[1] + noms[4]}$ is not a right triangle, so the diagonals are not perpendicular.<br>`\n          if (c2 === c3) texteCorr += `The diagonals are the same length. $${nom}$ is a parallelogram whose diagonals are of equal length, $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is therefore a rectangle')}.<br>`\n          else texteCorr += `In addition, they do not have the same length, so $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is not a particular parallelogram')}.<br>`\n        } else {\n          texteCorr += `The triangle $${noms[0] + noms[1] + noms[4]}$ is a right triangle, so the diagonals are perpendicular.<br>`\n          if (c2 === c3) texteCorr += `moreover the diagonals have the same length. $${nom}$ is a parallelogram whose diagonals are perpendicular and of the same length, $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is therefore a square')}.<br>`\n        }\n        texteCorr += '.<br>'\n        t1 = traceCompas(A, O, 20)\n        t2 = traceCompas(B, O, 20)\n        t3 = traceCompas(O, C, 30)\n        t4 = traceCompas(O, D, 30)\n\n        tri = polygoneAvecNom(A, B, O)\n        xm = Math.min(A.x, B.x, O.x) - 0.8\n        ym = Math.min(A.y, B.y, O.y) - 0.8\n        xM = Math.max(A.x, B.x, O.x) + 0.8\n        yM = Math.max(A.y, B.y, O.y) + 0.8\n        break\n      case 7:\n        A = point(0, 0, noms[0])\n        c1 = calculANePlusJamaisUtiliser(randint(30, 40) / 5) // AC\n        c2 = randint(25, 40)// angle OAB\n        c3 = randint(30, 45, c2) // angle OCB\n\n        C = pointAdistance(A, c1, randint(-30, 30), noms[2])\n        O = milieu(A, C, noms[4])\n        B = rotation(C, A, c2)\n        dd1 = droite(A, B)\n        D = rotation(A, C, -c3)\n        dd2 = droite(C, D)\n        B = pointIntersectionDD(dd1, dd2, noms[1])\n        D = rotation(B, O, 180, noms[3])\n        texte = `$${nom}$ is a parallelogram with center $${noms[4]}$ such that`\n        texte += `$${noms[0] + noms[2]}=${texNombre(c1)}$ cm.<br>$\\\\widehat{${noms[4] + noms[0] + noms[1]}}=${c2}\\\\degree$ counterclockwise.<br>$\\\\widehat{${noms[4] + noms[2] + noms[1]}}=${c3}\\\\degree$ clockwise.<br>`\n        objetsEnonce.push(tracePoint(A, C), labelPoint(A, C))\n        if (this.correctionDetaillee) {\n          texteCorr += `Since $${nom}$ is a parallelogram, its opposite sides are parallel.<br>`\n          texteCorr += `The diagonal $[${noms[0]}${noms[2]}]$ forms equal alternate-internal angles $\\\\widehat{${noms[4] + noms[0] + noms[1]}}$ and $\\\\widehat{${noms[4] + noms[2] + noms[3]}}$.<br>`\n          texteCorr += `Likewise the angles $\\\\widehat{${noms[4] + noms[0] + noms[3]}}$ and $\\\\widehat{${noms[4] + noms[2] + noms[1]}}$ are also alternate-internal equal.<br>`\n          texteCorr += `We deduce that $\\\\widehat{${noms[4] + noms[0] + noms[3]}}=\\\\widehat{${noms[4] + noms[2] + noms[1]}}=${miseEnEvidence(c3, 'red')}\\\\degree$ and that $\\\\widehat{${noms[4] + noms[0] + noms[1]}}=\\\\widehat{${noms[4] + noms[2] + noms[3]}}=${miseEnEvidence(c2, 'blue')}\\\\degree$.<br>`\n          texteCorr += `Let's first construct the triangle $${noms[0] + noms[1] + noms[2]}$`\n          texteCorr += `then the point $${noms[3]}$ symmetrical to $${noms[1]}$ with respect to $${noms[4]}$.<br>`\n        }\n\n        texteCorr += `The triangle $${noms[0] + noms[1] + noms[2]}$ is not an isosceles triangle because its angles are not equal.<br>`\n        texteCorr += `Moreover, in this triangle $${noms[0] + noms[1] + noms[2]}$, the angle $\\\\widehat{${noms[0] + noms[1] + noms[2]}}$ measures $${180 - c2 - c3}\\\\degree$ and is not straight therefore $${miseEnEvidence(nom)}$ ${texteEnCouleurEtGras('is not a particular parallelogram')}.<br>`\n        t1 = afficheMesureAngle(O, A, B, 'blue', 1, texNombre(c2) + '°')\n        t2 = afficheMesureAngle(O, C, B, 'red', 1, texNombre(c3) + '°')\n        t3 = traceCompas(O, D, 30)\n        t5 = tracePoint(O)\n        t5.style = '+'\n        objetsCorrection.push(t1, t2)\n        tri = polygoneAvecNom(A, B, C)\n        xm = Math.min(A.x, B.x, C.x) - 0.8\n        ym = Math.min(A.y, B.y, C.y) - 0.8\n        xM = Math.max(A.x, B.x, C.x) + 0.8\n        yM = Math.max(A.y, B.y, C.y) + 0.8\n        break\n    }\n    texte += `Construct the parallelogram $${nom}$ and specify if it is a particular parallelogram.<br>`\n\n    const p = polygoneAvecNom(A, B, C, D)\n\n    const xMin = Math.min(A.x, B.x, C.x, D.x) - 2\n    const yMin = Math.min(A.y, B.y, C.y, D.y) - 2\n    const xMax = Math.max(A.x, B.x, C.x, D.x) + 2\n    const yMax = Math.max(A.y, B.y, C.y, D.y) + 2\n\n    const cellule1 = celluleAlea(5)\n    const cellule2 = celluleAlea(5)\n    const cellule3 = celluleAlea(5)\n    const result1 = dansLaCibleCarree(B.x, B.y, 5, 0.3, cellule3)\n    const result2 = dansLaCibleCarree(C.x, C.y, 5, 0.3, cellule1)\n    const result3 = dansLaCibleCarree(D.x, D.y, 5, 0.3, cellule2)\n    const cible1 = cibleCarree({ x: result1[0], y: result1[1], rang: 5, num: '', taille: 0.4, color: 'gray' })\n    cible1.taille = 0.3\n    cible1.opacite = 0.7\n    const cible2 = cibleCarree({ x: result2[0], y: result2[1], rang: 5, num: '', taille: 0.4, color: 'gray' })\n    cible2.taille = 0.3\n    cible2.opacite = 0.7\n    const cible3 = cibleCarree({ x: result3[0], y: result3[1], rang: 5, num: '', taille: 0.4, color: 'gray' })\n    cible3.taille = 0.3\n    cible3.opacite = 0.7\n    dd1 = segment(O, A)\n    dd2 = segment(O, B)\n    const dd3 = segment(O, C)\n    const dd4 = segment(O, D)\n\n    switch (typeDeQuestion) {\n      case 1:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, objetsCorrection, t1, t2, tri[0], tri[1], afficheLongueurSegment(D, B)) + '<br>'\n        objetsEnonce.push(cible3, cible2)\n        objetsCorrection.push(p[0], p[1], t3)\n        objetsCorrection.push(cible3, cible2, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('|||', 'blue', B, O, O, D), afficheLongueurSegment(O, B))\n        break\n      case 2:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, codageSegments('||', 'red', A, O, O, C), t3, dd1, dd3, dd2, afficheMesureAngle(A, O, B, 'black', 1, alpha + '°'), tracePoint(A, O, C), labelPoint(A, O, C), texteParPosition('x', B.x - 0.5, B.y), afficheLongueurSegment(A, O), afficheLongueurSegment(O, C)) + '<br>'\n        objetsEnonce.push(cible3, cible2, cible1)\n        objetsCorrection.push(p[0], p[1], t3, afficheLongueurSegment(O, D))\n        objetsCorrection.push(cible3, cible2, cible1, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('|||', 'blue', B, O, O, D), afficheMesureAngle(A, O, B, 'black', 1, alpha + '°'))\n\n        break\n      case 3:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, objetsCorrection, tri[0], tri[1], codageAngleDroit(D, A, B)) + '<br>'\n        objetsEnonce.push(cible3, cible2)\n        objetsCorrection.push(p[0], p[1], t1, t3)\n        objetsCorrection.push(cible3, cible2, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('||', 'red', B, O, O, D))\n\n        break\n      case 4:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, objetsCorrection, tri[0], tri[1], afficheLongueurSegment(D, B), t2, traceCompas(A, B, 60), traceCompas(A, D, 60)) + '<br>'\n        objetsEnonce.push(cible3, cible2)\n        objetsCorrection.push(p[0], p[1], t3, afficheLongueurSegment(O, B))\n        objetsCorrection.push(codageAngleDroit(A, O, D), cible3, cible2, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('|||', 'blue', B, O, O, D))\n        break\n      case 5:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, tri[0], tri[1], demiDroite(A, B), demiDroite(A, D), afficheMesureAngle(B, A, D, 'black', 1, alpha + '°'), afficheLongueurSegment(A, B), afficheLongueurSegment(A, D)) + '<br>'\n        objetsEnonce.push(cible3, cible2)\n        objetsCorrection.push(p[0], p[1], t3)\n        objetsCorrection.push(cible3, cible2, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('|||', 'blue', B, O, O, D), afficheMesureAngle(B, A, D, 'black', 1, alpha + '°'), afficheLongueurSegment(B, A), afficheLongueurSegment(A, D), afficheLongueurSegment(C, B), afficheLongueurSegment(D, C))\n\n        break\n      case 6:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, objetsCorrection, tri[0], tri[1], afficheLongueurSegment(B, A), afficheLongueurSegment(O, B), afficheLongueurSegment(A, O), t1, t2, t5) + '<br>'\n        objetsEnonce.push(cible3, cible2)\n        objetsCorrection.push(p[0], p[1], t3, t4)\n        objetsCorrection.push(cible3, cible2, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('|||', 'blue', B, O, O, D))\n        break\n      case 7:\n        if (this.correctionDetaillee) texteCorr += mathalea2d({ xmin: xm, ymin: ym, xmax: xM, ymax: yM, pixelsParCm: 25, scale: 1 }, objetsCorrection, tri[0], tri[1], afficheLongueurSegment(C, O), afficheLongueurSegment(O, A), labelPoint(O), t5, codageSegments('||', 'red', A, O, O, C)) + '<br>'\n        objetsEnonce.push(cible3, cible1)\n        objetsCorrection.push(p[0], p[1], t3)\n        objetsCorrection.push(cible3, t1, t2, t3, cible1, dd1, dd2, dd3, dd4, labelPoint(O), codageSegments('||', 'red', A, O, O, C), codageSegments('|||', 'blue', B, O, O, D), afficheMesureAngle(O, A, D, 'red', 1, texNombre(c3) + '°'), afficheMesureAngle(O, C, D, 'blue', 1, texNombre(c2) + '°'))\n        break\n    }\n    texte += mathalea2d({ xmin: xMin, ymin: yMin, xmax: xMax, ymax: yMax, pixelsParCm: 25, scale: 1 }, objetsEnonce)\n    texteCorr += mathalea2d({ xmin: xMin, ymin: yMin, xmax: xMax, ymax: yMax, pixelsParCm: 25, scale: 1 }, objetsCorrection)\n\n    this.listeQuestions.push(texte)\n    this.listeCorrections.push(texteCorr)\n    listeQuestionsToContenu(this)\n  }\n  this.besoinFormulaireNumerique = ['Level of difficulty', 3, '1: Easy figures\\n2: More difficult figures\\n3: Combination']\n  // this.needFormular2BoxCheck = [\"With points on both 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