import{E as me,ag as ge,c as f,h as W,r as O,C as w,W as H,u as ee,I as Z,w as P,o as Q,aa as j,R as $e,b1 as fe,ai as Ie,T as ye,aH as ce,f as K,ay as he,m as R,aL as be,s as ke,bq as Oe,a as we,bI as Pe,l as ve}from"./index-XCg2QAX4.js";import{c as v}from"./aleatoires-C4JUoVmT.js";import{a as m,l as se,k as te,v as S,s as pe,C as T,b as ue,o as Ce,e as Me,n as Qe,t as je,g as ie,q as xe}from"./3d-n_2g9rXi.js";import{RedactionPythagore as G}from"./_pythagore-2u_lJort.js";import"./dateEtHoraires-1H8goc58.js";const We="Determining lengths in geometry in space",Ae=!0,Le="AMCHybride",Ee=!0,Be="mathLive",Ne="19/12/2022",Ke="13/11/2023",Re="57c70",Se="3G44";function Je(){me.call(this),this.titre=We,this.nbQuestions=4,this.sup2=1,this.sup=10,this.nouvelleVersion=function(){this.autoCorrection=[];const de=ge({saisie:this.sup,min:1,max:9,defaut:10,melange:10,nbQuestions:this.nbQuestions});this.listeQuestions=[],this.listeCorrections=[];const h=["km","hmm","dam","m","dm","cm","mm"];for(let D=0,I,o,C,a,e,n,A,t,z,re,b,$,i,g,l,s,k=[],y,p,L,x,_,oe,M,q,r,u,c,ne,ae,U,E,V,B,X,d,N,J,Y,le=0;D<this.nbQuestions&&le<50;){switch(this.autoCorrection[D]={},I="",o="",a=[],f.anglePerspective=W([-30,-60,30,60]),c=O(0,6),de[D]){case 1:e=m(0,0,0),u=O(5,10),n=m(u,0,0),t=m(0,0,u),q=W([u,-u]),z=m(0,q,0),l=v(8,"OQWX").join(""),i=ie(e,n,t,z,"blue",!0,l),k.push(["02","1"],["13","2"]),q>0?(k.push(["16","5"],["25","6"]),f.anglePerspective>0?k.push(["27","3"],["36","2"]):k.push(["14","5"],["05","1"])):(k.push(["07","3"],["34","0"]),f.anglePerspective<0?k.push(["27","3"],["36","2"]):k.push(["14","5"],["05","1"])),s=W(k),L=l[parseInt(s[0][0])],x=l[parseInt(s[0][1])],M=l[parseInt(s[1])],g=L+x,I+=`Knowing that the cube $${l}$ has edges of $${u}$ ${h[c]}, calculate the length $${g}$, rounded to the tenth of ${h[c]}.<br>`,b=w(i.sommets[parseInt(s[0][0])].c2d,i.sommets[parseInt(s[0][1])].c2d,"#f15929"),b.epaisseur=2,a.push(...i.c2d,b),I+=Q(Object.assign({},j(a),{scale:.7,style:"block"}),a)+"<br>",a.push(new T(i.sommets[parseInt(s[0][0])],i.sommets[parseInt(s[1])],i.sommets[parseInt(s[0][1])],"#f15929",2)),o+="<br>"+Q(Object.assign({},j(a),{scale:.7,style:"block"}),a)+"<br>",C=H(Math.sqrt(u**2+u**2),1),o+="<br>"+G(M,L,x,1,u,u,C,h[c])[0];break;case 2:u=O(5,10),l=v(8,"OQWX"),l=l.join(""),k=[["60","5","05","1"],["71","4","14","0"],["24","1","41","0"],["35","0","50","1"]],s=W(k),n=l[parseInt(s[0][0])],A=l[parseInt(s[0][1])],e=l[parseInt(s[1])],t=l[parseInt(s[3])],g=n+A,I+=`Knowing that the cube $${l}$ has edges of $${u}$ ${h[c]}, calculate the length $${g}$, rounded to the tenth of ${h[c]}.<br>`,i=xe(1,1,1,u,"blue","","","",!1,!0,l),b=w(i.sommets[parseInt(s[0][0])].c2d,i.sommets[parseInt(s[0][1])].c2d,"#f15929"),b.epaisseur=2,b.pointilles=2,a.push(...i.c2d,b),I+=Q(Object.assign({},j(a),{scale:.7,style:"block"}),a)+"<br>",a.push(new T(i.sommets[parseInt(s[0][0])],i.sommets[parseInt(s[1])],i.sommets[parseInt(s[0][1])],"#f15929",2)),$=w(i.sommets[parseInt(s[1])].c2d,i.sommets[parseInt(s[0][1])].c2d,"green"),$.epaisseur=2,$.pointilles=1,a.push($),a.push(new T(i.sommets[parseInt(s[1])],i.sommets[parseInt(s[3])],i.sommets[parseInt(s[0][1])],"green",2)),o+="<br>"+Q(Object.assign({},j(a),{scale:.7,style:"block"}),a)+"<br>",o+=`The triangle $${g+e}$ is right-angled in $${e}$ so according to the Pythagorean theorem, we have:`,o+=`$${g}^2=${e+n}^2+${e+A}^2$.`,o+=`<br> We cannot continue if we do not know the value of $${e+A}^2$. Let's find it.`,o+=`<br> ${K(10)}The triangle $${A+e+t}$ is right-angled in $${t}$ so according to the Pythagorean theorem, we have:`,o+=`$${e+A}^2=${e+t}^2+${t+A}^2$.`,re=P(he(Math.sqrt(u**2+u**2+u**2))),C=H(Math.sqrt(u**2+u**2+u**2),1),o+=`<br> ${K(10)}$${e+A}^2=${P(u)}^2+${P(u)}^2$`,o+=`<br> ${K(10)}$${R(e+A,"green")}^2=${R(P(u**2+u**2),"green")}$`,o+=`<br> ${K(10)} No need to find the value of $${e+A}$ because only its square interests us here.`,o+=`<br> Let's go back to $${g}^2=${e+n}^2+${e+A}^2$.`,o+=`<br> $${g}^2=${P(u)}^2+${R(P(u**2+u**2),"green")} = ${P(u**2+u**2+u**2)}$`,o+=`<br> $${g}=\\sqrt{${P(u**2+u**2+u**2)}}$`,o+=`<br> $${g}\\approx${R(re)}$ ${be(h[c])}`;break;case 3:y=O(5,20),r=O(5,20,[y]),p=O(5,20,[y,r]),e=m(0,0,0),n=m(y,0,0),t=m(0,0,r),q=W([p,-p]),z=m(0,q,0),l=v(8,"OQWX").join(""),i=ie(e,n,t,z,"blue",!0,l),k=[],k.push(["02","1",y,r],["13","2",r,y]),q>0?(k.push(["16","5",p,r],["25","6",p,r]),f.anglePerspective>0?k.push(["27","3",y,p],["36","2",y,p]):k.push(["14","5",p,y],["05","1",y,p])):(k.push(["07","3",r,p],["34","0",p,r]),f.anglePerspective<0?k.push(["27","3",y,p],["36","2",y,p]):k.push(["14","5",p,y],["05","1",y,p])),s=W(k),L=l[parseInt(s[0][0])],x=l[parseInt(s[0][1])],M=l[parseInt(s[1])],g=L+x,I+=`Knowing that in the right box $${l}$, $${l[0]}${l[1]} = ${y}$ ${h[c]}, $${l[0]}${l[3]} = ${r}$ ${h[c]} and $${l[0]}${l[4]} = ${p}$ ${h[c]}, calculate the length $${g}$, rounded to the tenth of ${h[c]}.<br>`,b=w(i.sommets[parseInt(s[0][0])].c2d,i.sommets[parseInt(s[0][1])].c2d,"#f15929"),b.epaisseur=2,a.push(...i.c2d,b),I+=Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a)+"<br>",a.push(new T(i.sommets[parseInt(s[0][0])],i.sommets[parseInt(s[1])],i.sommets[parseInt(s[0][1])],"#f15929",2)),o+="<br>"+Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a)+"<br>",C=H(Math.sqrt(s[2]**2+s[3]**2),1),o+="<br>"+G(M,L,x,1,s[2],s[3],C,h[c])[0];break;case 4:y=O(5,20),r=O(5,20,[y]),p=O(5,20,[y,r]),e=m(0,0,0),n=m(y,0,0),t=m(0,0,r),q=W([p,-p]),z=m(0,q,0),l=v(8,"OQWX").join(""),i=ie(e,n,t,z,"blue",!0,l),k=[["60","5","05","1",p,y,r],["71","4","14","0",p,y,r],["24","1","41","0",y,p,r],["35","0","50","1",y,p,r]],s=W(k),L=l[parseInt(s[0][0])],x=l[parseInt(s[0][1])],M=l[parseInt(s[1])],_=l[parseInt(s[3])],g=L+x,I+=`Knowing that in the right box $${l}$, $${l[0]}${l[1]} = ${y}$ ${h[c]}, $${l[0]}${l[3]} = ${r}$ ${h[c]} and $${l[0]}${l[4]} = ${p}$ ${h[c]}, calculate the length $${g}$, rounded to the tenth of ${h[c]}.<br>`,b=w(i.sommets[parseInt(s[0][0])].c2d,i.sommets[parseInt(s[0][1])].c2d,"#f15929"),b.epaisseur=2,b.pointilles=2,a.push(...i.c2d,b),I+=Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a)+"<br>",$=w(i.sommets[parseInt(s[1])].c2d,i.sommets[parseInt(s[0][1])].c2d,"green"),$.epaisseur=2,$.pointilles=1,a.push($),a.push(new T(i.sommets[parseInt(s[0][0])],i.sommets[parseInt(s[1])],i.sommets[parseInt(s[0][1])],"#f15929",2)),a.push(new T(i.sommets[parseInt(s[1])],i.sommets[parseInt(s[3])],i.sommets[parseInt(s[0][1])],"green",2)),o+="<br>"+Q(Object.assign({},j(a),{scale:.7,style:"block"}),a)+"<br>",o+=`The triangle $${g+M}$ is right-angled in $${M}$ so according to the Pythagorean theorem, we have:`,o+=`$${g}^2=${M+L}^2+${M+x}^2$.`,o+=`<br> We cannot continue if we do not know the value of $${M+x}^2$. Let's find it.`,o+=`<br> ${K(10)}The triangle $${x+M+_}$ is right-angled in $${_}$ so according to the Pythagorean theorem, we have:`,o+=`$${M+x}^2=${M+_}^2+${_+x}^2$.`,oe=P(he(Math.sqrt(s[4]**2+s[5]**2+s[6]**2))),C=H(Math.sqrt(s[4]**2+s[5]**2+s[6]**2),1),o+=`<br> ${K(10)}$${M+x}^2=${P(s[4])}^2+${P(s[5])}^2$`,o+=`<br> ${K(10)}$${R(M+x,"green")}^2=${R(P(s[4]**2+s[5]**2),"green")}$`,o+=`<br> ${K(10)} No need to find the value of $${M+x}$ because only its square interests us here.`,o+=`<br> Let's go back to $${g}^2=${M+L}^2+${M+x}^2$.`,o+=`<br> $${g}^2=${P(s[6])}^2+${R(P(s[4]**2+s[5]**2),"green")} = ${P(s[6]**2+s[4]**2+s[5]**2)}$`,o+=`<br> $${g}=\\sqrt{${P(s[6]**2+s[4]**2+s[5]**2)}}$`,o+=`<br> $${g}\\approx${R(oe)}$ ${be(h[c])}`;break;case 5:for(d=O(4,10),r=O(5,15,[d]),e=m(0,0,0,!0,v(1,"OQWX")[0],"left"),W([0,1,2])===0?(n=m(d,0,0),t=m(0,0,r,!0,v(1,"OQWX"+e.label)[0],"left")):W([0,1])===0?(n=m(0,d,0),t=m(r,0,0,!0,v(1,"OQWX"+e.label)[0],"left")):(n=m(0,0,d),t=m(0,r,0,!0,v(1,"OQWX"+e.label)[0],"left")),J=S(e,n),i=Qe(e,t,J,J,"blue",!1,!0,!0,"black"),ne=Ie(ye(2,i.pointsBase2.length-3,[16,17,18,19,20])),ae=0;ce(t.c2d,i.pointsBase2[ne[ae]])<ce(t.c2d,i.pointsBase2[0])/2;)ae++;E=se(je(t,J),te(e,S(e,t)),i.angleDepart+10*ne[ae]),E.c2d.nom=v(1,"OQWX"+e.label+t.label)[0],g=e.label+E.c2d.nom,b=w(e.c2d,E.c2d,"#f15929"),b.epaisseur=2,b.pointilles=2,a.push(...i.c2d,b,ee(E),Z(e,t,E.c2d)),I+=`In this cylinder of revolution, the radius of its bases (with respective centers $${e.label}$ and $${t.label}$) is $${d}$ ${h[c]} and its height is $${r}$ ${h[c]}. Knowing that the point $${E.c2d.nom}$ is on the base of center $${t.label}$, calculate the length $${g}$, rounded to the tenth of ${h[c]}.<br>`,I+="<br>"+Q(Object.assign({},j([...i.c2d]),{scale:f.isHtml?.7:.3,style:"block"}),a),a.push(new T(e,t,E,"green",2)),$=w(t.c2d,E.c2d,"green"),a.push($),$=w(t.c2d,e.c2d,"green"),a.push($),o+="<br>"+Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),C=H(Math.sqrt(r**2+d**2),1),o+="<br>"+G(t.label,e.label,E.c2d.nom,1,r,d,C,h[c])[0];break;case 6:d=O(4,10),r=O(5,15,[d]),e=m(0,0,0,!0,v(1,"OQWX")[0],"left"),n=m(d,0,0),t=m(0,0,r*1,!0,v(1,"OQWX"+e.label)[0],"left"),J=S(e,n),V=[n],B=W([3,5,6,7]);for(let F=1;F<B;F++)V.push(se(n,te(e,S(t,e)),F*360/B));p=ue(V,"blue"),i=Me(p,t,"blue",e,!0,"black",!0),X=O(0,B-1),b=i.aretesSommet[X].c2d,b.epaisseur=2,b.color=$e("#f15929"),g=t.label+p.listePoints2d[X].nom,I+=`$${i.nom}$ is a regular pyramid. The distance between $${e.label}$, the center of the base, and one of the vertices of the base is $${d}$ ${h[c]} and the height of this pyramid is $${r}$ ${h[c]}. Calculate the length $${g}$, rounded to the tenth of ${h[c]}.<br>`,I+=Q(Object.assign({},j([...i.c2d]),{scale:f.isHtml?.7:.3,style:"block"}),[...i.c2d,b]),a.push(...i.c2d,b,new T(t,e,p.listePoints[X],"green",2)),$=w(e.c2d,p.listePoints2d[X],"green"),a.push($),$=w(t.c2d,e.c2d,"green"),a.push($),o+="<br>"+Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),C=H(Math.sqrt(r**2+d**2),1),o+="<br>"+G(e.label,t.label,p.listePoints2d[X].nom,1,r,d,C,h[c])[0];break;case 7:d=O(4,10),r=O(5,15,[d]),e=m(0,0,0,!0,v(1,"OQWX")[0],"left"),n=m(d,0,0),t=m(0,0,r,!0,v(1,"OQWX"+e.label)[0],"left"),J=S(e,n),V=[n],B=36;for(let F=1;F<B;F++)V.push(se(n,te(e,S(t,e)),F*360/B));p=ue(V,"blue"),i=Ce(e,t,J,"blue",!0,"black",fe("gray",100)),X=O(1,Math.floor(B/2)-1),f.anglePerspective<0&&(X=(B-X)%B),b=w(t.c2d,p.listePoints2d[X],"#f15929"),b.epaisseur=2,I+=`In this cone of revolution, the radius of its base is $${d}$ ${h[c]} and its height is $${r}$ ${h[c]}. Calculate the length of a generatrix of this cone, rounded to the tenth of ${h[c]}.<br>`,I+=Q(Object.assign({},j([...i.c2d]),{scale:f.isHtml?.7:.3,style:"block"}),[...i.c2d,b]),U=p.listePoints2d[X],U.nom=v(1,"OQWX"+e.label+t.label)[0],U.positionLabel="below",g=t.label+U.nom,a.push(...i.c2d,b,new T(t,e,p.listePoints[X],"green",2)),$=w(e.c2d,p.listePoints2d[X],"green"),a.push($),$=w(t.c2d,e.c2d,"green"),a.push($),a.push(Z(U,t.c2d)),o+="<br>"+Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),C=H(Math.sqrt(r**2+d**2),1),o+="<br>"+G(e.label,t.label,U.nom,1,r,d,C,h[c])[0];break;case 8:f.anglePerspective=O(2,6)*W([10,-10]),d=O(4,10),q=W([1,2,4]),N=d*Math.cos(q/6*Math.PI/2),e=m(0,0,0,!0,v(1,"OQWX")[0],"left"),r=e.z+d*Math.sin(q/6*Math.PI/2),n=m(N,e.y,r,!0),n=se(n,te(e,S(0,0,1)),f.anglePerspective<0?30:-30),n.label=v(1,"OQWX"+e.label)[0],n.c2d.nom=n.label,n.c2d.positionLabel="below",t=m(0,e.y,r,!0,v(1,"OQWX"+e.label+n.label)[0]),g=n.label+e.label,b=w(e.c2d,n.c2d,"#f15929"),b.epaisseur=2,b.pointilles=2,$=w(t.c2d,n.c2d,"black"),$.pointilles=2,r=H(r,1),N=H(N,1),Y=pe(e,d,"green","blue",12,"lightgray",0,"black",!0),a=Y.c2d,a.push(b,$,ee(n,t,e),Z(e,t,n),new T(e,t,n,"black",1)),I+=`$${n.label}$ belongs to this sphere with center $${e.label}$. $${t.label}$ is a point inside the sphere, $${P(r)}$ ${h[c]} from $${e.label}$ and the distance between $${n.label}$ and $${t.label}$ is $${P(N)}$ ${h[c]}. Calculate the radius of the sphere, rounded to the tenth of ${h[c]}.<br>`,I+=Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),a=Y.c2d,$.epaisseur=2,$.color=$e("green"),a.push(b,$,ee(n,t,e),Z(e,t,n),new T(e,t,n,"green",1)),$=w(t.c2d,e.c2d,"green"),$.epaisseur=2,$.pointilles=2,a.push($),o+="<br>"+Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),C=H(Math.sqrt(r**2+N**2),1),o+="<br>"+G(t.label,e.label,n.label,1,r,N,C,h[c])[0];break;case 9:f.anglePerspective=O(2,6)*W([10,-10]),d=O(4,10),q=W([1,2,4]),N=d*Math.cos(q/6*Math.PI/2),e=m(0,0,0,!0,v(1,"OQWX")[0],"left"),r=e.z+d*Math.sin(q/6*Math.PI/2),n=m(N,e.y,r,!0),n=se(n,te(e,S(0,0,1)),f.anglePerspective<0?30:-30),n.label=v(1,"OQWX"+e.label)[0],n.c2d.nom=n.label,n.c2d.positionLabel="below",t=m(0,e.y,r,!0,v(1,"OQWX"+e.label+n.label)[0]),g=n.label+t.label,b=w(t.c2d,n.c2d,"#f15929"),b.epaisseur=2,b.pointilles=2,r=H(r,1),Y=pe(e,d,"green","blue",12,"lightgray",0,"black",!0),a=Y.c2d,a.push(b,ee(n,t,e),Z(e,t,n),new T(e,t,n,"black",1)),I+=`$${n.label}$ belongs to this sphere with center $${e.label}$ and radius $${P(d)}$ ${h[c]}. $${t.label}$ is a point inside the sphere, at $${P(r)}$ ${h[c]} of $${e.label}$. Calculate the distance between $${n.label}$ and $${t.label}$, rounded to the tenth of ${h[c]}.<br>`,I+=Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),a=Y.c2d,$=w(t.c2d,e.c2d,"green"),$.epaisseur=2,$.pointilles=2,a.push(b,$,ee(n,t,e),Z(e,t,n),new T(e,t,n,"green",1)),$=w(n.c2d,e.c2d,"green"),$.epaisseur=2,$.pointilles=2,a.push($),o+="<br>"+Q(Object.assign({},j(a),{scale:f.isHtml?.7:.3,style:"block"}),a),C=H(Math.sqrt(d**2-r**2),1),o+="<br>"+G(t.label,n.label,e.label,2,C,r,d,h[c])[0];break}ke(this,D,new Oe(C,h[c]),{formatInteractif:"units"}),this.interactif&&f.isHtml&&(I+=`<br>$${g}\\approx$`+we(this,D,"largeur25 inline units[lengths]")),f.isAmc&&(this.autoCorrection[D]={enonce:I,enonceAvant:this.sup2===2,options:{ordered:!1}},this.autoCorrection[D].propositions=[],this.sup2===1?this.autoCorrection[D].propositions.push({type:"AMCNum",propositions:[{reponse:{texte:I,valeur:[C],alignement:"center",param:{digits:Pe(C),decimals:1,signe:!1}}}]}):this.autoCorrection[D].propositions.push({type:"AMCOpen",propositions:[{texte:o,statut:4,enonce:I,sanscadre:!1,pointilles:!1}]})),this.questionJamaisPosee(D,l,d,r)&&(this.listeQuestions.push(I),this.listeCorrections.push(o),D++),le++}ve(this)},f.isAmc&&(this.besoinFormulaire2Numerique=["AMC exercise",2,`1: Open question
2: Numerical response`]),this.besoinFormulaireTexte=["Type of length to find",`Numbers separated by hyphens
1: Diagonal of a face of a cube
2: Diagonal of a cube
3: Diagonal of a face of a straight block
4: Diagonal of
 a straight block
5: In a cylinder
6: In a pyramid
7: In a cone
8: Radius of a sphere
9: Radius of a section of a sphere
10: Combination`]}export{Ae as amcReady,Le as amcType,Ke as dateDeModifImportante,Ne as dateDePublication,Je as default,Ee as interactifReady,Be as interactifType,Se as ref,We as titre,Re as uuid};
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