import{E as fe,ag as pe,F as ue,r as h,q as ge,W as I,h as te,B as Q,aF as Y,K as re,al as g,jK as Z,aN as _,G as W,jH as w,L as F,w as a,o as J,aA as y,ae as f,a as ie,s as se,c as le,bI as ne,bJ as ae,f as E,m as ee,jG as ce,l as de}from"./index-XCg2QAX4.js";import{a as me}from"./outilsMathjs-KwNosraq.js";import"./create-9DBTQ71r.js";const be="Enlarge or reduce figures, based on a situation of proportionality",Pe=!0,Ae="mathLive",Te=!0,De="AMCHybride",Me="13/03/2022",Ie="4c6e2",Ee="6P14";function Oe(){fe.call(this),this.titre=be,this.besoinFormulaireTexte=["Type of figures",["Numbers separated by hyphens","1: Equilateral triangle","2: Square","3: Triangle with reduction or enlargement coefficient","4: Triangle with initial length and final length","5: Rectangle with reduction or enlargement coefficient","6: Rectangle with initial length and final length","7: Combination"].join(`
ot`)],this.sup=7,this.nbQuestions=4,this.spacingCorr=1,this.spacing=2,this.nouvelleVersion=function(){this.listeQuestions=[],this.listeCorrections=[];const ye=pe({max:6,defaut:7,melange:7,nbQuestions:this.nbQuestions,saisie:this.sup}),D=[[" enlargement","e reduction"],["the requested expansion","the requested reduction"]];let N=0,C;for(let q=0,v,e,i,s,u,n,l,T,j,H,oe,m,k,t,P,M,U,z,b,x,R,K,$,p,O,d,L,$e,B,V,he,X,S,G,o,r,c,A,xe=0;q<this.nbQuestions&&xe<50;){switch(X=[],C=0,$=[],m=[1.5,2,3,5,.5,.25,.75],k=[new ue(1,2),new ue(1,4),new ue(3,4)],t=h(0,6),e=ge(0,0),x=t<5?h(5,11,[6,9]):2*h(4,7),ye[q]){case 1:r=I(m[t]*x,1),R=h(10,170),K=h(10,170,[R]),P=te([-1,1]),i=Q(e,x,P*R),n=Q(e,r,P*K),s=Q(e,x,P*(R+60)),l=Q(e,r,P*(K+60)),j=re(e,i,s),H=re(e,n,l),p=h(1,26,[4,5,15,23,24,25]),d=h(1,26,[4,5,15,23,24,25,p]),L=h(1,26,[4,5,15,23,24,25,p,d]),M=g(p)+g(d)+g(L),O=h(1,26,[4,5,15,23,24,25,p,d,L]),B=h(1,26,[4,5,15,23,24,25,p,d,O]),V=h(1,26,[4,5,15,23,24,25,p,d,O,B]),U=g(O)+g(B)+g(V),$.push(j,_("||","red",j.listePoints),W(P<0?e:i,P<0?i:e,"blue",.5,"",!0),Z(j,M)),G=`We decide to perform a ${D[0][t<4?0:1]} of coefficient $${a(m[t])}$ of the equilateral triangle ${M}. What will be the length of the side of the triangle to construct?`,v=G,S="<br>"+J({xmin:y(e.x,i.x,s.x)-1,ymin:y(e.y,i.y,s.y)-1,xmax:f(e.x,i.x,s.x)+1,ymax:f(e.y,i.y,s.y)+1,pixelsParCm:20,scale:.5},$),v+=S,this.interactif?(v+=ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,r)):le.isAmc?(X[C]={type:"AMCOpen",propositions:[{texte:"",statut:1,sanscadre:!0,enonce:S+`<br>Trace, on a white sheet, ${D[1][t<4?0:1]}.`,sanslignes:!0}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:G,valeur:[r],alignement:"center",param:{digits:ne(r),decimals:ae(r),signe:!1}}}]},C++):(v=`Plot a${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ of triangle ${M}.`,v+="<br>"+J({xmin:y(e.x,i.x,s.x)-1,ymin:y(e.y,i.y,s.y)-1,xmax:f(e.x,i.x,s.x)+1,ymax:f(e.y,i.y,s.y)+1,pixelsParCm:20,scale:.5},$)),o=`Performing a ${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ involves multiplying all lengths by this coefficient`,t>=4&&(o+=` or, since $${a(m[t])}=${k[t-4].texFraction}$, this amounts to dividing all lengths by $${k[t-4].den}$`,o+=t===6?" then multiply each of these results by 3":""),o+=".<br>",o+=`$${x} \\times ${m[t]}=${a(r)}$`,t===6?o+=` or $(${x} \\div 4) \\times 3=${a(I(x/4,1))} \\times 3=${a(r)}$`:t>=4&&(o+=` or $${x} \\div ${k[t-4].den}=${a(r)}$`),o+=`<br>The equilateral triangle resulting from a${D[0][t<4?0:1]} of the triangle ${M} with coefficient $${a(m[t])}$ therefore has sides of length $${ee(a(r))}$.`,o+="<br>Here is a creation below.",$=[],$.push(H,_("|||","blue",H.listePoints),W(P<0?e:n,P<0?n:e,"red",.5,"",!0),Z(H,U)),o+="<br>"+J({xmin:y(e.x,n.x,l.x)-1,ymin:y(e.y,n.y,l.y)-1,xmax:f(e.x,n.x,l.x)+1,ymax:f(e.y,n.y,l.y)+1,pixelsParCm:20,scale:.5},$);break;case 2:r=I(m[t]*x,1),R=h(10,170),K=h(10,170,[R]),P=te([-1,1]),i=Q(e,x,P*R),n=Q(e,r,P*K),s=Y(e,i,90),u=Y(i,e,-90),l=Y(e,n,90),T=Y(n,e,-90),j=re(e,i,s,u),H=re(e,n,l,T),p=h(1,26,[4,5,15,23,24,25]),d=h(1,26,[4,5,15,23,24,25,p]),L=h(1,26,[4,5,15,23,24,25,p,d]),$e=h(1,26,[4,5,15,23,24,25,p,d,L]),M=g(p)+g(d)+g(L)+g($e),O=h(1,26,[4,5,15,23,24,25,p,d,L]),B=h(1,26,[4,5,15,23,24,25,p,d,O]),V=h(1,26,[4,5,15,23,24,25,p,d,O,B]),he=h(1,26,[4,5,15,23,24,25,p,d,O,B,V]),U=g(O)+g(B)+g(V)+g(he),$.push(j,_("||","red",j.listePoints),W(e,i,"blue",.5,"",!0),Z(j,M)),$.push(F(e,i,s),F(u,s,i),F(e,u,s),F(i,e,u)),G=`We decide to perform a ${D[0][t<4?0:1]} of coefficient $${a(m[t])}$, of the square ${M}. What will be the length of the side of the square to be constructed?`,v=G,S="<br>"+J({xmin:y(e.x,i.x,s.x,u.x)-1,ymin:y(e.y,i.y,s.y,u.y)-1,xmax:f(e.x,i.x,s.x,u.x)+1,ymax:f(e.y,i.y,s.y,u.y)+1,pixelsParCm:20,scale:.5},$),v+=S,this.interactif?(v+=ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,r)):le.isAmc?(X[C]={type:"AMCOpen",propositions:[{texte:"",statut:1,sanscadre:!0,enonce:S+`<br>Trace, on a white sheet, ${D[1][t<4?0:1]}.`,sanslignes:!0}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:G,valeur:[r],alignement:"center",param:{digits:ne(r),decimals:ae(r),signe:!1}}}]},C++):(v=`Plot a${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ of the square ${M}.`,v+="<br>"+J({xmin:y(e.x,i.x,s.x,u.x)-1,ymin:y(e.y,i.y,s.y,u.y)-1,xmax:f(e.x,i.x,s.x,u.x)+1,ymax:f(e.y,i.y,s.y,u.y)+1,pixelsParCm:20,scale:.5},$)),o=`Performing a ${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ involves multiplying all lengths by this coefficient`,t>=4&&(o+=` or, since $${a(m[t])}=${k[t-4].texFraction}$, this amounts to dividing all lengths by $${k[t-4].den}$`,o+=t===6?" then multiply each of these results by 3":""),o+=".<br>",o+=`$${x} \\times ${m[t]}=${a(r)}$`,t===6?o+=` or $(${x} \\div 4) \\times 3=${a(I(x/4,1))} \\times 3=${a(r)}$`:t>=4&&(o+=` or $${x} \\div ${k[t-4].den}=${a(r)}$`),o+=`<br>The square resulting from a${D[0][t<4?0:1]} of the square ${M} with coefficient $${a(m[t])}$ therefore has sides of length $${ee(a(r))}$.`,o+="<br>Here is a creation below.",$=[],$.push(H,_("|||","blue",H.listePoints),W(e,n,"red",.5,"",!0),Z(H,U)),$.push(F(e,n,l),F(T,l,n),F(e,T,l),F(n,e,T)),o+="<br>"+J({xmin:y(e.x,n.x,l.x,T.x)-1,ymin:y(e.y,n.y,l.y,T.y)-1,xmax:f(e.x,n.x,l.x,T.x)+1,ymax:f(e.y,n.y,l.y,T.y)+1,pixelsParCm:20,scale:.5},$);break;case 3:b=t<4?h(5,11,[6,9,x]):2*h(4,7,[I(x/2,0)]),z=Object.values(me({absD:!0,test:`absD>0 and absD<${x+b} and ${x}<absD+${b} and ${b}<absD+${x}`},{valueOf:!0}))[0],r=I(m[t]*x,1),c=I(m[t]*b,1),A=I(m[t]*z,1),R=h(10,170),K=h(10,170,[R]),P=te([-1,1]),i=Q(e,x,P*R),n=Q(e,r,P*K),oe=h(1,2),j=ce(e,i,b,z,oe),s=j.listePoints[2],H=ce(e,n,c,A,oe),l=H.listePoints[2],p=h(1,26,[4,5,15,23,24,25]),d=h(1,26,[4,5,15,23,24,25,p]),L=h(1,26,[4,5,15,23,24,25,p,d]),M=g(p)+g(d)+g(L),O=h(1,26,[4,5,15,23,24,25,p,d,L]),B=h(1,26,[4,5,15,23,24,25,p,d,O]),V=h(1,26,[4,5,15,23,24,25,p,d,O,B]),U=g(O)+g(B)+g(V),$.push(j,Z(j,M)),$.push(W(w(s,e,i)>0?e:i,w(s,e,i)>0?i:e,"blue",.5,"",!0)),$.push(W(w(e,i,s)>0?i:s,w(e,i,s)>0?s:i,"blue",.5,"",!0)),$.push(W(w(i,s,e)>0?s:e,w(i,s,e)>0?e:s,"blue",.5,"",!0)),G=`We decide to perform a ${D[0][t<4?0:1]} of coefficient $${a(m[t])}$ of the triangle ${M}. What will be the respective lengths of each side of the triangle to be constructed?`,S="<br>"+J({xmin:y(e.x,i.x,s.x)-1,ymin:y(e.y,i.y,s.y)-1,xmax:f(e.x,i.x,s.x)+1,ymax:f(e.y,i.y,s.y)+1,pixelsParCm:20,scale:.5},$),this.interactif?(v=G,v+=S,v+="<br> In the new triangle, the smallest length will be:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,y(r,c,A)),N++,v+="<br> In the new triangle, the greatest length will be:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,f(r,c,A)),N++,v+="<br> In the new triangle, the last length will be:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,te([r,c,A],[y(r,c,A),f(r,c,A)]))):le.isAmc?(X[C]={type:"AMCOpen",propositions:[{texte:"",statut:1,sanscadre:!0,enonce:S+`<br>Trace, on a white sheet, ${D[1][t<4?0:1]}.`,sanslignes:!0}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:G+"<br> <br>In the new triangle, the smallest length will be:",valeur:[y(r,c,A)],alignement:"center",param:{digits:ne(y(r,c,A)),decimals:ae(y(r,c,A)),signe:!1}}}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:"In the new triangle, the greatest length will be:",valeur:[f(r,c,A)],alignement:"center",param:{digits:ne(f(r,c,A)),decimals:ae(f(r,c,A)),signe:!1}}}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:"In the new triangle, the last length will be:",valeur:[te([r,c,A],[y(r,c,A),f(r,c,A)])],alignement:"center",param:{digits:ne(te([r,c,A],[y(r,c,A),f(r,c,A)])),decimals:ae(te([r,c,A],[y(r,c,A),f(r,c,A)])),signe:!1}}}]},C++):(v=`Plot a${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ of triangle ${M}.`,v+="<br>"+J({xmin:y(e.x,i.x,s.x)-1,ymin:y(e.y,i.y,s.y)-1,xmax:f(e.x,i.x,s.x)+1,ymax:f(e.y,i.y,s.y)+1,pixelsParCm:20,scale:.5},$)),o=`Performing a ${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ involves multiplying all lengths by this coefficient`,t>=4&&(o+=` or, since $${a(m[t])}=${k[t-4].texFraction}$, this amounts to dividing all lengths by $${k[t-4].den}$`,o+=t===6?" then multiply each of these results by 3":""),o+=".<br>",o+=`$${x} \\times ${m[t]}=${a(r)}${E(10)}$`,o+=`$${b} \\times ${m[t]}=${a(c)}${E(10)}$`,o+=`$${z} \\times ${m[t]}=${a(A)}$`,t===6?(o+=`${E(10)} or ${E(10)}$(${x} \\div 4) \\times 3=${a(I(x/4,1))} \\times 3=${a(r)}$`,o+=`${E(10)}$(${b} \\div 4) \\times 3=${a(I(b/4,1))} \\times 3=${a(c)}$`,o+=`${E(10)}$(${z} \\div 4) \\times 3=${a(I(z/4,1))} \\times 3=${a(A)}$`):t>=4&&(o+=`${E(10)} or ${E(10)}$${x} \\div ${k[t-4].den}=${a(r)}$`,o+=`${E(10)}$${b} \\div ${k[t-4].den}=${a(c)}$`,o+=`${E(10)}$${z} \\div ${k[t-4].den}=${a(A)}$`),o+=`<br>The triangle resulting from a${D[0][t<4?0:1]} of the triangle ${M} with coefficient $${a(m[t])}$ therefore has sides of respective length $${ee(a(r))}$; $${ee(a(c))}$ and $${ee(a(A))}$.`,o+="<br>Here is a creation below.",$=[],$.push(H,Z(H,U)),$.push(W(w(l,e,n)>0?e:n,w(l,e,n)>0?n:e,"red",.5,"",!0)),$.push(W(w(e,n,l)>0?n:l,w(e,n,l)>0?l:n,"red",.5,"",!0)),$.push(W(w(n,l,e)>0?l:e,w(n,l,e)>0?e:l,"red",.5,"",!0)),o+="<br>"+J({xmin:y(e.x,n.x,l.x)-1,ymin:y(e.y,n.y,l.y)-1,xmax:f(e.x,n.x,l.x)+1,ymax:f(e.y,n.y,l.y)+1,pixelsParCm:20,scale:.5},$);break;case 4:b=t<4?h(5,11,[6,9,x]):2*h(4,7,[I(x/2,0)]),z=Object.values(me({absD:!0,test:`absD>0 and absD<${x+b} and ${x}<absD+${b} and ${b}<absD+${x}`},{valueOf:!0}))[0],r=I(m[t]*x,1),c=I(m[t]*b,1),A=I(m[t]*z,1),R=h(10,170),K=h(10,170,[R]),P=te([-1,1]),i=Q(e,x,P*R),n=Q(e,r,P*K),oe=h(1,2),j=ce(e,i,b,z,oe),s=j.listePoints[2],H=ce(e,n,c,A,oe),l=H.listePoints[2],p=h(1,26,[4,5,15,23,24,25]),d=h(1,26,[4,5,15,23,24,25,p]),L=h(1,26,[4,5,15,23,24,25,p,d]),M=g(p)+g(d)+g(L),O=h(1,26,[4,5,15,23,24,25,p,d,L]),B=h(1,26,[4,5,15,23,24,25,p,d,O]),V=h(1,26,[4,5,15,23,24,25,p,d,O,B]),U=g(O)+g(B)+g(V),$.push(j,Z(j,M)),$.push(W(w(s,e,i)>0?e:i,w(s,e,i)>0?i:e,"blue",.5,"",!0)),$.push(W(w(e,i,s)>0?i:s,w(e,i,s)>0?s:i,"blue",.5,"",!0)),$.push(W(w(i,s,e)>0?s:e,w(i,s,e)>0?e:s,"blue",.5,"",!0)),G=`We decide to perform a${D[0][t<4?0:1]} of the triangle ${M}, such that the length of the side associated with [${g(d)+g(L)}] will measure $${a(A)}$.<br>What will be the respective lengths of the two other sides of the triangle to construct?`,S="<br>"+J({xmin:y(e.x,i.x,s.x)-1,ymin:y(e.y,i.y,s.y)-1,xmax:f(e.x,i.x,s.x)+1,ymax:f(e.y,i.y,s.y)+1,pixelsParCm:20,scale:.5},$),this.interactif?(v=G,v+=S,v+="<br> In the new triangle, the smallest length to find will be:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,y(r,c)),N++,v+="<br> In the new triangle, the greatest length to find will be:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,f(r,c))):le.isAmc?(X[C]={type:"AMCOpen",propositions:[{texte:"",statut:1,sanscadre:!0,enonce:S+`<br>Trace, on a white sheet, ${D[1][t<4?0:1]}.`,sanslignes:!0}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:G+"<br> <br>In the new triangle, the smallest length to find will be:",valeur:[y(r,c)],alignement:"center",param:{digits:ne(y(r,c)),decimals:ae(y(r,c)),signe:!1}}}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:"In the new triangle, the greatest length to find will be:",valeur:[f(r,c)],alignement:"center",param:{digits:ne(f(r,c)),decimals:ae(f(r,c)),signe:!1}}}]},C++):(v=`Draw a ${D[0][t<4?0:1]} of the triangle ${M} such that the length of the side associated with [${g(d)+g(L)}] will measure $${a(A)}$.`,v+="<br>"+J({xmin:y(e.x,i.x,s.x)-1,ymin:y(e.y,i.y,s.y)-1,xmax:f(e.x,i.x,s.x)+1,ymax:f(e.y,i.y,s.y)+1,pixelsParCm:20,scale:.5},$)),o=`Performing a ${D[0][t<4?0:1]} involves multiplying all lengths by a proportionality coefficient. Let's find this coefficient.<br>`,o+=`To find this coefficient, divide the known length of the future triangle by its associated length in the current triangle: $${a(A)} \\div ${z} = ${m[t]}$. The proportionality coefficient is therefore $${m[t]}$.<br>`,o+=`Let's multiply all known lengths of the current triangle by $${m[t]}$`,t>=4&&(o+=`, or, since $${a(m[t])}=${k[t-4].texFraction}$, this amounts to dividing all lengths by $${k[t-4].den}$`,o+=t===6?" then multiply each of these results by 3":""),o+=`.<br>$${x} \\times ${m[t]}=${a(r)}${E(10)}$`,o+=`$${b} \\times ${m[t]}=${a(c)}$`,t===6?(o+=`${E(10)} or ${E(10)}$(${x} \\div 4) \\times 3=${a(I(x/4,1))} \\times 3=${a(r)}$`,o+=`${E(10)}$(${b} \\div 4) \\times 3=${a(I(b/4,1))} \\times 3=${a(c)}$`):t>=4&&(o+=`${E(10)} or ${E(10)}$${x} \\div ${k[t-4].den}=${a(r)}$`,o+=`${E(10)}$${b} \\div ${k[t-4].den}=${a(c)}$`),o+=`<br>The triangle resulting from a${D[0][t<4?0:1]} of the triangle ${M} with coefficient $${a(m[t])}$ therefore has sides of respective length $${a(A)}$; $${ee(a(c))}$ and $${ee(a(r))}$.`,o+="<br>Here is a creation below.",$=[],$.push(H,Z(H,U)),$.push(W(w(l,e,n)>0?e:n,w(l,e,n)>0?n:e,"red",.5,"",!0)),$.push(W(w(e,n,l)>0?n:l,w(e,n,l)>0?l:n,"red",.5,"",!0)),$.push(W(w(n,l,e)>0?l:e,w(n,l,e)>0?e:l,"red",.5,"",!0)),o+="<br>"+J({xmin:y(e.x,n.x,l.x)-1,ymin:y(e.y,n.y,l.y)-1,xmax:f(e.x,n.x,l.x)+1,ymax:f(e.y,n.y,l.y)+1,pixelsParCm:20,scale:.5},$);break;case 5:b=t<4?h(5,11,[6,9,x]):2*h(4,7,[I(x/2,0)]),z=t<4?h(5,11,[6,9,x,b]):2*h(4,7,[I(x/2,0),I(b/2,0)]),r=I(m[t]*x,1),c=I(m[t]*b,1),R=h(10,170),K=h(10,170,[R]),P=te([-1,1]),i=Q(e,x,P*R),n=Q(e,r,P*K),s=Y(Q(i,b,180+P*R),i,-90),l=Y(Q(n,c,180+P*K),n,-90),u=Y(Q(e,b,P*R),e,90),T=Y(Q(e,c,P*K),e,90),j=re(e,i,s,u),H=re(e,n,l,T),p=h(1,26,[4,5,15,23,24,25]),d=h(1,26,[4,5,15,23,24,25,p]),L=h(1,26,[4,5,15,23,24,25,p,d]),$e=h(1,26,[4,5,15,23,24,25,p,d,L]),M=g(p)+g(d)+g(L)+g($e),O=h(1,26,[4,5,15,23,24,25,p,d,L]),B=h(1,26,[4,5,15,23,24,25,p,d,O]),V=h(1,26,[4,5,15,23,24,25,p,d,O,B]),he=h(1,26,[4,5,15,23,24,25,p,d,O,B,V]),U=g(O)+g(B)+g(V)+g(he),$.push(j,Z(j,M)),$.push(_("||","red",e,i,s,u)),$.push(_("X","red",i,s,u,e)),$.push(W(w(i,s,u)>0?s:u,w(i,s,u)>0?u:s,"blue",.5,"",!0)),$.push(W(w(s,u,e)>0?u:e,w(s,u,e)>0?e:u,"blue",.5,"",!0)),$.push(F(e,i,s),F(u,s,i),F(e,u,s),F(i,e,u)),G=`We decide to perform a ${D[0][t<4?0:1]} of coefficient $${a(m[t])}$ of the rectangle ${M}. What will be the respective lengths of each side of the rectangle to be constructed?`,S="<br>"+J({xmin:y(e.x,i.x,s.x,u.x)-1,ymin:y(e.y,i.y,s.y,u.y)-1,xmax:f(e.x,i.x,s.x,u.x)+1,ymax:f(e.y,i.y,s.y,u.y)+1,pixelsParCm:20,scale:.5},$),this.interactif?(v=G,v+=S,v+="<br> In the new rectangle, the shorter side will have the length:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,y(r,c)),N++,v+="<br> In the new rectangle, the longest side will have the length:"+ie(this,q+N,"inline",{tailleExtensible:!0}),se(this,q+N,f(r,c))):le.isAmc?(X[C]={type:"AMCOpen",propositions:[{texte:"",statut:1,sanscadre:!0,enonce:S+`<br>Trace, on a white sheet, ${D[1][t<4?0:1]}.`,sanslignes:!0}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:G+"<br> <br>In the new rectangle, the shortest side will have the length:",valeur:[y(r,c)],alignement:"center",param:{digits:ne(y(r,c)),decimals:ae(y(r,c)),signe:!1}}}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:"In the new rectangle, the longest side will have the length:",valeur:[f(r,c)],alignement:"center",param:{digits:ne(f(r,c)),decimals:ae(f(r,c)),signe:!1}}}]},C++):(v=`Plot a${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ of the rectangle ${M}.`,v+="<br>"+J({xmin:y(e.x,i.x,s.x,u.x)-1,ymin:y(e.y,i.y,s.y,u.y)-1,xmax:f(e.x,i.x,s.x,u.x)+1,ymax:f(e.y,i.y,s.y,u.y)+1,pixelsParCm:20,scale:.5},$)),o=`Performing a ${D[0][t<4?0:1]} with coefficient $${a(m[t])}$ involves multiplying all lengths by this coefficient`,t>=4&&(o+=` or, since $${a(m[t])}=${k[t-4].texFraction}$, this amounts to dividing all lengths by $${k[t-4].den}$`,o+=t===6?" then multiply each of these results by 3":""),o+=".<br>",o+=`$${x} \\times ${m[t]}=${a(r)}${E(10)}$`,o+=`$${b} \\times ${m[t]}=${a(c)}$`,t===6?(o+=`${E(10)} or ${E(10)}$(${x} \\div 4) \\times 3=${a(I(x/4,1))} \\times 3=${a(r)}$`,o+=`${E(10)}$(${b} \\div 4) \\times 3=${a(I(b/4,1))} \\times 3=${a(c)}$`):t>=4&&(o+=`${E(10)} or ${E(10)}$${x} \\div ${k[t-4].den}=${a(r)}$`,o+=`${E(10)}$${b} \\div ${k[t-4].den}=${a(c)}$`),o+=`<br>The rectangle resulting from a${D[0][t<4?0:1]} of the rectangle ${M} with coefficient $${a(m[t])}$ therefore has sides of respective length $${ee(a(r))}$ and $${ee(a(c))}$.`,o+="<br>Here is a creation below.",$=[],$.push(H,Z(H,U)),$.push(_("|||","blue",e,n,l,T)),$.push(_("XX","blue",n,l,T,e)),$.push(W(w(l,e,n)>0?e:n,w(l,e,n)>0?n:e,"red",.5,"",!0)),$.push(W(w(e,n,l)>0?n:l,w(e,n,l)>0?l:n,"red",.5,"",!0)),$.push(F(e,n,l),F(T,l,n),F(e,T,l),F(n,e,T)),o+="<br>"+J({xmin:y(e.x,n.x,l.x,T.x)-1,ymin:y(e.y,n.y,l.y,T.y)-1,xmax:f(e.x,n.x,l.x,T.x)+1,ymax:f(e.y,n.y,l.y,T.y)+1,pixelsParCm:20,scale:.5},$);break;case 6:b=t<4?h(5,11,[6,9,x]):2*h(4,7,[I(x/2,0)]),z=t<4?h(5,11,[6,9,x,b]):2*h(4,7,[I(x/2,0),I(b/2,0)]),r=I(m[t]*x,1),c=I(m[t]*b,1),R=h(10,170),K=h(10,170,[R]),P=te([-1,1]),i=Q(e,x,P*R),n=Q(e,r,P*K),s=Y(Q(i,b,180+P*R),i,-90),l=Y(Q(n,c,180+P*K),n,-90),u=Y(Q(e,b,P*R),e,90),T=Y(Q(e,c,P*K),e,90),j=re(e,i,s,u),H=re(e,n,l,T),p=h(1,26,[4,5,15,23,24,25]),d=h(1,26,[4,5,15,23,24,25,p]),L=h(1,26,[4,5,15,23,24,25,p,d]),$e=h(1,26,[4,5,15,23,24,25,p,d,L]),M=g(p)+g(d)+g(L)+g($e),O=h(1,26,[4,5,15,23,24,25,p,d,L]),B=h(1,26,[4,5,15,23,24,25,p,d,O]),V=h(1,26,[4,5,15,23,24,25,p,d,O,B]),he=h(1,26,[4,5,15,23,24,25,p,d,O,B,V]),U=g(O)+g(B)+g(V)+g(he),$.push(j,Z(j,M)),$.push(_("||","red",e,i,s,u)),$.push(_("X","red",i,s,u,e)),$.push(W(w(i,s,u)>0?s:u,w(i,s,u)>0?u:s,"blue",.5,"",!0)),$.push(W(w(s,u,e)>0?u:e,w(s,u,e)>0?e:u,"blue",.5,"",!0)),$.push(F(e,i,s),F(u,s,i),F(e,u,s),F(i,e,u)),G=`We decide to perform a ${D[0][t<4?0:1]} of the rectangle ${M}, such that the length of the side associated with [${g(p)+g(d)}] will measure $${a(r)}$. What will be the other dimension of the rectangle to construct?`,S="<br>"+J({xmin:y(e.x,i.x,s.x,u.x)-1,ymin:y(e.y,i.y,s.y,u.y)-1,xmax:f(e.x,i.x,s.x,u.x)+1,ymax:f(e.y,i.y,s.y,u.y)+1,pixelsParCm:20,scale:.5},$),this.interactif?(v=G,v+=ie(this,q+N,"inline",{tailleExtensible:!0}),v+=S,se(this,q+N,c)):le.isAmc?(X[C]={type:"AMCOpen",propositions:[{texte:"",statut:1,sanscadre:!0,enonce:S+`<br>Trace, on a white sheet, ${D[1][t<4?0:1]}.`,sanslignes:!0}]},C++,X[C]={type:"AMCNum",propositions:[{texte:"",statut:"",reponse:{texte:G+"<br> <br>In the new rectangle, the second length will be:",valeur:[c],alignement:"center",param:{digits:ne(c),decimals:ae(c),signe:!1}}}]},C++):(v=`Draw a${D[0][t<4?0:1]} of the rectangle ${M} such that the length of the side associated with [${g(p)+g(d)}] will measure $${a(r)}$.`,v+="<br>"+J({xmin:y(e.x,i.x,s.x,u.x)-1,ymin:y(e.y,i.y,s.y,u.y)-1,xmax:f(e.x,i.x,s.x,u.x)+1,ymax:f(e.y,i.y,s.y,u.y)+1,pixelsParCm:20,scale:.5},$)),o=`Performing a ${D[0][t<4?0:1]} involves multiplying all lengths by a proportionality coefficient. Let's find this coefficient.<br>`,o+=`To find this coefficient, divide the known length of the future rectangle by its associated length in the current rectangle: $${a(r)} \\div ${x} = ${m[t]}$. The proportionality coefficient is therefore $${m[t]}$.<br>`,o+=`Let's multiply all known lengths of the current triangle by $${m[t]}$`,t>=4&&(o+=` or, since $${a(m[t])}=${k[t-4].texFraction}$, this amounts to dividing all lengths by $${k[t-4].den}$`,o+=t===6?" then multiply each of these results by 3":""),o+=".<br>",o+=`$${b} \\times ${m[t]}=${a(c)}$`,t===6?o+=`${E(10)} or ${E(10)}$(${b} \\div 4) \\times 3=${a(I(b/4,1))} \\times 3=${a(c)}$`:t>=4&&(o+=`${E(10)} or ${E(10)}$${b} \\div ${k[t-4].den}=${a(c)}$`),o+=`<br>The rectangle resulting from a${D[0][t<4?0:1]} of the rectangle ${M} with coefficient $${a(m[t])}$ therefore has sides of respective length $${ee(a(r))}$ and $${ee(a(c))}$.`,o+="<br>Here is a creation below.",$=[],$.push(H,Z(H,U)),$.push(_("|||","blue",e,n,l,T)),$.push(_("XX","blue",n,l,T,e)),$.push(W(w(l,e,n)>0?e:n,w(l,e,n)>0?n:e,"red",.5,"",!0)),$.push(W(w(e,n,l)>0?n:l,w(e,n,l)>0?l:n,"red",.5,"",!0)),$.push(F(e,n,l),F(T,l,n),F(e,T,l),F(n,e,T)),o+="<br>"+J({xmin:y(e.x,n.x,l.x,T.x)-1,ymin:y(e.y,n.y,l.y,T.y)-1,xmax:f(e.x,n.x,l.x,T.x)+1,ymax:f(e.y,n.y,l.y,T.y)+1,pixelsParCm:20,scale:.5},$);break}le.isAmc&&(this.autoCorrection[q]={enonce:"",options:{multicols:!0,barreseparation:!1},enonceCentre:!0,enonceAvant:!0,propositions:X}),this.questionJamaisPosee(q,v,x,t)&&(this.listeQuestions.push(v),this.listeCorrections.push(o),q++),xe++}de(this)}}export{Te as amcReady,De as amcType,Me as dateDePublication,Oe as default,Pe as interactifReady,Ae as interactifType,Ee as ref,be as titre,Ie as uuid};
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