DERIVE for Windows version 5.00 DfW file saved on 19 Dec 2000 Macrolugarcentros(a, b):=[[-1, 0; 1, 0; a, b; -1, 0], abx - y(a^2 - b + -1) + -b = 0] Macrovertices(a, b, t):=[t, 0; (a^2t + ab(t + 1) + b^2 - b(t + 1) + -t)/(a^2 + b^2 + -1), b(a(t - 1) + b(t + 1) + t + -1)/(a^2 + b^2 + -1); (a^2t + ab(1 - t) - b^2 + b(1 - t) + -t)/(a^2 + b^2 + -1), b(a(t + 1) + b(1 - t) - t + -1)/(a^2 + b^2 + -1); t, 0] Verticesbydycentroparat(a, b, t):=[(a^2t + ab(t + 1) + b^2 - b(t + 1) + -t)/(a^2 + b^2 + -1), b(a(t - 1) + b(t + 1) + t + -1)/(a^2 + b^2 + -1); (a^2t + ab(1 - t) - b^2 + b(1 - t) + -t)/(a^2 + b^2 + -1), b(a(t + 1) + b(1 - t) - t + -1)/(a^2 + b^2 + -1); (a^2t + ab + - t(b + 1))/(a^2 + b^2 + -1), b(at + b + -1)/(a^2 + b^2 + -1)] hCross:=APPROX(- 17/2) vCross:=APPROX(34318181818181821/10000000000000000) 6CTextObj 6{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Un lugar curioso. \par **************** \par En un tri\'e1ngulo hay muchos cuadrados que tienen tres de sus v\'e9rtices sobre cada uno de los lados del tri\'e1ngulo. Se trata de mirar c\'f3mo es el lugar de los centros. \par ***************** \par Para hacer las cuentas m\'e1s f\'e1ciles, tomamos el tri\'e1ngulo de v\'e9rtices \par } CExpnObj8User[[-1,0],[1,0],[a,b],[-1,0]] CDispOleObj# CDispItem"ࡱ>   !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryF಼۞iOle CONTENTSCompObj{ F5Imagen (Mapa de bits independiente del dispositivo ) StaticDib9q     BM(HCONTENTS o{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Tomamos un punto A(t,0) sobre la base y por el una recta y=m(x-t), que corta al lado (1,0)--(a,b) en B, cuyas coordenadas calculamos. \par } 8$User y=m*(x-t)80TUsery=b+b/(a-1)*(x-a)`r{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Hallamos la interseccion B \par } 8~Solve(User,[x,y])*SOLVE([y=m*(x-t),y=b+b/(a-1)*(x-a)],[x,y])HSimp(#4){Gz?5[x=(a*m*t-b-m*t)/(a*m-b-m) AND y=b*m*(t-1)/(a*m-b-m)]{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Asi el punto B es \par } 8H User-[(a*m*t-b-m*t)/(a*m-b-m),b*m*(t-1)/(a*m-b-m)],ࡱ>   !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryF(iOle CONTENTSCompObj{ F5Imagen (Mapa de bits independiente del dispositivo ) StaticDib9q      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~BM(CONTENTS"X{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Si el punto B es (p,q) y lo giramos 90 grados con centro en (t,0) obtenemos el punto (r,s) de coordenadas r=t-q, s=p-t. As\'ed obtenemos el vertice D del cuadrado de lado AB. As\'ed en nuestro caso, podemos calcular D, que es \par } 8dUser1[t-b*m*(t-1)/(a*m-b-m),(a*m*t-b-m*t)/(a*m-b-m)-t]{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Calculamos el lado (a,b)--(-1,0), que es \par } 8Usery=b/(a+1)*(x+1) {\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 y expresamos que D pertenece a el \par } 8$Sub(#8) =(a*m*t-b-m*t)/(a*m-b-m)-t=b/(a+1)*((t-b*m*(t-1)/(a*m-b-m))+1)0B{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 De aqu\'ed resulta el m adecuado para el t escogido \par } 8NPr Solve(#9,m) FSOLVE((a*m*t-b-m*t)/(a*m-b-m)-t=b/(a+1)*((t-b*m*(t-1)/(a*m-b-m))+1),m)~(Simp(Solve(#9,m))q= ףp? -m=(a*(t-1)+b*(t+1)+t-1)/(a*(t+1)+b*(1-t)-t-1)1{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 que substituimos en las coordenadas de D para obtenerlas en funcion de t \par } X Fctr(User)L7A`? Q[(a^2*t+a*b*(1-t)-b^2+b*(1-t)-t)/(a^2+b^2-1),b*(a*(t+1)+b*(1-t)-t-1)/(a^2+b^2-1)]&{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 y en las coordenadas de B \par } X2n Simp(User))\(? Q[(a^2*t+a*b*(t+1)+b^2-b*(t+1)-t)/(a^2+b^2-1),b*(a*(t-1)+b*(t+1)+t-1)/(a^2+b^2-1)]z({\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 El centro que buscamos tiene por coordenadas 1/2(B+D), es decir \par } P Simp(User){Gz?9[(a^2*t+a*b-t*(b+1))/(a^2+b^2-1),b*(a*t+b-1)/(a^2+b^2-1)]7{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 lo que pone en claro que se trata de una recta cuya ecuaci\'f3n podemos hallar \par } 8:User!x=(a^2*t+a*b-t*(b+1))/(a^2+b^2-1)0FSimp(Solve(#15,t)){Gz?t=(x*(a^2+b^2-1)-a*b)/(a^2-b-1)8Userb*(a*t+b-1)/(a^2+b^2-1)=y Simp(Solve(#17,t)){Gz?t=(y*(a^2+b^2-1)-b*(b-1))/(a*b)8(Usera*b*x-y*(a^2-b-1)-b=0*< {\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Podemos representarlo directamente \par } 8HhUserKMacrolugarcentros(a,b):=[[[-1,0],[1,0],[a,b],[-1,0]],a*b*x-y*(a^2-b-1)-b=0]{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Experimentamos \par } 8UserMacrolugarcentros(0.5,3)(2  Approx(#21)~jtx?0[[[-1,0],[1,0],[0.5,3],[-1,0]],1.5*x+3.75*y-3=0]>  ࡱ> \  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryF[a/iOle CONTENTSΪCompObj{ F5Imagen (Mapa de bits independiente del dispositivo ) StaticDib9q      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[BMΪ(CONTENTS . &{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Tambi\'e9n podemos representar A, B y D para un t determinado \par } 8:  UserMacrovertices(a,b,t):=[[t,0],[(a^2*t+a*b*(t+1)+b^2-b*(t+1)-t)/(a^2+b^2-1),b*(a*(t-1)+b*(t+1)+t-1)/(a^2+b^2-1)],[(a^2*t+a*b*(1-t)-b^2+b*(1-t)-t)/(a^2+b^2-1),b*(a*(t+1)+b*(1-t)-t-1)/(a^2+b^2-1)],[t,0]]  {\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}} {\colortbl\red0\green0\blue0;} \deflang1034\pard\plain\f3\fs24 Experimentamos \par } 8  UserMacrovertices(0.5,3,0.3)@ D  Approx(#24){Gzt?9[[0.3,0],[0.827272,1.03636],[-0.736363,0.527272],[0.3,0]]P ( ࡱ> d  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryFgiOle CONTENTSCompObj{ F5Imagen (Mapa de bits independiente del dispositivo ) StaticDib9q      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcBM(CONTENTS84 @ UserMacrovertices(0.5,3,0.7)8L   Approx(#26)9[[0.7,0],[0.718181,1.6909],[-0.990909,0.0181818],[0.7,0]]8  UserMacrovertices(0.5,3,0.1)@  Approx(#28)8[[0.1,0],[0.881818,0.70909],[-0.60909,0.781818],[0.1,0]] $ ࡱ> @  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryFiOle CONTENTSsCompObj{ F5Imagen (Mapa de bits independiente del dispositivo ) StaticDib9q      !"#$%&'()*+,-./0123456789:;<=>?BMs(rCONTENTS8 UserMacrovertices(0.5,3,2)@"v Approx(#30)4[[2,0],[0.363636,3.81818],[-1.81818,-1.63636],[2,0]]8User Macrovertices(0.5,3,-2)@ Approx(#32);On?!5[[-2,0],[1.45454,-2.72727],[0.727272,3.45454],[-2,0]]".5;