DERIVE for Windows version 5.00 DfW file saved on 05 Jan 2001
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   Segmento(©, r, t):=[-1 + y^2 + x^2 = 0, - r^2 + y^2 + (x - ©)^2 = 0, [© + r·(1 - t^2)/(1 + t^2), r·2·t/(1 + t^2); - r·(1 - t^4 + r^2·(t^4 - 1) - 2·©·r·(t^4 + 1) + ©^2·(t^4 - 1) + 2·t·‹(t^2 + 1)·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))·(-1 - 2·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4) + 2·©·r·(1 - t^4) + ©^2·(1 + 2·t^2 + t^4))/((t^2 + 1)·(- r^2·(t^2 + 1)·(-1 + 6·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4)) + 2·©·r·(t^2 - 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(t^4 + 1)) - ©^2·(t^2 + 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(3 - 4·t^2 + 3·t^4)) + 4·©^3·r·(t^2 + 1)·(t^4 - 1) - ©^4·(t^2 + 1)^3 + 4·r·t·‹(t^2 + 1)·(©·(t^2 + 1) + r·(1 - t^2))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))), - (x·((t^2 + 1)^(3/2)·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)) - 2·r·t·(©·(t^2 + 1) + r·(1 - t^2))) + r·((t^2 - 1)·‹(t^2 + 1)·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)) - 2·t·(-1 - t^2 + ©·r·(1 - t^2) + ©^2·(t^2 + 1))))/(-1 - 2·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4) + 2·©·r·(1 - t^4) + ©^2·(1 + 2·t^2 + t^4))]]
m(©, r, t):=(2·r·t·(©·(t^2 + 1) + r·(1 - t^2)) - (t^2 + 1)^(3/2)·‹(©^2·(t^2 + 1) + 2·©·r·(1 - t^2) + (r^2 - 1)·(t^2 + 1)))/(©^2·(t^4 + 2·t^2 + 1) + 2·©·r·(1 - t^4) + r^2·(t^4 - 2·t^2 + 1) - t^4 - 2·t^2 + -1)
s(©, r, t):=2·(2·r·t·‹(t^2 + 1)·‹(©^2·(t^2 + 1) + 2·©·r·(1 - t^2) + (r^2 - 1)·(t^2 + 1))·(©^2·(t^2 + 1)^2 + 2·©·r·(t^2 + 1)·(1 - t^2) + r^2·(t^2 - 1)^2 + t^4 + 2·t^2 + 1) - ©^5·(t^4 + 2·t^2 + 1)^2 + 4·©^4·r·(t^4 - 1)·(t^4 + 2·t^2 + 1) - ©^3·(t^4 + 2·t^2 + 1)·(6·r^2·(t^4 - 2·t^2 + 1) - t^4 - 2·t^2 + -1) + ©^2·r·(t^2 + 1)·(t^2 - 1)·(4·r^2·(t^2 - 1)^2 - t^4 - 2·t^2 + -1) - ©·r^2·(r^2·(t^4 - 2·t^2 + 1)^2 + t^8 + 8·t^6 + 14·t^4 + 8·t^2 + 1) + r·(t^2 + 1)·(t^2 - 1)·(r^2·(t^4 + 6·t^2 + 1) - (t^2 + 1)^2))/((t^2 + 1)·(4·r·t·‹(t^2 + 1)·(©·(t^2 + 1) + r·(1 - t^2))·‹(©^2·(t^2 + 1) + 2·©·r·(1 - t^2) + (r^2 - 1)·(t^2 + 1)) - ©^4·(t^2 + 1)^3 + 4·©^3·r·(t^2 + 1)·(t^4 - 1) - ©^2·(t^2 + 1)·(2·r^2·(3·t^4 - 4·t^2 + 3) - t^4 - 2·t^2 + -1) + 2·©·r·(t^2 - 1)·(2·r^2·(t^4 + 1) - t^4 - 2·t^2 + -1) + - r^2·(t^2 + 1)·(r^2·(t^4 - 2·t^2 + 1) - t^4 + 6·t^2 + -1))) - © + r·(1 - 2/(t^2 + 1))
segmento(©, r, t):=[-1 + y^2 + x^2 = 0, - r^2 + y^2 + (x - ©)^2 = 0, [© + r·(1 - t^2)/(1 + t^2), r·2·t/(1 + t^2)], [- r·(1 - t^4 + r^2·(t^4 - 1) - 2·©·r·(t^4 + 1) + ©^2·(t^4 - 1) + 2·t·‹(t^2 + 1)·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))·(-1 - 2·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4) + 2·©·r·(1 - t^4) + ©^2·(1 + 2·t^2 + t^4))/((t^2 + 1)·(- r^2·(t^2 + 1)·(-1 + 6·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4)) + 2·©·r·(t^2 - 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(t^4 + 1)) - ©^2·(t^2 + 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(3 - 4·t^2 + 3·t^4)) + 4·©^3·r·(t^2 + 1)·(t^4 - 1) - ©^4·(t^2 + 1)^3 + 4·r·t·‹(t^2 + 1)·(©·(t^2 + 1) + r·(1 - t^2))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))), r·(2·t·(t^2 + 1)·((t^2 + 1)^2·(1 + 2·t^2 + t^4) - 4·r^2·(t^2 + 1)^2·(1 - 2·t^2 + t^4) + 2·r^4·(1 - 2·t^2 + t^4)·(1 + 4·t^2 + t^4) + r^6·(t^2 - 1)^2·(1 - 2·t^2 + t^4)) + 2·©·r·t·(1 - t^2)·(- 7·(t^2 + 1)^2·(1 + 2·t^2 + t^4) + 3·r^2·(t^2 + 1)^2·(3 - 2·t^2 + 3·t^4) + 4·r^4·(1 - t^2 - t^6 + t^8)) + 2·©^2·t·(t^2 + 1)·(- 3·(t^2 + 1)^4 + 15·r^2·(t^2 + 1)^2·(1 - 2·t^2 + t^4) + r^4·(5 - 8·t^2 + 6·t^4 - 8·t^6 + 5·t^8)) + 2·©^3·r·t·(1 - t^2)·(1 + 2·t^2 + t^4)·(4·r^2·t^2 + 11·(t^2 + 1)^2) + 2·©^4·t·(t^2 + 1)^3·(3·(t^2 + 1)^2 - 5·r^2·(1 - 2·t^2 + t^4)) + 8·©^5·r·t·(t^2 - 1)·(t^2 + 1)^4 - 2·©^6·t·(t^2 + 1)^5 + (t^2 - 1)·‹(t^2 + 1)·((t^2 + 1)^2·(1 + 2·t^2 + t^4) - r^2·(t^2 + 1)^2·(3 - 10·t^2 + 3·t^4) + 2·r^4·(1 - 2·t^2 + t^4)·(1 + 4·t^2 + t^4) + 2·©·r·(1 - t^4)·(4·r^2·(1 + t^2 + t^4) - 3·(t^2 + 1)^2) + 3·©^2·(1 + 2·t^2 + t^4)·(4·r^2·(1 - t^2 + t^4) - (t^2 + 1)^2) - 8·©^3·r·(t^2 + 1)·(-1 - t^2 + t^4 + t^6) + 2·©^4·(t^2 + 1)·(1 + 3·t^2 + 3·t^4 + t^6))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))/((t^2 + 1)·(- r^2·(t^2 + 1)·(-1 + 6·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4)) + 2·©·r·(t^2 - 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(t^4 + 1)) - ©^2·(t^2 + 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(3 - 4·t^2 + 3·t^4)) + 4·©^3·r·(t^2 + 1)·(t^4 - 1) - ©^4·(t^2 + 1)^3 + 4·r·t·‹(t^2 + 1)·(©·(t^2 + 1) + r·(1 - t^2))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))·(-1 - t^2 + r·(1 - t^2) + ©·(t^2 + 1))·(1 + t^2 + r·(1 - t^2) + ©·(t^2 + 1)))], [- r·(1 - t^4 + r^2·(t^4 - 1) - 2·©·r·(t^4 + 1) + ©^2·(t^4 - 1) + 2·t·‹(t^2 + 1)·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))·(-1 - 2·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4) + 2·©·r·(1 - t^4) + ©^2·(1 + 2·t^2 + t^4))/((t^2 + 1)·(- r^2·(t^2 + 1)·(-1 + 6·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4)) + 2·©·r·(t^2 - 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(t^4 + 1)) - ©^2·(t^2 + 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(3 - 4·t^2 + 3·t^4)) + 4·©^3·r·(t^2 + 1)·(t^4 - 1) - ©^4·(t^2 + 1)^3 + 4·r·t·‹(t^2 + 1)·(©·(t^2 + 1) + r·(1 - t^2))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))), r·(2·t·(t^2 + 1)·((t^2 + 1)^2·(1 + 2·t^2 + t^4) - 4·r^2·(t^2 + 1)^2·(1 - 2·t^2 + t^4) + 2·r^4·(1 - 2·t^2 + t^4)·(1 + 4·t^2 + t^4) + r^6·(t^2 - 1)^2·(1 - 2·t^2 + t^4)) + 2·©·r·t·(1 - t^2)·(- 7·(t^2 + 1)^2·(1 + 2·t^2 + t^4) + 3·r^2·(t^2 + 1)^2·(3 - 2·t^2 + 3·t^4) + 4·r^4·(1 - t^2 - t^6 + t^8)) + 2·©^2·t·(t^2 + 1)·(- 3·(t^2 + 1)^4 + 15·r^2·(t^2 + 1)^2·(1 - 2·t^2 + t^4) + r^4·(5 - 8·t^2 + 6·t^4 - 8·t^6 + 5·t^8)) + 2·©^3·r·t·(1 - t^2)·(1 + 2·t^2 + t^4)·(4·r^2·t^2 + 11·(t^2 + 1)^2) + 2·©^4·t·(t^2 + 1)^3·(3·(t^2 + 1)^2 - 5·r^2·(1 - 2·t^2 + t^4)) + 8·©^5·r·t·(t^2 - 1)·(t^2 + 1)^4 - 2·©^6·t·(t^2 + 1)^5 + (t^2 - 1)·‹(t^2 + 1)·((t^2 + 1)^2·(1 + 2·t^2 + t^4) - r^2·(t^2 + 1)^2·(3 - 10·t^2 + 3·t^4) + 2·r^4·(1 - 2·t^2 + t^4)·(1 + 4·t^2 + t^4) + 2·©·r·(1 - t^4)·(4·r^2·(1 + t^2 + t^4) - 3·(t^2 + 1)^2) + 3·©^2·(1 + 2·t^2 + t^4)·(4·r^2·(1 - t^2 + t^4) - (t^2 + 1)^2) - 8·©^3·r·(t^2 + 1)·(-1 - t^2 + t^4 + t^6) + 2·©^4·(t^2 + 1)·(1 + 3·t^2 + 3·t^4 + t^6))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))/((t^2 + 1)·(- r^2·(t^2 + 1)·(-1 + 6·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4)) + 2·©·r·(t^2 - 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(t^4 + 1)) - ©^2·(t^2 + 1)·(-1 - 2·t^2 - t^4 + 2·r^2·(3 - 4·t^2 + 3·t^4)) + 4·©^3·r·(t^2 + 1)·(t^4 - 1) - ©^4·(t^2 + 1)^3 + 4·r·t·‹(t^2 + 1)·(©·(t^2 + 1) + r·(1 - t^2))·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))·(-1 - t^2 + r·(1 - t^2) + ©·(t^2 + 1))·(1 + t^2 + r·(1 - t^2) + ©·(t^2 + 1))); © + r·(1 - t^2)/(1 + t^2), r·2·t/(1 + t^2)]]
segmentotangente(©, r, t):=[x^2 + y^2 + -1 = 0, (x - ©)^2 + y^2 + - r^2 = 0, [© + r·(1 - t^2)/(1 + t^2), r·2·t/(1 + t^2); s(©, r, t), u(©, r, t)]]
tangente(©, r, t):=[-1 + y^2 + x^2 = 0, - r^2 + y^2 + (x - ©)^2 = 0, [© + r·(1 - t^2)/(1 + t^2), r·2·t/(1 + t^2)], y = r·2·t/(1 + t^2) + (2·r·t·(©·(t^2 + 1) + r·(1 - t^2)) - (t^2 + 1)^(3/2)·‹((r^2 - 1)·(t^2 + 1) + 2·©·r·(1 - t^2) + ©^2·(t^2 + 1)))/(-1 - 2·t^2 - t^4 + r^2·(1 - 2·t^2 + t^4) + 2·©·r·(1 - t^4) + ©^2·(1 + 2·t^2 + t^4))·(x - (© + r·(1 - t^2)/(1 + t^2)))]
u(©, r, t):=r·2·t/(1 + t^2) + m(©, r, t)·(s(©, r, t) - (© + r·(1 - t^2)/(1 + t^2)))
hCross:=APPROX(- 6659090909090909/500000000000000)
vCross:=APPROX(45238095238095237/10000000000000000)
©:=
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\deflang1034\pard\plain\f3\fs24 Un ensayo para hacer los c\'e1lculos de Poncelet.
\par 
\par Tomamos la circunferencia de radio 1 y centro (0,0) como circunferencia interior y la de radio r y centro (\'a9,0) como la circunferencia exterior (r habr\'e1 de ser mayor que 1+\'a9 para que la circunferencia unitaria sea interior).
\par Un punto gen\'e9rico de la exterior ser\'e1
\par }
ÿÿ   CExpnObj8   –     Ò    User      ð¿      '[alpha+r*(1-t^2)/(1+t^2),r*2*t/(1+t^2)]€   Þ        ÿ={\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Trazamos una tangente a la circunferencia interior. Primero una recta de pendiente m
\par }
€8     x  J   User      ð¿      .y=r*2*t/(1+t^2)+m(x-(alpha+r*(1-t^2)/(1+t^2)))€   V    h   ÿ&{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Hallamos m para que sea tangente a la circunferencia interior
\par }
€8   t     Œ   User      ð¿      	x^2+y^2=1€8   ˜  Ø  Ô   User      ð¿      8x^2+(r*2*t/(1+t^2)+m(x-(alpha+r*(1-t^2)/(1+t^2))))^2-1=0€   à    ò   ÿ{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Esta es la ecuaci\'f3n que resulta
\par }
€8   þ     ^   User      ð¿      ¿x^2*m(+1)^2*(t^2+1)^2+2*m(x)*(t^2+1)*(r*(m(t^2-1)+2*t)-alpha*m(t^2+1))+alpha^2*(m(t^2+1)^2)^2-2*alpha*m(r)*(t^2+1)*(m(t^2-1)+2*t)+(m(r)^2)^2*(t^2-1)^2+4*m(r)^2*t*(t^2-1)+4*r^2*t^2-(t^2+1)^2=0€   j        ÿ·{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Obtenemos su discriminante, que anulamos y as\'ed, resolviendo la ecuaci\'f3n en m que resulta, obtenemos los dos valores de m que hacen tangentes las rectas desde el punto correspondiente al par\'e1metro t
\par }
€8   ¬        User      ð¿      Ã(2*m(t^2+1)*(r*(m(t^2-1)+2*t)-alpha*m(t^2+1)))^2-4*((m(+1)^2*(t^2+1)^2)*(alpha^2*(m(t^2+1)^2)^2-2*alpha*m(r)*(t^2+1)*(m(t^2-1)+2*t)+(m(r)^2)^2*(t^2-1)^2+4*m(r)^2*t*(t^2-1)+4*r^2*t^2-(t^2+1)^2))=0€       *   ÿ"{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Elegimos la primera pendiente, por ejemplo, y comprobamos
\par }
€   6    H   ÿ7{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 La recta tangente desde el punto correspondiente al par\'e1metro t es entonces
\par }
€8   T         Sub(#2)      ð¿      Ýy=r*2*t/(1+t^2)+(2*r*t*(alpha*(t^2+1)+r*(1-t^2))-(t^2+1)^(3/2)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1)))/(alpha^2*(t^4+2*t^2+1)+2*alpha*r*(1-t^4)+r^2*(t^4-2*t^2+1)-t^4-2*t^2-1)*(x-(alpha+r*(1-t^2)/(1+t^2)))€   ,    >   ô{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Comprobamos
\par }
€8   J     ^   User      ð¿      ÿ>tangente(alpha,r,t):=[x^2+y^2-1=0,(x-alpha)^2+y^2-r^2=0,[alpha+r*(1-t^2)/(1+t^2),r*2*t/(1+t^2)],y=r*2*t/(1+t^2)+(2*r*t*(alpha*(t^2+1)+r*(1-t^2))-(t^2+1)^(3/2)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1)))/(alpha^2*(t^4+2*t^2+1)+2*alpha*r*(1-t^4)+r^2*(t^4-2*t^2+1)-t^4-2*t^2-1)*(x-(alpha+r*(1-t^2)/(1+t^2)))]€8   j  À   v   User      ð¿	      tangente(2,5,1)€`   ‚  Ø  š  
Approx(#9){®Gázt?
      =[x^2+y^2-1=0,x^2-4*x+y^2-21=0,[2,5],y=0.0016469*(953*x+1130)]€8   ¦  À   ²   User      ð¿      tangente(3,7,2)€8   ¾  ø  Ö  Approx(#11)ú~j¼t“ˆ?      E[x^2+y^2-1=0,x^2-6*x+y^2-40=0,[-1.2,5.6],y=-0.00355871*(7893*x+7898)]€   â    <   ÿø{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fswiss\fprq2 System;}{\f3\fmodern\fcharset2 DfW5 Printer;}}
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\deflang1034\pard\plain\f3\fs24 Se ve que va bien por ahora.
\par 
\par Tratamos a continuaci\'f3n de obtener la otra intersecci\'f3n de esta recta con la circunferencia exterior. Como conocemos una intersecci\'f3n, la otra se obtiene f\'e1cilmente. Substituimos en la ecuaci\'f3n de la circunferencia
\par }
€8   H  è   `   User      ð¿      (x-alpha)^2+y^2-r^2=0€   l    ~   ÿ {\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 la y por la de la recta
\par }
€8   Š     V   User      ð¿      Ýy=r*2*t/(1+t^2)+(2*r*t*(alpha*(t^2+1)+r*(1-t^2))-(t^2+1)^(3/2)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1)))/(alpha^2*(t^4+2*t^2+1)+2*alpha*r*(1-t^4)+r^2*(t^4-2*t^2+1)-t^4-2*t^2-1)*(x-(alpha+r*(1-t^2)/(1+t^2)))€   b    t   ÿ{\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 Resulta la ecuaci\'f3n en x
\par }
€8   €     H	   User      ð¿      ÿBx^2*(t^2+1)*(4*r*t*SQRT(t^2+1)*(alpha*(t^2+1)+r*(1-t^2))*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1))-alpha^4*(t^2+1)^3+4*alpha^3*r*(t^2+1)*(t^4-1)-alpha^2*(t^2+1)*(2*r^2*(3*t^4-4*t^2+3)-t^4-2*t^2-1)+2*alpha*r*(t^2-1)*(2*r^2*(t^4+1)-t^4-2*t^2-1)-r^2*(t^2+1)*(r^2*(t^4-2*t^2+1)-t^4+6*t^2-1))-2*x*(2*r*t*SQRT(t^2+1)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1))*(alpha^2*(t^2+1)^2+2*alpha*r*(t^2+1)*(1-t^2)+r^2*(t^2-1)^2+t^4+2*t^2+1)-alpha^5*(t^4+2*t^2+1)^2+4*alpha^4*r*(t^4-1)*(t^4+2*t^2+1)-alpha^3*(t^4+2*t^2+1)*(6*r^2*(t^4-2*t^2+1)-t^4-2*t^2-1)+alpha^2*r*(t^2+1)*(t^2-1)*(4*r^2*(t^2-1)^2-t^4-2*t^2-1)-alpha*r^2*(r^2*(t^4-2*t^2+1)^2+t^8+8*t^6+14*t^4+8*t^2+1)+r*(t^2+1)*(t^2-1)*(r^2*(t^4+6*t^2+1)-(t^2+1)^2))+(4*r*t*(t^2+1)^(3/2)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1))-alpha^5*(t^2+1)^3+3*alpha^4*r*(t^6+t^4-t^2-1)+alpha^3*(t^2+1)*((t^2+1)^2-2*r^2*(t^4-4*t^2+1))+alpha^2*r*(1-t^2)*(2*r^2*(t^4+4*t^2+1)-(t^2+1)^2)+alpha*r^2*(t^2+1)*(3*r^2*(t^4-2*t^2+1)-5*t^4-2*t^2-5)+r*(1-t^2)*(r^4*(t^4-2*t^2+1)-3*r^2*(t^2+1)^2+2*(t^2+1)^2))*(alpha*(t^2+1)+r*(1-t^2))=0€   T	    f	   ÿ{\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 Una de sus soluciones es la que conocemos
\par }
€8   r	  Ø   ®	   User      ð¿      x=alpha+r*(1-t^2)/(1+t^2)€   º	    Ì	   ô{\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 Comprobamos
\par }
€ˆ  Ø	  °  ä	  
Simp(User)š™™™™™Ù?      0=0€   ð	    
   ÿ0{\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 La otra soluci\'f3n s para x la obtenemos de la suma x+s=-b/a, es decir
\par }
€   
     
   ÿ{\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 Ahora hallamos la y correspondiente a esta s
\par }
€8   ,
  x  h
   User      ð¿      .y=r*2*t/(1+t^2)+m(x-(alpha+r*(1-t^2)/(1+t^2)))€8   t
     `   User      ð¿      ÿøs(alpha,r,t):=2*(2*r*t*SQRT(t^2+1)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1))*(alpha^2*(t^2+1)^2+2*alpha*r*(t^2+1)*(1-t^2)+r^2*(t^2-1)^2+t^4+2*t^2+1)-alpha^5*(t^4+2*t^2+1)^2+4*alpha^4*r*(t^4-1)*(t^4+2*t^2+1)-alpha^3*(t^4+2*t^2+1)*(6*r^2*(t^4-2*t^2+1)-t^4-2*t^2-1)+alpha^2*r*(t^2+1)*(t^2-1)*(4*r^2*(t^2-1)^2-t^4-2*t^2-1)-alpha*r^2*(r^2*(t^4-2*t^2+1)^2+t^8+8*t^6+14*t^4+8*t^2+1)+r*(t^2+1)*(t^2-1)*(r^2*(t^4+6*t^2+1)-(t^2+1)^2))/((t^2+1)*(4*r*t*SQRT(t^2+1)*(alpha*(t^2+1)+r*(1-t^2))*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1))-alpha^4*(t^2+1)^3+4*alpha^3*r*(t^2+1)*(t^4-1)-alpha^2*(t^2+1)*(2*r^2*(3*t^4-4*t^2+3)-t^4-2*t^2-1)+2*alpha*r*(t^2-1)*(2*r^2*(t^4+1)-t^4-2*t^2-1)-r^2*(t^2+1)*(r^2*(t^4-2*t^2+1)-t^4+6*t^2-1)))-alpha+r*(1-2/(t^2+1))€8   l     8   User      ð¿      ½m(alpha,r,t):=(2*r*t*(alpha*(t^2+1)+r*(1-t^2))-(t^2+1)^(3/2)*SQRT(alpha^2*(t^2+1)+2*alpha*r*(1-t^2)+(r^2-1)*(t^2+1)))/(alpha^2*(t^4+2*t^2+1)+2*alpha*r*(1-t^4)+r^2*(t^4-2*t^2+1)-t^4-2*t^2-1)€8   D  `  €   User      ð¿      Qu(alpha,r,t):=r*2*t/(1+t^2)+m(alpha,r,t)*(s(alpha,r,t)-(alpha+r*(1-t^2)/(1+t^2)))€8   Œ     @   User      ð¿      †segmentotangente(alpha,r,t):=[x^2+y^2-1=0,(x-alpha)^2+y^2-r^2=0,[[alpha+r*(1-t^2)/(1+t^2),r*2*t/(1+t^2)],[s(alpha,r,t),u(alpha,r,t)]]]€8   L     X   User      ð¿      segmentotangente(2,5,3)€8   d     p   User      ð¿      segmentotangente(3,5,7)€   |    ²   ÿÃ{\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 Por los experimentos va bien. Lo que ahora hace falta es trazar la tangente desde el punto que hemos obtenido y luego la otra y expresar que se cierra el tri\'e1ngulo. De esas condiciones eliminar la t. Un buen l\'edo.
\par }
         ÿÿÿ      ð              