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Geometria2D`Conicas`
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KiepertHyperbola::usage="Kiepert's hyperbola passes through the three vertices of the triangle, passes through the barycenter, the orthocenter and many other important points. References: Wolfram Math World: Kiepert Hyperbola; Personal. ";KiepertHyperbola[tri={{x11_,y11_},{x2_,y2_},{x3_,y3_}},{X0_,X1_},{Y0_,Y1_}]:=Module{},DEP=Det[{{x11,y11,1},{x2,y2,1},{x3,y3,1}}];IfDEP==0||X0≥X1||Y0≥Y1,Print[PasteButton["Incorrect data or points do not form a triangle"]],ρ=TriangleCenter[tri,"Orthocenter"];xo=ρ[[1]];yo=ρ[[2]];α=TriangleCenter[tri,"Centroid"];xb=α[[1]];yb=α[[2]];xm1=;ym1=;xm2=;ym2=;xm3=;ym3=;tri1={{xm1,ym1},{xm2,ym2},{xm3,ym3}};ϕ=TriangleCenter[tri1,"Circumcenter"];xce=ϕ[[1]];yce=ϕ[[2]];ϕ1=TriangleMeasurement[tri1,"Circumradius"];rce=ϕ1;Print[""];s=NSolve[ae1x11^2+x11y11+ce1y11^2+de1x11+ee1y11+fe1==0, ae1x2^2+x2y2+ce1y2^2+de1x2+ee1y2+fe1==0, ae1x3^2+x3y3+ce1y3^2+de1x3+ee1y3+fe1==0, ae1xb^2+xbyb+ce1yb^2+de1xb+ee1yb+fe1==0, ae1xo^2+xoyo+ce1yo^2+de1xo+ee1yo+fe1==0,{ae1,ce1,de1,ee1,fe1}]; a11=ae1/.s;c11=ce1/.s;d11=de1/.s;e11=ee1/.s;f11=fe1/.s; at1=Dot[a11,{1}]; bt1=1; ct1=Dot[c11,{1}]; dt1=Dot[d11,{1}]; et1=Dot[e11,{1}]; ft22=Dot[f11,{1}];μ=ResourceFunction["ConicProperties"][{at1,bt1,ct1,dt1,et1,ft22}.{,xy,,x,y,1}==0,{x,y}];p=μ[[6,1]];p1=μ[[6,2]];Print[PasteButton["The equation of Kiepert's hyperbola is:"]];Print[" "];Print[{at1,bt1,ct1,dt1,et1,ft22}.{,xy,,x,y,1}==0];Print[""];Conicas[at1,bt1,ct1,dt1,et1,ft22,{X0,X1},{Y0,Y1}];Print[PasteButton["The center of the Kiepert's hyperbola () is:"]];Print[{hi,ki}];g1=Line[{{x11,y11},{x2,y2},{x3,y3},{x11,y11}}];g2=Point[{xo,yo}];g3=Point[{xb,yb}];g4=Point[{x11,y11}];g5=Point[{x2,y2}];g6=Point[{x3,y3}];g8=Point[{hi,ki}];g7=ContourPlot[at1x^2+bt1xy+ct1y^2+dt1x+et1y+ft22==0,{x,X0,X1},{y,Y0,Y1}];g9=ContourPlot[p,{x,X0,X1},{y,Y0,Y1},ContourStyle->{Gray,Dashed}];g10=ContourPlot[p1,{x,X0,X1},{y,Y0,Y1},ContourStyle->{Gray,Dashed}];g11=Circle[{xce,yce},rce];SwatchLegend[{Gray,Purple,Green,Brown,Orange},{"Asymptotes","Circumference of the 9 points","Orthocenter","Barycenter","Center ()"}]Show[{Graphics[{Green,PointSize[0.01],g2}],Graphics[{Blue,g1}],Graphics[{Brown,PointSize[0.01],g3}],Graphics[{Black,PointSize[0.01],g4}],Graphics[{Black,PointSize[0.01],g5}],Graphics[{Black,PointSize[0.01],g6}],Graphics[{Orange,PointSize[0.01],g8}],g7,g9,g10,Graphics[{Purple,g11}]},AxesTrue,AxesOrigin{0,0},ImageSize900,AspectRatioAutomatic,PlotLabelStyle[Framed["Kiepert's Hyperbola and detail of the triangle"],16,Blue,BackgroundYellow]]
x11+x2
2
y11+y2
2
x2+x3
2
y2+y3
2
x3+x11
2
y3+y11
2
2
x
2
y
2
x
2
y
X
115
X
115
In the command below, type the new coordinates of the points. You can also change the extension to X and Y. Evaluate the notebook.
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KiepertHyperbola[tri={{-2,5},{7,12},{9,-4}},{-20,20},{-20,20}]
52.4667-4.35556x-0.377778-11.8222y+xy+0.3777780
2
x
2
y
Los valores de a, b, c, y e son:
LL´
3.8653 | 3.8653 | 5.46635 | 7.73059 | 1.41421 |
Eje transverso en x''
Después de la rotación la ecuación es:
-52.4667+12.5244-0.626671+1.36913+0.626671= 0
´
x
´2
x
´
y
´2
y
El ángulo de rotación es:
θ = 63.5365°
Después de la traslación la ecuación es:
9.36278-0.626671+0.626671= 0
´´2
x
´´2
y
Las coordenadas del nuevo origen son:
´´
O
{5.431,8.45898}
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