Equal Parallelians Point

Author

Eric W. Weisstein
November 16, 2005

This notebook downloaded from http://mathworld.wolfram.com/notebooks/PlaneGeometry/EqualParalleliansPoint.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/EqualParalleliansPoint.html.

©2005 Wolfram Research, Inc. except for portions noted otherwise

Derivation

d=Simplify[(Distance@@@Parallelians[{α,β,γ}])^2]

Subtract@@@Partition[Numerator[d],2,1]

(2ToSidelengths[Area])^2-(((a α+b β+c γ))^2)

gb=Factor/@GroebnerBasis[Append[Subtract@@@Partition[Numerator[d],2,1], (a+b-c) (a-b+c) (-a+b+c) (a+b+c)-4(((a α+b β+c γ))^2)],α,{β,γ},MonomialOrder->EliminationOrder]

FullSimplify[Solve[#==0,α],TriangleQ[]]&/@gb

Factor//@%

CyclicTrilinears[((-a b-a c+b c)/a)]

Trilinears[X[192]]

Name[X[192]]

Lengths

Factor[Simplify[Distance@@@Parallelians[Trilinears[X[192]]],TriangleQ[]&&Area>0]]

Figure

Factor[Transpose[{{a/c,b/a,c/b},{a/b,b/c,c/a}}]List@@@Simplify[Parallelians[Trilinears[X[192]]]]]

t=Triangle[{{0,0},{1,0},{.3,.7}}];
p=X[192][t];
Show[Graphics[{
t,
{Red,Parallels[t,p]},
PointLabels[Join[
Transpose[{Point/@Vertices[t],{"A","B","C"}}],
Append[#,Blue]&/@Transpose[{Point/@Flatten[Coordinates/@Parallels[t,p],1],{"(Subscript[P, CB])","(Subscript[P, BC])","(Subscript[P, CA])","(Subscript[P, AC])","(Subscript[P, BA])","(Subscript[P, AB])"}}],
{{X[192][t],"(Subscript[X, 192])",Red}}
],{LabelOffset->-.04,PointSize[.03]}
]
}],AspectRatio->Automatic,PlotRange->All]

Ellipse

Trilinears[X[192]]

pts=Flatten[List@@@Parallelians[Trilinears[X[192]]],1]

Factor[Conic[Take[pts,5]]]

conic=Collect[Last[%],{α^2,β^2,γ^2,α β,β γ,γ α},Factor]

Expand[conic/.Thread[{α,β,γ}->#]&/@pts]

ctr=ToAngles[Symmetrize[Factor[Center[conic]]]]

KimberlingLookup[ctr]

t=Triangle[{{0,0},{1,0},{.3,.7}}];
Module[{α0,β0,γ0},
{α0,β0,γ0}=Trilinears[X[192]]/.Thread[{a,b,c}->SideLengths[Evaluate[t]]];
Show[Graphics[{
t,
{Green,TrilinearCurve[t,conic]},
{Red,
Cartesian[#[t]]&/@Parallelians[{α0,β0,γ0}]
},
PointLabels[Join[
{{Cartesian[{α0,β0,γ0}[t]],"P",Red}},
Transpose[{VertexPoints[t],{"A","B","C"}}],
Append[#,Blue]&/@Transpose[{Flatten[Map[Point,Coordinates[Cartesian[#[t]]]&/@Parallelians[{α0,β0,γ0}],{2}]],{"(Subscript[P, CB])","(Subscript[P, BC])","(Subscript[P, AC])","(Subscript[P, CA])","(Subscript[P, BA])","(Subscript[P, AB])"}}],
{{Cartesian[ctr[t]],"(Subscript[O, P])",Green}}
],LabelOffset->-.04,PointSize[.03]
]
}
],AspectRatio->Automatic,PlotRange->All]
]

Perspective Center of Incentral Triangle and Anticomplementary triangle

t=Triangle[{{0,0},{1,0},{.4,.6}}];
Show[Graphics[{
t,
{Blue,
ct=CevianTriangle[t,Incenter[t]],
Dashing[{.015}],
Cevians[t,Incenter[t]]
},
{Red,
at=AnticevianTriangle[t,Centroid[t]],
Dashing[{.015}],
Anticevians[t,Centroid[t]]
},
{Purple,
LongestLine[{Sequence@@#,Coordinates[X[192][t]]}]&/@Transpose[Vertices/@{at,ct}]
},
PointLabels[Join[
Transpose[{Point/@Vertices[t],{"A","B","C"}}],
Append[#,Red]&/@Transpose[{Point/@Vertices[at],{"A'","B'","C'"}}],
Append[#,Blue]&/@Transpose[{Point/@Vertices[ct],{"(Subscript[A, I])","(Subscript[B, I])","(Subscript[C, I])"}}],
{
{Incenter[t],"I",Blue},
{Centroid[t],"G",Red},
{X[192][t],"(Subscript[X, 192])",Purple}
}
],{LabelOffset->-.08,PointSize[.03]}
]
}],AspectRatio->Automatic,PlotRange->All]

Trilinears

Together[1/a(-1/a+1/b+1/c)]