Inner Vecten Triangle

Author

Eric W. Weisstein
November 8, 2005

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

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

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

Figure

t=Triangle[{{0,0},{1,0},{3/10,8/10}}];
Show[Graphics[{
{Blue,ErectPolygons[Vertices@t,4,1]},
t,
{Red,InnerVectenTriangle[t]},
PointLabels[Join[
Append[#,Blue]&/@Transpose[{VertexPoints[t],{"A","B","C"}}],
Append[#,Red]&/@Transpose[{VertexPoints[InnerVectenTriangle[t]],{"(Subscript[I, C])","(Subscript[I, A])","(Subscript[I, B])"}}]
],PointSize[.025]]
},AspectRatio->Automatic]
]

Trilinear Vertex Matrix

TrilinearVertexMatrix[InnerVectenTriangle]//TraditionalForm

Area

FullSimplify[Factor[Together//@ToSidelengths[Area[InnerVectenTriangle]]/.Sqrt[x_]Sqrt[y_]:>FullSimplify[Sqrt[x y],TriangleQ[]]/.Sqrt[x_]:>FullSimplify[Sqrt[x],TriangleQ[]]],TriangleQ[]]

%/.(Sqrt[(a+b-c) (a-b+c) (-a+b+c) (a+b+c)])->4Area

TraditionalForm[%]

Side Lengths

FullSimplify[Together//@ToSidelengths[SideLengths[InnerVectenTriangle]]/.Sqrt[x_]Sqrt[y_]:>FullSimplify[Sqrt[x y],TriangleQ[]]/.Sqrt[x_]:>FullSimplify[Sqrt[x],TriangleQ[]],TriangleQ[]]

TraditionalForm[%/.(Sqrt[-((a-b-c) (a+b-c) (a-b+c) (a+b+c))])->Area]

Inner Vecten Point

Simplify[Intersections[{{1,0,0},m[[1]]},{{0,1,0},m[[2]]}]]4c

Simplify[Intersections[{{1,0,0},m[[1]]},{{0,0,1},m[[3]]}]]4b

%/(-a b c(Cos[A]-Sin[A])(Cos[B]-Sin[B]) (Cos[C]-Sin[C]) )

FullSimplify[%]

Simplify[TrigExpand[Sec[Pi/4+A]]]

Show[Graphics[{
{Red,
ErectPolygons[Vertices@t,4,1]
},
t,
{Blue,
LongestLine/@Transpose[{Vertices[s],RotateRight[Vertices[t]],Table[Coordinates[X[486][t]]//Simplify,{3}]}]
},
PointLabels[Join[
Append[#,Red]&/@Transpose[{VertexPoints[t],{"A","B","C"}}],
Append[#,Blue]&/@Transpose[{VertexPoints[s],{"(Subscript[I, C])","(Subscript[I, A])","(Subscript[I, B])"}}],
{{X[486][t],"(Subscript[X, 486])",Purple}}
],PointSize[.025]]
},AspectRatio->Automatic]
]

Derivation

t=Triangle[{{0,0},{1,0},{3/10,8/10}}];
s=Triangle[ErectedPolygonCenters[Vertices@t,4,1]];

ExactTrilinears[CartesianToTrilinear[t,VertexPoints[s][[2]]]]/.Thread[{a,b,c}->SideLengths[Evaluate[t]]]//Simplify

SideLengths[Evaluate[t]]

As expected, (Subscript[α, 1])= a/2

The side B C and line from the midpoint of B C to the center of the square are perpendicular, so

Collect[LineEquation[{{0,c,b},{a/2,β1,γ1}}],{α,β,γ}]

l1=(l ll+m mm+n nn)-(m nn+mm n)Cos[A]-(n ll+nn l)Cos[B]-(l mm+ll m)Cos[C]/.Thread[{l,m,n}->{1,0,0}]/.Thread[{ll,mm,nn}->{(-b β1+c γ1),(1/2) a b,-((a c )/2)}]//FullSimplify

soln=Simplify[Solve[{
l1==0,
Distance[{a/2,β1,γ1},{0,c,b}]==a/2
},{β1,γ1}],c>0]

coords=FullSimplify[DistanceTrilinears[Last/@soln[[2]]],TriangleQ[]]

Area[]

coords=Prepend[Simplify[LawOfCosines[{((-(a^2)-(b^2)+(c^2)+4Δ)/(4 b)),((-(a^2)+(b^2)-(c^2)+4Δ)/(4 c))}]],(a/2)]

MapIndexed[RotateRight[#1,#2[[1]]-1]&,CyclicTrilinears[coords]]//TraditionalForm

m=(({{(a/2), (1/2) a (Sin[C]- cos(C)), (1/2) a (Sin[B]- cos(B))}, {(1/2) b (Sin[C]-cos(C)), (b/2), (1/2) b (Sin[A]- cos(A))}, {(1/2) c (Sin[B]- cos(B)), (1/2) c (Sin[A]- cos(A)), (c/2)}}))

t=Triangle[{{0,0},{1,0},{3/10,8/10}}];
s=Triangle[ErectedPolygonCenters[Vertices@t,4,1]];

ps=TrilinearToCartesian[t,
#/.Δ->Area[]/.Thread[{A,B,C}->Angles[t]]/.Thread[{a,b,c}->SideLengths[Evaluate[t]]]]&/@MapIndexed[RotateRight[#1,#2[[1]]-1]&,CyclicTrilinears[coords]];

Show[Graphics[{
{Red,
ErectPolygons[Vertices@t,4,1]
},
t,
{PointSize[.04],Green,Point/@ps},
PointLabels[Join[
Append[#,Red]&/@Transpose[{VertexPoints[t],{"A","B","C"}}],
Append[#,Blue]&/@Transpose[{VertexPoints[s],{"(Subscript[O, C])","(Subscript[O, A])","(Subscript[O, B])"}}]
],PointSize[.025]]
},AspectRatio->Automatic]
]

Centers

(pts={#,KimberlingLookup[#[InnerVectenTriangle]]}&/@Take[KimberlingCenters,100]);//Timing

{#[[1]],Name[#[[1]]],X[#[[2,1]]],Name[X[#[[2,1]]]]}&/@Cases[pts,_?(Last[#]=!={}&)]//TableForm