Incentral Triangle

Author

Eric W. Weisstein
November 11, 2005

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

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

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

Graphic

t=Triangle[{{0,0},{1,0},{.2,.5}}];
Show[Graphics[{
t,
{
Red,
in=IncentralTriangle[t],
Dashing[{.02}],
Incircle[t]
},
{
Blue,
Line/@Transpose[Vertices/@{t,in}]
},
PointLabels[Join[
Transpose[{Point/@Vertices[t],{"A","B","C"}}],
Append[#,Red]&/@Transpose[{Point/@Vertices[in],{"(Subscript[I, A])","(Subscript[I, B])","(Subscript[I, C])"}}],
{
{Incenter[t],"I",Blue}
}
],{PointSize[.025],LabelOffset->.03}]
}],AspectRatio->Automatic,PlotRange->All]

Vertices

CevianTriangle[Incenter]//MatrixForm

Side Lengths

abc=SideLengths[CevianTriangle[Incenter]]

ToAngles[Factor//@FullSimplify[ToSidelengths[abc],TriangleQ[]]]

Equilateral

abc/.{a->1,b->1,c->1,A->60Degree,B->60Degree,C->60Degree}

2((1/2)(a+b+c)-a)Sin[A/2]/.{a->1,b->1,c->1,A->60Degree,B->60Degree,C->60Degree}

Right Isosceles

{Abs[((a b c (Sqrt[3+2 (-Cos[A]+Cos[B]+Cos[C])]))/((a+b) (a+c)))],2((1/2)(a+b+c)-a)Sin[A/2]}/.{a->1,b->1,c->(Sqrt[2]),A->45Degree,B->45Degree,C->90Degree}//FunctionExpand//FullSimplify

N[%]

Random

{Abs[((a b c (Sqrt[3+2 (-Cos[A]+Cos[B]+Cos[C])]))/((a+b) (a+c)))],2((1/2)(a+b+c)-a)Sin[A/2]}/.RandomTriangleRules[]

Sum

Timing[sum=FullSimplify[Plus@@abc,TriangleQ[]]]

Product

Timing[prod=FullSimplify[Times@@abc,TriangleQ[]]]

Area

FullSimplify[Area[CevianTriangle[Incenter]] Δ/Area[],TriangleQ[]]

Centers

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

l=Map[First,DeleteCases[pts,_?(Last[#]==={}&)],{2}]

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