Tucker Circles

Author

Eric W. Weisstein
February 9, 2006

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

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

©2006 Wolfram Research, Inc. except for portions noted otherwise

Code

TuckerCircleLambda[c_Circle]:=Module[{λ},λ/.Solve[CircleRadius[c]^2==Circumradius^2((λ^2)+(((1-λ))^2)((Tan[BrocardAngle])^2)),λ]
]
TuckerCircleLambda[c_]:=TuckerCircleLambda[Circle[c]]/;MemberQ[CentralCircles,c]

Graphics

t=Triangle[{{0,0},{1,0},{.3,.7}}];
Show[GraphicsArray[Partition[Table[
Show[Graphics[{
t,
TuckerCircle[t,r],
{Thickness[.01],TuckerHexagon[t,r]}
}],
AspectRatio->Automatic,DisplayFunction->Identity],
{r,.05,.95,(.95-.05)/8}],4]]]

t=Triangle[{{0,0},{1,0},{.3,.7}}];
Show[GraphicsArray[Partition[Table[
Show[Graphics[{
Triangle[t],
TuckerCircle[t,r],
{Thickness[.01],TuckerHexagon[t,r]}
}],
AspectRatio->Automatic,DisplayFunction->Identity],
{r,.05,.95,(.95-.05)/8}],4]]]

Special Cases

Apollonius Circle

This needs some work

Circle[ApolloniusCircle]

Factor[ToSidelengths[CircleRadius[ApolloniusCircle]]]/. 1/Sqrt[x_/(a+b+c)]:>(a+b+c)/Sqrt[x(a+b+c)]

circ=TuckerCircle[-2ArcTan[Inradius/Semiperimeter]];

Center

Symmetrize[ToSidelengths[TrigExpand[Center[circ]]]/.Sqrt[x_]Sqrt[y_]:>FullSimplify[Sqrt[x y],TriangleQ[]]]

Radius

Factor[Together//@ToSidelengths[TrigExpand[CircleRadius[circ]]]/.Sqrt[x_/(a+b+c)]:>Sqrt[x(a+b+c)]/(a+b+c)]

Lambda

Factor[Factor//@ToSidelengths[TuckerCircleLambda[ApolloniusCircle]]/.Sqrt[x_]:>Refine[Sqrt[x],TriangleQ[]]]

Factor[ToSidelengths[((((Semiperimeter^2)-(Inradius^2))Semiperimeter)/(a b c))]]

Circumcircle

circ=TuckerCircle[0]

Circle[Circumcircle]

solns=λ/.Solve[CircleRadius[Circumcircle]^2==Circumradius^2((λ^2)+(((1-λ))^2)((Tan[BrocardAngle])^2)),λ]

Cosine Circle

Circle[CosineCircle]

FullSimplify[Together//@ToSidelengths[TuckerCircle[Pi/2]],TriangleQ[]]/.Circle[ctr_List,r_]:>Circle[Symmetrize[ctr],r]

solns=λ/.Solve[CircleRadius[CosineCircle]^2==Circumradius^2((λ^2)+(((1-λ))^2)((Tan[BrocardAngle])^2)),λ]

Simplify[#>=0,TriangleQ[]]&/@solns

First Lemoine Circle

Circle[FirstLemoineCircle]

Refine[ToAngles[Simplify[ToSidelengths[(1/2) Circumradius Sec[BrocardAngle]],TriangleQ[]]],Area>0]/. 1/Sqrt[1/x_]:>Sqrt[x]

Simplify[ToSidelengths[TrigExpand//@TuckerCircle[BrocardAngle]],TriangleQ[]]/.Circle[ctr_List,r_]:>Circle[Symmetrize[ctr],r]

Factor//@Refine[ToAngles[%],Area>0]/. 1/Sqrt[1/x_]:>Sqrt[x]

solns=λ/.Solve[CircleRadius[FirstLemoineCircle]^2==Circumradius^2((λ^2)+(((1-λ))^2)((Tan[BrocardAngle])^2)),λ]

Simplify[#>=0,TriangleQ[]]&/@solns

Simplify[#>=0,TriangleQ[]&&AcuteQ[]]&/@solns

Simplify[#>=0,TriangleQ[]&&!(AcuteQ[])]&/@solns

Gallatly Circle

Circle[GallatlyCircle]

MapAt[Cancel[((2a b c(Sqrt[(a^2) (b^2)+(a^2) (c^2)+(b^2) (c^2)]))/(Sqrt[-(a-b-c) (a+b-c) (a-b+c) (a+b+c)]))#]&,Factor//@((Together//@ToSidelengths[TuckerCircle[Pi/2-BrocardAngle]/.Sin[x_]:>TrigExpand[Sin[x]]])/.{1/Sqrt[x_]:>FullSimplify[1/Sqrt[x],TriangleQ[]],
Sqrt[x_]:>FullSimplify[Sqrt[x],TriangleQ[]]}),1]

solns=λ/.Solve[CircleRadius[GallatlyCircle]^2==Circumradius^2((λ^2)+(((1-λ))^2)((Tan[BrocardAngle])^2)),λ];

Union[Factor[Together//@ToSidelengths[solns]]]

Factor[FullSimplify[ToSidelengths[Sin[BrocardAngle]],TriangleQ[]]^2]

Kenmotu Circle

Circle[KenmotuCircle]

circ=TuckerCircle[Pi/4];

Radius

CircleRadius[circ]

Center

Symmetrize[TrigExpand[Center[circ]]]

Lambda

λ=Together//@(TuckerCircleLambda[KenmotuCircle]/.t_Csc:>TrigExpand[t])

Simplify[ToSidelengths[(S/(SW+S))]]

r=RandomTriangleRules[]

((Sqrt[(a+b-c) (a-b+c) (-a+b+c) (a+b+c)])/((a^2)+(b^2)+(c^2)+(Sqrt[(a+b-c) (a-b+c) (-a+b+c) (a+b+c)])))/.r

ToSidelengths[λ]/.r

Taylor Circle

Circle[TaylorCircle]

Factor//@Simplify[Together//@(ToSidelengths[Circle[TaylorCircle]/.Cos[x_]:>TrigExpand[Cos[x]]]/.Sqrt[x_]Sqrt[y_]:>Simplify[Sqrt[x y],TriangleQ[]]),TriangleQ[]]/.Circle[ctr_List,r_]:>Circle[Symmetrize[ctr],r]

Factor//@((Together//@(ToSidelengths[TuckerCircle[ArcTan[-Tan[A]Tan[B]Tan[C]]]/.f:(_Cos|_Csc):>TrigExpand[f]]))/.Sqrt[s_]Sqrt[t_]:>Simplify[Sqrt[Expand[s t]],TriangleQ[]]/.Sqrt[s_]:>Simplify[Sqrt[s],TriangleQ[]])/.Circle[ctr_List,r_]:>Circle[Symmetrize[ctr],r]

solns=λ/.Solve[CircleRadius[TaylorCircle]^2==Circumradius^2((λ^2)+(((1-λ))^2)((Tan[BrocardAngle])^2)),λ]

Together//@Simplify[solns,TriangleQ[]]

Map[Factor,Simplify[Together//@ToSidelengths[%],TriangleQ[]],-1]

solns2=FullSimplify[%4,TriangleQ[]]

poly=Factor[(b^2) (c^2) ((((b^2)-(c^2)))^2)+(a^6) ((b^2)+(c^2))+(a^2) ((((b^2)-(c^2)))^2) ((b^2)+(c^2))-(a^4) (2 (b^4)+(b^2) (c^2)+2 (c^4))]

Factor[solns2/.Abs[x_]:>x]

FullSimplify[#>0,TriangleQ[]&&AcuteQ[]]&/@%45

Factor[solns2/.Abs[x_]:>-x]

FullSimplify[#>0,TriangleQ[]&&AcuteQ[]]&/@%40

Table

l={
{ApolloniusCircle,(((a^3)+(a^2) b+a (b^2)+(b^3)+(a^2) c+4 a b c+(b^2) c+a (c^2)+b (c^2)+(c^3))/(4 a b c))},
{Circumcircle,1},
{CosineCircle,0},
{FirstLemoineCircle,1/2},
{GallatlyCircle,Sin[BrocardAngle]^2},
{TaylorCircle,-Cos[A]Cos[B]Cos[C]}
};

Do[Print[{l[[i,1]],FullSimplify[Together//@ToSidelengths[Subtract@@{CircleRadius[l[[i,1]]],Circumradius√(((l[[i,2]])^2)+(((1-l[[i,2]]))^2)((Tan[BrocardAngle])^2))}],TriangleQ[]]//Timing}],
{i,Length[l]}]