Stammler Triangle

Author

Eric W. Weisstein
November 8, 2005

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

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

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

Equilateral

EquilateralQ[StammlerTriangle[RandomTriangle[20]]]

Simplify[#^2&/@EquilateralQ[StammlerTriangle]/.C->π-A-B]

Figure

<<Graphics`Colors`

t=Triangle[{{0,0},{1,0},{.15,.7}}];
Show[Graphics[{
{Purple,Sides[t,2]},
t,
{Blue,circs=StammlerCircles[t,{1,2,3}]},
{Green,ct=CircumtangentialTriangle[t]},{Red,StammlerTriangle[t]},
{DarkGreen,Circumcircle[t]},
PointLabels[Join[
Transpose[{VertexPoints[t],{"A","B","C"},{LabelOffset->-.1{1,1},LabelOffset->.1{1,-1},{}}}],
Append[#,Green]&/@Transpose[{VertexPoints[ct],{"A'","B'","C'"}}],
Append[#,Red]&/@Transpose[{Center/@circs,{"(Subscript[O, A])","(Subscript[O, B])","(Subscript[O, C])"}}]
]]
}],AspectRatio->Automatic,PlotRange->{{-1.5,2.5},{-1.9,2.2}}]

Trilinear Vertex Matrix

Vertices[Triangle[StammlerTriangle]]//TraditionalForm

Area

rules1={Cos[x_]:>TrigExpand[Cos[x]],Sin[x_]:>TrigExpand[Sin[x]]};
rules2={Cos[x_/3]:>((1+(((Cos[x]+I Sin[x]))^(2/3)))/(2 (((Cos[x]+I Sin[x]))^(1/3)))),Sin[x_/3]:>1+4(((((1+(((Cos[x]+I Sin[x]))^(2/3)))/(2 (((Cos[x]+I Sin[x]))^(1/3))))))^2)};

area=Area[StammlerTriangle]/.rules1/.(Sqrt[(a+b-c) (a-b+c) (-a+b+c) (a+b+c)]):>sqrt;

area2=Together//@area;

Denominator[area2]

Gröbner

eqns=Join[
{poly-Denominator[area2]},
TrigExpand[Cos[3#]]-Cos[3#]&/@{A,B,C}/.{Cos[x_]:>Cos[x/3],Sin[x_]:>Sin[x/3]},
TrigExpand[Sin[3#]]-Sin[3#]&/@{A,B,C}/.{Cos[x_]:>Cos[x/3],Sin[x_]:>Sin[x/3]},
Cos[#/3]^2+Sin[#/3]^2-1&/@{A,B,C},
Cos[#]^2+Sin[#]^2-1&/@{A,B,C},
Cos[A#]^2+Cos[B#]^2+Cos[C#]^2+2Cos[A#]Cos[B#]Cos[C#]-1&/@{1,1/3}
]/.{
Cos[A/3]:>cosa,Cos[B/3]:>cosb,Cos[C/3]:>cosc,
Sin[A/3]:>sina,Sin[B/3]:>sinb,Sin[C/3]:>sinc,
Cos[A]:>cos3a,Cos[B]:>cos3b,Cos[C]:>cos3c,
Sin[A]:>sin3a,Sin[B]:>sin3b,Sin[C]:>sin3c
}

(gb=Factor/@GroebnerBasis[eqns,poly,{cosa,cosb,cosc,sina,sinb,sinc},MonomialOrder->EliminationOrder])//Timing

...hangs...

TrigToExp

area2=Factor//@(Together//@TrigToExp[area])

ToSidelengths[area2]

Together//@(TrigToExp//@%)

Simplify[(area/.rules1)/.rules2]

Together//@ToSidelengths[(area/.rules1)/.rules2]

FullSimplify[area]

Side Lengths

SideLengths[Evaluate[Triangle[Center/@Circles[StammlerCircles]]]]/.Thread[{a,b,c}->SideLengths[Evaluate[t]]]/.Thread[{A,B,C}->Angles[t]]

sides=SideLengths[StammlerTriangle];

Simplify[sides]

Solve[-3y+4y^3==x,y]

((1+(((x+(Sqrt[-1+(x^2)])))^(2/3)))/(2 (((x+(Sqrt[-1+(x^2)])))^(1/3))))/.x->Sin[x]

{((1+(((Cos[x]+(Sqrt[-1+((Cos[x])^2)])))^(2/3)))/(2 (((Cos[x]+(Sqrt[-1+((Cos[x])^2)])))^(1/3)))),Cos[x/3]}/.x->.1234

FullSimplify[((1+(((Cos[x]+I Sin[x]))^(2/3)))/(2 (((Cos[x]+I Sin[x]))^(1/3)))),0<x<π]

FullSimplify[((Sin[x])/(-1+4(((((1+(((Cos[x]+I Sin[x]))^(2/3)))/(2 (((Cos[x]+I Sin[x]))^(1/3))))))^2)))]

FullSimplify[%,0<x<π]

Together//@ToSidelengths[(sides[[1]]//.

Radical Circle

Center/@Circles[StammlerCircles]

?RadicalCenter

?CircleFunction

CircleFunction[TrilinearEquation[Circle[{Cos[A]-2 Cos[((B-C)/3)],Cos[B]+2 Cos[(1/3) (B+2 C)],Cos[C]+2 Cos[(1/3) (2 B+C)]},r1]]]

Centers

(pts={#,KimberlingLookup[#[StammlerTriangle]]}&/@Take[KimberlingCenters,10]);//Timing

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