Johnson Triangle Circumcircle

Author

Eric W. Weisstein
February 13, 2006

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

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

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

Figure

t=Triangle[{{0,0},{1,0},{.7,.8}}];
ctrs=Center/@JohnsonCircles[t];
Show[Graphics[{
t,
{Purple,Circumcircle[t]},
{Blue,JohnsonCircles[t]},
{Red,JohnsonTriangleCircumcircle[t]},
{Green,JohnsonTriangle[t]},
PointLabels[Join[
Transpose[{VertexPoints[t],{"A","B","C"}}],
Append[#,Red]&/@Transpose[{ctrs,{"(Subscript[J, A])","(Subscript[J, B])","(Subscript[J, C])"}}],
{
{Circumcenter[t],"O",{Purple}},
{Orthocenter[t],"H",{Red}}
}
],LabelOffset->.06]
}],AspectRatio->Automatic]

Incident Centers

(ctrs=IncidentCenters[KimberlingCenters,JohnsonTriangleCircumcircle])//Timing

{#,Name[#]}&/@ctrs//ColumnForm

Name[X[#]]&/@{3,125}

Derivation

circ=Circumcircle[JohnsonTriangle];

Radius

Factor[FullSimplify[Factor//@ToSidelengths[CircleRadius[circ]^2],TriangleQ[]]]

Refine[Sqrt[ToAngles[%7]],TriangleQ[]&&Area>0]

Refine[ToAngles[Circumradius//ToSidelengths],TriangleQ[]&&Area>0]

Center

Symmetrize[Simplify[Factor[ToSidelengths[Center[circ]]/.x_^(5/2):>x^2 Sqrt[x]],TriangleQ[]]]

Symmetrize[ToAngles[%%]]

Circle Function

ToAngles[Factor[ToSidelengths[CircleFunction[Circle[{Sec[A],Sec[B],Sec[C]},Circumradius]]]]]

Simplify[ToAngles[%22]]

ToAngles[Circumradius^2//ToSidelengths]

((-(((a^2) b c (1+2 Cos[2 A]))/(16 (Area^2))))/((2 Area (1+2 Cos[2 A]))/(b c)))((R^2)/(((a^2) (b^2) (c^2))/(16 (Area^2))))

FullSimplify[Together//@ToSidelengths[TrigExpand[((-(((a^2) (-(a^2)+(b^2)-b c+(c^2)) (-(a^2)+(b^2)+b c+(c^2)))/(16 (Area^2) b c)))/(-((R^2)/(2 Area))Sin[3A]))/.R->Circumradius]],TriangleQ[]]

KimberlingLookup[%22]

Name[X[50]]

CenterFunction[X[50]]

Simplify[ToAngles[Factor//@Simplify[Together//@ToSidelengths[TrigExpand[Sin[3A]]],TriangleQ[]]],Area>0]