Conway Circle

Author

Eric W. Weisstein
November 30, 2004

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

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

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

Circle Function

CircleFunction[ConwayCircle]

IncidentCenters[KimberlingCenters,CyclicTrilinears[CircleFunction[ConwayCircle]]]//Timing

Through[{CenterName,CenterFunction}[X[213]]]

Circle Radius

FullSimplify[Together//@DistanceTrilinears[CircleRadius[ConwayCircle]],TriangleQ[]]

FullSimplify[Inradius[]^2+Semiperimeter[]^2]

Factor[%]

Incident Centers

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

{#,CenterName[#]}&/@ctrs//TableForm

Check

Short/@(FullSimplify[DistanceTrilinears[TrigExpand[eqn/.Thread[{α,β,γ}->Trilinears[#]]]],
TriangleQ[]]&/@ctrs)//Timing

Figure

t=Triangle[{{0,0},{1,0},{.1,.5}}];
sides=Thread[{a,b,c}->SideLengths[Evaluate[t]]];
Show[Graphics[{
{Blue,Sides[t,1.4]},
t,
MapIndexed[EqualSideTicks[Coordinates[#1],#2[[1]],.05,.06]&,Sides[t]],
{Red,ConwayCircle[t]},
PointLabels[Join[
Transpose[{{aa,bb,cc}=VertexPoints[t],{"A","B","C"},{LabelOffset->{0,-.2},{},{}}}],
{{Incenter[t],"I",{Red,LabelOffset->{-.05,-.2}}}},
Append[#,{Red,LabelOffset->{0,.15}}]&/@{
{ab=Point[TrilinearToCartesian[t,{c(b+a),0,-(a^2)}/.sides]],"(Subscript[A, b])"},
{ac=Point[TrilinearToCartesian[t,{b(c+a),-(a^2),0}/.sides]],"(Subscript[A, c])"},
{ba=Point[TrilinearToCartesian[t,{0,c(a+b),-(b^2)}/.sides]],"(Subscript[B, a])"},
{bc=Point[TrilinearToCartesian[t,{-(b^2),a(c+b),0}/.sides]],"(Subscript[B, c])"},
{ca=Point[TrilinearToCartesian[t,{0,-(c^2),b(a+c)}/.sides]],"(Subscript[C, a])"},
{cb=Point[TrilinearToCartesian[t,{-(c^2),0,a(b+c)}/.sides]],"(Subscript[C, b])"}
}
],PointSize[.025]],
{
Blue,
EqualSideTicks[Coordinates/@{aa,#},1,.05,.06]&/@{ab,ac},
EqualSideTicks[Coordinates/@{bb,#},2,.05,.06]&/@{ba,bc},
EqualSideTicks[Coordinates/@{cc,#},3,.05,.06]&/@{ca,cb}
}
}],AspectRatio->Automatic,ImageSize->300]

Check that Points are Concyclic

rs=Factor/@(FullSimplify[(Together//@Distance[Trilinears[X[1]],#]^2)&/@{
{c(b+a),0,-(a^2)},
{b(c+a),-(a^2),0},
{0,c(a+b),-(b^2)},
{-(b^2),a(c+b),0},
{0,-(c^2),b(a+c)},
{-(c^2),0,a(b+c)}
},TriangleQ[]])

Equal@@rs

Derivation Given Center and Radius

CenterName[X[1]]

Factor[Simplify[Together//@DistanceTrilinears[CircleEquation[X[1],Sqrt[Inradius[]^2+Semiperimeter[]^2]]],TriangleQ[]]]

Cancel[CircleFunction[((a^3) b (α^2)+(a^3) c (α^2)+2 (a^2) (b^2) α β+(a^2) b c α β+a (b^2) c α β+a b (c^2) α β+a (b^3) (β^2)+(b^3) c (β^2)+(a^2) b c α γ+a (b^2) c α γ+2 (a^2) (c^2) α γ+a b (c^2) α γ+(a^2) b c β γ+a (b^2) c β γ+a b (c^2) β γ+2 (b^2) (c^2) β γ+a (c^3) (γ^2)+b (c^3) (γ^2))]]

Derivation Given Points

Factor[Together//@CircleEquation[{{c(b+a),0,-(a^2)},{b(c+a),-(a^2),0},{0,c(a+b),-(b^2)}}]]

eqn=((a^3) b (α^2)+(a^3) c (α^2)+2 (a^2) (b^2) α β+(a^2) b c α β+a (b^2) c α β+a b (c^2) α β+a (b^3) (β^2)+(b^3) c (β^2)+(a^2) b c α γ+a (b^2) c α γ+2 (a^2) (c^2) α γ+a b (c^2) α γ+(a^2) b c β γ+a (b^2) c β γ+a b (c^2) β γ+2 (b^2) (c^2) β γ+a (c^3) (γ^2)+b (c^3) (γ^2));

Center

center=Together//@DistanceTrilinears[CircleCenter[eqn]]

Equal@@center

Symmetrize/@center

Radius

Simplify[Together//@DistanceTrilinears[CircleRadius[eqn]],TriangleQ[]]

Factor//@%

(Sqrt[Factor[Inradius[]^2+Semiperimeter[]^2]])