Garfunkel's Inequality

Author

Eric W. Weisstein
October 25, 2004

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

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

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

Graphic

Show[Graphics[{
t=Triangle[{{0,0},{1,0},{.1,.7}}],
l=Line[Coordinates/@{Incenter[t],#}]&/@VertexPoints[t];
p=Line[Coordinates/@{Centroid[t],#}]&/@VertexPoints[t];
t1=(Intersections[circ,#]&/@l)[[Sequence@@#]]&/@{{1,2},{2,2},{3,1}};
t2=(Intersections[circ,#]&/@p)[[Sequence@@#]]&/@{{1,2},{2,2},{3,1}};
{Red,{Dashing[{.02}],l},
Triangle[Coordinates/@t1]
},
{Blue,{Dashing[{.02}],p},
Triangle[Coordinates/@t2]
},
PointSize[.02],
{
Purple,
circ=Incircle[t]
},
PointLabels[Join[
Transpose[{VertexPoints[t],{"A","B","C"}}],
Append[#,Red]&/@Transpose[{
t1,{"U","V","W"}}],
Append[#,{Blue,LabelOffset->{0,-.05}}]&/@Transpose[{
t2,{"D","E","F"}}],
{
{Incenter[t],"I",Red},
{Centroid[t],"G",Purple}
}
],{LabelOffset->.04,PointSize[.025]}]
}],AspectRatio->Automatic,PlotRange->All]

Vertex-Incenter-Incircle Triangle

FullSimplify[Solve[
{
LineEquation[{Trilinears[Incenter],#}]==0,
CircleEquation[Incircle]==0,
γ==1
},{α,β,γ}],TriangleQ[]]&/@IdentityMatrix[3]

imat={{(((((b-c))^2)-2 (Sqrt[b (a+b-c) c (a-b+c)])-a (b+c))/(a (a-b-c))),1,1},
{1,(((a^2)-b c+(c^2)-2 (Sqrt[-a (a-b-c) (a+b-c) c])-a (b+2 c))/(b (-a+b-c))),1},
{((c (-(((a-b))^2)+(a+b) c-2 (Sqrt[a b (-(((a-b))^2)+(c^2))])))/((((a-b))^2) (a+b-c))),((c (-(((a-b))^2)+(a+b) c-2 (Sqrt[a b (-(((a-b))^2)+(c^2))])))/((((a-b))^2) (a+b-c))),1}};

pts=TrilinearToCartesian[t,#]/.Thread[{a,b,c}->SideLengths[Evaluate[t]]]&/@imat;

Show[Graphics[{
t,
circ,
l,
PointSize[.02],Point/@pts
}],AspectRatio->Automatic]

Perimeter

(isides=Simplify[ExpandAll/@SideLengths[Evaluate[imat]],TriangleQ[]])//Timing

isides/.Sqrt[x_]:>Simplify[Sqrt[LawOfCosines[Expand[x]]],TriangleQ[]]

Area

(iarea=FullSimplify[ExpandAll/@Area[Evaluate[imat]],TriangleQ[]])//Timing

FullSimplify[LawOfCosines[Expand[(a-b-c) (a+b-c)]]]

Vertex-Centroid-Incircle Triangle

FullSimplify[Solve[
{
LineEquation[{Trilinears[Centroid],#}]==0,
CircleEquation[Incircle]==0,
γ==1
},{α,β,γ}],TriangleQ[]]&/@IdentityMatrix[3]

gmat={
{-((2 c (a+(Sqrt[(a+b-c) (a-b+c)])))/(a (a-b-c))),(c/b),1},
{(c/a),((2 c (b+(Sqrt[-(a^2)+(b^2)+2 a c-(c^2)])))/(b (a-b+c))),1},
{(((a+b-c) c (c-(Sqrt[-(((a-b))^2)+(c^2)])))/(2 a (((a-b))^2))),(((a+b-c) c (c-(Sqrt[-(((a-b))^2)+(c^2)])))/(2 b (((a-b))^2))),1}
};

pts=TrilinearToCartesian[t,#]/.Thread[{a,b,c}->SideLengths[Evaluate[t]]]&/@gmat;

Show[Graphics[{
t,
circ,
p,
PointSize[.02],Point/@pts
}],AspectRatio->Automatic]

Perimeter

(gsides=Simplify[ExpandAll/@SideLengths[Evaluate[gmat]],TriangleQ[]])//Timing

Area

(garea=FullSimplify[ExpandAll/@Area[Evaluate[gmat]],TriangleQ[]])//Timing