Heilbronn Triangle Problem

Author

Eric W. Weisstein
January 23, 2009

This notebook downloaded from http://mathworld.wolfram.com/notebooks/ComputationalGeometry/HeilbronnTriangleProblem.nb.

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

©2009 Wolfram Research, Inc. except for portions noted otherwise

Additional Authors

David Cantrell
Ed Pegg, Jr.

Code

MinArea[v_]:=Min[RootReduce[TriangleArea[#]]&/@Subsets[v,{3}]]
TriangleArea[v_List?MatrixQ]:=Abs[Det[Append[#,1]&/@v]]/2/;Dimensions[v]=={3,2}

Square

Numerical

Heilbronn[Square,n_,opts___]:=Module[{x,v},
v=Array[x,{n,2}];
{First[#],Partition[Last/@#[[2]],2]}&@NMaximize[{Min[TriangleArea[#]&/@Subsets[v,{3}]],
And@@(0≤#≤1&/@Flatten[v])},
Flatten[v],
opts]
]

(Subscript[H, 3])

Heilbronn[Square,3]

(Subscript[H, 4])

Heilbronn[Square,4]

(Subscript[H, 5])

t=Table[Print[i," ",Timing[h=Heilbronn[Square,5,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

(Subscript[H, 6])

t=Table[Print[i," ",Timing[h=Heilbronn[Square,6,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

(Subscript[H, 7])

t=Table[Print[i," ",Timing[h=Heilbronn[Square,7,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

Yang et al. showed it must be ≥1/12 = 0.08333…

Analytic

triangles={
{{0,0},{1,1},{0,1}},
{{0,0},{1,0},{1,1},{0,1}},
{{0,0},{0,2/3},{Sqrt[3]/3,1},{1,Sqrt[3]/3},{2/3,0}},
{{0,1/2},{1/4,1},{3/4,1},{1,1/2},{3/4,0},{1/4,0}}
};

MinArea/@triangles

N[%]

N[{1/12,(2-Sqrt[3])/4,1/21,(3Sqrt[17]-11)/32,1/27,1/33}]

Old Figure

HeilbronnDiagram[y_,v_,Square]:=Module[{lines=Subsets[v,{2}],tris=Subsets[v,{3}],areas,t,u=1.3},
areas=Simplify[TriangleArea[#]&/@tris];
t=Flatten[Position[areas,Min[areas]]];
Table[
{
Line[{u(i-1),-u y}+#&/@{{0,0},{1,0},{1,1},{0,1},{0,0}}],
{Blue,Polygon[{u(i-1),-u y}+#&/@tris[[t[[i]]]]]},
Function[l,Line[{u(i-1),-u y}+#&/@l]]/@lines,
{Red,PointSize[.015],Point[{u(i-1),-u y}+#]&/@v}
},{i,Length[t]}
]
]

Show[Graphics[
MapIndexed[HeilbronnDiagram[#2[[1]],#1,Square]&,triangles]
],AspectRatio->Automatic]

New Figure

Heilbronn3 = {{0,0},{0,1},{1,1}};
Heilbronn4 = {{0,0},{0,1},{1,1},{1,0}};
Heilbronn5 = {{0.336124,1},{1,0.336140},{0,0.420215},{1,1},{0.420225,0}};
Heilbronn6 = {{0,0.497844},{1,0.498730},{0.377357,0},{0.879367,0},{0.620047,1},{0.121426,1}};
Heilbronn7 =With[{z=Root[1-5 #1+5 ((#1)^2)+((#1)^3)&,2]},{{-50/19z-17/38 z^2+37/38,0},{1,0},{0,z},{9/19+1/19z^2+7/19z,z},{40/19z^2+223/19 z-58/19,-1+6z+z^2},{58/19z-15/19+11/19z^2, 1},{1,1}}];
Heilbronn8=With[{z=Sqrt[13]},{{0,0},{(1+z)/6, 0},{1,(7-z)/18},{(5-z)/6,(7-z)/9}, {(1+z)/6,(2+z)/9},{0,(11+z)/18},{(5-z)/6,1},{1,1}}];
Heilbronn9=With[{z=Sqrt[65]},{{(10-z)/10,0},{(25+z)/40,0},{0,(15-z)/40},{1,(15-z)/40},{(15-z)/20,(5+z)/20},{0,(35+3z)/80},{1,z/10},{(45-3z)/80,1},{(25+z)/40,1}}];
Heilbronn10 =With[{x=Root[-1+10 #1-27 ((#1)^2)+24 ((#1)^3)&,1],y=Root[-3+2 #1+21 ((#1)^2)+72 ((#1)^3)&,1],z=Root[-4+20 #1-27 ((#1)^2)+12 ((#1)^3)&,1]},{{x, 0},{1-y, 0},{0 ,x},{1 ,y},{1-z, z},{z ,1-z},{0,1-y},{1,1-x},{y ,1},{1-x, 1}}];
Heilbronn11 = {{1/3,0},{2/3,0},{0,2/9},{1,2/9},{1/3,4/9},{2/3,4/9},{0,2/3},{1,2/3},{1/2,7/9},{1/6,1},{5/6,1}};
Heilbronn12 = With[{x = 1+ Root[1-2 #1+4 ((#1)^3)&,1],y =Root[-1+6 #1-4 ((#1)^2)+8 ((#1)^3)&,1] },{{x, 0},{1-x, 0},{0, x},{1, x},{1/2, y},{y ,1/2},{1-y ,1/2},{1/2,1-y},{0,1-x},{1,1-x},{x ,1},{1-x, 1}}];
Heilbronn13 = {{0,0.7567698216633331`},{1,0.7567698216633331`},{0.1753163783941777`,0},{0.8246836216058223`,0},{0.007742638743361058`,0.09683785018436035`},{0.992257361256639`,0.09683785018436035`},{0.3073866312393005`,0.24231400660377114`},{0.6926133687606995`,0.24231400660377114`},{0.2500860333763238`,0.6499433318585518`},{0.7499139666236763`,0.6499433318585518`},{0.07879960827922133`,0.936615842880567`},{0.9212003917207787`,0.936615842880567`},{(1/2),1}};
Heilbronn14 = With[{v=Root[-17+162 #1-288 ((#1)^2)+40 ((#1)^3)&,1],w=Root[-191+1248 #1-2124 ((#1)^2)+280 ((#1)^3)&,2],x = Root[1-12 #1-12 ((#1)^2)+8 ((#1)^3)&,2],y =Root[-7+54 #1-96 ((#1)^2)+40 ((#1)^3)&,1], z= Root[3+6 #1-72 ((#1)^2)+56 ((#1)^3)&,2]},{{x,0},{1-x,1},{1-x,0},{x,1},{0,y},{1,1-y},{1,y},{0,1-y},{z,w},{1-z,1-w},{1-z,w},{z,1-w},{1/2,v},{1/2,1-v}}];
Heilbronn15={{(2/31),(2/45)},{(29/31),1},{(23/31),0},{0,(16/45)},{1,(31/45)},{1,(2/45)},{0,1},{(8/31),(2/5)},{(23/31),(29/45)},{(10/31),(4/45)},{(21/31),(43/45)},{(21/31),(14/45)},{(10/31),(11/15)},{(29/31),(16/45)},{(2/31),(31/45)}};
Heilbronn16 = {{0,10/33},{0,31/33},{2/31,0},{2/31,7/11},{8/31,4/11},{8/31,1},{10/31,2/33},{10/31,23/33},{21/31,10/33},{21/31,31/33},{23/31,0},{23/31,7/11},{29/31,4/11},{29/31,1},{1,2/33},{1,23/33}};

GraphicsGrid[Partition[Graphics[{Black,AbsoluteThickness[.25], Line/@Tuples[#,{2}],Black,Thick,Line[{{0,0},{1,0},{1,1},{0,1},{0,0}}],Red,Disk[#,.04]&/@#}]&/@{Heilbronn5,Heilbronn6,Heilbronn7,Heilbronn8,Heilbronn9,Heilbronn10,Heilbronn11,Heilbronn12,Heilbronn13,Heilbronn14,Heilbronn15,Heilbronn16},4],ImageSize-> 400]

Quiet[TableForm[MapIndexed[{#2[[1]]+4,TraditionalForm[#1[[2]]],#1[[1]],InputForm[#1[[2]]]}&,(First[Sort[Map[{Abs[N[1/2 Det[Map[Join[#,{1}]&,#]]]],RootReduce[Abs[1/2 Det[Map[Join[#,{1}]&,#]]]]}&,Subsets[#,{3}]]]]&/@{Heilbronn5,Heilbronn6,Heilbronn7,Heilbronn8,Heilbronn9,Heilbronn10,Heilbronn11,Heilbronn12,Heilbronn13,Heilbronn14,Heilbronn15,Heilbronn16})]]/.#1-> x]

Disk

Numerical

Heilbronn[Disk,n_,opts___]:=Module[{x,v},
v=Array[x,{n,2}];
{First[#],Partition[Last/@#[[2]],2]}&@NMaximize[{Min[TriangleArea[#]&/@Subsets[v,{3}]],
And@@(#.#≤1&/@v)},
Flatten[v],
opts]
]

(Subscript[H, 3])

Heilbronn[Disk,3]

#.#&/@{{-0.11120569556281418`,0.9937973998382287`},{-0.8050510115533529`,-0.5932055897915388`},{0.9162565455456642`,-0.4005919692274154`}}

(Subscript[H, 4])

Heilbronn[Disk,4]

(Subscript[H, 5])

t=Table[Print[i," ",Timing[h=Heilbronn[Disk,5,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

(Subscript[H, 6])

t=Table[Print[i," ",Timing[h=Heilbronn[Disk,6,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

Old Figure

HeilbronnDiagram[y_,v_,Disk]:=Module[{lines=Subsets[v,{2}],tris=Subsets[v,{3}],areas,t,u=2.4},
areas=RootReduce[TriangleArea[#]&/@tris];
t=Flatten[Position[areas,Min[areas]]];
Table[
{
Circle[{u(i-1),-u y},1],
{Blue,Polygon[{u(i-1),-u y}+#&/@tris[[t[[i]]]]]},
Function[l,Line[{u(i-1),-u y}+#&/@l]]/@lines,
{Red,PointSize[.018],Point[{u(i-1),-u y}+#]&/@v}
},{i,Length[t]}
]
]

disktriangles={
Through[{Sin,Cos}[2Pi #/3]]&/@Range[0,2],
Through[{Sin,Cos}[2Pi #/4]]&/@Range[0,3],
Through[{Sin,Cos}[2Pi #/5]]&/@Range[0,4],
Through[{Sin,Cos}[2Pi #/6]]&/@Range[0,5]
};

ToRadicals[MinArea/@disktriangles/Pi]

N[%]

Show[Graphics[
MapIndexed[HeilbronnDiagram[#2[[1]],#1,Disk]&,disktriangles]
],AspectRatio->Automatic]

New Figure

Unit Circles

HeilbronnCircle = {{{(1/2) (Sqrt[((5+(Sqrt[5]))/(2 π))]),((-1+(Sqrt[5]))/(4 (Sqrt[π])))},{(1/2) (Sqrt[((5-(Sqrt[5]))/(2 π))]),((-1-(Sqrt[5]))/(4 (Sqrt[π])))},{-(1/2) (Sqrt[((5-(Sqrt[5]))/(2 π))]),((-1-(Sqrt[5]))/(4 (Sqrt[π])))},{-(1/2) (Sqrt[((5+(Sqrt[5]))/(2 π))]),((-1+(Sqrt[5]))/(4 (Sqrt[π])))},{0,(1/(Sqrt[π]))}},

{{((Sqrt[(3/π)])/2),(1/(2 (Sqrt[π])))},{((Sqrt[(3/π)])/2),-(1/(2 (Sqrt[π])))},{0,-(1/(Sqrt[π]))},{-((Sqrt[(3/π)])/2),-(1/(2 (Sqrt[π])))},{-((Sqrt[(3/π)])/2),(1/(2 (Sqrt[π])))},{0,(1/(Sqrt[π]))}},

{{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,5])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,3])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,6])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,2])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,4])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,1])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,3])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,1])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,1])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,2])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,2])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,3])/(Sqrt[π]))},{0,(1/(Sqrt[π]))}},

{{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,5])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,3])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,6])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,2])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,4])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,1])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,3])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,1])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,1])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,2])/(Sqrt[π]))},{((Root[-7+56 ((#1)^2)-112 ((#1)^4)+64 ((#1)^6)&,2])/(Sqrt[π])),((Root[-1-4 #1+4 ((#1)^2)+8 ((#1)^3)&,3])/(Sqrt[π]))},{0,(1/(Sqrt[π]))},{0,0}},

{{0,0.1150062412`},{0.2649178125`,0.4981249269`},{0.5188437929`,0.221610044`},{0.1640523245`,-0.5398117464`},{0.5083974895`,-0.2026588249`},{-0.2649178125`,0.4981249269`},{-0.5188437929`,0.221610044`},{-0.1640523245`,-0.5398117464`},{-0.5083974895`,-0.2026588249`}},

{{((Sin[((2 π)/9)])/(Sqrt[π])),-((Cos[((2 π)/9)])/(Sqrt[π]))},{((Cos[(π/18)])/(Sqrt[π])),-((Sin[(π/18)])/(Sqrt[π]))},{((Sqrt[(3/π)])/2),(1/(2 (Sqrt[π])))},{((Sin[(π/9)])/(Sqrt[π])),((Cos[(π/9)])/(Sqrt[π]))},{-((Sin[(π/9)])/(Sqrt[π])),((Cos[(π/9)])/(Sqrt[π]))},{-((Sqrt[(3/π)])/2),(1/(2 (Sqrt[π])))},{-((Cos[(π/18)])/(Sqrt[π])),-((Sin[(π/18)])/(Sqrt[π]))},{-((Sin[((2 π)/9)])/(Sqrt[π])),-((Cos[((2 π)/9)])/(Sqrt[π]))},{0,-(1/(Sqrt[π]))},{0,0}},

{{0,0.1167113202429419`},{0.4407136731028269`,0.3522518197596789`},{0.19882337391708982`,0.5279954092300574`},{0.14949642644566685`,-0.5440227060185687`},{0.47119940320144693`,-0.31029181201957445`},{0.5635932759415178`,-0.025932710951593926`},{-0.4407136731028269`,0.3522518197596789`},{-0.19882337391708982`,0.5279954092300574`},{-0.14949642644566685`,-0.5440227060185687`},{-0.47119940320144693`,-0.31029181201957445`},{-0.5635932759415178`,-0.025932710951593926`}},

{{0,(1/(Sqrt[π]))},{0.3379176623`,0.451798118289014`},{-0.3379176623`,0.451798118289014`},{0,0.2325606754`},{-((Sqrt[(3/π)])/2),-(1/(2 (Sqrt[π])))},{-0.5602274789702928`,0.06674622079474407`},{-0.22230981667029287`,-0.5185443390837581`},{-0.20140345281766675`,-0.1162803377`},{((Sqrt[(3/π)])/2),-(1/(2 (Sqrt[π])))},{0.22230981667029287`,-0.5185443390837581`},{0.5602274789702928`,0.06674622079474407`},{0.20140345281766675`,-0.1162803377`}},

{{-0.21504982224067237`,0.5215970284980982`},{-0.15466490450029988`,-0.5425759426105288`},{0.5057608253448745`,0.25003174544497747`},{0.5641339568912458`,0.00792242803796609`},{-0.2267901367842186`,-0.1745702049221536`},{-0.4291205355721967`,-0.3662860250323806`},{0,0.30984036406200965`},{0.21504982224067237`,0.5215970284980982`},{0.15466490450029988`,-0.5425759426105288`},{-0.5057608253448745`,0.25003174544497747`},{-0.5641339568912458`,0.00792242803796609`},{0.2267901367842186`,-0.1745702049221536`},{0.4291205355721967`,-0.3662860250323806`}},

{{0.1178970343800088`,0.5517337903991287`},{0.5595334113412647`,0.07233427801946847`},{0.296888883180507`,-0.47975710232118707`},{0.4220467297220386`,0.3309254629566527`},{0.5202575727603586`,-0.20504415468139423`},{0.178991848800498`,-0.07014352589173518`},{-0.1178970343800088`,0.5517337903991287`},{-0.5595334113412647`,0.07233427801946847`},{-0.296888883180507`,-0.47975710232118707`},{-0.4220467297220386`,0.3309254629566527`},{-0.5202575727603586`,-0.20504415468139423`},{-0.178991848800498`,-0.07014352589173518`},{0,0.1421202139696768`},{0,-(1/(Sqrt[π]))}},

{{0.2730219342456227`,0.4870921567028016`},{0.45033724486238264`,0.3398621074399868`},{0.541455619680651`,0.012739157675487556`},{0.4877059016670484`,-0.2836420978326772`},{0.30231308639185206`,-0.47635772690282174`},{0.14670632656297244`,0.1878967979719508`},{0.15671724542124976`,-0.1413841589751206`},{-0.2730219342456226`,0.4870921567028016`},{-0.4503372448623826`,0.33986210743998685`},{-0.541455619680651`,0.012739157675487667`},{-0.48770590166704847`,-0.28364209783267713`},{-0.30231308639185217`,-0.47635772690282174`},{-0.1467063265629724`,0.1878967979719508`},{-0.1567172454212498`,-0.14138415897512058`},{0,-0.5501033678313902`}},

{{(1/2),Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]},{Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2],0},{Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],(1/2)},{Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],-(1/2)},{Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1]},{0,Root[-3+8 #1+84 ((#1)^2)-156 ((#1)^3)+60 ((#1)^4)+8 ((#1)^5)&,2]},{(1/2),Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1]},{-(1/2),Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1]},{Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1],Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1]},{Root[-3+8 #1+84 ((#1)^2)-156 ((#1)^3)+60 ((#1)^4)+8 ((#1)^5)&,2],0},{Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1],-(1/2)},{Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1],(1/2)},{Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1],Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{0,Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2]},{-(1/2),Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]}} ((1/Sqrt[Pi])/(√((1/4)+((Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3])^2))))};

GraphicsGrid[Partition[Graphics[{Black,AbsoluteThickness[.25], Line/@Tuples[#,{2}],Black,Thick,Circle[{0,0},(1/(Sqrt[π]))],Red,Disk[#,.04]&/@#}]&/@HeilbronnCircle,4],ImageSize-> 400]

Quiet[TableForm[MapIndexed[{#2[[1]]+4,TraditionalForm[RootReduce[FullSimplify[#1[[2]]]]],#1[[1]],InputForm[FullSimplify[#1[[2]]]]}&,(First[Sort[Map[{Abs[N[1/2 Det[Map[Join[#,{1}]&,#]]]],RootReduce[Abs[1/2 Det[Map[Join[#,{1}]&,#]]]]}&,Subsets[#,{3}]]]]&/@( HeilbronnCircle) )]]/.#1-> x]

In this second notebook, I look at the Heilbronn triangle problem in a circle.

Initialize:

tri[a_,b_,c_,rd_,gr_,bl_]:={{RGBColor[rd,gr,bl],Polygon[{ptlist[[a]],ptlist[[b]],ptlist[[c]]}]},GrayLevel[0],Line[{ptlist[[a]],ptlist[[b]],ptlist[[c]],ptlist[[a]]}]}

N = 3..7, area = (1/π)Sin[((2π)/N)](1-Cos[((2π)/N)]), complete symmetry

The points are just the vertices of a circumscribed regular N-gon. If you wanted to show one example, it could be

n=7;Clear[x,y,z];Do[z[j]=((-I)/(Sqrt[π]))(E^((2π I j)/n));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n}];ptlist=Table[{x[j],y[j]},{j,1,n}];ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[6,7,1,1,0,0],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False]

with area (1/π)Sin[((2π)/7)](1-Cos[((2π)/7)]) = 0.093700+.

N = 8, area = (1/(2π))Sin[(π/7)] = 0.069054+, complete symmetry

Note that a configuration using the vertices of a regular inscribed octagon is not optimal, the smallest triangles having area =(((Sqrt[2])-1)/(2 π)) = 0.065924+.

n=8;Clear[x,y,z];Do[z[j]=((-I)/(Sqrt[π]))(E^((2π I j)/(n-1)));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n-1}];{x[n],y[n]}={0,0};ptlist=Table[{x[j],y[j]},{j,1,n}];ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,5,8,0,1,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False]

N = 9, area = 0.055310+, horizontal symmetry

This configuration is better than the one which I'd mentioned in the AZPC group. That former configuration had assumed that all but one of the points were on the circle. Under that assumption, my former configuration was optimal, I suppose. But if you look closely at the new configuration, you'll see that two of the points are close to, but not on, the circle.

ptlist={{0,0.1150062412`},{0.2649178125`,0.4981249269`},{0.5188437929`,0.221610044`},{0.1640523245`,-0.5398117464`},{0.5083974895`,-0.2026588249`},{-0.2649178125`,0.4981249269`},{-0.5188437929`,0.221610044`},{-0.1640523245`,-0.5398117464`},{-0.5083974895`,-0.2026588249`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,9,0,0,1],tri[1,2,8,0,1,0],tri[2,3,5,0,1,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,3,9,1,0,0],tri[1,3,7,1,1,0],tri[4,8,5,1,1/2,0],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

N = 10, area = (1/π)Sin[((2π)/9)](1-Cos[((2π)/9)]) = 0.047868+, complete symmetry

This simple configuration is like the one for N = 8, except that the smallest triangles don't have a vertex at the center here. It's better than the one which I'd mentioned in the AZPC group; I don't know how I overlooked this before.

n=10;Clear[x,y,z];Do[z[j]=((-I)/(Sqrt[π]))(E^((2π I j)/(n-1)));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n-1}];{x[n],y[n]}={0,0};ptlist=Table[{x[j],y[j]},{j,1,n}]

n=10;Clear[x,y,z];Do[z[j]=((-I)/(Sqrt[π]))(E^((2π I j)/(n-1)));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n-1}];{x[n],y[n]}={0,0};ptlist=Table[{x[j],y[j]},{j,1,n}];ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[8,9,1,0,3/4,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False]

N = 11, area = 0.034941+, horizontal symmetry

ptlist={{0,0.1167113202429419`},{0.4407136731028269`,0.3522518197596789`},{0.19882337391708982`,0.5279954092300574`},{0.14949642644566685`,-0.5440227060185687`},{0.47119940320144693`,-0.31029181201957445`},{0.5635932759415178`,-0.025932710951593926`},{-0.4407136731028269`,0.3522518197596789`},{-0.19882337391708982`,0.5279954092300574`},{-0.14949642644566685`,-0.5440227060185687`},{-0.47119940320144693`,-0.31029181201957445`},{-0.5635932759415178`,-0.025932710951593926`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,3,9,0,0,1],tri[3,7,8,0,1,0],tri[7,10,11,0,1,1],tri[2,3,6,1,1,0],tri[4,5,6,1,0,0],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,11,1,1/2,0],tri[4,9,10,1/2,1/2,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

N = 12, area = 0.033395+, symmetry of an equilateral triangle

In the literature I've seen, there's not much about Heilbronn in a circle. However, Goldberg does show a configuration for N = 12, with area 0.02442. His configuration looks a bit like mine, but his points on the circle are the vertices of a regular enneagon.

ptlist={{0,(1/(Sqrt[π]))},{0.3379176623`,0.451798118289014`},{-0.3379176623`,0.451798118289014`},{0,0.2325606754`},{-((Sqrt[(3/π)])/2),-(1/(2 (Sqrt[π])))},{-0.5602274789702928`,0.06674622079474407`},{-0.22230981667029287`,-0.5185443390837581`},{-0.20140345281766675`,-0.1162803377`},{((Sqrt[(3/π)])/2),-(1/(2 (Sqrt[π])))},{0.22230981667029287`,-0.5185443390837581`},{0.5602274789702928`,0.06674622079474407`},{0.20140345281766675`,-0.1162803377`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,7,8,0,1,1],tri[1,4,12,0,1,0],tri[2,10,12,0,0,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

N = 13, area = 0.0856+, horizontally symmetric

ptlist={{-0.21504982224067237`,0.5215970284980982`},{-0.15466490450029988`,-0.5425759426105288`},{0.5057608253448745`,0.25003174544497747`},{0.5641339568912458`,0.00792242803796609`},{-0.2267901367842186`,-0.1745702049221536`},{-0.4291205355721967`,-0.3662860250323806`},{0,0.30984036406200965`},{0.21504982224067237`,0.5215970284980982`},{0.15466490450029988`,-0.5425759426105288`},{-0.5057608253448745`,0.25003174544497747`},{-0.5641339568912458`,0.00792242803796609`},{0.2267901367842186`,-0.1745702049221536`},{0.4291205355721967`,-0.3662860250323806`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[7,8,11,1,1,0],tri[6,7,8,0,1,0],tri[3,5,6,0,0,1],tri[2,4,12,1,0,0],tri[2,9,13,1/2,0,1],tri[1,10,11,1,1/2,0],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->225]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[3,4,13,0,1,1],tri[1,2,5,1,0,1/2],tri[12,13,7,1/2,1,1/2],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->225]

N = 14, area = 0.0758+, horizontal symmetry

IIRC, in my previous Heilbronn submissions, the triangle area I gave was for a configuration in a unit-area circle. But knowing that your site deals with unit-radius circles, the area shown above matches your convention:

0.024146112951410725` π

tri[a_,b_,c_,rd_,gr_,bl_]:={{RGBColor[rd,gr,bl],Polygon[{ptlist[[a]],ptlist[[b]],ptlist[[c]]}]},GrayLevel[0],Line[{ptlist[[a]],ptlist[[b]],ptlist[[c]],ptlist[[a]]}]};ptlist={{0.1178970343800088`,0.5517337903991287`},{0.5595334113412647`,0.07233427801946847`},{0.296888883180507`,-0.47975710232118707`},{0.4220467297220386`,0.3309254629566527`},{0.5202575727603586`,-0.20504415468139423`},{0.178991848800498`,-0.07014352589173518`},{-0.1178970343800088`,0.5517337903991287`},{-0.5595334113412647`,0.07233427801946847`},{-0.296888883180507`,-0.47975710232118707`},{-0.4220467297220386`,0.3309254629566527`},{-0.5202575727603586`,-0.20504415468139423`},{-0.178991848800498`,-0.07014352589173518`},{0,0.1421202139696768`},{0,-(1/(Sqrt[π]))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,4,0,0,1],tri[1,3,6,0,1,0],tri[1,9,12,0,1,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,12,13,1,0,0],tri[5,6,12,1,1,0],tri[5,6,13,1,1/2,0],tri[2,4,5,2/3,1/2,0],tri[12,14,10,0,1/2,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,6,11,1,0,1/2],tri[4,11,13,1/2,1,0],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,9,13,1,0,1],tri[3,6,13,1,1,0],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False,ImageSize->200]

N = 15, area = 0.022294+, horizontal symmetry

ptlist={{0.2730219342456227`,0.4870921567028016`},{0.45033724486238264`,0.3398621074399868`},{0.541455619680651`,0.012739157675487556`},{0.4877059016670484`,-0.2836420978326772`},{0.30231308639185206`,-0.47635772690282174`},{0.14670632656297244`,0.1878967979719508`},{0.15671724542124976`,-0.1413841589751206`},{-0.2730219342456226`,0.4870921567028016`},{-0.4503372448623826`,0.33986210743998685`},{-0.541455619680651`,0.012739157675487667`},{-0.48770590166704847`,-0.28364209783267713`},{-0.30231308639185217`,-0.47635772690282174`},{-0.1467063265629724`,0.1878967979719508`},{-0.1567172454212498`,-0.14138415897512058`},{0,-0.5501033678313902`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,3,0,0,1],tri[3,4,5,1,0,0],tri[1,6,7,0,1,0],tri[1,10,13,1,1,0],tri[5,12,15,1/2,0,1],tri[13,12,14,0,1,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,3,4,0,1,0],tri[9,6,13,1,1/2,0],tri[4,10,14,1/2,0,1],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,10,13,0,1/2,1/2],tri[7,14,11,1,0,0],tri[2,7,15,1,1,0],tri[4,5,15,1/2,1,1/2],Circle[{0,0},(1/(Sqrt[π]))]},PlotRange->All,Axes->False]

N = 16, area = 0.021051+, symmetry of a square

You may have noticed that, for Heilbronn in circles, I haven't given many precise coordinates or areas. Being precise throughout would have taken much longer and things would have been very messy. But here at the end of the notebook, I will be precise. The area of the smallest triangles is exactly Root[-9+103 #1+452 ((#1)^2)+476 ((#1)^3)+3776 ((#1)^4)+976 ((#1)^5)&,3]/π.

BTW, there's an interesting story behind this configuration. I began thinking about Goldberg's configuration for 16 in a square (since its points can be nicely inscribed in a circle):

Some discrepancy between his work and mine seemed to be lurking. Finally, I uncovered the problem. Goldberg chose the 4 small triangles shown and equated their areas to get his configuration. Unfortunately, he made one bad choice, shown in red above. That caused his area to be (-5+(Sqrt[13])+2 (Sqrt[-122+34 (Sqrt[13])]))/8 = 0.017517+. If he had instead chosen the triangle corresponding to the one in red in my configuration, he would have gotten a configuration with area Root[-9+356 #1+8496 ((#1)^2)+23104 ((#1)^3)+32000 ((#1)^4)+1024 ((#1)^5)&,3]= 0.017561+.

ptlist={{(1/2),Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]},{Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2],0},{Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],(1/2)},{Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],-(1/2)},{Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1]},{0,Root[-3+8 #1+84 ((#1)^2)-156 ((#1)^3)+60 ((#1)^4)+8 ((#1)^5)&,2]},{(1/2),Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1]},{-(1/2),Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1]},{Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1],Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1]},{Root[-3+8 #1+84 ((#1)^2)-156 ((#1)^3)+60 ((#1)^4)+8 ((#1)^5)&,2],0},{Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1],-(1/2)},{Root[1+3 #1-48 ((#1)^2)-64 ((#1)^3)-16 ((#1)^4)+16 ((#1)^5)&,1],(1/2)},{Root[-9-180 #1-724 ((#1)^2)-1024 ((#1)^3)-560 ((#1)^4)+64 ((#1)^5)&,1],Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{0,Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2]},{-(1/2),Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[9,16,10,0,0,1],tri[16,7,11,1,1,0],tri[8,15,14,1,0,0],tri[1,2,4,0,1,0],Circle[{0,0},√((1/4)+((Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3])^2))]},PlotRange->All,Axes->False]

Equilateral Triangle

#1{1,Sqrt[3]}/2+#2{1,0}

Simplify[Solve[y-y1==(y2-y1)/(x2-x1)(x-x1)/.{x1->1,y1->0,x2->1/2,y2->Sqrt[3]/2},y]]

p=Sqrt[3]/3Through[{Cos,Sin}[2Pi(#-1/4)/3]]&/@Range[0,2]

p[[2]]-p[[3]]

Numerical

Heilbronn[Triangle,n_,opts___]:=Module[{x,v},
v=Array[x,{n,2}];
{First[#]4/Sqrt[3],Partition[Last/@#[[2]],2]}&@NMaximize[{
Min[
Function[t,TriangleArea[#1{1,Sqrt[3]}/2+#2{1,0}&@@@t]]/@Subsets[v,{3}]
],
And@@(0≤#≤1&/@Flatten[v])&&And@@(#[[2]]<Sqrt[3](1-#[[1]])&/@(#1{1,Sqrt[3]}/2+#2{1,0}&@@@v))},
Flatten[v],
opts]
]

(Subscript[H, 3])

Heilbronn[Triangle,3]

(Subscript[H, 4])

t=Table[Print[Timing[h=Heilbronn[Triangle,4,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,10}];

(Subscript[H, 5])

t=Table[Print[Timing[h=Heilbronn[Triangle,5,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

3-2Sqrt[2.]

Heilbronn[Triangle,5,Method->{"NelderMead","RandomSeed"->3}]

{Min[#],#}&[4/Sqrt[3]Function[t,area[#1{1,Sqrt[3]}/2+#2{1,0}&@@@t]]/@KSubsets[%[[2]],3]]

Heilbronn[Triangle,5,Method->{"NelderMead","RandomSeed"->3},WorkingPrecision->20]//Timing

eqns={TriangleArea[{{x,0},{1,0}+y{-(1/2),((Sqrt[3])/2)},{1,0}+(1-y){-(1/2),((Sqrt[3])/2)}}],
TriangleArea[{{0,0},{1,0}+(1-y){-(1/2),((Sqrt[3])/2)},x{1,Sqrt[3]}/2}]
}

FullSimplify[Solve[Equal@@Append[eqns,(3-2Sqrt[2])Sqrt[3]/4],{x,y}]]

N[%]

{{1/Sqrt[2],1-1/Sqrt[2]},{0,2-(Sqrt[2])},{2-(Sqrt[2]),0},{1-1/Sqrt[2],1/Sqrt[2]},{0,0}}

(Subscript[H, 6])

t=Table[Print[i," ",Timing[h=Heilbronn[Triangle,6,Method->{"NelderMead","RandomSeed"->i}]]];h,{i,20}];

Sort[t][[-1]]

Heilbronn[Triangle,6,Method->{"NelderMead","RandomSeed"->6}]

Heilbronn[Triangle,6,Method->{"NelderMead","RandomSeed"->6}]

{Min[#],#}&[4/Sqrt[3]Function[t,TrianleArea[#1{1,Sqrt[3]}/2+#2{1,0}&@@@t]]/@KSubsets[%[[2]],3]]

f[x_,y_,z_?NumericQ]:=Min[4/Sqrt[3]Function[t,TriangleArea[#1{1,Sqrt[3]}/2+#2{1,0}&@@@t]]/@KSubsets[{{0,0},{y,0},{x,x},{1-x,x},{0,z},{x,1-x}},3]]

FindMaximum[f[x,y,z],{x,.227,.226,.235},{y,.607,.606,.65},{z,.710,.709,.711},WorkingPrecision->30,PrecisionGoal->10]

FindMaximum[f[x,y,z],{x,.227,.226},{y,.607,.606},{z,.710,.709},WorkingPrecision->30,PrecisionGoal->10]

With[{x=1/4,y=3/4,z=2/3},
Min[4/Sqrt[3]Function[t,TriangleArea[#1{1,Sqrt[3]}/2+#2{1,0}&@@@t]]/@KSubsets[{{0,0},{y,0},{x,x},{1-x,x},{0,z},{x,1-x}},3]]
]

With[{x=1/4,y=3/4,z=2/3},
{{0,0},{y,0},{x,x},{1-x,x},{0,z},{x,1-x}}
]

Old Figures

HeilbronnDiagram[y_,v_,Triangle]:=Module[{lines=Subsets[v,{2}],tris=Subsets[v,{3}],areas,t,u=1.2},
areas=RootReduce[TriangleArea[#]&/@tris];
t=Flatten[Position[areas,_?(Abs[#-Min[areas]]<.01&),{1}]];
Table[
{
Line[{u(i-1),-u y}+{1/2,Sqrt[3]/6}+#&/@(Sqrt[3]/3Through[{Cos,Sin}[2Pi(#-1/4)/3]]&/@Range[0,3])],
{Blue,Polygon[{u(i-1),-u y}+#&/@tris[[t[[i]]]]]},
Function[l,Line[{u(i-1),-u y}+#&/@l]]/@lines,
{Red,PointSize[.015],Point[{u(i-1),-u y}+#]&/@v}
},{i,Length[t]}
]
]

tritriangles=Map[#[[1]]{1,Sqrt[3]}/2+#[[2]]{1,0}&,
{
{{1,0},{0,0},{0,1}},
{{0,1},{0,0},{1/3,1/3},{1,0}},
{{1/Sqrt[2],1-1/Sqrt[2]},{0,2-(Sqrt[2])},{2-(Sqrt[2]),0},{1-1/Sqrt[2],1/Sqrt[2]},{0,0}},
{{0,0},{(3/4),0},{(1/4),(1/4)},{(3/4),(1/4)},{0,(2/3)},{(1/4),(3/4)}}
},
{2}];

FullSimplify[4/Sqrt[3] MinArea/@tritriangles]

N[%]

Show[Graphics[
MapIndexed[HeilbronnDiagram[#2[[1]],#1,Triangle]&,tritriangles]
],AspectRatio->Automatic,PlotRange->All]

New Figures

Unit Triangles

HeilbronnTriangle = {{{(-2 + Sqrt[2])/3^(1/4), (-4 + 3*Sqrt[2])/3^(3/4)}, {(1 - Sqrt[2])/3^(1/4), -3^(-3/4)}, {-((-2 + Sqrt[2])/3^(1/4)), (-4 + 3*Sqrt[2])/3^(3/4)}, {(-1 + Sqrt[2])/3^(1/4), -3^(-3/4)},  {0, 2/3^(3/4)}},

{{1/(2*3^(1/4)),-3^(-3/4)},{-1/(2*3^(1/4)),-3^(-3/4)},{1/(4*3^(1/4)),5/(4*3^(3/4))},{3^(3/4)/4,-1/(4*3^(3/4))},{-3^(3/4)/4,-1/(4*3^(3/4))},{-1/(4*3^(1/4)),5/(4*3^(3/4))}},

{{2/(3*3^(1/4)),-3^(-3/4)},{3^(3/4)/4,-1/(4*3^(3/4))},{1/(6*3^(1/4)),3^(1/4)/2},{-1/(4*3^(1/4)),5/(4*3^(3/4))},{-5/(6*3^(1/4)),-1/(2*3^(3/4))},{-1/(2*3^(1/4)),-3^(-3/4)},{0,0}},

{{-3^(-1/4),-3^(-3/4)},{0.5236274599,-3^(-3/4)},{-0.0846397964,0.7307822477},{-0.2450656014,-0.2912911357},{0.6747341394,-0.2912911357},{0.1371228657,0.0815218223},{0.2123263582,0.5096226351},{-0.3101032349,0.1394695068}},

{{0.597013737,-3^(-3/4)},{0.6648562315,-0.2741820973},{0.1628219297,0.1253403637},{0.08141097432579625,0.7363747314756971},{-0.09497945422530327,0.7128734349933876},{-0.1899589039337809,0.07833774556340398},{-0.6784247113257963,-0.2976833938248663},{-0.5698767772746967,-3^(-3/4)},{0.02713697423378089,-0.203678109263404}},

{{-0.4239530851`,-(1/(3^(3/4)))},{0.6572305907`,-(1/(3^(3/4)))},{0.5113077447`,-0.008228316822194603`},{-0.4499289558`,0.09808286405964994`},{-0.5904406817`,-0.3381065411`},{0.1915018448`,0.3406624472`},{-0.0940212592`,0.3808833609`},{0.0006283394`,-0.0419999715`},{0.2720503627`,-0.3381065411`},{0,(2/(3^(3/4)))}},

{{0.3627654674443981,-3^(-3/4)},{0.5452866147601714,-0.06708144615019251},{0.11345362658894519,0.6808752297466613},{0.156604360381382,0.19756294694637877},{0.6058832138845257,-0.33919819056486644},{-0.3627654674443981,-3^(-3/4)},{-0.5452866147601714,-0.06708144615019251},{-0.11345362658894519,0.6808752297466613},{-0.156604360381382,0.19756294694637877},{-0.6058832138845257,-0.33919819056486644},{0,-0.13327173403398437}},

{{-0.3828761953531324`,-(1/(3^(3/4)))},{-0.5713559405023625`,-0.11223484285473073`},{-0.1540971234828158`,0.0889680157241507`},{0,0.7154254571730748`},{0.5713559405023625`,-0.11223484285473068`},{0.38287619535313244`,-(1/(3^(3/4)))},{0,-0.1779360314483014`},{-0.6195766204259787`,-0.3577127285865374`},{-0.18847974514923008`,0.5509261805055614`},{0.18847974514923005`,0.5509261805055615`},{0.1540971234828158`,0.0889680157241507`},{0.6195766204259787`,-0.3577127285865374`}},

{{-0.19260874`,-(1/(3^(3/4)))},{0.0942069387`,-0.3014895661`},{0.1552028182`,0.6085635087`},{0.6074897898`,-0.3278596045`},{0.47622221282579624`,0.05254160699450344`},{0.21399415386854767`,0.23233038517696336`},{-0.6046328674503922`,-0.16987217104986216`},{-0.01981014852828117`,0.6900313927564687`},{-0.2836134728257963`,0.38614973065632735`},{-0.30820109256854766`,0.06915918092303663`},{0.4494300492503922`,-0.4386913376501378`},{-0.5876796412717189`,-0.3621717882564688`},{0,0}},

{{-0.3904545987`,-(1/(3^(3/4)))},{-0.1904885733`,-0.1532488072`},{0.6524136764`,-0.378782794`},{-0.0064065584`,-0.3433769196`},{0.5531430703`,-0.0806892263`},{0.2614131862`,-0.0452096335`},{0.0325763013`,0.8209588662`},{0.2754746774`,0.2740947036`},{-0.5691154992`,-0.1083542847`},{-0.1403958289`,0.6342099665`},{-0.0250107551`,0.320761482`},{-0.6620957954`,-0.3803161242`},{-0.2852346791`,0.202926372`},{0.4241299371`,-(1/(3^(3/4)))}},

{{-0.38761748842791954`,-(1/(3^(3/4)))},{-0.18760767830514943`,-0.16578440610915707`},{0,0.7685627508192401`},{0,-0.352595053943903`},{0.38761748842791954`,-(1/(3^(3/4)))},{0.5737265870397561`,-0.11634092310428357`},{0.23737734639442082`,-0.07958081230269956`},{-0.6655948666119113`,-0.38428137540962004`},{0.3053562739641645`,0.1762975269719515`},{0.1861090986118365`,0.5550322607551144`},{-0.1861090986118365`,0.5550322607551144`},{-0.04976966808927137`,0.24536521841185663`},{0.6655948666119113`,-0.38428137540962004`},{-0.3053562739641645`,0.1762975269719515`},{-0.5737265870397561`,-0.11634092310428357`}},

{{0.5855611067622228`,-(1/(3^(3/4)))},{-0.37458685982327644`,-0.02344929700189427`},{0.3435082898333592`,-0.0765241545484251`},{-0.08914493930189583`,0.7229791111931344`},{0.0005649738876899437`,-0.4078864888841451`},{0.08713728944468485`,0.7264564627496322`},{0.20760111681616494`,-0.3126770880298508`},{-0.10548228307461696`,0.3357489826804494`},{-0.5815458070478009`,-(1/(3^(3/4)))},{0.3529575742902637`,0.20443252618128688`},{-0.6726983962069077`,-0.28776512509880137`},{0.1669857430071115`,0.336126385031745`},{-0.23802600675874222`,-0.2592248281320243`},{0.6706907463496967`,-0.2842877735423036`},{-0.35352254817795364`,0.2034539627028582`},{0,0}}

};

unitareatriangle=Line[{{(1/(3^(1/4))),-((3^(1/4))/3)},{-(1/(3^(1/4))),-((3^(1/4))/3)},{0,((2(3^(1/4)))/3)},{(1/(3^(1/4))),-((3^(1/4))/3)}}];

GraphicsGrid[Partition[Graphics[{Black,AbsoluteThickness[.25], Line/@Tuples[#,{2}],Black,Thick,unitareatriangle,Red,Disk[#,.04]&/@#}]&/@HeilbronnTriangle,4],ImageSize-> 400]

Quiet[TableForm[MapIndexed[{#2[[1]]+4,TraditionalForm[RootReduce[FullSimplify[#1[[2]]]]],#1[[1]],InputForm[FullSimplify[#1[[2]]]]}&,(First[Sort[Map[{Abs[N[1/2 Det[Map[Join[#,{1}]&,#]]]],RootReduce[Abs[1/2 Det[Map[Join[#,{1}]&,#]]]]}&,Subsets[#,{3}]]]]&/@( HeilbronnTriangle) )]]/.#1-> x]

In this third notebook, I look at the Heilbronn triangle problem in a triangle.

Initialize:

unitareatriangle=Line[{{(1/(3^(1/4))),-((3^(1/4))/3)},{-(1/(3^(1/4))),-((3^(1/4))/3)},{0,((2(3^(1/4)))/3)},{(1/(3^(1/4))),-((3^(1/4))/3)}}];tri[a_,b_,c_,rd_,gr_,bl_]:={{RGBColor[rd,gr,bl],Polygon[{ptlist[[a]],ptlist[[b]],ptlist[[c]]}]},GrayLevel[0],Line[{ptlist[[a]],ptlist[[b]],ptlist[[c]],ptlist[[a]]}]}

N = 4, area = (1/3) = 0.333333+, symmetry of an equilateral triangle

Trivial.

ptlist={{0,((2(3^(1/4)))/3)},{-(1/(3^(1/4))),-((3^(1/4))/3)},{(1/(3^(1/4))),-((3^(1/4))/3)},{0,0}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,3,4,1,0,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 5, area = 3-2(Sqrt[2]) = 0.171572+

I have not seen the paper by Yang, Zhang and Zeng in Acta Math. Sinica 37 (1994). Perhaps they mention everything here and for N = 6.

Symmetric configuration 1, horizontal symmetry

This is the one already shown at MathWorld.

ptlist={{((-2+(Sqrt[2]))/(3^(1/4))),((-4+3 (Sqrt[2]))/(3^(3/4)))},{((1-(Sqrt[2]))/(3^(1/4))),-(1/(3^(3/4)))},{-((-2+(Sqrt[2]))/(3^(1/4))),((-4+3 (Sqrt[2]))/(3^(3/4)))},{((-1+(Sqrt[2]))/(3^(1/4))),-(1/(3^(3/4)))},{0,(2/(3^(3/4)))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,3,2,0,0,1],tri[1,2,5,1,0,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

Symmetric configuration 2, horizontal symmetry

ptlist={{((-1+(Sqrt[2]))/(3^(1/4))),((5-3 (Sqrt[2]))/(3^(3/4)))},{(1/(3^(1/4))),-(1/(3^(3/4)))},{-((-1+(Sqrt[2]))/(3^(1/4))),((5-3 (Sqrt[2]))/(3^(3/4)))},{0,((8-6 (Sqrt[2]))/(3^(3/4)))},{-(1/(3^(1/4))),-(1/(3^(3/4)))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,3,2,0,0,1],tri[4,2,5,0,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[3,4,1,1,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[5,4,1,1,0,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

An infinite family

The configuration above is the only symmetric member of an infinite family. (Of course, the red and blue triangles above are congruent. I showed them both just to facilitate this discussion.) Rather than showing an animation this time, I'll describe: The yellow and red triangles above share a vertex which lies on a side of the equilateral triangle; let's call that vertex P. Now let P move up a little along the side of the equilateral triangle. The area of the yellow triangle stays constant, while that of the red triangle increases, meaning that it then ceases to be a minimum-area triangle. But the area of another triangle, which had not been a minimum-area triangle, decreases until it finally does become one, shown in orange below.

The configurations shown above and below are, thus, the extreme members of the family.

ptlist={{-(1/(3^(1/4))),-(1/(3^(3/4)))},{-((-2+(Sqrt[2]))/(2 (3^(1/4)))),((-2+3 (Sqrt[2]))/(2 (3^(3/4))))},{0,((8-6 (Sqrt[2]))/(3^(3/4)))},{-((-1+(Sqrt[2]))/(3^(1/4))),((5-3 (Sqrt[2]))/(3^(3/4)))},{(1/(3^(1/4))),-(1/(3^(3/4)))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,3,5,0,0,1],tri[1,3,5,0,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,3,2,1,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,1,2,1,1/2,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 6, area = (1/8) = 0.125

A configuration with the symmetry of an equilateral triangle

ptlist={{(1/(2 (3^(1/4)))),-(1/(3^(3/4)))},{-(1/(2 (3^(1/4)))),-(1/(3^(3/4)))},{(1/(4 (3^(1/4)))),(5/(4 (3^(3/4))))},{((3^(3/4))/4),-(1/(4 (3^(3/4))))},{-((3^(3/4))/4),-(1/(4 (3^(3/4))))},{-(1/(4 (3^(1/4)))),(5/(4 (3^(3/4))))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,4,0,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

An infinite family

Since this is somewhat like what we found for N = 5, I'll not go into much detail here. There are an infinite number of symmetric members in the family, one of which is shown at MathWorld. Below I show a typical (i.e., not an extreme) one with horizontal symmetry. There are also infinitely many asymmetric members in the family.

ptlist={{0,(2/(3^(3/4)))},{-0.55`,-0.07524526886122096`},{-(1/(2 (3^(1/4)))),-(1/(3^(3/4)))},{0,(1/(2 (3^(3/4))))},{0.55`,-0.07524526886122096`},{(1/(2 (3^(1/4)))),-(1/(3^(3/4)))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,3,1,1,0,0],tri[5,6,4,1,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 7, area = (7/72) = 0.097222+, 120° rotational symmetry

ptlist={{(2/(3 (3^(1/4)))),-(1/(3^(3/4)))},{((3^(3/4))/4),-(1/(4 (3^(3/4))))},{(1/(6 (3^(1/4)))),((3^(1/4))/2)},{-(1/(4 (3^(1/4)))),(5/(4 (3^(3/4))))},{-(5/(6 (3^(1/4)))),-(1/(2 (3^(3/4))))},{-(1/(2 (3^(1/4)))),-(1/(3^(3/4)))},{0,0}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,7,1,1,1/2,0],tri[2,3,1,1,0,0],tri[5,6,7,1,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 8, area = 0.067789+, no symmetry

Minimum-area triangles come in 11 different shapes. Just for completeness, I've shown them all; I don't suppose you'll want to, Erich.

ptlist={{-(1/(3^(1/4))),-(1/(3^(3/4)))},{0.5236274599`,-(1/(3^(3/4)))},{-0.0846397964`,0.7307822477`},{-0.2450656014`,-0.2912911357`},{0.6747341394`,-0.2912911357`},{0.1371228657`,0.0815218223`},{0.2123263582`,0.5096226351`},{-0.3101032349`,0.1394695068`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[3,8,1,0,0,1],tri[8,7,1,0,1,0],tri[4,6,1,1,1,0],tri[4,5,1,1,0,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[3,8,4,1,0,1/2],tri[3,6,2,0,1/2,1],tri[3,7,5,1/2,1,1/2],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,4,5,1,1,0],tri[4,6,7,1/2,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,5,6,1,0,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[5,6,8,0,0,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 9, area = (43/784) = 0.0548+, 120° rotationally symmetric

ptlist={{0.597013737`,-(1/(3^(3/4)))},{0.6648562315`,-0.2741820973`},{0.1628219297`,0.1253403637`},{0.08141097432579625`,0.7363747314756971`},{-0.09497945422530327`,0.7128734349933876`},{-0.1899589039337809`,0.07833774556340398`},{-0.6784247113257963`,-0.2976833938248663`},{-0.5698767772746967`,-(1/(3^(3/4)))},{0.02713697423378089`,-0.203678109263404`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[3,6,9,1,0,0],tri[2,7,9,0,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,6,9,1/2,0,1],tri[4,5,6,1,1,0],tri[8,6,7,0,1,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 10, area = 0.0433+, asymmetric

This is an improved configuration; 
the previous symmetric one had area = 0.0420+.

ptlist={{-0.4239530851`,-(1/(3^(3/4)))},{0.6572305907`,-(1/(3^(3/4)))},{0.5113077447`,-0.008228316822194603`},{-0.4499289558`,0.09808286405964994`},{-0.5904406817`,-0.3381065411`},{0.1915018448`,0.3406624472`},{-0.0940212592`,0.3808833609`},{0.0006283394`,-0.0419999715`},{0.2720503627`,-0.3381065411`},{0,(2/(3^(3/4)))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,4,5,1/2,0,1/2],tri[1,7,10,0,1,0],tri[6,1,8,1,0,0],tri[2,3,6,0,0,1],tri[2,1,9,1,1,0],tri[2,8,9,1/2,1/2,1/2],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,5,0,1,1],tri[2,5,9,1,0,0],tri[2,4,8,0,1,0],tri[3,4,8,0,1/2,1/2],tri[3,6,7,1,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,9,5,1/2,0,0],tri[6,10,9,1,0,0],tri[4,5,10,0,1,1/2],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

0.0433767435`/ConvexHullArea[ptlist]

N = 11, area = 0.036092+, horizontal symmetry

ptlist={{0.3627654674443981`,-(1/(3^(3/4)))},{0.5452866147601714`,-0.06708144615019251`},{0.11345362658894519`,0.6808752297466613`},{0.156604360381382`,0.19756294694637877`},{0.6058832138845257`,-0.33919819056486644`},{-0.3627654674443981`,-(1/(3^(3/4)))},{-0.5452866147601714`,-0.06708144615019251`},{-0.11345362658894519`,0.6808752297466613`},{-0.156604360381382`,0.19756294694637877`},{-0.6058832138845257`,-0.33919819056486644`},{0,-0.13327173403398437`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[10,7,8,1,1,0],tri[8,9,6,0,1,0],tri[8,4,1,0,0,1],tri[8,4,5,1,0,0],tri[1,9,11,0,1,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[10,6,1,1,1/2,0],tri[10,2,11,1,0,1/2],tri[7,2,11,1/2,0,1],tri[1,2,5,0,1,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 12, area = 0.031004+, symmetry of an equilateral triangle

ptlist={{-0.3828761953531324`,-(1/(3^(3/4)))},{-0.5713559405023625`,-0.11223484285473073`},{-0.1540971234828158`,0.0889680157241507`},{0,0.7154254571730748`},{0.5713559405023625`,-0.11223484285473068`},{0.38287619535313244`,-(1/(3^(3/4)))},{0,-0.1779360314483014`},{-0.6195766204259787`,-0.3577127285865374`},{-0.18847974514923008`,0.5509261805055614`},{0.18847974514923005`,0.5509261805055615`},{0.1540971234828158`,0.0889680157241507`},{0.6195766204259787`,-0.3577127285865374`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,8,1,1,0],tri[1,4,3,0,1,0],tri[1,7,11,0,0,1],tri[1,6,12,1,0,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 13, area = 0.0245+, 120 ° rotationally symmetric

0.0245642594`/ConvexHullArea[ptlist]

ptlist={{-0.19260874`,-(1/(3^(3/4)))},{0.0942069387`,-0.3014895661`},{0.1552028182`,0.6085635087`},{0.6074897898`,-0.3278596045`},{0.47622221282579624`,0.05254160699450344`},{0.21399415386854767`,0.23233038517696336`},{-0.6046328674503922`,-0.16987217104986216`},{-0.01981014852828117`,0.6900313927564687`},{-0.2836134728257963`,0.38614973065632735`},{-0.30820109256854766`,0.06915918092303663`},{0.4494300492503922`,-0.4386913376501378`},{-0.5876796412717189`,-0.3621717882564688`},{0,0}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,6,13,1,1,0],tri[1,3,13,1,0,0],tri[3,6,11,0,1,0],tri[1,9,10,0,0,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[2,4,7,0,1,0],tri[2,4,12,1,0,1/2],tri[1,12,11,1/2,0,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 14, area = 0.0237+, asymmetric

0.0237757742`/ConvexHullArea[ptlist]

There are 23 different triangles having that minimum area.

ptlist={{-0.3904545987`,-(1/(3^(3/4)))},{-0.1904885733`,-0.1532488072`},{0.6524136764`,-0.378782794`},{-0.0064065584`,-0.3433769196`},{0.5531430703`,-0.0806892263`},{0.2614131862`,-0.0452096335`},{0.0325763013`,0.8209588662`},{0.2754746774`,0.2740947036`},{-0.5691154992`,-0.1083542847`},{-0.1403958289`,0.6342099665`},{-0.0250107551`,0.320761482`},{-0.6620957954`,-0.3803161242`},{-0.2852346791`,0.202926372`},{0.4241299371`,-(1/(3^(3/4)))}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,2,11,1,0,0],tri[1,2,8,0,1,0],tri[1,4,5,1,1,0],tri[1,10,13,0,1/2,1],tri[3,4,14,1/2,1/2,1],tri[9,10,12,1,1/2,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,12,14,1,1/2,0],tri[2,3,9,0,1/2,1/2],tri[2,5,6,1/2,1/2,1/2],tri[3,4,12,1/2,0,1],tri[5,8,10,1,0,1],tri[6,11,10,0,1,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[4,9,14,1,1,0],tri[3,5,8,1/2,1/2,0],tri[3,6,11,1,0,0],tri[9,13,11,0,0,1],tri[7,10,13,0,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[3,7,8,1,1,0],tri[4,7,11,1/2,1/2,1],tri[2,4,13,1,0,0],tri[7,12,13,0,1,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

I know your page is set up to handle four figures, at most. For completeness, the two triangles I was not clever enough to fit into the four figures above are:

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[6,10,14,1,1,1],tri[8,11,13,1,1,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 15, area = 0.0210+, 120 ° rotationally symmetric

0.0210907714`/ConvexHullArea[ptlist]

ptlist={{-0.38761748842791954`,-(1/(3^(3/4)))},{-0.18760767830514943`,-0.16578440610915707`},{0,0.7685627508192401`},{0,-0.352595053943903`},{0.38761748842791954`,-(1/(3^(3/4)))},{0.5737265870397561`,-0.11634092310428357`},{0.23737734639442082`,-0.07958081230269956`},{-0.6655948666119113`,-0.38428137540962004`},{0.3053562739641645`,0.1762975269719515`},{0.1861090986118365`,0.5550322607551144`},{-0.1861090986118365`,0.5550322607551144`},{-0.04976966808927137`,0.24536521841185663`},{0.6655948666119113`,-0.38428137540962004`},{-0.3053562739641645`,0.1762975269719515`},{-0.5737265870397561`,-0.11634092310428357`}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[8,10,12,1,0,0],tri[1,5,13,1,1,0],tri[1,4,6,0,1,0],tri[3,8,14,0,0,1],tri[1,2,12,0,1,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[9,12,14,1,1/2,1/2],tri[6,7,15,1/2,0,1/2],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

N = 16, area = 0.0179+, 120 ° rotationally symmetric

0.0179762759`/ConvexHullArea[ptlist]

ptlist={{0.5855611067622228`,-(1/(3^(3/4)))},{-0.37458685982327644`,-0.02344929700189427`},{0.3435082898333592`,-0.0765241545484251`},{-0.08914493930189583`,0.7229791111931344`},{0.0005649738876899437`,-0.4078864888841451`},{0.08713728944468485`,0.7264564627496322`},{0.20760111681616494`,-0.3126770880298508`},{-0.10548228307461696`,0.3357489826804494`},{-0.5815458070478009`,-(1/(3^(3/4)))},{0.3529575742902637`,0.20443252618128688`},{-0.6726983962069077`,-0.28776512509880137`},{0.1669857430071115`,0.336126385031745`},{-0.23802600675874222`,-0.2592248281320243`},{0.6706907463496967`,-0.2842877735423036`},{-0.35352254817795364`,0.2034539627028582`},{0,0}};ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[1,5,9,1,0,0],tri[1,7,13,1,1,0],tri[1,2,7,0,0,1],tri[1,15,16,0,1/2,1/2],tri[1,3,12,0,1,1],tri[8,15,12,1,1/2,0],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

ListPlot[ptlist,PlotStyle->PointSize[0.025],AspectRatio->Automatic,Prolog->{tri[15,16,14,1/2,1,0],tri[4,12,14,1/2,0,1],tri[2,13,15,1,0,1],unitareatriangle},PlotRange->{{-0.78,0.78},{-0.46,0.9}},Axes->False,ImageSize->200]

Ultimate Heilbronn

By David Cantrell

Initialize:

<<Statistics`MultiDescriptiveStatistics`

tri[a_,b_,c_,rd_,gr_,bl_]:={{RGBColor[rd,gr,bl],Polygon[{ptlist[[a]],ptlist[[b]],ptlist[[c]]}]},{GrayLevel[0],Line[{ptlist[[a]],ptlist[[b]],ptlist[[c]],ptlist[[a]]}]},{PointSize[0.025],Point[ptlist[[a]]],Point[ptlist[[b]]],Point[ptlist[[c]]]}};pt[a_]:={PointSize[0.025],Point[ptlist[[a]]]}

Heilbronn in unit-area convex regions

I mentioned symmetries below mainly because that's consistent with what's been done for the other Heilbronn categories. But I'm not sure that you want to mention symmetries here, for two reasons: 
1) The figures make the symmetries reasonably obvious. 
2) The symmetries are just artifacts of the particular way that the configuration happens to be shown here. (See, for example, my comment for N = 8.)

Speaking of symmetries, a good question is:
Is there some N for which the optimal configuration in a convex region cannot be rendered so that it is symmetric? (Obviously, I didn't find any such N, but that doesn't mean much.)

N = 3, A = 1, completely symmetric

n=3;Do[z[j]=(E^((2π I( j-1/4))/n));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n}];ptlist=Table[{x[j],y[j]},{j,1,n}];PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,2,3,1,0,0]},LabelPoints->False]

N = 4, A = (1/2) = 0.5, completely symmetric

n=4;Do[z[j]=(E^((2π I( j-1/2))/n));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n}];ptlist=Table[{x[j],y[j]},{j,1,n}];PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,2,3,1,0,0],pt[1]},LabelPoints->False]

N = 5, A = ((5-(Sqrt[5]))/10) = 0.2763+, completely symmetric

n=5;Do[z[j]=(E^((2π I( j+1/4))/n));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n}];ptlist=FunctionExpand[Table[{x[j],y[j]},{j,1,n}]];PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,2,3,1,0,0],pt[1],pt[5]},LabelPoints->False]

N = 6, A = (1/6) = 0.1666+, completely symmetric

n=6;Do[z[j]=(E^((2π I( j))/n));x[j]=Re[z[j]];y[j]=Im[z[j]],{j,1,n}];ptlist=Table[{x[j],y[j]},{j,1,n}];PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,5,3,1,0,0],pt[1],pt[2],pt[6]},LabelPoints->False]

N = 7, A = (1/9) = 0.1111+, completely symmetric

There is an alternative optimal configuration: Just take my configuration for N = 7 in an equilateral triangle and trim off the parts lying outside the convex hull of the 7 points. The result has 9 smallest triangles and 120° rotational symmetry; by contrast, this configuration has 6 smallest triangles and complete symmetry.

ptlist={{((-1+(Sqrt[3]))/(2 (Sqrt[2]))),((1+(Sqrt[3]))/(2 (Sqrt[2])))},{-((-1+(Sqrt[3]))/(2 (Sqrt[2]))),((1+(Sqrt[3]))/(2 (Sqrt[2])))},{((1+(Sqrt[3]))/(2 (Sqrt[2]))),-((-1+(Sqrt[3]))/(2 (Sqrt[2])))},{-((1+(Sqrt[3]))/(2 (Sqrt[2]))),-((-1+(Sqrt[3]))/(2 (Sqrt[2])))},{(1/(Sqrt[2])),-(1/(Sqrt[2]))},{-(1/(Sqrt[2])),-(1/(Sqrt[2]))},{0,0}};PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,2,7,0,1,0],tri[3,4,7,1,0,0],pt[5],pt[6]},LabelPoints->False]

N = 8, A = 0.0800+, horizontally symmetric

As you might guess from the ptlist below, the area is not a simple 8/100. Rather, it is exactly
Root[-1+26 #1-240 ((#1)^2)+1064 ((#1)^3)-2332 ((#1)^4)+2060 ((#1)^5)&,1]
which is 0.080000139329466... Henceforth, I will not bother to give areas exactly when they are this messy.

Also, I should mention somewhere, and this is probably as good a place as any, that, since we're dealing with general convex regions, an optimal configuration can be presented in many different forms. For example, the asymmetric set of 8 points

is equivalent to the symmetric form I use below. In general, I've tried to present each configuration in as visually attractive a way as possible.

ptlist={{1,(1/8) (7-3 Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]-22 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)-6 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)+7 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)-3 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5))},{(1/2) (-2-Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]+13 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)+5 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)-5 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)+2 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5)),(1/2) (3-3 Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]-8 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)-((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)+2 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)-((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5))},{Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1],(1/8) (-9-Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]+50 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)+14 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)-17 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)+7 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5))},{-1,(1/8) (7-3 Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]-22 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)-6 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)+7 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)-3 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5))},{(1/2) (2+Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]-13 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)-5 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)+5 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)-2 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5)),(1/2) (3-3 Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]-8 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)-((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)+2 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)-((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5))},{-Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1],(1/8) (-9-Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1]+50 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^2)+14 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^3)-17 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^4)+7 ((Root[-1-#1+4 ((#1)^2)+4 ((#1)^3)-3 ((#1)^4)+((#1)^5)&,1])^5))},{0,1},{0,0}};PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[3,6,7,0,1,1],tri[1,4,8,1,1,0],pt[5],pt[2]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,6,7,0,1,0],tri[3,5,8,1,0,0],pt[1],pt[2]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,5,8,0,0,1],tri[1,2,3,1,1/2,0],pt[6],pt[7]},LabelPoints->False]

N = 9, A = 0.0640+, horizontally symmetric

ptlist={{0,0.29657672069554386`},{0.4349303574500163`,1},{1,0.4880335182903961`},{0.2573230436263183`,-1},{0.8859980920105824`,-0.25596722818629963`},{-0.4349303574500163`,1},{-1,0.4880335182903961`},{-0.2573230436263183`,-1},{-0.8859980920105824`,-0.25596722818629963`}};PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,3,9,0,0,1],tri[1,3,7,1,0,0],tri[4,9,8,0,1,0],tri[3,4,5,1,1,0],pt[6],pt[2]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,2,9,1,0,1/2],tri[1,2,8,0,1,1/2],pt[3],pt[4],pt[5],pt[6],pt[7]},LabelPoints->False]

N = 10, A = ((154+33 (Sqrt[6])-(Sqrt[14346+6756 (Sqrt[6])]))/1136) = 0.0519+,
symmetries of a rectangle

Sometimes the best configuration in a general convex region can be obtained merely by "trimming" the previously known best configuration in a square, circle or triangle. Trivial examples of that include trimming the best configurations in a circle for N = 3..6 to get the configurations given above. I mention this here because, if we trim the best configuration for N = 10 in a circle, we get

with A = 0.051990+. That's not optimal, but surprisingly close; for the configuration below, A = 0.051993+.

ptlist={{(1/12) (3+(Sqrt[6])-(Sqrt[-9+6 (Sqrt[6])])),-1},{1,(1/4) (-2+(Sqrt[6])-(Sqrt[-6+4 (Sqrt[6])]))},{(1/12) (-3-(Sqrt[6])+(Sqrt[-9+6 (Sqrt[6])])),-1},{1,(1/4) (2-(Sqrt[6])+(Sqrt[-6+4 (Sqrt[6])]))},{(1/12) (-3-2 (Sqrt[6])+3 (Sqrt[9+4 (Sqrt[6])])),0},{(1/4)+(1/(Sqrt[6]))-(1/4) (Sqrt[9+4 (Sqrt[6])]),0},{-1,(1/4) (-2+(Sqrt[6])-(Sqrt[-6+4 (Sqrt[6])]))},{(1/12) (3+(Sqrt[6])-(Sqrt[-9+6 (Sqrt[6])])),1},{-1,(1/4) (2-(Sqrt[6])+(Sqrt[-6+4 (Sqrt[6])]))},{(1/12) (-3-(Sqrt[6])+(Sqrt[-9+6 (Sqrt[6])])),1}};PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[3,4,5,0,1,1],tri[1,2,3,1/2,1,0],tri[4,5,7,1,1,0],pt[6],pt[8],pt[9],pt[10]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[6,4,5,1/2,0,1],pt[1],pt[2],pt[3],pt[9],pt[7],pt[8],pt[10]},LabelPoints->False]

N = 11, A = (2/47) = 0.0425+, horizontally symmetric

This may be obtained by trimming Goldberg's configuration in a square, and I have presented the configuration in that way. (Alternatively, it may be obtained by omitting the top point in the next configuration, for N = 12. But if you want to see it presented in exactly that way, then AspectRatio->((Sqrt[3])/2) should be used in the following figures.)

ptlist={{(1/3),0},{(2/3),0},{0,(2/9)},{1,(2/9)},{(1/3),(4/9)},{(2/3),(4/9)},{0,(2/3)},{1,(2/3)},{(1/2),(7/9)},{(1/6),1},{(5/6),1}};

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[6,4,5,1/2,0,1],tri[1,4,2,1/2,1,1],tri[6,8,5,1,0,1/2],tri[1,10,5,0,1,0],tri[3,10,7,1,1/2,0],tri[5,11,9,1,1,0]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,9,5,1,0,0],tri[1,6,8,0,0,1],tri[8,10,9,0,1,0],tri[3,9,5,1,1,0],pt[2],pt[4],pt[7],pt[11]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[3,8,5,0,1,0],tri[3,6,8,0,1/2,1],tri[3,5,11,1,1,0],tri[3,9,11,1/2,0,1],pt[1],pt[7],pt[2],pt[4],pt[10]},LabelPoints->False]

N = 12, A = (2/51) = 0.0392+, completely symmetric

Since the previous configuration is just like this one but with its top point removed, why are so few different triangles shown here as compared to N = 11? Answer: Much greater symmetry here! I have always tried to show all triangles having the same minimum area, except that I do not show multiple triangles which, from the stated symmetry, can be deduced to be congruent. (Of the 14 triangles shown above for N = 11, there are indeed many congruencies, but they cannot be deduced merely from horizontal symmetry.)

ptlist={{0,1},{(1/(Sqrt[3])),(2/3)},{-(1/(Sqrt[3])),(2/3)},{0,(1/3)},{-((Sqrt[3])/2),-(1/2)},{-((Sqrt[3])/2),(1/6)},{-(1/(2 (Sqrt[3]))),-(5/6)},{-(1/(2 (Sqrt[3]))),-(1/6)},{((Sqrt[3])/2),-(1/2)},{(1/(2 (Sqrt[3]))),-(5/6)},{((Sqrt[3])/2),(1/6)},{(1/(2 (Sqrt[3]))),-(1/6)}};

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[5,7,10,0,1,0],tri[1,4,12,1,1,0],tri[10,2,12,0,1,1],pt[3],pt[6],pt[8],pt[9],pt[11]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[5,8,2,1,0,0],pt[1],pt[2],pt[3],pt[4],pt[6],pt[7],pt[9],pt[10],pt[11],pt[12]},LabelPoints->False]

N = 13, A = 0.0306+, horizontally symmetric

ptlist={{0.3344454904506242`,0.27847892630838433`},{0.24875582782436484`,-0.14991673439270953`},{(1/2),-0.052296067446687744`},{-0.09556830467281317`,(1/2)},{-0.463450351885976`,-0.2350218499076573`},{-0.07989486474500392`,-(1/2)},{0,0.19664563836728338`},{-0.33444549045062416`,0.2784789263083844`},{-0.24875582782436484`,-0.1499167343927095`},{-(1/2),-0.052296067446687695`},{0.09556830467281323`,(1/2)},{0.463450351885976`,-0.23502184990765737`},{0.07989486474500385`,-(1/2)}};

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,7,13,1,0,0],tri[4,7,2,1,1,0],tri[6,8,9,1,0,1],tri[5,10,9,0,1,1],tri[1,3,12,0,0,1],tri[1,4,11,0,1,0]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[4,8,10,1,1/2,0],tri[1,5,9,1,0,0],tri[1,7,10,1,1,0],tri[2,5,9,0,1/2,0],tri[12,6,13,1/2,1/2,1],pt[3],pt[11]},LabelPoints->False]

N = 14, A = 0.0277+, horizontally symmetric

ptlist={{(1/2),0.04148940941975993`},{0.19960240861396789`,0.18467307322033563`},{-0.44380312022592455`,-0.37080741303043135`},{-0.4243887975524955`,0.29901658941442594`},{0.1203297002844073`,-0.1811348618000502`},{-0.17666061015642245`,-(1/2)},{0.1431529415261378`,(1/2)},{-(1/2),0.04148940941975993`},{-0.19960240861396789`,0.18467307322033563`},{0.44380312022592455`,-0.37080741303043135`},{0.4243887975524955`,0.29901658941442594`},{-0.1203297002844073`,-0.1811348618000502`},{0.17666061015642245`,-(1/2)},{-0.1431529415261378`,(1/2)}};

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,3,12,1,1,0],tri[3,4,8,1,0,1],tri[2,4,9,0,0,1],tri[3,6,13,0,1,0],tri[2,11,12,1,0,0],pt[5],pt[7],pt[10],pt[14]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[10,5,12,1,1,0],tri[8,7,9,0,1/2,0],tri[2,5,7,1,1/2,0],tri[2,7,10,1,0,0],pt[1],pt[2],pt[3],pt[4],pt[5],pt[8],pt[9],pt[10],pt[11],pt[6],pt[13],pt[14]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[14,6,12,0,1,0],tri[6,7,12,1,0,1/2],tri[11,6,5,1/2,0,1],pt[1],pt[7],pt[2],pt[3],pt[4],pt[5],pt[8],pt[9],pt[10],pt[11],pt[12],pt[13],pt[14]},LabelPoints->False]

N = 15, A = 0.0244+, horizontally symmetric

n=15;ptlist={{0.2730219342456227`,0.4870921567028016`},{0.45033724486238264`,0.3398621074399868`},{0.541455619680651`,0.012739157675487556`},{0.4877059016670484`,-0.2836420978326772`},{0.30231308639185206`,-0.47635772690282174`},{0.14670632656297244`,0.1878967979719508`},{0.15671724542124976`,-0.1413841589751206`},{-0.2730219342456226`,0.4870921567028016`},{-0.4503372448623826`,0.33986210743998685`},{-0.541455619680651`,0.012739157675487667`},{-0.48770590166704847`,-0.28364209783267713`},{-0.30231308639185217`,-0.47635772690282174`},{-0.1467063265629724`,0.1878967979719508`},{-0.1567172454212498`,-0.14138415897512058`},{0,-0.5501033678313902`}};PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[1,2,3,0,0,1],tri[3,4,5,1,0,0],tri[1,6,7,0,1,0],tri[1,10,13,1,1,0],tri[5,12,15,1/2,0,1],tri[13,12,14,0,1,1],Table[pt[i],{i,1,n}]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[2,3,4,0,1,0],tri[9,6,13,1,1/2,0],tri[4,10,14,1/2,0,1],Table[pt[i],{i,1,n}]},LabelPoints->False]

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[2,10,13,0,1/2,1/2],tri[7,14,11,1,0,0],tri[2,7,15,1,1,0],tri[4,5,15,1/2,1,1/2],Table[pt[i],{i,1,n}]},LabelPoints->False]

N = 16, A = 0.0222+, completely symmetric

ptlist={{0,Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2]},{Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],(1/2)},{(1/2),Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]},{Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{-Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2],0},{-(1/2),Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]},{-Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],(1/2)},{-Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{0,-Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2]},{-Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],-(1/2)},{-(1/2),-Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]},{-Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],-Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]},{Root[3+8 #1-84 ((#1)^2)-156 ((#1)^3)-60 ((#1)^4)+8 ((#1)^5)&,2],0},{(1/2),-Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3]},{Root[-1+3 #1+48 ((#1)^2)-64 ((#1)^3)+16 ((#1)^4)+16 ((#1)^5)&,3],-(1/2)},{Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3],-Root[9-180 #1+724 ((#1)^2)-1024 ((#1)^3)+560 ((#1)^4)+64 ((#1)^5)&,3]}};

PlanarGraphPlot[ptlist,ConvexHull[ptlist],Epilog->{tri[7,4,2,0,1,0],tri[1,4,11,1,1,0],tri[10,11,12,0,0,1],tri[13,3,9,1,0,1/2],Table[pt[i],{i,1,n}]},LabelPoints->False]

Summary

By David Cantrell

Notes:
1.  Values are truncated, not rounded. Normally 4 decimal places are shown. 
2.  Best values are shown with a cyan background. Of course, the convex row is always best, and the circle row can never be best.
3.  Worst values are shown with a yellow background. Although, as I said, the circle can never be the best, it is also never the worst except for N = 3 and 4. Of the regions considered here, the triangle seems to be the clear winner in the "worst" category. I'm guessing that trend will continue for larger N but with occassional exceptions (like we had at N = 7).

The graph below shows values for the best configurations now known. Square, circle and triangle symbols are used in the obvious way; red dots are used for values for convex regions. Upper and lower bounds in black and blue, resp., are adapted from forms of bounds given in the literature. The red curve is my invention, intended to approximate the values for convex regions. It's fairly accurate considering the simplicity of my formula:

((Log[n-2]+1)/((((n-2))^2)-(2/3)(n-2)+(2/3)))