(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 724810, 20522]*) (*NotebookOutlinePosition[ 725752, 20555]*) (* CellTagsIndexPosition[ 725708, 20551]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "Applying ", StyleBox["GroebnerBasis", FontFamily->"Courier", FontWeight->"Plain"], " to Three Problems in Geometry" }], "Title", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "Mathematica's highly optimized Gr\[ODoubleDot]bner basis reduction function \ is put to interesting geometric purposes."], "Subsubtitle"], Cell["by Michael Trott", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The new version of ", StyleBox["GroebnerBasis", FontFamily->"Courier"], " is extremely powerful. We will demonstrate three applications: a proof of \ Lester's recent circle theorem, a solution to the three-dimensional version \ of the tangency problem of Apollonius, and a calculation of 28 bitangents to \ a plane quartic curve." }], "Text"], Cell[CellGroupData[{ Cell["A Proof of Lester's Circle Theorem", "Section"], Cell["\<\ J. Lester [Kimberling 1996] recently came up with a beautiful \ theorem in elementary geometry.\ \>", "Text"], Cell["\<\ The circumcenter, the nine-point center, and the first and second \ Fermat points of a triangle lie on a circle.\ \>", "Text", CellFrame->True, FontWeight->"Bold"], Cell["We will prove and illustrate this theorem.", "Text"], Cell[CellGroupData[{ Cell["Definition of circumcenter", "Subsubsection"], Cell["\<\ The circumcenter of a triangle is the center of the circle through \ the vertices of the triangle (the circumcircle). The circumcenter can be \ constructed as the point of intersection of the perpendicular bisectors of \ the sides of the triangle. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Definition of nine-point center", "Subsubsection"], Cell["\<\ The nine-point center is the midpoint of the circle through the \ midpoints of the sides of the triangle. \ \>", "Text"], Cell["\<\ The other six points that lie on the nine-point circle are the feet \ of the altitudes of the triangle and the midpoints of the line segments from \ the orthocenter to the vertices of the triangle. (The orthocenter is the \ point of intersection of the altitudes.) The nine-point circle is also called Feuerbach's circle. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Definition of the first and second Fermat points", "Subsubsection"], Cell[TextData[{ "Construct equilateral triangles on the sides of a triangle, going ", StyleBox["outwards", FontSlant->"Italic"], ". Each equilateral triangle has a vertex that was not a vertex of the \ original triangle. Draw lines from those vertices to the opposite vertices of \ the original triangle. These three lines meet in a single point, the first \ Fermat point. " }], "Text"], Cell[TextData[{ "The second Fermat point is constructed similarly, except that the \ equilateral triangles are constructed on the sides of the triangle going ", StyleBox["inwards", FontSlant->"Italic"], ". " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["The intersection of two lines", "Subsubsection"], Cell[TextData[ "The function \[ScriptCapitalI] calculates the intersection of two lines. \ Each line is given in the form of a point and a direction."], "Text"], Cell[BoxData[ \(\[ScriptCapitalI][{point1_, dir1_}, {point2_, dir2_}]\ := \n Module[{p, \ d}, \n \t\t{p\_x1, \ p\_y1} = point1; {p\_x2, \ p\_y2} = point2; \n \t\t{d\_x1, \ d\_y1} = dir1; {d\_x2, \ d\_y2} = dir2; \n \ \t{p\_x1, \ p\_y1} + \ {d\_x1\ , d\_y1} \(d\_x2\ \((p\_y2 - p\_y1)\) + d\_y2\ \((p\_x1 - p\_x2\ )\)\)\/\(\(d\_x2\) d\_y1 - \(d\_x1\) d\_y2\)]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Perpendicular direction", "Subsubsection"], Cell[TextData[ "The function \[ScriptCapitalN] gives the direction perpendicular to the \ given one."], "Text"], Cell[BoxData[ \(\(\[ScriptCapitalN][dir_] := \ {\(-1\), 1} Reverse[dir]; \)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Calculating the circumcenter", "Subsubsection"], Cell[TextData[{ "Let the three points of the original triangle be ", Cell[BoxData[ \(TraditionalForm\`P\_A\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(P\_B, \)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`P\_C\)]], ". " }], "Text"], Cell[BoxData[ \(P\_A = \ {P\_Ax, P\_Ay}; P\_B = \ {P\_Bx, P\_By}; P\_C = \ {P\_Cx, P\_Cy}; \)], "Input"], Cell["\<\ The circumcenter can now be calculated as the point of intersection \ of two of the perpendicular bisectors of the sides of the triangle. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(P\_1 = Together\ @ \[ScriptCapitalI][{\(P\_A + P\_B\)\/2, \[ScriptCapitalN][P\_A - P\_B]}, \ {\(P\_B + P\_C\)\/2, \ \[ScriptCapitalN][P\_B - P\_C]}]\)], "Input"], Cell[BoxData[ \({\(( \(-P\_Ay\)\ P\_Bx\%2 + P\_Ax\%2\ P\_By + P\_Ay\%2\ P\_By - P\_Ay\ P\_By\%2 + P\_Ay\ P\_Cx\%2 - P\_By\ P\_Cx\%2 - P\_Ax\%2\ P\_Cy - P\_Ay\%2\ P\_Cy + P\_Bx\%2\ P\_Cy + P\_By\%2\ P\_Cy + P\_Ay\ P\_Cy\%2 - P\_By\ P\_Cy\%2)\)/ \((2\ \(( \(-P\_Ay\)\ P\_Bx + P\_Ax\ P\_By + P\_Ay\ P\_Cx - P\_By\ P\_Cx - P\_Ax\ P\_Cy + P\_Bx\ P\_Cy)\))\), \((P\_Ax\%2\ P\_Bx + P\_Ay\%2\ P\_Bx - P\_Ax\ P\_Bx\%2 - P\_Ax\ P\_By\%2 - P\_Ax\%2\ P\_Cx - P\_Ay\%2\ P\_Cx + P\_Bx\%2\ P\_Cx + P\_By\%2\ P\_Cx + P\_Ax\ P\_Cx\%2 - P\_Bx\ P\_Cx\%2 + P\_Ax\ P\_Cy\%2 - P\_Bx\ P\_Cy\%2)\)/ \((2\ \(( P\_Ay\ P\_Bx - P\_Ax\ P\_By - P\_Ay\ P\_Cx + P\_By\ P\_Cx + P\_Ax\ P\_Cy - P\_Bx\ P\_Cy)\))\)}\)], "Output"] }, Open ]], Cell["Here is a sketch of the construction for a sample triangle.", "Text"], Cell[BoxData[ \(CircumcenterSketch[{a_, \ b_, \ c_}, opts___] := \n Module[{r, m = P\_1\ /. \n Flatten[Thread\ /@ \ {{P\_Ax, P\_Ay} \[Rule] a, {P\_Bx, P\_By} \[Rule] b, {P\_Cx, P\_Cy} \[Rule] \ c}]}, \n r = Sqrt[\(#.#&\)[m - a]]; \n Show[Graphics[{ \n{GrayLevel[0], Thickness[0.005], Line[{a, b, c, a}]}, \n{RGBColor[1, 0, 0], Thickness[0.001], Circle[m, r]}, \n{GrayLevel[0.5], PointSize[0.02], Point[m]}, \n{RGBColor[1, 0, 0], Thickness[0.001], \ \ Line[{\(a + b\)\/2, m}], Line[{\(a + c\)\/2, m}], Line[{\(b + c\)\/2, m}]}}], \n\ PlotRange \[Rule] All, AspectRatio \[Rule] Automatic, \ opts]]\)], "Input"], Cell[CellGroupData[{ Cell["CircumcenterSketch[{{0,0},{1,0},{3/4,1}}];", "Input", AspectRatioFixed->True], 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Ool2O00/Ool002ioo`03O01oogoo03moo`05001oogooOol00005Ool2O00^Ool002moo`Al03ioo`05 001oogoo001oo`04O00`Ool003=oo`9l03eoo`800003Ooml07oo03=oo`00=Goo0W`0?7oo0W`0=Goo 000gOol2O00hOol2O00gOol003Uoo`=l035oo`Al03Uoo`00?7oo17`0:Goo17`0?Goo0010Ool4O00Q Ool4O011Ool004Aoo`Al01Uoo`Al04Eoo`00B7oo6G`0BGoo002ZOol00:Yoo`00ZWoo002ZOol00001 \ \>"], ImageRangeCache->{{{0, 169.375}, {169.375, 0}} -> {-0.176454, -0.270204, 0.00798765, 0.00798765}}] }, Open ]], Cell["\<\ We can prove that the three lines really meet in one point by \ calculating the point of intersection of a different pair of lines.\ \>", "Text"], Cell[BoxData[ \(\(P\&^\_1 = Together\ @ \[ScriptCapitalI][{\(P\_A + P\_B\)\/2, \[ScriptCapitalN][P\_A - P\_B]}, \ {\(P\_C + P\_A\)\/2, \ \[ScriptCapitalN][P\_C - P\_A]}]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(P\&^\_1 - P\_1\ // \ Together\)], "Input"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Calculating the nine-point center", "Subsubsection"], Cell["\<\ The nine-point center can be calculated as the circumcenter of the \ triangle formed by the midpoints of the edges of the original triangle.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(P\_2 = Together\ @ \[ScriptCapitalI][{\(\(P\_A + P\_B\)\/2 + \(P\_B + P\_C\)\/2\)\/2, \[ScriptCapitalN][\(P\_A + P\_B\)\/2 - \(P\_B + P\_C\)\/2]}, \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\(\(P\_B + P\_C\)\/2 + \(P\_C + P\_A\)\/2\)\/2, \[ScriptCapitalN][\(P\_B + P\_C\)\/2 - \(P\_C + P\_A\)\/2]}]\)], "Input"], Cell[BoxData[ \({\(( \(-2\)\ P\_Ax\ P\_Ay\ P\_Bx - P\_Ay\ P\_Bx\%2 + P\_Ax\%2\ P\_By - P\_Ay\%2\ P\_By + 2\ P\_Ax\ P\_Bx\ P\_By + P\_Ay\ P\_By\%2 + 2\ P\_Ax\ P\_Ay\ P\_Cx - 2\ P\_Bx\ P\_By\ P\_Cx + P\_Ay\ P\_Cx\%2 - P\_By\ P\_Cx\%2 - P\_Ax\%2\ P\_Cy + P\_Ay\%2\ P\_Cy + P\_Bx\%2\ P\_Cy - P\_By\%2\ P\_Cy - 2\ P\_Ax\ P\_Cx\ P\_Cy + 2\ P\_Bx\ P\_Cx\ P\_Cy - P\_Ay\ P\_Cy\%2 + P\_By\ P\_Cy\%2)\)/ \((4\ \(( \(-P\_Ay\)\ P\_Bx + P\_Ax\ P\_By + P\_Ay\ P\_Cx - P\_By\ P\_Cx - P\_Ax\ P\_Cy + P\_Bx\ P\_Cy)\))\), \((\(-P\_Ax\%2\)\ P\_Bx + P\_Ay\%2\ P\_Bx + P\_Ax\ P\_Bx\%2 - 2\ P\_Ax\ P\_Ay\ P\_By + 2\ P\_Ay\ P\_Bx\ P\_By - P\_Ax\ P\_By\%2 + P\_Ax\%2\ P\_Cx - P\_Ay\%2\ P\_Cx - P\_Bx\%2\ P\_Cx + P\_By\%2\ P\_Cx - P\_Ax\ P\_Cx\%2 + P\_Bx\ P\_Cx\%2 + 2\ P\_Ax\ P\_Ay\ P\_Cy - 2\ P\_Bx\ P\_By\ P\_Cy - 2\ P\_Ay\ P\_Cx\ P\_Cy + 2\ P\_By\ P\_Cx\ P\_Cy + P\_Ax\ P\_Cy\%2 - P\_Bx\ P\_Cy\%2)\)/ \((4\ \(( P\_Ay\ P\_Bx - P\_Ax\ P\_By - P\_Ay\ P\_Cx + P\_By\ P\_Cx + P\_Ax\ P\_Cy - P\_Bx\ P\_Cy)\))\)}\)], "Output"] }, Open ]], Cell["\<\ Here is a sketch of the construction of the nine-point circle for a \ sample triangle.\ \>", "Text"], Cell[BoxData[ \(FeuerbachCircleSketch[{a_, \ b_, \ c_}, opts___] := \n Module[{r, m = P\_2\ /. \n Flatten[Thread\ /@ \ {{P\_Ax, P\_Ay} \[Rule] a, {P\_Bx, P\_By} \[Rule] b, {P\_Cx, P\_Cy} \[Rule] c}]}, \n r = Sqrt[\(#.#&\)[m - \(a + b\)\/2]]; \ \n Show[Graphics[{ \n{GrayLevel[0], Thickness[0.005], Line[{a, b, c, a}]}, \n{GrayLevel[0], Thickness[0.001], Line[{\(a + b\)\/2, \(b + c\)\/2, \(c + a\)\/2, \(a + b\)\/2}]}, \n{RGBColor[0, 1, 0], Thickness[0.001], \ Circle[m, \ r]}, \n{GrayLevel[0.5], \ PointSize[0.02], \ Point[m]}, \n{RGBColor[0, 1, 0], Thickness[0.001], \ \ Line[{\(a + b\)\/4 + c\/2, m}], Line[{\(a + c\)\/4 + b\/2, m}], Line[{\(b + c\)\/4 + a\/2, m}]}}], \n\ PlotRange \[Rule] All, AspectRatio \[Rule] Automatic, opts]]\)], "Input"], Cell[CellGroupData[{ Cell["FeuerbachCircleSketch[{{0,0},{1,0},{3/4,1}}];", "Input", AspectRatioFixed->True], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: 1.02524 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics /Courier findfont 10 scalefont setfont % Scaling calculations 0.0238095 0.952381 0.0484511 0.952381 [ [ 0 0 0 0 ] [ 1 1.02524 0 0 ] ] MathScale % Start of Graphics 1 setlinecap 1 setlinejoin newpath 0 g 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0003h0?P049oo`030n1oogoo00Aoo`03001oogoo01Uoo`00BWoo00@007ooOol3h41oo`030n1oogoo 00Eoo`03001oogoo01Uoo`00Bgoo00<007ooOol00P?P?7oo0P?P27oo00<007ooOol06Goo0017oo0P?P2Goo00<007ooOol06Woo001=Ool00`00Oomoo`02Ool20n0dOol20n0; Ool00`00Oomoo`0JOol004eoo`03001oogoo00Aoo`030n1oogoo031oo`030n1oogoo00]oo`03001o ogoo01Yoo`00CWoo00<007ooOol017oo0P?P;Woo0P?P3Woo00<007ooOol06Woo001?Ool00`00Oomo o`05Ool20n0ZOol20n0?Ool00`00Oomoo`0KOol0051oo`03001oogoo00Ioo`<3h2Aoo`<3h15oo`03 001oogoo01]oo`00D7oo00<007ooOol02Goo0`?P7goo0P?P57oo00<007ooOol06goo001AOol00`00 Oomoo`0;Ool50n0DOol60n0FOol00`00Oomoo`0KOol0059oo`03001oogoo00moo`D3h0Yoo`D3h1]o o`03001oogoo01aoo`00Dgoo00<007ooOol04goo2P?P87oo00<007ooOol077oo001COol00`00Oomo o`0mOol00`00Oomoo`0LOol005Aoo`03001oogoo03aoo`03001oogoo01aoo`00EGoo00<007ooOol0 >Woo00<007ooOol07Goo001FOol00`00Oomoo`0iOol00`00Oomoo`0MOol005Ioo`03001oogoo03Uo o`03001oogoo01eoo`00Egoo00<007ooOol0>7oo00<007ooOol07Goo001HOol00`00Oomoo`0fOol0 0`00Oomoo`0NOol005Uoo`03001oogoo03Eoo`03001oogoo01ioo`00FGoo00<007ooOol0=Goo00<0 07ooOol07Woo001JOol00`00Oomoo`0dOol00`00Oomoo`0NOol005]oo`03001oogoo039oo`03001o ogoo01moo`00G7oo00<007ooOol0"], ImageRangeCache->{{{0, 177.5}, {182, 0}} -> {-0.0250052, -0.0509379, 0.00591555, 0.00591555}}] }, Open ]], Cell["\<\ Again, by choosing a different pair of lines we get the same point \ of intersection.\ \>", "Text"], Cell[BoxData[ \(\(P\&^\_2 = Together\ @ \[ScriptCapitalI][{\(\(P\_A + P\_B\)\/2 + \(P\_B + P\_C\)\/2\)\/2, \[ScriptCapitalN][\(P\_A + P\_B\)\/2 - \(P\_B + P\_C\)\/2]}, \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\(\(P\_C + P\_A\)\/2 + \(P\_A + P\_B\)\/2\)\/2, \[ScriptCapitalN][\(P\_C + P\_A\)\/2 - \(P\_A + P\_B\)\/2]}]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(P\&^\_2 - P\_2 // Together\)], "Input"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Calculating the first Fermat point", "Subsubsection"], Cell["\<\ The first Fermat point is the point of intersection of the three \ lines given in the definition at the beginning. Only two of the three lines \ are needed for the calculation. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(P\_3 = \[ScriptCapitalI][{P\_A, \(P\_B + P\_C\)\/2\ - \ \@3\/2\ \[ScriptCapitalN][P\_C - P\_B] - P\_A}, \ \t\t\ \ \ \ \ \ \ \ \ \ \ \ \n \t\t\t\ \ \ \ \ \ {P\_B, \(P\_C + P\_A\)\/2\ - \ \@3\/2\ \[ScriptCapitalN][P\_A - P\_C] - P\_B}]\ // Together\)], "Input"], Cell[BoxData[ \({\(( \@3\ P\_Ax\%2\ P\_Bx - 4\ P\_Ax\ P\_Ay\ P\_Bx + \@3\ P\_Ay\%2\ P\_Bx + \@3\ P\_Ax\ P\_Bx\%2 - P\_Ay\ P\_Bx\%2 + P\_Ax\%2\ P\_By - 3\ P\_Ay\%2\ P\_By + 4\ P\_Ax\ P\_Bx\ P\_By + \@3\ P\_Ax\ P\_By\%2 + 3\ P\_Ay\ P\_By\%2 + \@3\ P\_Ax\%2\ P\_Cx + 4\ P\_Ax\ P\_Ay\ P\_Cx + \@3\ P\_Ay\%2\ P\_Cx - 6\ \@3\ P\_Ax\ P\_Bx\ P\_Cx + \@3\ P\_Bx\%2\ P\_Cx - 2\ \@3\ P\_Ay\ P\_By\ P\_Cx - 4\ P\_Bx\ P\_By\ P\_Cx + \@3\ P\_By\%2\ P\_Cx + \@3\ P\_Ax\ P\_Cx\%2 + P\_Ay\ P\_Cx\%2 + \@3\ P\_Bx\ P\_Cx\%2 - P\_By\ P\_Cx\%2 - P\_Ax\%2\ P\_Cy + 3\ P\_Ay\%2\ P\_Cy - 2\ \@3\ P\_Ay\ P\_Bx\ P\_Cy + P\_Bx\%2\ P\_Cy - 2\ \@3\ P\_Ax\ P\_By\ P\_Cy - 3\ P\_By\%2\ P\_Cy - 4\ P\_Ax\ P\_Cx\ P\_Cy + 4\ P\_Bx\ P\_Cx\ P\_Cy + \@3\ P\_Ax\ P\_Cy\%2 - 3\ P\_Ay\ P\_Cy\%2 + \@3\ P\_Bx\ P\_Cy\%2 + 3\ P\_By\ P\_Cy\%2)\)/ \((2\ \(( \@3\ P\_Ax\%2 + \@3\ P\_Ay\%2 - \@3\ P\_Ax\ P\_Bx - 3\ P\_Ay\ P\_Bx + \@3\ P\_Bx\%2 + 3\ P\_Ax\ P\_By - \@3\ P\_Ay\ P\_By + \@3\ P\_By\%2 - \@3\ P\_Ax\ P\_Cx + 3\ P\_Ay\ P\_Cx - \@3\ P\_Bx\ P\_Cx - 3\ P\_By\ P\_Cx + \@3\ P\_Cx\%2 - 3\ P\_Ax\ P\_Cy - \@3\ P\_Ay\ P\_Cy + 3\ P\_Bx\ P\_Cy - \@3\ P\_By\ P\_Cy + \@3\ P\_Cy\%2)\))\), \((3\ P\_Ax\%2\ P\_Bx - P\_Ay\%2\ P\_Bx - 3\ P\_Ax\ P\_Bx\%2 + \@3\ P\_Ay\ P\_Bx\%2 + \@3\ P\_Ax\%2\ P\_By + 4\ P\_Ax\ P\_Ay\ P\_By + \@3\ P\_Ay\%2\ P\_By - 4\ P\_Ay\ P\_Bx\ P\_By + P\_Ax\ P\_By\%2 + \@3\ P\_Ay\ P\_By\%2 - 3\ P\_Ax\%2\ P\_Cx + P\_Ay\%2\ P\_Cx - 2\ \@3\ P\_Ay\ P\_Bx\ P\_Cx + 3\ P\_Bx\%2\ P\_Cx - 2\ \@3\ P\_Ax\ P\_By\ P\_Cx - P\_By\%2\ P\_Cx + 3\ P\_Ax\ P\_Cx\%2 + \@3\ P\_Ay\ P\_Cx\%2 - 3\ P\_Bx\ P\_Cx\%2 + \@3\ P\_By\ P\_Cx\%2 + \@3\ P\_Ax\%2\ P\_Cy - 4\ P\_Ax\ P\_Ay\ P\_Cy + \@3\ P\_Ay\%2\ P\_Cy - 2\ \@3\ P\_Ax\ P\_Bx\ P\_Cy + \@3\ P\_Bx\%2\ P\_Cy - 6\ \@3\ P\_Ay\ P\_By\ P\_Cy + 4\ P\_Bx\ P\_By\ P\_Cy + \@3\ P\_By\%2\ P\_Cy + 4\ P\_Ay\ P\_Cx\ P\_Cy - 4\ P\_By\ P\_Cx\ P\_Cy - P\_Ax\ P\_Cy\%2 + \@3\ P\_Ay\ P\_Cy\%2 + P\_Bx\ P\_Cy\%2 + \@3\ P\_By\ P\_Cy\%2)\)/ \((2\ \(( \@3\ P\_Ax\%2 + \@3\ P\_Ay\%2 - \@3\ P\_Ax\ P\_Bx - 3\ P\_Ay\ P\_Bx + \@3\ P\_Bx\%2 + 3\ P\_Ax\ P\_By - \@3\ P\_Ay\ P\_By + \@3\ P\_By\%2 - \@3\ P\_Ax\ P\_Cx + 3\ P\_Ay\ P\_Cx - \@3\ P\_Bx\ P\_Cx - 3\ P\_By\ P\_Cx + \@3\ P\_Cx\%2 - 3\ P\_Ax\ P\_Cy - \@3\ P\_Ay\ P\_Cy + 3\ P\_Bx\ P\_Cy - \@3\ P\_By\ P\_Cy + \@3\ P\_Cy\%2)\))\)}\)], "Output"] }, Open ]], Cell["A sketch reveals that these three lines meet in one point.", "Text"], Cell[BoxData[ \(FirstFermatPointSketch[{a_, b_, c_}, opts___] := \n Module[{m = P\_3 /. \n Flatten[Thread\ /@ \ {{P\_Ax, P\_Ay} \[Rule] a, {P\_Bx, P\_By} \[Rule] b, {P\_Cx, P\_Cy} \[Rule] c}]}, \n a\_1\ = \ \(b + c\)\/2 - \@3\/2\ \[ScriptCapitalN][c - b]; \n b\_1\ = \ \(c + a\)\/2 - \@3\/2\ \[ScriptCapitalN][a - c]; \n c\_1\ = \ \(a + b\)\/2 - \@3\/2\ \[ScriptCapitalN][b - a]; \n Show[Graphics[{ \n{GrayLevel[0], Thickness[0.005], Line[{a, b, c, a}]}, \n{GrayLevel[0.5], Thickness[0.001], \ Line[{a, c\_1, b}], Line[{b, a\_1, c}], Line[{c, b\_1, a}]}, \n{GrayLevel[0.5], PointSize[0.02], Point[m]}, \n{RGBColor[0, 0, 1], Thickness[0.001], \ \ Line[{a, a\_1}], Line[{b, b\_1}], Line[{c, c\_1}]}}], \nPlotRange \[Rule] All, AspectRatio \[Rule] Automatic, opts]]\)], "Input"], Cell[CellGroupData[{ Cell["FirstFermatPointSketch[{{0,0},{1,0},{3/4,1}}];", "Input", AspectRatioFixed->True], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .903 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics /Courier findfont 10 scalefont setfont % Scaling calculations 0.233322 0.426684 0.391019 0.426684 [ [ 0 0 0 0 ] [ 1 .903 0 0 ] ] MathScale % Start of Graphics 1 setlinecap 1 setlinejoin newpath 0 g .005 w [ ] 0 setdash .23332 .39102 m .66001 .39102 L .55334 .8177 L .23332 .39102 L s .5 g .001 w .23332 .39102 m .44666 .0215 L .66001 .39102 L s .66001 .39102 m .97619 .69674 L .55334 .8177 L s .55334 .8177 m .02381 .8815 L .23332 .39102 L s .02 w .51151 .50551 Mdot 0 0 1 r .001 w .23332 .39102 m .97619 .69674 L s .66001 .39102 m .02381 .8815 L s .55334 .8177 m .44666 .0215 L s 0 0 m 1 0 L 1 .903 L 0 .903 L closepath clip newpath % End of Graphics MathPictureEnd \ \>"], "Graphics", ImageSize->{182.875, 165}, ImageMargins->{{35.5625, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, ImageCache->GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHgOol01cg_Oomoogoo01moocg_065oo`00CGoo00Ool00`0OOomoo`08Ool00cg_Oomoo`1DOol0049oo`03?Nmoogoo00moo`0301mo ogoo00Qoo`03?Nmoogoo05Aoo`00@Woo00Ool003]oo`03?Nmo ogoo01Qoo`0301moogoo00eoo`03?Nmoogoo04eoo`00>goo00Ool00cg_Oomoo`1Goo00Ool00`00 Oomoo`0GOol201l;Ool00`0OOomoo`04Ool00`0OOomoo`0=Ool00`00Oomoo`0AOol00cg_Oomoo`0/ Ool002=oo`03?Nmoogoo011oo`03001oogoo01Qoo`<07`Qoo`0301moogoo00=oo`0301moogoo00io o`03001oogoo019oo`03?Nmoogoo02]oo`008goo00Ool00`0OOomoo`0>Ool00`00Oomoo`06Ool301lNOol00cg_Oomo o`0JOol001aoo`03?Nmoogoo02=oo`03001oogoo00Moo`0301moogoo00moo`0301moogoo00ioo`03 001oogoo00Uoo`807aeoo`03?Nmoogoo01Uoo`0077oo00Ool001Moo`03?Nmoogoo02=oo`807`aoo`03001oogoo019oo`0301moogoo00Uoo`03001o ogoo02Moo`807`ioo`03?Nmoogoo00eoo`005goo00Ool00`00Oomoo`0AOol00`0OOomoo`09Ool00`00Oomoo`0/Ool201l;Ool0 0cg_Oomoo`0;Ool001Ioo`03?Nmoogoo025oo`0301moogoo011oo`03001oogoo011oo`0301moogoo 00Uoo`03001oogoo02ioo`<07`Uoo`03?Nmoogoo00Yoo`005Woo00Ool00`0OOomoo`08Ool00`00Oomoo`0dOol301l6 Ool00cg_Oomoo`08Ool001Eoo`03?Nmoogoo01ioo`0301moogoo01Ioo`03001oogoo00ioo`0301mo ogoo00Qoo`03001oogoo03Moo`807`Eoo`03?Nmoogoo00Moo`0057oo00Goo0P0O17oo00Ool0015oo`03?Nmoogoo01Ioo`0301moogoo02Uoo`03 001oogoo00Uoo`0301moogoo00Aoo`03001oogoo03Eoo`@mka5oo`004Goo00Ool00cg_Oomoo`0? Ool00`0OOomoo`0iOol00`00Oomoo`04Ool01@0OOomoogoo00007Goo13g_;Goo000=Ool00cg_Oomo o`0>Ool201lmOol00`00Oomoo`03Ool0100OOomoo`006goo0cg_7oo000goo000;Ool00cg_Oomoo`0; Ool201m5Ool01P00Oomoo`0OOol000ioo`"], ImageRangeCache->{{{0, 181.875}, {164, 0}} -> {-0.548504, -0.916424, 0.0129045, 0.0129045}}] }, Open ]], Cell["\<\ Taking a second pair from the set of three lines shows that all \ three lines meet in the same point.\ \>", "Text"], Cell[BoxData[ \(\(P\&^\_3 = \[ScriptCapitalI][{P\_A, \(P\_B + P\_C\)\/2\ - \ \@3\/2\ \[ScriptCapitalN][P\_C - P\_B] - P\_A}, \ \t\t\ \ \ \ \ \ \ \ \ \ \n \t\t\t\ \ \ \ \ \ {P\_C, \(P\_A + P\_B\)\/2\ - \ \@3\/2\ \[ScriptCapitalN][P\_B - P\_A] - P\_C}]\ // Together; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(P\&^\_3 - P\_3 // Together\)], "Input"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Calculating the second Fermat point", "Subsubsection"], Cell["\<\ The construction of the second Fermat point is very similar to the \ that of the first Fermat point. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(P\_4 = \[ScriptCapitalI][{P\_A, \(P\_B + P\_C\)\/2\ + \ \@3\/2\ \[ScriptCapitalN][P\_C - P\_B] - P\_A}, \ \t\t\ \ \ \ \ \ \ \ \ \ \n \t\t\t\ \ \ \ \ \ {P\_B, \(P\_C + P\_A\)\/2\ + \ \@3\/2\ \[ScriptCapitalN][P\_A - P\_C] - P\_B}]\ // Together\)], "Input"], Cell[BoxData[ \({\(( \@3\ P\_Ax\%2\ P\_Bx + 4\ P\_Ax\ P\_Ay\ P\_Bx + \@3\ P\_Ay\%2\ P\_Bx + \@3\ P\_Ax\ P\_Bx\%2 + P\_Ay\ P\_Bx\%2 - P\_Ax\%2\ P\_By + 3\ P\_Ay\%2\ P\_By - 4\ P\_Ax\ P\_Bx\ P\_By + \@3\ P\_Ax\ P\_By\%2 - 3\ P\_Ay\ P\_By\%2 + \@3\ P\_Ax\%2\ P\_Cx - 4\ P\_Ax\ P\_Ay\ P\_Cx + \@3\ P\_Ay\%2\ P\_Cx - 6\ \@3\ P\_Ax\ P\_Bx\ P\_Cx + \@3\ P\_Bx\%2\ P\_Cx - 2\ \@3\ P\_Ay\ P\_By\ P\_Cx + 4\ P\_Bx\ P\_By\ P\_Cx + \@3\ P\_By\%2\ P\_Cx + \@3\ P\_Ax\ P\_Cx\%2 - P\_Ay\ P\_Cx\%2 + \@3\ P\_Bx\ P\_Cx\%2 + P\_By\ P\_Cx\%2 + P\_Ax\%2\ P\_Cy - 3\ P\_Ay\%2\ P\_Cy - 2\ \@3\ P\_Ay\ P\_Bx\ P\_Cy - P\_Bx\%2\ P\_Cy - 2\ \@3\ P\_Ax\ P\_By\ P\_Cy + 3\ P\_By\%2\ P\_Cy + 4\ P\_Ax\ P\_Cx\ P\_Cy - 4\ P\_Bx\ P\_Cx\ P\_Cy + \@3\ P\_Ax\ P\_Cy\%2 + 3\ P\_Ay\ P\_Cy\%2 + \@3\ P\_Bx\ P\_Cy\%2 - 3\ P\_By\ P\_Cy\%2)\)/ \((2\ \(( \@3\ P\_Ax\%2 + \@3\ P\_Ay\%2 - \@3\ P\_Ax\ P\_Bx + 3\ P\_Ay\ P\_Bx + \@3\ P\_Bx\%2 - 3\ P\_Ax\ P\_By - \@3\ P\_Ay\ P\_By + \@3\ P\_By\%2 - \@3\ P\_Ax\ P\_Cx - 3\ P\_Ay\ P\_Cx - \@3\ P\_Bx\ P\_Cx + 3\ P\_By\ P\_Cx + \@3\ P\_Cx\%2 + 3\ P\_Ax\ P\_Cy - \@3\ P\_Ay\ P\_Cy - 3\ P\_Bx\ P\_Cy - \@3\ P\_By\ P\_Cy + \@3\ P\_Cy\%2)\))\), \((\(-3\)\ P\_Ax\%2\ P\_Bx + P\_Ay\%2\ P\_Bx + 3\ P\_Ax\ P\_Bx\%2 + \@3\ P\_Ay\ P\_Bx\%2 + \@3\ P\_Ax\%2\ P\_By - 4\ P\_Ax\ P\_Ay\ P\_By + \@3\ P\_Ay\%2\ P\_By + 4\ P\_Ay\ P\_Bx\ P\_By - P\_Ax\ P\_By\%2 + \@3\ P\_Ay\ P\_By\%2 + 3\ P\_Ax\%2\ P\_Cx - P\_Ay\%2\ P\_Cx - 2\ \@3\ P\_Ay\ P\_Bx\ P\_Cx - 3\ P\_Bx\%2\ P\_Cx - 2\ \@3\ P\_Ax\ P\_By\ P\_Cx + P\_By\%2\ P\_Cx - 3\ P\_Ax\ P\_Cx\%2 + \@3\ P\_Ay\ P\_Cx\%2 + 3\ P\_Bx\ P\_Cx\%2 + \@3\ P\_By\ P\_Cx\%2 + \@3\ P\_Ax\%2\ P\_Cy + 4\ P\_Ax\ P\_Ay\ P\_Cy + \@3\ P\_Ay\%2\ P\_Cy - 2\ \@3\ P\_Ax\ P\_Bx\ P\_Cy + \@3\ P\_Bx\%2\ P\_Cy - 6\ \@3\ P\_Ay\ P\_By\ P\_Cy - 4\ P\_Bx\ P\_By\ P\_Cy + \@3\ P\_By\%2\ P\_Cy - 4\ P\_Ay\ P\_Cx\ P\_Cy + 4\ P\_By\ P\_Cx\ P\_Cy + P\_Ax\ P\_Cy\%2 + \@3\ P\_Ay\ P\_Cy\%2 - P\_Bx\ P\_Cy\%2 + \@3\ P\_By\ P\_Cy\%2)\)/ \((2\ \(( \@3\ P\_Ax\%2 + \@3\ P\_Ay\%2 - \@3\ P\_Ax\ P\_Bx + 3\ P\_Ay\ P\_Bx + \@3\ P\_Bx\%2 - 3\ P\_Ax\ P\_By - \@3\ P\_Ay\ P\_By + \@3\ P\_By\%2 - \@3\ P\_Ax\ P\_Cx - 3\ P\_Ay\ P\_Cx - \@3\ P\_Bx\ P\_Cx + 3\ P\_By\ P\_Cx + \@3\ P\_Cx\%2 + 3\ P\_Ax\ P\_Cy - \@3\ P\_Ay\ P\_Cy - 3\ P\_Bx\ P\_Cy - \@3\ P\_By\ P\_Cy + \@3\ P\_Cy\%2)\))\)}\)], "Output"] }, Open ]], Cell["\<\ Here is a sketch of the construction of the second Fermat \ point.\ \>", "Text"], Cell[BoxData[ \(SecondFermatPointSketch[{a_, b_, c_}, opts___] := \n Module[{m = P\_4 /. \n Flatten[Thread\ /@ \ {{P\_Ax, P\_Ay} \[Rule] a, {P\_Bx, P\_By} \[Rule] b, {P\_Cx, P\_Cy} \[Rule] c}]}, \n a\_1\ = \ \(b + c\)\/2 + \@3\/2\ \[ScriptCapitalN][c - b]; \n b\_1\ = \ \(c + a\)\/2 + \@3\/2\ \[ScriptCapitalN][a - c]; \n c\_1\ = \ \(a + b\)\/2 + \@3\/2\ \[ScriptCapitalN][b - a]; \n Show[Graphics[{ \n{GrayLevel[0], Thickness[0.005], Line[{a, b, c, a}]}, \n{GrayLevel[0.5], Thickness[0.001], \ Line[{a, c\_1, b}], Line[{b, a\_1, c}], Line[{c, b\_1, a}]}, \n{GrayLevel[0.5], \ PointSize[0.02], \ Point[m]}, \n{RGBColor[1, 1, 0], Thickness[0.001], \ \ Line[{a, a\_1, m}], \ Line[{b, b\_1, m}], Line[{c, c\_1, m}]}}], \n PlotRange\ \[Rule] \ All, AspectRatio \[Rule] Automatic, opts]]\)], "Input"], Cell[CellGroupData[{ Cell["SecondFermatPointSketch[{{0,0},{1,0},{3/4,1}}];", "Input", AspectRatioFixed->True], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .92627 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics /Courier findfont 10 scalefont setfont % Scaling calculations 0.0238095 0.767415 0.136797 0.767415 [ [ 0 0 0 0 ] 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00]oo`03On1oogoo025oo`03?Nmoogoo009oo`03?Nmoogoo00ioo`03001oogoo02aoo`03?Nmoogoo 01ioo`03001oogoo00]oo`03?Nmoogoo03Yoo`002Goo00=oh7ooOol02Goo0WoP9Goo00DmkgooOomo ocg_015oo`03001oogoo02Yoo`03?Nmoogoo01moo`03001oogoo00Yoo`03?Nmoogoo03]oo`002Goo 00=oh7ooOol01goo0WoP:7oo0Sg_0Woo00goo0009Ool00goPOomoo`06Ool00goPOomoo`0ZOol00cg_Oolm k`0AOol00`00Oomoo`0XOol00cg_Oomoo`0OOol00`00Oomoo`0;Ool00cg_Oomoo`0kOol000Uoo`03 On1oogoo00Aoo`9oh2ioo`03?Nmoocg_015oo`03001oogoo02Moo`03?Nmoogoo01moo`03001oogoo 00Yoo`03?Nmoogoo03aoo`002Goo00=oh7ooOol00goo00=oh7ooOol0;goo00Woo00Ool0 0cg_Oomoo`0;Ool2?Nl8Ool00`00Oomoo`0ROol00`00Oomoo`04Ool00cg_Oomoo`1Ool005=oo`9oh0Moo`03?Nmoogoo00Eoo`03?Nmoogoo00Ioo`03?Nmoogoo00Eoo`03001oogoo 01eoo`03001oogoo009oo`03?Nmoogoo04moo`00EGoo0WoP1Goo00aoo`00k7oo003/ Ool00>aoo`00\ \>"], ImageRangeCache->{{{0, 235.562}, {218.062, 0}} -> {-0.0314264, -0.178263, 0.00691896, 0.00691896}}] }, Open ]], Cell["\<\ Yet again, here is a proof that all three lines meet in one \ point.\ \>", "Text"], Cell[BoxData[ \(\(\n\t\t \(P\&^\_4 = \[ScriptCapitalI][{P\_A, \(P\_B + P\_C\)\/2\ + \ \@3\/2\ \[ScriptCapitalN][P\_C - P\_B] - P\_A}, \ \t\t\ \ \ \ \ \ \ \ \ \n \t\t\t\ \t\t\t\ \ \ {P\_C, \(P\_A + P\_B\)\/2\ + \ \@3\/2\ \[ScriptCapitalN][P\_B - P\_A] - P\_C}]\ // Together; \)\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(P\&^\_4 - P\_4 // Together\)], "Input"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["A traditional proof of the theorem", "Subsubsection"], Cell["\<\ We calculate the center of the Lester circle explicitly using three \ of the four points.\ \>", "Text"], Cell[BoxData[ \(\(M\_\(1\ \) = \[ScriptCapitalI][{\(P\_1 + P\_2\)\/2, \[ScriptCapitalN][P\_1 - P\_2]}, {\(P\_2 + P\_3\)\/2, \[ScriptCapitalN][P\_2 - P\_3]}] /. \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {P\_Ax \[Rule] 0, P\_Ay \[Rule] 0, P\_By \[Rule] 0}; \)\)], "Input"], Cell["\<\ Choosing another triple gives the same center, proving the theorem. \ \ \>", "Text"], Cell[BoxData[ \(\(M\&^\_\(1\ \) = \[ScriptCapitalI][{\(P\_1 + P\_2\)\/2, \[ScriptCapitalN][P\_1 - P\_2]}, {\(P\_2 + P\_4\)\/2, \[ScriptCapitalN][P\_2 - P\_4]}] /. \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {P\_Ax \[Rule] 0, P\_Ay \[Rule] 0, P\_By \[Rule] 0}; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(M\&^\_\(1\ \) - M\_\(1\ \) // Together\)], "Input"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["An algebraic proof of the theorem", "Subsubsection"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`{X\_0, Y\_0}\)]], " and ", Cell[BoxData[ \(TraditionalForm\`R\)]], " be the center and radius of the Lester circle. \nIf the four points ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(P\_1\), "TraditionalForm"], ",", " ", FormBox[\(P\_2\), "TraditionalForm"], ",", " ", FormBox[\(P\_3\), "TraditionalForm"], ",", " ", FormBox[\(P\_4\), "TraditionalForm"]}], TraditionalForm]]], " actually lie on one circle, they will satisfy the equations ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(\(({X\_0, Y\_0} - P\_i)\).\(({\)\), "TraditionalForm"], FormBox[\({X\_0, Y\_0} - P\_i)\), "TraditionalForm"]}], "\[Equal]", \(R\^2\)}], TraditionalForm]]], " simultaneously. " }], "Text"], Cell[TextData[{ "Without loss of generality we assume the vertex ", Cell[BoxData[ \(TraditionalForm\`P\_A\)]], " of the triangle to be at the origin and the point ", Cell[BoxData[ \(TraditionalForm\`P\_B\)]], " to be on the ", StyleBox["x", FontSlant->"Italic"], "-axis. This makes the expressions to be manipulated much smaller and the \ calculations much faster. 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This property can be checked by \ calculating a ", StyleBox["Gr\[ODoubleDot]bner Basis", "InlineInput"], " for \[ScriptCapitalE]." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({1} =!= GroebnerBasis[\[ScriptCapitalE], \ {X\_0, Y\_0, R}, \n \t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tMonomialOrder\ \[Rule] \ DegreeLexicographic, \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tCoefficientDomain\ \[Rule] \ RationalFunctions]\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[TextData[{ "The last test actually proves only that the four points lie on a circle in \ ", StyleBox["complex", FontSlant->"Italic"], " space, where the coordinates of points might be complex numbers. To prove \ that the theorem holds in ", StyleBox["real", FontSlant->"Italic"], " space we have to consider the corresponding radical ideal. 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