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A bitangent is a line which is tangent to two points \ of a given curve at the same time. Here we will explicitly calculate all 28 \ bitangents for the quartic\n\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(12\^2\ \((x\^4 + y\^4)\) - 15\^2\ \((x\^2 + y\^2)\) + 350 x\^2\ y\^2 + 81\), "TraditionalForm"], "=", "0"}], TraditionalForm]]], ". \n\nThis quartic is symmetric with respect to ", Cell[BoxData[ \(TraditionalForm\`x\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y\)]], ". In the following we will make use of the symmetry ", Cell[BoxData[ \(TraditionalForm \`under the exchange of variables x \[LeftRightArrow] \ \(y.\)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(p[x_, y_] = 12\^2\ \((x\^4 + y\^4)\) - 15\^2\ \((x\^2 + y\^2)\) + 350 x\^2\ y\^2 + 81\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(81 + 350\ x\^2\ y\^2 - 225\ \((x\^2 + y\^2)\) + 144\ \((x\^4 + y\^4)\)\)], "Output"] }, Open ]], Cell["\<\ We implement the calculation of the discriminant of a polynomial, \ defined as the resultant of a polynomial and its derivative. We will later on \ make use of the obvious fact that a polynomial has a double root if and only \ if its discriminant vanishes.\ \>", "Text"], Cell[BoxData[ \(Discriminant[\[ScriptP]_, x_] := Factor[Resultant[\[ScriptP], \[PartialD]\_x \[ScriptP], x]] /; PolynomialQ[\[ScriptP], x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Using the function ", StyleBox["ContourPlot", "Input", FontWeight->"Plain"], " we can visualize the quartic under interest. Its shape resembles two \ intersecting ellipses. This is not accidental. If one takes an ellipse ", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)]], "+", Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)]], "-1 = 0 and its image after rotation by ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ", namely ", Cell[BoxData[ \(TraditionalForm\`x\^2\/b\^2\)]], "+", Cell[BoxData[ \(TraditionalForm\`y\^2\/a\^2\)]], "-1=0, multiplies these two implicit representations and adds a constant ", Cell[BoxData[ \(TraditionalForm\`C\)]], " to make the resulting polynomial irreducible (adding the constant ", Cell[BoxData[ \(TraditionalForm\`C\)]], " means, geometrically speaking, smoothing out the cusps at the points of \ intersection of the two ellipses) one obtains\n \n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"(", RowBox[{ FormBox[\(x\^2\/a\^2\), "TraditionalForm"], "+", FormBox[\(y\^2\/b\^2\), "TraditionalForm"], "-", "1"}], ")"}], RowBox[{"(", RowBox[{\(x\^2\/b\^2\), "+", FormBox[\(y\^2\/a\^2\), "TraditionalForm"], "-", "1"}], ")"}]}], "-", "C"}], 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Each of these pairs of \ regions will have four bitangents, two on the outside boundaries and two \ crossing ones. This makes 24 bitangents. Additionally there are 4 bitangents \ on the inner side of every region itself parallel to the ", Cell[BoxData[ \(TraditionalForm\`x\)]], "- and ", Cell[BoxData[ \(TraditionalForm\`y\)]], "-axes. Taking them all together results in 28 real bitangents." }], "Text"], Cell[CellGroupData[{ Cell["Exact Calculation of the 28 Bitangents", "Subsubsection"], Cell["\<\ In the following we will in a heuristic manner calculate most of \ these tangents. At the end we give a general method of calculation.\ \>", "Text"], Cell[TextData[{ "Looking at the quartic ", StyleBox["p[x,y],", "Input", FontWeight->"Plain"], " it becomes obvious that one class of bitangents will be parallel to the \ ", Cell[BoxData[ \(TraditionalForm\`x\)]], "- and ", Cell[BoxData[ \(TraditionalForm\`y\)]], "-axes. Assuming ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], " we get the following equation for the intersection of the line ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], " with our quartic." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(r\_1 = Factor[Resultant[p[x, y], x - a, x]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(81 - 225\ a\^2 + 144\ a\^4 - 225\ y\^2 + 350\ a\^2\ y\^2 + 144\ y\^4\)], "Output"] }, Open ]], Cell[TextData[{ "The condition that this intersection be a tangent requires the last \ equation to have a double root for ", Cell[BoxData[ \(TraditionalForm\`y\)]], ", so using the above defined function ", StyleBox["Discriminant", "Input", FontWeight->"Plain"], " we get:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(d\_1 = Discriminant[r\_1, y]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(2985984\ \((\(-1\) + a)\)\ \((1 + a)\)\ \((\(-3\) + 4\ a)\)\ \((3 + 4\ a)\)\ \((3969 - 27900\ a\^2 + 39556\ a\^4)\)\^2\)], "Output"] }, Open ]], Cell[TextData[{ "Solving for ", Cell[BoxData[ \(TraditionalForm\`\(a, \)\)]], " we have the following result:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(s\_1 = Union[Solve[d\_1 == 0, a]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({{a \[Rule] \(-1\)}, {a \[Rule] \(-\(3\/4\)\)}, {a \[Rule] 3\/4}, { a \[Rule] 1}, {a \[Rule] \(-3\)\ \@\(\(775 - 12\ \@806\)\/19778\)}, { a \[Rule] 3\ \@\(\(775 - 12\ \@806\)\/19778\)}, { a \[Rule] \(-3\)\ \@\(\(775 + 12\ \@806\)\/19778\)}, { a \[Rule] 3\ \@\(\(775 + 12\ \@806\)\/19778\)}}\)], "Output"] }, Open ]], Cell["\<\ We sort out all imaginary solutions and all lines which do not have \ exactly two common points with our quartic (the condition for a bitangent).\ \ \>", "Text"], Cell[BoxData[ \(goodSolutionQ[sol_] := \n \(\((Length[Union[#]] === 2 && Union[Im[y /. #]] == {0})\)&\)[\ \t\t\n \t\t\ \ \ \ \ \ \ \ \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solve[\((p[a, y] /. sol)\) \[Equal] 0, y]]\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(s\_1 = Select[s\_1, goodSolutionQ]\)], "Input"], Cell[BoxData[ \({{a \[Rule] \(-3\)\ \@\(\(775 - 12\ \@806\)\/19778\)}, { a \[Rule] 3\ \@\(\(775 - 12\ \@806\)\/19778\)}, { a \[Rule] \(-3\)\ \@\(\(775 + 12\ \@806\)\/19778\)}, { a \[Rule] 3\ \@\(\(775 + 12\ \@806\)\/19778\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "The points of intersection corresponding to these values of ", Cell[BoxData[ \(TraditionalForm\`a\)]], " are given by:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(t\_1 = Union[Flatten[ Function[s, \((Point[{a /. s, #}]&)\)/@ \((y /. Solve[\((p[a, y] /. s)\) == 0, y])\)]/@s\_1]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({Point[{\(-3\)\ \@\(\(775 - 12\ \@806\)\/19778\), \(-\(5\/2\)\)\ \@\(\(3\ \((372 + 7\ \@806)\)\)\/19778\)}], Point[{\(-3\)\ \@\(\(775 - 12\ \@806\)\/19778\), 5\/2\ \@\(\(3\ \((372 + 7\ \@806)\)\)\/19778\)}], Point[{3\ \@\(\(775 - 12\ \@806\)\/19778\), \(-\(5\/2\)\)\ \@\(\(3\ \((372 + 7\ \@806)\)\)\/19778\)}], Point[{3\ \@\(\(775 - 12\ \@806\)\/19778\), 5\/2\ \@\(\(3\ \((372 + 7\ \@806)\)\)\/19778\)}], Point[{\(-3\)\ \@\(\(775 + 12\ \@806\)\/19778\), \(-\(5\/2\)\)\ \@\(\(3\ \((372 - 7\ \@806)\)\)\/19778\)}], Point[{\(-3\)\ \@\(\(775 + 12\ \@806\)\/19778\), 5\/2\ \@\(\(3\ \((372 - 7\ \@806)\)\)\/19778\)}], Point[{3\ \@\(\(775 + 12\ \@806\)\/19778\), \(-\(5\/2\)\)\ \@\(\(3\ \((372 - 7\ \@806)\)\)\/19778\)}], Point[{3\ \@\(\(775 + 12\ \@806\)\/19778\), 5\/2\ \@\(\(3\ \((372 - 7\ \@806)\)\)\/19778\)}]}\)], "Output"] }, Open ]], Cell[TextData[{ "Here we show the four bitangents that are parallel to the ", Cell[BoxData[ \(TraditionalForm\`y\)]], "-axis, as well as the equivalent ones which are parallel to the ", Cell[BoxData[ \(TraditionalForm\`x\)]], "-axis (here we use the above mentioned ", Cell[BoxData[ \(TraditionalForm\`x \[LeftRightArrow] y\)]], " symmetry of ", StyleBox["p[x,y]", "Input", FontWeight->"Plain"], ")." }], "Text"], Cell[BoxData[ \(\(opts\ = \ {PlotRange \[Rule] {{\(-1.25\), 1.25}, {\(-1.25\), 1.25}}, Frame \[Rule] False, Axes \[Rule] False}; \)\)], "Input"], Cell[BoxData[ \(x\_min = \(-\(3\/2\)\); \ \ x\_max = 3\/2; \)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Show", "[", RowBox[{ RowBox[{"{", RowBox[{"quartic", ",", "\n", "\t\t ", RowBox[{\(B\_1\), "=", RowBox[{"Graphics", "[", RowBox[{"{", RowBox[{ RowBox[{"Thickness", "[", StyleBox["0.0001", StyleBoxAutoDelete->True, PrintPrecision->1], "]"}], ",", \(PointSize[0.012]\), ",", "\n", "\t\t\t\t\t", \(\((Line[{{#, x\_min}, {#, x\_max}}]&)\)/@ \((a /. s\_1)\)\), ",", 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0`00Oomoo`1FOol00`00Oomoo`0HOol00`00Oomoo`0dOol003Eoo`03001oogoo01Qoo`03001oogoo 05Ioo`03001oogoo01Qoo`03001oogoo03Aoo`00=Goo00<007ooOol067oo00<007ooOol0EWoo00<0 07ooOol067oo00<007ooOol0=7oo000eOol00`00Oomoo`0HOol00`00Oomoo`1FOol00`00Oomoo`0H Ool00`00Oomoo`0dOol003Eoo`03001oogoo01Qoo`03001oogoo05Ioo`03001oogoo01Qoo`03001o ogoo03Aoo`00=Goo00<007ooOol067oo00<007ooOol0EWoo00<007ooOol067oo00<007ooOol0=7oo 000eOol00`00Oomoo`0HOol00`00Oomoo`1FOol00`00Oomoo`0HOol00`00Oomoo`0dOol003Eoo`03 001oogoo01Qoo`03001oogoo05Ioo`03001oogoo01Qoo`03001oogoo03Aoo`00=Goo00<007ooOol0 67oo00<007ooOol0EWoo00<007ooOol067oo00<007ooOol0=7oo000eOol00`00Oomoo`0HOol00`00 Oomoo`1FOol00`00Oomoo`0HOol00`00Oomoo`0dOol00001\ \>"], ImageRangeCache->{{{0, 250}, {250, 0}} -> {-1.25001, -1.25001, 0.0125001, 0.0125001}}] }, Open ]], Cell[TextData[{ "The second class of bitangents will be the ones crossing between opposite \ segments of the quartic. Because of the symmetry of the quartic under \ consideration, these lines will go through the origin. Proceeding as above, \ we calculate the intersection of the quartic with a line ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["y", FontSlant->"Italic"], "=", RowBox[{ StyleBox["ax", FontSlant->"Italic"], "."}]}], TraditionalForm]]], " We have:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(r\_2 = Factor[Resultant[p[x, y], y - a x, x]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(81\ a\^4 - 225\ a\^2\ y\^2 - 225\ a\^4\ y\^2 + 144\ y\^4 + 350\ a\^2\ y\^4 + 144\ a\^4\ y\^4\)], "Output"] }, Open ]], Cell["\<\ Again requiring that for a tangent we have a double zero, we get:\ \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(d\_2 = Discriminant[r\_2, y]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(34012224\ a\^12\ \((49 - 150\ a\^2 + 49\ a\^4)\)\^2\ \((72 + 175\ a\^2 + 72\ a\^4)\)\^2\)], "Output"] }, Open ]], Cell[TextData[{ "We can easily solve for possible ", Cell[BoxData[ \(TraditionalForm\`a\)]], "'s." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(s\_2 = Union[Solve[d\_2 == 0, a]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({{a \[Rule] 0}, {a \[Rule] \(-\(1\/7\)\)\ \@\(75 - 2\ \@806\)}, { a \[Rule] 1\/7\ \@\(75 - 2\ \@806\)}, { a \[Rule] \(-\(1\/7\)\)\ \@\(75 + 2\ \@806\)}, { a \[Rule] 1\/7\ \@\(75 + 2\ \@806\)}, { a \[Rule] \(-\(1\/12\)\)\ I\ \@\(175 - \@9889\)}, { a \[Rule] 1\/12\ I\ \@\(175 - \@9889\)}, { a \[Rule] \(-\(1\/12\)\)\ I\ \@\(175 + \@9889\)}, { a \[Rule] 1\/12\ I\ \@\(175 + \@9889\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "Selecting again the right ", Cell[BoxData[ \(TraditionalForm\`a'\)]], "s we finally arrive at the following values for ", Cell[BoxData[ \(TraditionalForm\`a\)]], ":" }], "Text"], Cell[BoxData[ \(goodSolutionQ[sol_] := \(Im[a /. sol] === 0 && \n \((Length[Union[#]] === 2 && \ Union[Im[x /. #]] === {0})\)&\)[\n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \t Solve[\((p[x, a x] /. sol)\) \[Equal] 0, x]]\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(s\_2 = Select[s\_2, goodSolutionQ]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({{a \[Rule] \(-\(1\/7\)\)\ \@\(75 - 2\ \@806\)}, { a \[Rule] 1\/7\ \@\(75 - 2\ \@806\)}, { a \[Rule] \(-\(1\/7\)\)\ \@\(75 + 2\ \@806\)}, { a \[Rule] 1\/7\ \@\(75 + 2\ \@806\)}}\)], "Output"] }, Open ]], Cell["\<\ The points of intersection of these bitangents with the quartic \ are:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(t\_2 = Union[Flatten[ Function[s, \((Point[{#, a # /. s}]&)\)/@ \((x /. Solve[\((p[a x, x] /. s)\) \[Equal] 0, x])\)]/@s\_2]] \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({Point[{\(-\(3\/5\)\)\ \@\(1\/62\ \((62 - \@806)\)\), \(-\(3\/35\)\)\ \@\(1\/62\ \((62 - \@806)\)\ \((75 + 2\ \@806)\)\)}], Point[{\(-\(3\/5\)\)\ \@\(1\/62\ \((62 - \@806)\)\), 3\/35\ \@\(1\/62\ \((62 - \@806)\)\ \((75 + 2\ \@806)\)\)}], Point[{3\/5\ \@\(1\/62\ \((62 - \@806)\)\), \(-\(3\/35\)\)\ \@\(1\/62\ \((62 - \@806)\)\ \((75 + 2\ \@806)\)\)}], Point[{3\/5\ \@\(1\/62\ \((62 - \@806)\)\), 3\/35\ \@\(1\/62\ \((62 - \@806)\)\ \((75 + 2\ \@806)\)\)}], Point[{\(-\(3\/5\)\)\ \@\(1\/62\ \((62 + \@806)\)\), \(-\(3\/35\)\)\ \@\(1\/62\ \((75 - 2\ \@806)\)\ \((62 + \@806)\)\)}], Point[{\(-\(3\/5\)\)\ \@\(1\/62\ \((62 + \@806)\)\), 3\/35\ \@\(1\/62\ \((75 - 2\ \@806)\)\ \((62 + \@806)\)\)}], Point[{3\/5\ \@\(1\/62\ \((62 + \@806)\)\), \(-\(3\/35\)\)\ \@\(1\/62\ \((75 - 2\ \@806)\)\ \((62 + \@806)\)\)}], Point[{3\/5\ \@\(1\/62\ \((62 + \@806)\)\), 3\/35\ \@\(1\/62\ \((75 - 2\ \@806)\)\ \((62 + \@806)\)\)}]}\)], "Output"] }, Open ]], Cell["Here we show the four bitangents that go through the origin:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(Show[{quartic, \n\t\t B\_2 = Graphics[{Thickness[0.0001], PointSize[0.015], \((Line[{{x\_min, \(x\_min\) #}, {x\_max, \(x\_max\) #}}]&)\)/@ \((a /. s\_2)\), t\_2}]}, opts]; \)\)], "Input", AspectRatioFixed->True], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: 1 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 0.5 0.4 0.5 0.4 [ [ 0 0 0 0 ] [ 1 1 0 0 ] ] MathScale % Start of Graphics 1 setlinecap 1 setlinejoin newpath 0 0 m 1 0 L 1 1 L 0 1 L closepath clip newpath 1 g -0.1 1.1 m 1.1 1.1 L 1.1 -0.1 L -0.1 -0.1 L F .5 g .2038 .32211 m .21236 .32532 L .21494 .34051 L .21465 .3557 L .21183 .37089 L .20934 .38608 L .20709 .40127 L .20511 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third class of bitangents\[LongDash]the outer ones for \ adjacent regions\[LongDash]are parallel to the lines ", Cell[BoxData[ \(TraditionalForm\`y = x\)]], " or ", Cell[BoxData[ \(TraditionalForm\`y = \(-x\)\)]], " . So this time we look for bitangents of the form ", Cell[BoxData[ \(TraditionalForm\`y = x + m\)]], ". The resulting equation for ", StyleBox["m", FontSlant->"Italic"], " is:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(r\_3 = Factor[Resultant[p[x, y], y - x - m, x]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(81 - 225\ m\^2 + 144\ m\^4 + 450\ m\ y - 576\ m\^3\ y - 450\ y\^2 + 1214\ m\^2\ y\^2 - 1276\ m\ y\^3 + 638\ y\^4\)], "Output"] }, Open ]], Cell[TextData[{ "We have a tangency condition if ", StyleBox["m", FontSlant->"Italic"], " satisfies the following equation:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(d\_3 = Discriminant[r\_3, y]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(1055058048\ \((648 - 900\ m\^2 + 319\ m\^4)\)\ \((117 - 1550\ m\^2 + 992\ m\^4)\)\^2\)], "Output"] }, Open ]], Cell[TextData[{ "This gives the following possible values for ", StyleBox["m", FontSlant->"Italic"], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(s\_3 = Union[Solve[d\_3 == 0, m]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({{m \[Rule] \(-3\)\ \((\(-1\))\)\^\(1/4\)\ \@\(2\/319\ \((\(-25\)\ I + \@13)\)\)}, { m \[Rule] 3\ \((\(-1\))\)\^\(1/4\)\ \@\(2\/319\ \((\(-25\)\ I + \@13)\)\)}, { m \[Rule] \(-3\)\ \((\(-1\))\)\^\(3/4\)\ \@\(2\/319\ \((25\ I + \@13)\)\)}, { m \[Rule] 3\ \((\(-1\))\)\^\(3/4\)\ \@\(2\/319\ \((25\ I + \@13)\)\)}, { m \[Rule] \(-\(1\/4\)\)\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}, { m \[Rule] 1\/4\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}, { m \[Rule] \(-\(1\/4\)\)\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}, { m \[Rule] 1\/4\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}}\)], "Output"] }, Open ]], Cell["Selecting the real values we have:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(s\_3 = Select[s\_3, Im[m /. #] == 0&]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({{m \[Rule] \(-\(1\/4\)\)\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}, { m \[Rule] 1\/4\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}, { m \[Rule] \(-\(1\/4\)\)\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}, { m \[Rule] 1\/4\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}}\)], "Output"] }, Open ]], Cell["\<\ The corresponding points where these lines meet the quartic are \ given by:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(t\_3 = Union[Flatten[ Function[s, \((Point[{#, # + m /. s}]&)\)/@ \((x /. Solve[\((p[x + m, x] /. s)\) \[Equal] 0, x])\)]/@ s\_3]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({Point[{ \(-\(1\/2\)\)\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\), \(-\(1\/2\)\)\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}], Point[{1\/2\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\), 1\/2\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}], Point[{\(-\(1\/2\)\)\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\), \(-\(1\/2\)\)\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}], Point[{1\/2\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\), 1\/2\ \@\(25\/32 + 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 - 7\ \@9889)\)\)}], Point[{\(-\(1\/2\)\)\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\), \(-\(1\/2\)\)\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}], Point[{1\/2\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\), 1\/2\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}], Point[{\(-\(1\/2\)\)\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\), \(-\(1\/2\)\)\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}], Point[{1\/2\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) + 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\), 1\/2\ \@\(25\/32 - 1799\/\(32\ \@9889\)\) - 1\/8\ \@\(1\/62\ \((775 + 7\ \@9889)\)\)}]}\)], "Output"] }, Open ]], Cell["Here are these eight bitangents:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(Show[{quartic, \n\t\t B\_3 = Graphics[{Thickness[0.0001], PointSize[0.015], \((Line[{{x\_min, \ \ \ x\_min + #}, {x\_max, \ x\_max + #}}]&) \)/@\((m /. s\_3)\), \((Line[{{x\_min, \(-x\_min\) - #}, {x\_max, \(-x\_max\) - #}}]&)\)/@\((m /. s\_3)\), t\_3, 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07ooOol0FGoo00<007oo0000Fgoo00<007ooOol06goo000LOol00`00Oomoo`1IOol01@00Oomoogoo 0000Fgoo00<007ooOol06Woo000KOol00`00Oomoo`1IOol00`00Oomoo`03Ool00`00Oomoo`1IOol0 0`00Oomoo`0IOol00001\ \>"], ImageRangeCache->{{{0, 244.75}, {244.75, 0}} -> {-1.25001, -1.25001, 0.0127683, 0.0127683}}] }, Open ]], Cell[TextData[{ "Now comes the final class of bitangents, the crossing ones between \ adjacent regions of the quartic. This time we have to take into account a \ general form of a line ", Cell[BoxData[ \(TraditionalForm\`y = m\ x\ + \ b\)]], ". The resulting lines will have nonobvious values for ", Cell[BoxData[ \(TraditionalForm\`m\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b\)]], ".\n\nThe condition of being a bitangent means that there are two points ", Cell[BoxData[ \(TraditionalForm\`\((x\_1, y\_1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((x\_2, y\_2)\)\)]], " which are are tangent and that the slope of the line through ", Cell[BoxData[ \(TraditionalForm\`\((x\_1, y\_1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((x\_2, y\_2)\)\)]], " has the same slope as the slope at ", Cell[BoxData[ \(TraditionalForm\`\((x\_1, y\_1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((x\_2, y\_2)\)\)]], ". Additionally we want the two points ", Cell[BoxData[ \(TraditionalForm\`\((x\_1, y\_1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((x\_2, y\_2)\)\)]], " to be distinct. In the following ", StyleBox["GroebnerBasis", "Input", FontWeight->"Plain"], " calculation we can force this by using the equation ", Cell[BoxData[ \(TraditionalForm \`1 - z\ \((\((x\_1 - x\_2)\)\^2 + \((y\_1 - y\_2)\)\^2)\)\)]], " where ", Cell[BoxData[ \(TraditionalForm\`z\)]], " is a new variable. These conditions translate into the following set of \ equations:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(Numerator\ [Together[#]]&\)[\n \t\t\t{\(y\_2 - y\_1\)\/\(x\_2 - x\_1\) + \[PartialD]\_\(x\_1\)p[x\_1, y\_1]\/\[PartialD]\_\(y\_1\)p[x\_1, y\_1], \(y\_2 - y\_1\)\/\(x\_2 - x\_1\) + \[PartialD]\_\(x\_2\)p[x\_2, y\_2]\/\[PartialD]\_\(y\_2\)p[x\_2, y\_2], \n\t\t\t\tp[x\_1, y\_1], p[x\_2, y\_2], \n\t\t\t 1 - z \((\((x\_1 - x\_2)\)\^2 + \((y\_1 - y\_2)\)\^2)\)}] /. { y\_1 \[Rule] \ m\ x\_1 + b, \ y\_2 \[Rule] m\ x\_2 + b}\)], "Input"], Cell[BoxData[ \({\(-225\)\ x\_1\%2 + 288\ x\_1\%4 - 225\ \((b + m\ x\_1)\)\^2 + 700\ x\_1\%2\ \((b + m\ x\_1)\)\^2 + 288\ \((b + m\ x\_1)\)\^4 + 225\ x\_1\ x\_2 - 288\ x\_1\%3\ x\_2 - 350\ x\_1\ \((b + m\ x\_1)\)\^2\ x\_2 + 225\ \((b + m\ x\_1)\)\ \((b + m\ x\_2)\) - 350\ x\_1\%2\ \((b + m\ x\_1)\)\ \((b + m\ x\_2)\) - 288\ \((b + m\ x\_1)\)\^3\ \((b + m\ x\_2)\), \(-225\)\ x\_1\ x\_2 + 225\ x\_2\%2 + 288\ x\_1\ x\_2\%3 - 288\ x\_2\%4 - 225\ \((b + m\ x\_1)\)\ \((b + m\ x\_2)\) + 350\ \((b + m\ x\_1)\)\ x\_2\%2\ \((b + m\ x\_2)\) + 225\ \((b + m\ x\_2)\)\^2 + 350\ x\_1\ x\_2\ \((b + m\ x\_2)\)\^2 - 700\ x\_2\%2\ \((b + m\ x\_2)\)\^2 + 288\ \((b + m\ x\_1)\)\ \((b + m\ x\_2)\)\^3 - 288\ \((b + m\ x\_2)\)\^4, 81 - 225\ x\_1\%2 + 144\ x\_1\%4 - 225\ \((b + m\ x\_1)\)\^2 + 350\ x\_1\%2\ \((b + m\ x\_1)\)\^2 + 144\ \((b + m\ x\_1)\)\^4, 81 - 225\ x\_2\%2 + 144\ x\_2\%4 - 225\ \((b + m\ x\_2)\)\^2 + 350\ x\_2\%2\ \((b + m\ x\_2)\)\^2 + 144\ \((b + m\ x\_2)\)\^4, 1 - z\ x\_1\%2 - z\ \((b + m\ x\_1)\)\^2 + 2\ z\ x\_1\ x\_2 - z\ x\_2\%2 + 2\ z\ \((b + m\ x\_1)\)\ \((b + m\ x\_2)\) - z\ \((b + m\ x\_2)\)\^2}\)], "Output"] }, Open ]], Cell[TextData[{ "We now eliminate the variables ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`z\)]], " to obtain a set of equations for ", Cell[BoxData[ \(TraditionalForm\`m\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(gb = Factor/@\ GroebnerBasis[%, {m, b}, {x\_1, x\_2, z}, MonomialOrder \[Rule] EliminationOrder]\)], "Input"], Cell[BoxData[ \({\(-b\)\ m\ \(( 114498 - 406250\ b\^2 + 316448\ b\^4 - 77175\ m\^2 - 88200\ b\^2\ m\^2)\), 4557999600 - 37518881832\ b\^2 + 83937261600\ b\^4 - 54600254368\ b\^6 - 13953060000\ m\^2 + 78626961675\ b\^2\ m\^2 - 39636024950\ b\^4\ m\^2 + 4557999600\ m\^4 + 16532892414\ b\^2\ m\^4, \(-b\)\ m\ \(( 146250 - 494450\ b\^2 + 316448\ b\^4 - 77175\ m\^2 - 31752\ m\^4)\), \(-6419878138800\) + 62535682114200\ b\^2 - 186346184911200\ b\^4 + 173485171560800\ b\^6 + 26943693734574\ m\^2 - 29827082580975\ b\^2\ m\^2 - 92562567048902\ b\^4\ m\^2 - 28739282897700\ m\^4 + 7291005554574\ m\^6, b\ \((\(-16079117544\) + 218368245600\ b\^2 - 900736637856\ b\^4 + 1049847884800\ b\^6 + 604603602225\ m\^2 - 1250249651650\ b\^2\ m\^2 - 501287458662\ m\^4)\), \(-m\)\ \(( 194481 - 547101\ b\^2 + 386012\ b\^4 - 1366976\ b\^6 - 595350\ m\^2 + 3120075\ b\^2\ m\^2 + 194481\ m\^4)\), b\ \((1555848 - 10936800\ b\^2 + 15505952\ b\^4 + 16284123\ m\^2 - 26146950\ b\^2\ m\^2 + 8227648\ b\^4\ m\^2 - 15040746\ m\^4)\)}\)], "Output"] }, Open ]], Cell["mbPolys=Last /@ gb;", "Input"], Cell[TextData[{ "By choosing to eliminate ", Cell[BoxData[ \(TraditionalForm\`b\)]], " from all possible combinations of the above equations, and by keeping \ equations with real solutions for ", Cell[BoxData[ \(TraditionalForm\`m\)]], ", we get the following set of equations." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Select[ Union[Flatten[ Table[Cases[ Factor[Resultant[ mbPolys\[LeftDoubleBracket]i\[RightDoubleBracket], mbPolys\[LeftDoubleBracket]j\[RightDoubleBracket], b]], _Plus?\((MemberQ[#, m, \[Infinity]]&)\), \[Infinity]], {i, Length[mbPolys]}, {j, i + 1, Length[mbPolys]}]]], And@@\(\((Im[#] \[Equal] 0&)\)/@ \((m /. {ToRules[Roots[# \[Equal] 0, m, Quartics \[Rule] False]]}) \)\)&]\)], "Input"], Cell[BoxData[ \({\(-1\) + m, 1 + m, \(-12722\) + 8575\ m\^2, \(-7397773571\) + 730406250\ m\^2, \(-31713190336453491520186\) + 24043094452364188653511\ m\^2, 49 - 150\ m\^2 + 49\ m\^4, 936 - 17503\ m\^2 + 936\ m\^4, 1786568616 - 67178178025\ m\^2 + 55698606518\ m\^4, 126873875960526137202889092493299618660 - 2631597114629323377892806551479423805775\ m\^2 + 7174372731304595472168502558756635026384\ m\^4}\)], "Output"] }, Open ]], Cell[TextData[{ "Within this approach we find all finite values for ", Cell[BoxData[ \(TraditionalForm\`b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`m\)]], ", and we also obtain the equations ", Cell[BoxData[ \(TraditionalForm\`m = 1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`m = \(-1\)\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`49 - 150\ m\^2 + 49\ m\^4\)]], "from above. After checking the explicit numerical values, we recognize \ that the last equation for ", Cell[BoxData[ \(TraditionalForm\`\(m\ \)\)]], "we are looking for is given by ", Cell[BoxData[ \(TraditionalForm\`936 - 17503\ m\^2 + 936\ m\^4\)]], ". In a similar manner one can obtain the equation for ", Cell[BoxData[ \(TraditionalForm\`b\)]], ", namely ", Cell[BoxData[ \(TraditionalForm\`279\ - \ 775 b\^2\ + \ 104 b\^4 = 0\)]], ". (The additional equations we got for ", Cell[BoxData[ \(TraditionalForm\`b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`m\)]], " would lead to complex ", Cell[BoxData[ \(TraditionalForm\`\((x\_1, y\_1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((x\_2, y\_2)\)\)]], " and are not of interest here.)" }], "Text"], Cell[TextData[{ "A much easier and faster way than using algebraic techniques like ", StyleBox["Resultant", "Input", FontWeight->"Plain"], " and ", StyleBox["GroebnerBasis", "Input", FontWeight->"Plain"], " is to use a numerical approch to calculate the remaining bitangents \ numerically.\n\nFor the two points ", Cell[BoxData[ \(TraditionalForm\`\((x\_1, y\_1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((x\_2, y\_2)\)\)]], " itself we get following system \[ScriptCapitalS] of polynomial \ equations." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[ScriptCapitalS] = Numerator\ @ \(Together@\n \t\t\t{\(y\_2 - y\_1\)\/\(x\_2 - x\_1\) + \[PartialD]\_\(x\_1\)p[x\_1, y\_1]\/\[PartialD]\_\(y\_1\)p[ x\_1, y\_1], \(y\_2 - y\_1\)\/\(x\_2 - x\_1\) + \[PartialD]\_\(x\_2\)p[x\_2, y\_2]\/\[PartialD]\_\(y\_2\)p[ x\_2, y\_2], p[x\_1, y\_1], p[x\_2, y\_2]}\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({\(-225\)\ x\_1\%2 + 288\ x\_1\%4 + 225\ x\_1\ x\_2 - 288\ x\_1\%3\ x\_2 - 225\ y\_1\%2 + 700\ x\_1\%2\ y\_1\%2 - 350\ x\_1\ x\_2\ y\_1\%2 + 288\ y\_1\%4 + 225\ y\_1\ y\_2 - 350\ x\_1\%2\ y\_1\ y\_2 - 288\ y\_1\%3\ y\_2, \(-225\)\ x\_1\ x\_2 + 225\ x\_2\%2 + 288\ x\_1\ x\_2\%3 - 288\ x\_2\%4 - 225\ y\_1\ y\_2 + 350\ x\_2\%2\ y\_1\ y\_2 + 225\ y\_2\%2 + 350\ x\_1\ x\_2\ y\_2\%2 - 700\ x\_2\%2\ y\_2\%2 + 288\ y\_1\ y\_2\%3 - 288\ y\_2\%4, 81 - 225\ x\_1\%2 + 144\ x\_1\%4 - 225\ y\_1\%2 + 350\ x\_1\%2\ y\_1\%2 + 144\ y\_1\%4, 81 - 225\ x\_2\%2 + 144\ x\_2\%4 - 225\ y\_2\%2 + 350\ x\_2\%2\ y\_2\%2 + 144\ y\_2\%4}\)], "Output"] }, Open ]], Cell[TextData[{ "Using the function ", StyleBox["FindRoot", "Input", FontWeight->"Plain"], " with starting values appropriate for a bitangent which touches the lower \ part of the quartic at the right side and the right side of the quartic at \ the upper end, we find these tangency points." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(s\_4 = FindRoot[Evaluate[\(# \[Equal] 0&\)/@\[ScriptCapitalS]], \n \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {x\_1, 0.7}, { y\_1, 0.4}, {x\_2, 0.4}, {y\_2, \(-0.7\)}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({x\_1 \[Rule] 0.714378711347535944`, y\_1 \[Rule] 0.425333573442719892`, x\_2 \[Rule] 0.44341038329793978`, y\_2 \[Rule] \(-0.744740038545805393`\)}\)], "Output"] }, Open ]], Cell[TextData[{ "Using these points we can calculate the parameters ", StyleBox["m", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({m, b} = {\(y\_2 - y\_1\)\/\(x\_2 - x\_1\) /. s\_4, y\_1 - \(\(y\_2 - y\_1\)\/\(x\_2 - x\_1\)\) x\_1 /. s\_4}\)], "Input",\ AspectRatioFixed->True], Cell[BoxData[ \({4.31811946588223882`, \(-2.65943904603894409`\)}\)], "Output"] }, Open ]], Cell[TextData[{ "These values are in agreement with the roots of the above derived \ equations for ", Cell[BoxData[ \(TraditionalForm\`m\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Module[{m, b}, \n \t{\(NRoots[936 - 17503\ m\^2 + 936\ m\^4 == 0, m] \)\[LeftDoubleBracket]\(-1\), 2\[RightDoubleBracket], \(NRoots[279 - 775\ b\^2 + 104\ b\^4 == 0, b]\)\[LeftDoubleBracket]1, 2\[RightDoubleBracket]}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({4.31811949439249342`, \(-2.65943906650009465`\)}\)], "Output"] }, Open ]], Cell["Thus we have found the following eight bitangents:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(Show[{quartic, \n\t\t B\_4 = Graphics[{Thickness[0.0001], PointSize[0.015], \n\t With[{twoTangents = N@{Line[{{x\_min, \(x\_min\) m + b}, {x\_max, \(x\_max\) m + b}}], 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