Manuel Kauers Hagenberg, November 24, 2002 RISC-Linz

Algorithmic Algebraic Geometry

Solution 4

The following lines are exactly the bitangents to the plane algebraic curve

2x^{4} + 2y^{4} + 2 − 3x^{2}y^{2} − 3x^{2} − 3y^{2} = 0.

Note that in this case all real lines are expressible over ℚ(\sqrt{2}).

t ↦ \binom{0}{0} + t\binom{−1 + \sqrt{2}}{1}  t ↦ \binom{0}{0} + t\binom{−1 − \sqrt{2}}{1},
t ↦ \binom{0}{1/2\sqrt{2}} + t\binom{1}{−1},  t ↦ \binom{0}{−1/2\sqrt{2}} + t\binom{1}{−1},
t ↦ \binom{0}{1} + t\binom{1}{−1/2\sqrt{2}},  t ↦ \binom{0}{1} + t\binom{1}{1/2\sqrt{2}},
t ↦ \binom{0}{−1} + t\binom{1}{−1},  t ↦ \binom{0}{−1 + \sqrt{2}} + t\binom{1}{0},
t ↦ \binom{0}{−1 − \sqrt{2}} + t\binom{1}{0},  t ↦ \binom{0}{−1} + t\binom{1}{−1/2\sqrt{2}},
t ↦ \binom{0}{−1} + t\binom{1}{1/2\sqrt{2}},  t ↦ \binom{0}{−\sqrt{2}} + t\binom{1/2\sqrt{2}}{1},
t ↦ \binom{0}{\sqrt{2}} + t\binom{−1/2\sqrt{2}}{1},  t ↦ \binom{0}{1} + t\binom{1}{−1},
t ↦ \binom{0}{−\sqrt{2}} + t\binom{−1/2\sqrt{2}}{1},  t ↦ \binom{0}{\sqrt{2}} + t\binom{1/2\sqrt{2}}{1},
t ↦ \binom{0}{0} + t\binom{1 − \sqrt{2}}{1},  t ↦ \binom{0}{0} + t\binom{1 + \sqrt{2}}{1},
t ↦ \binom{0}{1/2\sqrt{2}} + t\binom{1}{1},  t ↦ \binom{0}{−1/2\sqrt{2}} + t\binom{1}{1},
t ↦ \binom{0}{−1} + t\binom{1}{1},  t ↦ \binom{0}{1 − \sqrt{2}} + t\binom{1}{0},
t ↦ \binom{0}{1 + \sqrt{2}} + t\binom{1}{0},  t ↦ \binom{−1 + \sqrt{2}}{0} + t\binom{0}{1},
t ↦ \binom{−1 − \sqrt{2}}{0} + t\binom{0}{1},  t ↦ \binom{1 − \sqrt{2}}{0} + t\binom{0}{1},
t ↦ \binom{1 + \sqrt{2}}{0} + t\binom{0}{1},  t ↦ \binom{0}{1} + t\binom{1}{1}.