双纽线

hbghlyj

给定四点A,B,C,D,$\sin\angle APB\cos\angle CPD+\cos\angle BPC\sin\angle DPA=0$,求P的轨迹

hbghlyj

$\cot\angle APB\cot\angle BPC=\cot\angle CPD\cot\angle DPA$
化简后是直线AC与直线BD与四次曲线$\Gamma$与四次曲线$\Omega$的并
$\Gamma:x^4-x_1 x^3-x_2 x^3-x_3 x^3-x_4 x^3+2 y^2 x^2+x_1 x_2 x^2+x_1 x_3 x^2+x_2 x_3 x^2+x_1 x_4 x^2+x_2 x_4 x^2+x_3 x_4 x^2-y y_1 x^2-y y_2 x^2+y_1 y_2 x^2-y y_3 x^2-y_1 y_3 x^2+y_2 y_3 x^2-y y_4 x^2+y_1 y_4 x^2-y_2 y_4 x^2+y_3 y_4 x^2-y^2 x_1 x-y^2 x_2 x-y^2 x_3 x-x_1 x_2 x_3 x-y^2 x_4 x-x_1 x_2 x_4 x-x_1 x_3 x_4 x-x_2 x_3 x_4 x+2 y x_3 y_1 x+2 y x_4 y_2 x-x_3 y_1 y_2 x-x_4 y_1 y_2 x+2 y x_1 y_3 x+x_2 y_1 y_3 x+x_4 y_1 y_3 x-x_1 y_2 y_3 x-x_4 y_2 y_3 x+2 y x_2 y_4 x-x_2 y_1 y_4 x-x_3 y_1 y_4 x+x_1 y_2 y_4 x+x_3 y_2 y_4 x-x_1 y_3 y_4 x-x_2 y_3 y_4 x+y^4+y^2 x_1 x_2-y^2 x_1 x_3+y^2 x_2 x_3+y^2 x_1 x_4-y^2 x_2 x_4+y^2 x_3 x_4+x_1 x_2 x_3 x_4-y^3 y_1-y x_2 x_3 y_1+y x_2 x_4 y_1-y x_3 x_4 y_1-y^3 y_2+y x_1 x_3 y_2-y x_1 x_4 y_2-y x_3 x_4 y_2+y^2 y_1 y_2+x_3 x_4 y_1 y_2-y^3 y_3-y x_1 x_2 y_3-y x_1 x_4 y_3+y x_2 x_4 y_3+y^2 y_1 y_3-x_2 x_4 y_1 y_3+y^2 y_2 y_3+x_1 x_4 y_2 y_3-y y_1 y_2 y_3-y^3 y_4-y x_1 x_2 y_4+y x_1 x_3 y_4-y x_2 x_3 y_4+y^2 y_1 y_4+x_2 x_3 y_1 y_4+y^2 y_2 y_4-x_1 x_3 y_2 y_4-y y_1 y_2 y_4+y^2 y_3 y_4+x_1 x_2 y_3 y_4-y y_1 y_3 y_4-y y_2 y_3 y_4+y_1 y_2 y_3 y_4=0$
$\Omega$过于复杂，就不贴了

hbghlyj

当A,B,C,D是圆周c上的调和分割点时,$\Gamma$与c重合.
当A,B重合时,$\Gamma$是以C,D为直径的圆
当A,B,C,D分别是(±cos22.5°,±sin22.5°)时,$\Gamma$是双纽线$(x^2+y^2)^2=\sqrt2(x^2-y^2)$

IMO会徽

（有点类似于IMO会徽）
下面是ggb文件.Enjoy!
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