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Degree-4 Cayley-Bacharach theoremThree quartic curves pass through the same thirteen points, no five of which are collinear, no nine of which are
conconic and no twelve of which are concubic. The three quartics then share a further three points.
moda-ch07
23. 让八边形的八个顶点位于一条圆锥曲线上，并交替将边着色为红色和蓝色。证明：异色边（的延伸）的其余八个交点共圆锥曲线。
size(200);
path conic = slant(.2)*unitcircle;pen p;
draw(conic);
real[] rel ={.2,.35,.5,.6,.7,.8,0};
pair[] vertices={point(conic,0.2)};
for (int i=0;i<7;++i) {
vertices.push(relpoint(conic,rel[i]));
}
for (int i=0;i<8;++i) {
if(i%2==0){p=red;}else{p=blue;}
draw(vertices[(i+1)%8]--vertices[i],p);
pair a=extension(vertices[i%8],vertices[(i+1)%8],vertices[(i+3)%8],vertices[(i+4)%8]);
dot(a);
draw(vertices[(i+1)%8]--a--vertices[(i+3)%8],dashed);
}
Let $Q_1$ be the union of red lines, $Q_2$ be the union of blue lines, and $Q_3$ be the union of the main conic with the conic passing through five of the other eight heterochromatic intersections. By the quartic version of Cayley-Bacharach, $Q_3$ must pass through the other three intersections. |
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