Set the intersection points equal, the equations equal to each other and solve.

学：使交点相等，方程式相等并求解。

If there are no objects between the intersection point and a light source, then that means that the point on the blue stripe is directly illuminated by that light source.

如果交点和光源之间没有其他物体，那么这意色条纹上的那个点是直接被该光源照亮的。

The one that I landed on just because I feel like both of these teams are just at interesting intersection points.

我最终择这个，是因为我觉得这两支球队正处在有趣的交汇点上。

So the total number of vertices here  would not be N but N plus N choose four total intersection points.

所以这里的顶点总数不是N，而是N加上n4的总交点数。

Then compare these two and find the intersection point, which is the determined opportunity point.

然后比较这两者，找出它们的交集点，这个交集点就是确定的机会点。

This light is called indirect illumination, and in order to calculate it, we take the intersection point from the primary ray and generate a secondary ray that bounces off it.

这种光线被称为间接照明，为了计算它， 我们从主光线的交点个二次光线， 该二次光线从该交点弹出。

I claim that that line also passes through the intersection point between the external tangents of the circles that we care about.

我声称那条线也经过我们关心的圆的外切线的交点。

Every intersection point corresponds  with some quadruplet of points.

每个交点对应于某四个点的组合。

Now that we have a set of intersection points on separate planes, we find whether the point is inside each corresponding triangle.

现在我们有组位于不同平面上的交点，我们需要判断这些点是否在各自对应的三角形内部。

But the key is to treat this as a new graph  where those intersection points are them vertices.

但关键是要将这视为个新的图，其中那些交点就是顶点。

Counting the number of intersection points in  the diagram is a little bit trickier.

计算图中的交点数量要棘手些。

But of course that can't happen, whether you're looking at it from one side, or looking at it from the other, those intersection points are all the same.

但当然这种情况不会发，无论是从哪边看，这些交点都是样的。

The claim, the theorem, the thing you need to prove is that those three intersection points of the external tangents always fall on the same line.

你需要证明的命题是那些三条外切线的交点总是落在同条直线上。

You might mentally imagine turning them with a little knob to find the sweet spot angle where all those intersection points line up.

你可能会在心里想象用个小旋钮来调整它们，找到个所有交点都对齐的最佳角度。

Maybe if you're a little bit lucky,   here's one thing you might notice; every intersection point is uniquely associated  with a quadruplet of points on the exterior.

也许如果你有点幸运， 你会注意到件事；每个交点都唯地与圆外的四个点相关联。

If it was the case that no bending happened, then because the wavelength is smaller in there, when you look at all those intersection points, they would have to be closer together.

如果确实没有发弯折，那么由于那里的波长更小，当你观察所有这些交点时，它们就必须靠得更近。

For example, look at this simple diagram where  we have three lines and two intersection points.

例如，看这个简单的示意图，我们有三条线和两个交点。

For example, if you calculate 4 choose 4,   everything cancels and you just get one and indeed, there is a single intersection point  in this diagram.

例如，如果你计算44， 所有项都会约掉， 最后得到1，确实，在这个图中只有个交点。

So, in effect,   each intersection point adds two more edges.

因此，每个交点实际上增加了两条边。

And then similarly, if you look at the lines passing through the tips of any other pair of cones, those lines will correspond to another one of those intersection points.

然后类似地，如果你看任意另对圆锥顶点所在的直线，这些直线将对应另个交点。