Shortest path between two points via two disks

Hallo everybody,

I have the following problem regarding shortest paths in $R^2$.

Suppose you are given two points $p$ and $q$ and two unit disks, as in the picture. I am looking for a path from $p$ to $q$ through a point $c_1$ in the first disk and $c_2$ in the second disk such that the sum $\overline{p c_1}+\overline{c_1 c_2}+\overline{c_2 q}$ is minimum.

I know how to find a path if there is only one disk, via reflection properties of ellipses. However, the case for two disks eludes me. I was hoping that you could have some suggestions, or some pointers to something to read.

• Do you really mean a point in the disk, or in the circle? Aug 28, 2014 at 13:19
• I mean closed disk, that is, inside and on the boundary. Not that you want to put anything inside anyway! Aug 28, 2014 at 13:23
• Sounds like the funicular polygon in engineering statics. If $O_1,O_2$ are centers of the circles, forces applied through $c_1O_1,c_2 O_2$ should be angular bisector of each string at $O_1,O_2.$ Aug 25, 2015 at 23:59

Let's name the circles $S_1,S_2$ so that $c_i \in S_i$. Using calculus you can prove that if the minimum is achieved with the configuration that you have shown then at each of the two points $c_i$, the two angles of incidence must be equal, meaning the angles that the two arcs make with the circle $S_i$ at the point $c_i$.
Consider the case where $$c_1$$ is fixed. If center of cicle containing point $$c_i$$ is $$O_i$$, then the minimum is attained when $$Oc_2$$ bisects the angle $$\angle c_1c_2q$$.
Now assume that $$c_i$$ are movable : For some $$c_i$$ if minimum of total length $$|pc_1|+|c_1c_2|+|c_2q|$$ is obtained and if we assume that there are not two bisections, then we can assume that there is no bisection at a point $$c_2$$. Then we fixes $$c_1$$ and we move $$c_2$$ so that there is bisection. Hence we have length decreasing which is a contradiction. Hence minimum of total length can happen when there are two bisections.