How to determine whether three ellipses have at least one common intersection point or not?

Question

How to establish a criterion described in equation so that it is easy to determine whether three ellipses have common intersection area (point) or not?

Update

Answer 1

We need oracles that

(a) test for any point ${\bf p}\in{\mathbb R}^2$ and any ellipse $E$ whether ${\bf p}\in {\rm int}(E)$,

(b) compute for any ellipse $E$ its center ${\bf m}_E$,

(c) compute for any two ellipses $E_1$ and $E_2$ the set $\partial E_1\cap\partial E_2$, consisting of $0$, $2$, or $4$ points.

(Degenerate cases will not be considered.)

Let $A$, $B$, $C$ be the three given ellipses.

Compute all intersection points of two of the three boundary ellipses $\partial A$, $\partial B$, $\partial C$, using oracle (c).

If $\partial A\cap\partial B$ is empty check whether ${\bf m}_A\in B$ or ${\bf m}_B\in A$. If ${\bf m}_A\in B$ we know that $A\subset B$, so it remains to check the ellipses $A$ and $C$ versus each other. (I leave this to the OP.) – Similarly when ${\bf m}_B\in A$.

Similarly when $\partial B\cap\partial C$ or $\partial C\cap\partial A$ is empty.

Assume now that any two boundary ellipses $\partial A$, $\partial B$, $\partial C$ intersect. Then we apply the following Lemma and oracle (a):

Lemma. The set $\Omega:={\rm int}(A\cap B\cap C)$ is nonempty iff at least one intersection point of two boundary ellipses lies in the interior of the third ellipse.

Proof. Any neighborhood of ${\bf r}\in\partial A\cap\partial B$ contains points of ${\rm int}(A\cap B)$. When ${\bf r}$ lies in ${\rm int}(C)$ the set $\Omega$ is nonempty.

For the proof of the converse consider the following "Gedankenexperiment": We pick a point ${\bf p}\in \Omega$ and draw a ray from ${\bf p}$ to infinity that does not go through any intersection point of two boundary ellipses. This ray will hit a first boundary ellipse, say $\partial A$, in a point ${\bf q}$. The point ${\bf q}$ is still in the interior of both $B$ and $C$. Now walk from ${\bf q}$ along $\partial A$ until you hit for the first time a point of $\partial A\cap\partial B$ or $\partial A\cap\partial C$. Assume this is the point ${\bf r}\in\partial A\cap\partial B$, say. This point ${\bf r}$ is still in the interior of $C$.

Answer 2

Inspired by @ Semiclassical comment, I got one answer:

1. suppose three ellipse are $a$, $b$ and $c$; so there are three combinations if we draw two from the three: $(a,b),(a,c),(b,c)$

2. test the number of interaction points between $(a,b), (a,c),\text{ or } (b,c)$, if the number is $0$, there is no common point between the three; if the number is $1$, then if the only point is not in or on the third ellipse, there is no common point between the three;

3. for the cases when interaction points between all the combinations are more than or equal to $2$ ($\leqslant 4$), test if there is any intersection point in/on the third ellipse; if all are no, then they have no common point; if any is yes, then the three ellipses have common interaction points.

This still has exception case:

Update

1. suppose three ellipse are $a$, $b$ and $c$; so there are three combinations if we draw two from the three: $(a,b),(a,c),(b,c)$

2. test the number of interaction points between $(a,b), (a,c),\text{ or } (b,c)$,

   2.1 if the number is $0$, check whether one ellipse is totally contained by another; if no, there is no common point between the three; if yes, the problem becomes determine whether the small ellipse and the third have common point or not; check the existence of intersection point and the containing relationship between the two;

   2.2 if the number is $1$, then if the only point is not in or on the third ellipse, there is no common point between the three;

3. for the cases when interaction points between all the combinations are more than or equal to $2$ ($\leqslant 4$), test if there is any intersection point in/on the third ellipse; if all are no, then they have no common point; if any is yes, then the three ellipses have common interaction points.