Let $A$, $B$, $C$ be finite subsets of the plane such that $A \cap B$, $B \cap C$, and $C \cap A$ are all empty. Let $S=A \cup B \cup C$. Assume that no three points of $S$ are collinear. Also, assume that each of $A$, $B$ and $C$ has at least $3$ points. Which of the following statements is always true?
(a) There exists a triangle having a vertex from each of $A$, $B$ and $C$ that doesn't contain any point of $S$ in its interior.
(b) There exists a triangle having a vertex from each of $A$, $B$ and $C$ that contains all the remaining points of $S$ in its interior.
(c) Any triangle having a vertex from each $A$, $B$ and $C$ must contain a point of $S$ in its interior.
(d) There exists $2$ triangles both having a vertex from each of $A$, $B$ and $C$ such that the two triangles do not intersect.
In my opinion, Option (a) must always be true as we can always choose choose three points on the edge is set but I am not sure how to write the solution.
The following figure gives a counter example for the items (b), (c) and (d).
Here, sets $A$, $B$, and $C$ are formed by red, blue and green dots, respectively.
(b) Notice that the convex hull of $S$ is an hexagon, so there is no triangle with vertices in S that contains all the remaining points of S in its interior.
(c) Notice that the triangle $\triangle XYZ$ doesn't contain any point of $S$ in its interior.
(d) Notice that any triangle having a vertex from each of $A$, $B$ and $C$ contains the black dot, so there are not $2$ such triangles such that the two do not intersect.
Therefore the correct answer must be (a). To check it out, just follow the tip given by Mike Earnest in the comment.