Suppose I have 3 points colored blue, red, and green resp. forming a triangle $T$ in $\mathbb R^2$. Suppose I have 3 more points colored blue, red, green resp. (possibly overlapping) in the interior of $T$. A rainbow triangle is one whose vertices are all different colors. A proper rainbow triangle is one properly contained within $T$.
Conjecture ($d=2$): the union of all proper rainbow triangles covers (the interior of) $T$.
More generally, suppose I have $d+1$ points colored distinct colors $c_1,\ldots, c_{d+1}$ resp. forming a $d$-simplex $T$ in $\mathbb R^d$. Suppose I have $d+1$ more points colored $c_1,\ldots, c_{d+1}$ resp. (possibly overlapping) in the interior of $T$. A rainbow simplex is one whose vertices are all different colors. A proper rainbow simplex is one properly contained within $T$.
Conjecture: the union of all proper rainbow simplices covers (the interior of) $T$.
For $d=2$, I drew some pictures and am convinced of its truth. But I only have a decent proof in the case that none of the sub/proper rainbow simplices overlap:
In this case, every rainbow $(d-1)$-dimensional interior face (i.e. every line segment between 2 different colored points lying in the interior of $T$) has exactly 2 rainbow $d$-simplices having it as a face; so the outward normals point in opposite directions.
So when we add simplices together, all the interior faces cancel, leaving only the faces on the boundary of $T$, since those are the only ones that only have 1 $d$-simplex containing it as a face.
Thus, when we add up all the proper rainbow $d$-simplices, the resulting boundary is the boundary of $T$. I'm not sure how to 100% rigorously conclude that the sum must be exactly $T$, but still this seems like mostly the right route.
P.S. I came across this conjecture while working on a proof of the strong Caratheodory theorem. The proof of that linked in the linked blog post uses topology, so I've been trying to find a proof using tools as elementary as possible.