I ran into the following problem on point equality in convex hulls.
Let $S$ be a subset of $d$-dimensional Euclidean space. Suppose that points $p_1,p_2$ do not lie in the convex hull of $S$, but $p_i$ lies in the convex hull of $S \cup \{ p_{3-i} \}$ for $i=1,2$. Show that $p_1=p_2$.
For this problem, it seems to be true that $p_i$ is the convex combination of a point in the convex hull of $S$ and the point $p_{3-i}$, but I'm not sure how to use this to obtain the needed result. I've tried for a direct way to prove this using the convex combination equalities, but those don't seem to lead anywhere nontrivial.
I'm looking for an elegant solution method possibly using the fact I mentioned.
First, note that without loss of generality we can replace $S$ with its convex hull and assume that $S$ is convex.
Assume $p,q\notin S$ but $p\in \text{conv}(\{q\}\cup S)$ and $q\in \text{conv}(\{p\}\cup S)$. We can write $$p=sq+(1-s)A$$ and $$q=tp+(1-t)B$$ for some $A,B\in S$ and $s,t\in [0,1]$. Assume that $s<1$. Then $1-st\in [0,1]$, and \begin{align*} p& =sq+(1-s)A=s\bigl[tp+(1-t)B\bigr]+(1-s)A \\ & = stp+s(1-t)B+(1-s)A,\end{align*} so $$(1-st)p=s(1-t)A+(1-s)B.$$ Therefore $$p=\frac{1-st}{1-st}\cdot p = \frac{s(1-t)}{1-st}A+\frac{1-s}{1-st}B\in \text{conv}(S), $$ a contradiction. Therefore it must be that $s=1$, which means $p=q$.