Given two Convex polygon $A$, $B$, denote their number of extreme points as $N(A), N(B)$.
One can obtain a bigger convex polygon by union $A, B$, and find the convex hull denote $C:= CH(A\cup B)$.
what's the relationship between $N(C)$ vs $N(A) + N(B)$, my gut feeling is $N(C) \leq N(A) +N(B)$ ? and what's the lower bound of $N(C)$ in the sense of $N(A), N(B)$?
and do the results generalize to non-convex polygon? and also do the result generalizes to convex polytope?
We are going to show $N(C) \le N(A) + N(B)$ in general.
For any $p \in C$, we have either $p \in A \cup B$ or $\not\in A \cup B$.
Case I - $p \not\in A\cup B$.
In this case, I claim
To see why this is true, consider the set of points $$C' \stackrel{def}{=} \{ \alpha a + (1-\alpha) b : \alpha \in [0,1], a \in A, b \in B \}$$
It is not hard to see
Since by definition $C$ is the smallest convex set containing $A$ and $B$, this forces $C = C'$.
This implies we can find an $\alpha \in [0,1], a \in A, b \in B$ such that $p = \alpha a + (1-\alpha) b$. Now $\alpha$ cannot be $0$ because $p \not\in B$ and cannot be $1$ because $p \not\in A$. This justifies above claim.
Next, choose a small positive $\epsilon < \min( \alpha, 1 - \alpha )$, we find $p$ is the mid point of two points $(\alpha \pm \epsilon ) a + (1 - \alpha \mp \epsilon ) b \in C$. This means $p$ cannot be an extreme point.
Case II - $p \in A \cup B \implies p \in A \lor p \in B$.
If $p \in A$ but not an vertex of $A$, it is not an extreme point of $A$ and hence cannot be an extreme point of the larger set $C$. By a similar argument, when $p \in B$ but not an vertex of $B$, $p$ cannot be an extreme point of $C$.
Combine these two cases, we find there are at most $N(A) + N(B)$ candidates of extreme points of $C$. Namely, the vertices of $A$ and $B$. From this, we can deduce $$N(C) \le N(A) + N(B)$$