Harvard-MIT Mathematics Tournament

Let $H = A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}$ be the hexagon, and for all $1 \leq i \leq 6$, let points $A_{i}'$ be considered such that $A_{i}A_{i}' < 1$. Let $H' = A_{1}'A_{2}'A_{3}'A_{4}'A_{5}'A_{6}'$, and consider all indices modulo $6$. For any point $P$ in the plane, let $D(P)$ denote the unit disk $\{ Q \mid PQ < 1 \}$ centered at $P$; it follows that $A_{i}' \in D(A_{i})$.
Let $X$ and $X'$ be points on line $A_{1}A_{6}$, and let $Y$ and $Y'$ be points on line $A_{3}A_{4}$ such that $A_{1}X = A_{1}X' = A_{3}Y = A_{3}Y' = 1$ and $X$ and $X'$ lie on opposite sides of $A_{1}$ and $Y$ and $Y'$ lie on opposite sides of $A_{3}$. If $X'$ and $Y'$ lie on segments $A_{1}A_{6}$ and $A_{3}A_{4}$, respectively, then segment $A_{1}'A_{3}'$ lies between the lines $XY$ and $X'Y'$. Note that $\frac{x}{2}$ is the distance from $A_{2}$ to $A_{1}A_{3}$.

If $\frac{x}{2} \geq 2$, then $C(A_{2})$ cannot intersect line $XY$, since the distance from $XY$ to $A_{1}A_{3}$ is $1$ and the distance from $XY$ to $A_{2}$ is at least $1$. Therefore, $A_{1}'A_{3}'$ separates $A_{2}'$ from the other $3$ vertices of the hexagon. By analogous reasoning applied to the other vertices, we may conclude that $H'$ is convex.
If $\frac{x}{2} < 2$, then $C(A_{2})$ intersects $XY$, so by choosing $A_{1}' = X$ and $A_{3}' = Y$, we see that we may choose $A_{2}'$ on the opposite side of $XY$, in which case $H'$ will be concave. Hence the answer is $4$, as desired.