Let $H$ be a regular hexagon of side length $x$. Call a hexagon in the same plane a "distortion" of $H$ if and only if it can be obtained from $H$ by translating each vertex of $H$ by a distance strictly less than 1. Determine the smallest value of $x$ for which every distortion of $H$ is necessarily convex.
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 $A_{i}^{\prime}$ be considered such that $A_{i} A_{i}^{\prime}<1$. Let $H^{\prime}=A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime} A_{5}^{\prime} A_{6}^{\prime}$, and consider all indices modulo 6 . For any point $P$ in the plane, let $D(P)$ denote the unit disk $\{Q \mid P Q<1\}$ centered at $P$; it follows that $A_{i}^{\prime} \in D\left(A_{i}\right)$. Let $X$ and $X^{\prime}$ be points on line $A_{1} A_{6}$, and let $Y$ and $Y^{\prime}$ be points on line $A_{3} A_{4}$ such that $A_{1} X=$ $A_{1} X^{\prime}=A_{3} Y=A_{3} Y^{\prime}=1$ and $X$ and $X^{\prime}$ lie on opposite sides of $A_{1}$ and $Y$ and $Y^{\prime}$ lie on opposite sides of $A_{3}$. If $X^{\prime}$ and $Y^{\prime}$ lie on segments $A_{1} A_{6}$ and $A_{3} A_{4}$, respectively, then segment $A_{1}^{\prime} A_{3}^{\prime}$ lies between the lines $X Y$ and $X^{\prime} Y^{\prime}$. Note that $\frac{x}{2}$ is the distance from $A_{2}$ to $A_{1} A_{3}$. If $\frac{x}{2} \geq 2$, then $C\left(A_{2}\right)$ cannot intersect line $X Y$, since the distance from $X Y$ to $A_{1} A_{3}$ is 1 and the distance from $X Y$ to $A_{2}$ is at least 1. Therefore, $A_{1}^{\prime} A_{3}^{\prime}$ separates $A_{2}^{\prime}$ from the other 3 vertices of the hexagon. By analogous reasoning applied to the other vertices, we may conclude that $H^{\prime}$ is convex. If $\frac{x}{2}<2$, then $C\left(A_{2}\right)$ intersects $X Y$, so by choosing $A_{1}^{\prime}=X$ and $A_{3}^{\prime}=Y$, we see that we may choose $A_{2}^{\prime}$ on the opposite side of $X Y$, in which case $H^{\prime}$ will be concave. Hence the answer is 4 , as desired. $\fbox{4}$.