Optimal unions of planar convex regions

Question

This post continues Optimal intersections between planar convex regions.

Question: Given two planar convex polygonal regions $C_1$ and $C_2$, how does one algorithmically find how to place and orient them so that the largest convex planar region that can be covered by the union of $C_1$ and $C_2$ is of maximum area?

Simple Examples: If $C_1$ and $C_2$ are identical squares, they can be put with a side coincident and a $2\times 1$ rectangle can be covered. If they are identical circular discs, one needs to make them partially overlap to maximize the largest convex region the two can together cover.

Further question: Which is the planar convex shape $C$ of unit area (special case: $C$ is centrally symmetric) such that the largest convex region coverable by two copies of $C$ has the least area? Is it the circular disk?

Answer 1

For the last question, the answer is no: there are unit-area shapes such that the greatest convex area in the union of two of them is less than the greatest convex area in the union of two unit-area circles.

Consider three shapes:

• Let $C$ be a circle of unit area.
• Let $D$ be the slightly larger area enclosed by $C$ and its tangents at $-1^\circ$, $+1^\circ$, $179^\circ$ and $181^\circ$.
• Let $E$ be similar to $D$ but of unit area.

The two copies of $C$ can cover a convex region of area at most $\sim\!\frac32$: If the circles are placed symmetrically, and $x$ is the central angle subtending their intersection, the precise area is $\frac1\pi(x+3\sin x)$, which is maximized at $x=\arccos(-\frac13)$.

The two copies of $D$ together have four pointy regions outside the circles, but they can cover a convex region of area at most $\sim\!\frac32$ plus two of the pointy regions.

So the two copies of $E$ can cover a convex region of area at most twice a weighted average of $\sim\!\frac34$ and $\frac12$, weighted by the areas of $C$ and $D-C$ respectively, and this area will be less than the maximal area covered by two copies of $C$.