Pigeonholing Polygons: Can two rigid regions fit in twice the space needed?

Question

This is a tweak of Henry Segerman's question Can an arbitrary collection of circles of total area 1/2 fit into a circle of area 1? , but restricted to the point of possibly having a proof in the literature.

If one takes the question above and restricts it to ask for a pair of circular regions with total area 1/2 to fit into a region of area 1, one has the answer yes except in the case the circles are the same size. In such a case, one needs a common point of the two regions, as well as each circle intersecting the outer circle in a point, if one wants the interiors of the small circles to be disjoint and also a subset of the interior of the enclosing region.

I will coopt the term "pigeonholing" from combinatorics and use it for geometrical packing. In this problem it will apply to two regions of the same shape and possibly of the same size, but in general it will refer to forming a packing using translates and scalings of a given region, but rotation and reflection are not allowed. Also, in a pigeonhole packing of closed objects inside a region, the objects occupy disjoint point sets in the space and are disjoint from the pointset that is the boundary of the enclosing region.

So to use this term, one can pigeonhole pack a pair of closed circles of combined area 1/2 inside a closed circle of area 1, unless the circles in the pair are the same size.

I don't have a proof, but it seems the same holds true if I replace the word circle with the word square. Or equilateral triangle. Possibly regular polygon.

This brings us to the main question. For what convex shapes C is the following true:

"You can pigeonhole pack a pair of closed C of combined area 1/2 into a region C of area 1, unless the two C in the pair are the same size." ?

I suspect the answer is all convex compact bodies in the plane, and that this is a consequence of some combinatorial geometric result. I am not a geometer, so I hope someone who knows geometry will answer and generalize this.

Bonus points for related info like more dimensions, nonconvex shapes, nonpigeonhole packings, looking at triples instead of pairs, and other small tweaks to the question. Improvements are also welcome.

Answer 1

Suppose you want to pigeonhole pack $aC$ and $bC$ into $C$, $a^2+b^2=1/2$, and $a\neq b$. Let $c=\tfrac12(a+(1-b))$, then $a<c<(1-b)$. Pick any direction $u\in S^1$, and let $w=w_u(C)$ be the width of $C$. There are two parallel lines $L_1$ and $L_2$ perpendicular to $u$ that bound $C$ and are a distance $w$ apart. Add another line $L_3$ a distance $cw$ from $L_1$. This line slices $C$ into two regions $A$ and $B$ such that $w_u(A)=c w>aw$ and $w_u(B)=(1-c)w>bw$. $A$ is the maximal subset of $C$ that contains $C\cap L_1$ and has width no greater than $cw$. Therefore it also contains a dilation of $C$ by a factor $c$ about a point in $C\cap L_1$. We now have immediately that $A$ contains a translate of $aC$ in its interior. The same argument works for $B$ containing $bC$.

For the "unless" part, the line separating the two pigeonhole packed copies of $\tfrac12C$ slices $C$ into two regions. The width of each slice is more than half the width of $C$, so you get a contradiction.