Sufficient condition to inscribe a polygon inside another one

Let $P$ be any convex polygon in the plane $\mathbb{R}^2$ with vertices $x_1,\dots,x_n$, $n\ge 4$. Let $P'$ be another convex polygon with vertices $x_1',\dots,x_n'$ (same number of vertices of $P$).

Question 1:

Is the hypothesis $|x_i'x_j'|\le|x_ix_j|$ for all $i,j=1,\dots,n$ enough to conclude that $P'$ can be inscribed inside $P$ (i.e. moved by isometries in such a way to be entirely contained inside $P$)?

If the answer to this question is yes, then: the answer would still be yes without the assumption of $P'$ being convex? What about the assumption of $P$ being convex?

Question 2:

Are there any other known sufficient conditions to inscribe a general polygon inside another one with the same number of vertices?

Thank you

Edit on Question 1: can $|x_i'x_j'|\le|x_ix_j|$ for all $i,j=1,\dots,n$ be a sufficient condition for $P'\subset P$ in case both $P'$ and $P$ are "particular" polygons in the following sense:

• $P'$ and $P$ have even number of sides

• sides of $P$ are pairwise parallel and parallel sides of $P$ also have equal length. There is a correspondence between sides of $P'$ and sides of $P$ in the sense that parallel sides of $P$ correspond to parallel sides of $P'$. Parallel sides of $P'$ also have equal length (which can be different from the length of corresponding parallel sides of $P$)

Solution 1:

The answer to the first question is no even if both polygons are convex:

In the picture above, all sides and diagonals of the blue polygon are at least slightly smaller than all sides and diagonals of the red polygon. (I've added a dashed arc to indicate where the blue vertex has to fit for the only case in which the comparison isn't obvious.)

But the only way to try to fit the blue polygon inside the red one is to match up their isosceles-right-triangle parts, after which the other half of the blue polygon goes in the wrong direction.

The answer to your second question is "yes, but they're mostly paywalled or written in German". A paper by Jiazu Zhou gives some conditions (you want to take the "nonnegative curvature" results and set $\epsilon=0$). I don't think these are quite the sort of conditions you're looking for (and they apply to sets more general than "polygons"). But the citations in the introduction can lead you to other places, including some papers by Hadwiger from 1941 that contain the first results in this direction.

I'm not completely sure if this counts as a counterexample to your revised question 1 in the non-convex case (since parallel sides of $P$ are not always equal), but it's worth keeping around:

Here, the blue polygon is just a scaled-down, slightly smaller version of the red one, so the inequalities still hold; but making the gap in the top smaller means that it doesn't fit inside the red polygon. (This is not a formal proof, but there should be a way to write one.)

It's possible that the revised version is true for convex polygons......