TITLE: Problem on distances in a polygon
QUESTION [6 upvotes]: In $\mathbb{R}^2$ consider a square (call it $S$) and three triangles (one acute $T_2$ and two obtuse $T_1$ and $T_3$) such that each triangle shares one different side with the square and the triangles and the square are disposed exactly as in the following picture. 

Define $P:=S\cup T_1\cup T_2\cup T_3$. 
Call $x_i$ the vertex of $T_i$ opposed to the side of $T_i$ shared with the square $S$. Choose any point $p$ inside the square $S$ such that all three segments $\overline{px_i}$ are entirely contained in $P$.
Now move all the vertexes of $P$ in a continuous way in such a manner that all the following lengths are not increased:

the lengths of all the sides of the square and of the three triangles
the lengths of the two diagonals of the square and the lengths of the segments from each $x_i$ to the vertexes of the square which are contained in $P$.

In the following picture I've drawn in blue all the segments whose lengths are not increased moving the vertexes of $P$:

Call $P':=S'\cup T_1'\cup T_2'\cup T_3'$ the polygon obtained in such a manner and $x_i'$ the vertexes of the new triangles. 
For any $p'\in S'$ define $d(x_i',p')$ in the following way:

If  $\overline{x_i'p'}\subset P'$, then $d(p',x_i'):=|p'x_i'|$ 
If  $\overline{x_i'p'}\not\subset P'$, then $d(p',x_i'):=$ minimum of $|v'p'|+|v'x_i'|$ for $v'$ which varies between the vertexes of $S'\cap T_i$

I'm wondering if it's always possible to find $p'\in S'$ such that $d(p',x_i')\le |px_i|$ for $i=1,2,3$. 
So my question is of course how to prove the existence of $p'$. I'm trying to consider all possible cases in which the vertexes of $P$ could move (given the bonds of the lengths), but it's quite complicated. Do you thing there's a better way to proceed?

REPLY [4 votes]: Let me prove a bit more general statement.

Let $P=[v_1\dots v_n]$ and $P'=[v_1'\dots v_n']$ be two solid polygons such that if $[vw]$ is a side of $P$ or a diagonal which lies in $P$ completely then $|v-w|\ge |v'-w'|$. Then there is a short map $P\to P'$ which sends any vertex $v$ of $P$ to the corresponding vertex $v'$ of $P'$.

Note that if there is one diagonal $[vw]$ such that $|v-w|=|v'-w'|$ then we can cut $P$ in two polygons and reduce the question to a simpler case.
First note that we can assume that $|v-w|=|v'-w'|$ for any side of $[vw]$ of $P$. 
If this is note the case choose a vertex $x'$ so that the triangle $[x'v'w']$ lies in $P'$.
Put joints in these vertexes and increase $|v'-w'|$ up to $|v-w|$ so that the rest of diagonals stay the same or increase. The new poygon $P''$ admits a short map to $P'$ so we can make each side to be the same as in $P$ one by one and reduce the question to this case.
Note that the convex angles get smaller in $P'$. 
Therefore one concave angle, say $a$, should go up, in particular it stays concave.
Choose 4 vertexes: $a$, its neighbors $x$ and $y$ and a vertex $b$ visible from $a$.
Put joints in these vertexes.
We can move the polygon straighting the angle at $a$, keeping the distances $|a-x|$, $|a-y|$, $|b-x|$ and $|b-y|$ and nondecreasing the rest of diagonals. 
Repeating if necessary, we will get one more diagonal fixed. 
P.S. Doing these operations, you can end in a generalized solid polygon --- a flat disc with polygonal boundary, when it is mapped to the plane you can get overlaps.