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. 

[![enter image description here][1]][1]

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 $P$:

[![enter image description here][2]][2]

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. 

**I'm wondering if it's always possible to find $p'\in S'$ such that $|p'x_i'|\le |px_i|$ 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?


  [1]: https://i.sstatic.net/IjCcb.jpg
  [2]: https://i.sstatic.net/9QBok.jpg