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?

Yes. $\endgroup$ – Joseph O'Rourke Apr 7 '17 at 15:59