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 $S'$

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?

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?

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.

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?

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 $S'$

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?