Isometrically embed a pentagon into $\mathbb{R}^2$ In this post, the regular pentagon is the metric space $M=\{1,2,3,4,5\}$ with
$$d(1,2) = d(2,3) = d(3,4) = d(4,5) = d(5,1) = 1,$$
$$d(1,3) = d(2,4) = d(3,5) = d(4,1) = d(5,2) = 2.$$
Questions:


*

*Is there a norm $\|\cdot\|$ on $\newcommand{\R}{\mathbb{R}}\R^2$ such that the regular pentagon can be isometrically embedded into $(\R^2,\|\cdot\|)$?

*If not, what is the minimal dimension $n$ that admits a norm $\|\cdot\|$ on $\R^n$ such that the regular pentagon is isometrically embedded into $(\R^n,\|\cdot\|)$?
What I know already:


*

*Since any metric space with $k$ elements can be isometrically embedded into $\R^k$ with $l^\infty$-norm, there is at least an embedding into $\R^5$ (e.g. by the Kuratowski embedding).

*For illustration: The square in the above sense (regular 4-gon) can be isometrically embedded into $\R^2$ with $\|\cdot\|_1$-norm (Manhattan distance) as $(0,0),(0,1),(1,1),(1,0)$.
For a generalization of this question, see Isometrically embed $K$-gon into $\mathbb{R}^N$
 A: Notation : $f: M\rightarrow (\mathbb{R}^n,\|\
\|),\ f(i)=x_i$
$[xy]$ line segment
$[xy)$ ray starting at $x$ and passing through $y$
$(xy)$ line containing $x,\ y$
$|xy|$ distance between $x$ and $y$
EXE : $n\neq 2$
Proof : Step 1 : Here $\Delta =[ x_1x_3x_4]$ is a
triangle. We will prove that a point $x_2$ is not in the triangle
$\Delta$.
If $x_2\in \Delta$, then let $$ [x_4x_2)\bigcap [x_1x_3] =\{z\}$$
By convexity of distance function $d(x)= |x_4x|$, since
$|x_4x_1|,\ |x_4x_3|\leq 2$, then $|x_4z|\leq 2$. By assumption, $z=
x_2$ and $|x_4z|=2$.
By assumption of $x_i$, $z$ is a midpoint of $[x_1x_3]$. If $z'$ is a
 midpoint of $[x_1x_4]$, then
$$ 2=|zx_4| \leq |z'x_4|+|zz'|=\frac{3}{2}$$ so that it is a
contraction.
Step 2 : $(x_ix_j)$ where $x_i\in
\{x_1,x_3,x_4\}$ divides $\mathbb{R}^2$ into 7 regions
$R_i$ is truncated cone whose vertex is $x_i$ and $S_i$ is a cone
whose vertex is $x_i$.
1) $x_2$ is not in $S_1,\ S_3,\ S_4$ and their boundaries :
If $x_2$ is in $S_1$, then we will use the convexity of the function
$d(x)=|xx_3|$ Recall that $d(x_1)=2,\ d(x_4)=1$ If we define
$$ [x_2x_3]\bigcap (x_1x_4) =\{z\} $$
then $$ d(x_2)\geq d(z)\geq d(x_1)\geq 2 $$
For $S_3,\ S_4$, we do the same argument.
2) $x_2$ is not in $R_3$ : If $x_2$ is in $R_3$, then define
$[x_2x_3]\bigcap [x_1x_4]=\{z\}$. Then $$ 2=|x_4x_2| \leq |zx_4| +
|zx_2| $$
so that $|zx_4| =1+\epsilon$. Hence $$ 2=|x_1x_3|\leq |zx_1| +|zx_3|
<2
$$ since $|zx_1|=1-\epsilon$. Contradiction.
3) $x_2$ is not in $R_1$ : Now, assume that $x_2\in R_1$.
If $[x_1 x_2]\bigcap [x_3x_4]=\{z\}$, then $|x_1x_3|\leq |x_1z|
+|zx_3| <2$. Contradiction.
4) $x_2$ is not in $R_4$ :
If $z,\ z'$ are mid points in $[x_1x_3],\ [x_1x_4]$, then $$2=|x_2x_4|\leq |x_2z| +|zz'| +|z'x_4|=|x_2z| +\frac{3}{2}$$
so that $|x_2z|\geq \frac{1}{2}$.
Since $|x_1x_2|,\ |x_1z|,\ |x_1z'|$ are 1, so $x_2$ is in a closed
half plane containing $(zz'),\ x_1$.
If $[zx_4]\bigcap [z'x_3]=\{w\}$, then $[wzz'],\ [wx_4x_3]$ are
similar with ratio 2.
Since $|zx_4|\leq \frac{3}{2}$, so $$ |zw|\leq \frac{1}{2},\
|wx_4|\leq 1 $$
If $x_2x_1x_2'x_3$ is parallelogram, then $x_2'$ is not in $[zz'w]$.
In further, $|x_3x_2'|=1$, then $x_2'$ is not in $[zx_3x_4]$. For
$x_5$ we have $x_5'$ so that $[zx_2']\bigcap [z'x_5'] $ is in
$[zz'w]$.
This contradicts to the fact that $[x_2x_2']$ is in the boundary of
unit ball whose center is $x_1$.
EXE : In $\mathbb{R}^3$, there exists an example
:
$$ (0,1,1),\ (-1,1,1),\ (-1,0,1),\ (-\frac{1}{2},0,\frac{3}{2} ),\
(0,\frac{1}{2},\frac{3}{2} ) $$ in $\|\ \|_1$
EXE : In $\mathbb{R}^4$, there exists an example :
$$ (0,0,0,0),\ (1,0,1,1),\ (2,-1,0,0),\ (2,0,0,-1),\
(1,1,-1,0) $$ in $\|\ \|_\infty$
