Given a unit disk (regular hexagon), how to tell the formula of the norm to show that it is not derived from a scalar product? In $\mathbb{R}^2$ there is a norm $\left\|\cdot\right\|$; the unit disk in this norm, $B = \{(x, y) \in \mathbb{R}^2 :\left\|(x,y)\right\| \leq 1\}$, has the shape of a regular hexagon with side length $1$ and one vertex at $(1, 0)$.
Show that the norm is not derived from a scalar product.
Now, I would like to demonstrate the failure of the parallelogram law holding. That is how I usually did such exercises. However here I am not given any formula for $\left\|\cdot\right\|$ , just the shape of unit circle.
So here comes my question: how to tell the formula of the norm if I'm given a unit disk
 A: Given any closed, bounded and convex subset $D$ of $\Bbb{R}^2$ that is symmetric about the origin, i.e., for all $v$, $v \in D$ implies $-v \in D$, we can define a norm $\|\cdot\|$ whose unit disc, i.e., the set $\{v \in \Bbb{R}^2\mid \|v\| \le 1\}$, is $D$. $\|\cdot\|$ can be defined by:
$$
\| v \| = \inf \{\lambda \in \Bbb{R_{\ge0}} \mid v \in \lambda D\}
$$
Equivalently $\|v\|$ is the unique $\lambda \ge 0$ such that $v \in \lambda S$, where $S$ is the boundary of $D$. As the unit disc of any norm is closed, bounded, convex and symmetric about the origin, all norms on $\Bbb{R}^2$ arise in this way. (This all generalises easily to highher dimensions.)
The points lying on and inside a regular hexagon centred on the origin meet the above criteria to be the unit disc of a norm. To see that the paralelogram identity is not satisfied, let the vertices of the hexagon be $v_1, v_2, \ldots v_6$, and consider any parallelogram $P$ with vertices $w_1 = (0, 0)$ $w_2 = v_1 + v_2$, $w_3$ and $w_4$ where $w_3$ and $w_4$ lie on the edge $[v_1, v_2]$.  Then each side of $P$ has length $1$ and the long diagonal $[w_1, w_2]$ has length $2$, but different choices of $w_3$ and $w_4$ let us vary the length of the short diagonal $[w_3, w_4]$ from $0$ to $1/2$. So the parallelogram identify cannot hold.
Norms that are induced by a scalar product are called Euclidean. The unit disc in a Euclidean norm is strictly convex, which mean that its boundary contains no line segments. Norms with a strictly convex unit disc are called strict. Another way of seeing that the norm with a hexagon as its unit circle is not Euclidean is to note that it is not strict. This approach applies to lots of interesting norms, although there are strict norms that are not Euclidean.
