Let $\|\cdot\|$ be a norm on $\mathbb R^2$. Show that the perimeter of the closed unit ball is between $4$ and $12$. Let $\|\cdot\|$ be a norm on $\mathbb R^2$. The length of a polygon line is given by this norm. Let $A$ be a non-empty bounded convex part of  $\mathbb R^2$. The perimeter of $A$ is defined as the upper bound of the perimeters of the convex polygons included in $A$. Show that the perimeter of the closed unit ball is between $4$ and $12$.
Not very familiar with geometry, I do not not see how to deal with this problem from an oral exam (if not to frame the unit ball with polygons). If you have an idea ...
Note that the norm is not necessarily euclidean (i.e. issue from a scalar product).
 A: Let $u$ be a point on the boundary $\partial B$ of the unit ball $B$ of a norm on $\mathbb R^2$. Then $\partial B$ and the boundary $u+\partial B$ of the unit ball with center $u$ intersects (for instance by the Intermediate Value Theorem). Let $v$ be a point in the intersection. Then $u$, $v$, $v-u$, $-u$, $-v$, $u-v$ are 6 points in this order on the boundary, with successive points at distance $1$ apart. They subdivide $\partial B$ into 6 arcs, each of length (in the norm) of at least $1$. This shows that the length of $\partial B$ is at least $6$.
Next, among all $u,v\in\partial B$, choose two for which the area $\det(u,v)$ of the parallelogram with vertex set $\{o, u, v, u+v\}$ is maximized.
Then the line through $u$ parallel to $v$ supports $B$ at $u$ (otherwise $\det(u,v)$ could be increased) and the line through $v$ parallel to $u$ supports $B$ at $v$. It follows that the parallelogram $P$ with vertices $\pm u\pm v$ contains $B$ and subdivides $B$ into $4$ parts. One of the parts is the arc of $\partial B$ from $u$ to $v$, which is contained in the convex hull of $o, u, u+v, v$. It can be shown (using the triangle inequality) that the length of the arc of $\partial B$ from $u$ to $v$ is at most the length of the corresponding arc of $\partial P$, which consists of the two segments $[u,u+v]$ and $[u+v,v]$, and thus has length $2$. It follows that the length of $\partial B$ is at most $8$.
Remarks


*

*The above two bounds were first proved by Stanislaw Golab in 1932. It can also be found in Chapter 4 of Thompson's book Minkowski Geometry.

*The result can also be shown by choosing $P$ to be a parallelogram of minimum area containing $B$.

*There is, up to isometry, only one norm where the length of the unit ball is $6$, namely when $B$ is an affine regular hexagon.

*There is, up to isometry, only one norm where the length of the unit ball is $8$, namely when $B$ is a parallelogram.

*See my survey with Martini and Weiß for an elementary proof from the triangle inequality that if a convex arc $\gamma$ from $a$ to $b$ is contained in the triangle $\triangle abc$, then the length of $\gamma$ is at most $\|a-c\|+\|b-c\|$. Thompson proves it indirectly using an area argument.


References
S. Golab, Some metric problems in the geometry of Minkowski (Polish. French summary), Prace Akademii Górniczej w Krakowie 6 (1932), 1-79.
H. Martini, K. J. Swanepoel, and G. Weiß, The geometry of Minkowski spaces — a survey. Part I, Expositiones Mathematicae 19 (2001), 97-142. 
A. C. Thompson, Minkowski Geometry, Cambridge University Press, Cambridge, 1996.
A: w.l.g. normalize the norm such that $||(1,0)|| = 1$, and take $a,b>0$ such that $||(1/2,a)|| = 1$ and $||(0,b)||=1$, then we can consider the following paralelograms (with sides paralels to the axes):
Polygon $P_1$: such that its corners are $\{(1/2,a), (-1/2,a), (-1/2,-a), (1/2,-a)\}$.
Polygon $P_2$: such that its corners are $\{(1,b), (-1,b), (-1,-b), (1,-b)\}$. 

Then notice that the ball of radius one $\mathbb{B} \subset \mathbb{R}^2$ satisfies 
$$ P_1 \subset \mathbb{B} \subset P_2$$
To prove this, just use the convexity of the sets.
Notice that since $||(1/2,a)|| = 1$ and $||(1/2,0)|| = 1/2$, employing the triangular inequality we can show that $1/2 \leq ||(0,a)||$. Therefore the perimeter of $P_1$, that is composed of 4 strokes, can be estimated as:
$$\text{Perim} (P_1) = 4 ||(1/2,0)|| + 4||(0,a)|| \geq 4\cdot 1/2 + 4\cdot 1/2 = 4$$
Similarly with $P_2$, its perimeter can be computed as
$$\text{Perim}(P_2) = 4||(1,0)|| + 4||(0,b)|| = 8$$.
So we proved that 
$$ 4 \leq \text{Perim}(\mathbb{B}) \leq 8 \leq 12 $$
