Can we build a geometry in which $\pi = 42$? More precisely, can we build a norm $N$ on $\mathbb{R}^2$, such that the ratio circumference / diameter (computed with norm $N$) of a standard circle is $42$?
(By standard circle, I mean a circle defined with the Euclidean norm, i.e. $x^2+y^2 = r^2$, and not a circle defined with the norm $N$).

Thus, this question is different to this one and this one.
 A: It's hopeless to get anything greater than $\pi$ with your definition:
Let $N$ be a norm on $\mathbb{R}^2$. By the positive homogeneity of the norm, it's enough to address only the case of the Euclidean circle $C$ centered at $(0,0)$ of radius $1$.


*

*On the one hand, the $N$-perimeter of $C$ is:
$$P_N=\int_0^{2\pi}N(-\sin t,\cos t)\,\mathrm{d}t=\int_0^{2\pi}N(\cos t,\sin t)\,\mathrm{d}t.$$
For the last equality we only used the $2\pi$-periodicity of $t\mapsto N(\cos t,\sin t)$.

*On the other hand, the $N$-diameter of $C$ is:
$$d_N=2\max_{t\in[0,2\pi]}N(\cos t,\sin t).$$
To see this: the definition of the diameter is:
$$d_N=\max_{u,v\in D}N(u-v),$$
where $D$ denotes the set
$$D=\bigl\{(x,y)\in\mathbb{R}^2\bigm\vert x^2+y^2\leq1\bigr\}.$$
Note that it's a maximum since $D$ is closed and bounded with respect to the $\lVert\cdot\rVert_2$ (hence compact) and the norm $N$ is continuous with respect to the topology induced by $\lVert\cdot\rVert_2$ in virtue of the equivalence of all the norms on $\mathbb{R}^2$ (since it's a finite-dimensional vector space). Now, if the max defining $d_N$ is attained for $u,v\in D$ such that $u\neq-v$ (and of course $u\neq v$) then $\lVert u-v\rVert_2<2$ hence taking $u'=(u-v)/\lVert u-v\rVert_2$ and $v'=(v-u)/\lVert u-v\rVert_2$ yields $u',v'\in D$ and
$$N(u'-v')=N\bigl(2(u-v)/\lVert u-v\rVert_2\bigr)=N(u-v)\frac2{\lVert u-v\rVert_2}>N(u-v).$$
Hence the maximum for $d_N$ is attained for opposite vectors. Hence
$$d_N=2\max_{u\in D}N(u).$$
By a similar argument, it is easy to show that the max is attained for a vector $u$ on $C=\partial D$, hence the result.


Now we clearly have
$$P_N\leq\pi d_N,$$
hence
$$\pi_N=\frac{P_N}{d_N}\leq\pi.$$

The mistake in Yves Daoust's answer is in point 2 when he says by symmetry we can assume $x=y$. This is wrong for $p>2$: the vectors on $C$ that lie on the line $x=y$ will have a minimal norm, so the diameter is not attained there. It's attained on the axes, so that $r_p=1$ when $p\geq2$.
A: A few notes on @YvesDaoust's answer (my edit was not accepted, so I post the notes as an additional answer).

Let $\mathcal{C}$ be the euclidean unit circle in $\mathbb{R}^2$ defined by $x^2 + y^2 = 1$.
1) Using a $p$-norm $||\cdot||_p$, the circumference of $\mathcal{C}$ will be $$h_p=2\int_0^{\pi}\sqrt[p]{|\cos t |^p+|\sin t|^p}\,dt,$$
which grows to infinity for $p\to0$. But for it to be a norm, one must have $p\ge1$.
From this integral we have
$$h_{10} \approx 5.68...$$
$$h_2=2 \pi \approx 6.28...$$
$$h_{1.5} \approx 6.74...$$
$$h_1=8$$ and can't achieve higher value than $h_1$ with a $p$-norm.
2) Let the diameter and radius be defined by
$$d_p = \max\limits_{a, b \in \mathcal{C}} ||a-b||_{p}$$
$$r_p = \max\limits_{x^2+y^2=1} \{ (|x|^p + |y|^p)^{1/p} \}$$
When $1 \leq p \leq 2$, by symmetry, we can find that
$$r_p=\sqrt[p]{\frac1{\sqrt2^p}+\frac1{\sqrt2^p}}=\frac{\sqrt[p]2}{\sqrt2}=2^{1/p - 1/2}.$$
As noted by user @gniourf_gniourf, this is no more true when $p \geq 2$.
When $p \geq 2$, $r_p = 1$.
If we evaluate $d_p$ in a similar fashion, we see that $d_p = 2\ r_p = 2^{1/p + 1/2}$ is always true for $p \geq 1$.
3) Then the circumference divided by the diameter
$$\pi_p = \frac{h_p}{2\ r_p}= \frac{h_p}{d_p}$$
satisfies (according  to numerical integration)
$$\pi_1=\frac{8}{2^{3/2}} = 2\sqrt2 \approx 2.828... $$
$$\pi_{1.5} \approx 3.00... $$
$$\pi_{2} = \frac{2 \pi}{2} = \pi \approx 3.14... $$
$$\pi_{3} \approx 2.96968... $$
$$\pi_{10} \approx 2.84... $$
Deeper analysis is required to find the maximum value, which seems to be reached for $p = 2$.
This is in agreement with @gniourf_gniourf's answer.
