**Whoever finds a norm for which $\pi=42$ is crowned nerd of the day!**

Can the principle of $\pi$ in euclidean space be generalized to 2-dimensional metric/normed spaces in a reasonable way?

For Example, let $(X,||.||)$ be a 2-dimensional normed vector space with a induced metric $d(x,y):=\|x-y\|$. Define the unit circle as $$\mathbb{S}^1 := \{x\in X|\;\|x\|=1\}$$ And define the outer diameter of a set $A\in X$ as $$d(A):=\sup_{x,y\in A}\{d(x,y)\}=\sup_{x,y\in A}\{\|x-y\|\}$$ Now choose a continuous Path $\gamma:[0,1]\rightarrow X$ for which the image $\gamma([0,1])=\mathbb{S}^1$. Using the standard definition of the length of a continuous (not necessarily rectificable) path given by $$ L(\gamma):=\sup\bigg\{\sum_{i=1}^nd(\gamma(t_i),\gamma(t_{i+1}))|n\in\mathbb{N},0\le t_0\lt t_1\lt ... \lt t_n\le 1\bigg\}$$ we can finally define $\pi$ in $(X;\|.\|)$ by $$\pi_{(X,\|.\|)}:=\frac{L(\gamma)}{d(\mathbb{S}^1)}$$ (This is way more well-defined than the old definition (check the rollbacks))

Examples:

- For the euclidean $\mathbb{R}^2$, $\pi_{\mathbb{R}^2}=3.141592...$
- For taxicab/infinity norms, $\pi_{(\mathbb{R}^2,\|.\|_1)}=\pi_{(\mathbb{R}^2,\|.\|_\infty)}=4$
- For a norm that has a n-gon as a unit circle, we have $\pi_{(\mathbb{R}^2,\|.\|)}=??$ (TODO: calculate)

While trying to calculate values for $\pi$ for interesting unit circles, I have defined a norm induced by a unit circle: let $\emptyset\neq D\subset X$ be star-shaped around $0\in D$. Define $\lambda D:=\{\lambda d|d\in D\}$. Now the (quasi-)norm in $X$ is defined as $\|x\|:=\min\{\lambda\mid x\in \lambda D\}$.

In other words: the scaling factor required to make x a part of the border of D. This allows us to easily find norms for most geometric shapes, that have exactly that Geometric shape as a unit circle and have the property, that the choice of the radius for the definition of $\pi$ is insignificant (for example $\pi=3$ is calculated for regular triangles with (0,0) in the centroid).

This allows for the following identity: let $x\in\mathbb{R}^2$, and $S\in\partial D$ be the Intersection of the Line $\overline{x0}$ with the border of D. Then we have $$\|x\|_D = \frac{\|x\|_p}{\|S\|_p}$$ where $\|.\|p$ is any (quasi)-norm in $\mathbb{R}^2$ (This follows out of the positive scalability of norms)

**EDIT:** I have been thinking about this a bit more. However I found out that my induced norm defined earlier is not even a norm... It violates the subadditivity axiom: Let the unit circle be a equilateral triangle where the centroid marks the point (0,0). Here we find $d(\mathbb{S}^1)=3$ which violates the triangle equation, as $d$ is measured in between 2 points of the unit sphere; Therefore, we have $\|a\|=1$ and $\|b\|=1$, but $\|a+b\|=3\nleq1+1=\|a\|+\|b\|$. Instead, we have a Quasinorm.

This indeed allows for $\pi=42$, if the unit circle is, for example, graph of a function $\varphi\mapsto(r(\varphi),\varphi)$ in Polar coordinates where $r(\varphi)=a*\sin(2\pi k)$.

**Questions**
Any other interesting norms?
Is this definition reasonable, and is there any practical use to this?
Feel free to share your thoughts. Mind me if I made some formal mistakes.
*And especially*, how do I define a norm with $\pi=42$?

**About the $\pi=42$**

Prince Alis Answer below shows that there can be no such p-norm, for which $\pi_p=42$, In fact this holds true for every norm. However, you can easily define a quasi-norm that has any arbitrary $\pi_{\|.\|}=\kappa\gt\pi$. For example,

$r(\theta)=1 + \frac{1}{a} \sin(b \theta)$

Defines a quasinorm with $\pi_{\|.\|_{a,b}}=\kappa$ for every $\kappa>\pi$. $\kappa$ can be increased by increasing a and b. The only thing left to do is find a and b for which we finally have $\pi_{\|.\|_{a,b}}=42$.

According to Mathematica, the Length of the boundary curve is 33.4581 (Took ~10 minutes to calculate) and the diameter is 4.5071, resulting in $\pi=7.42342$ for the norm given above ($a=b=10$). I doubt I will be able to easily find a solution for $\pi=42$ using this method... (Testing manually, I got exemplary values $a=9.95$, $b=175$ with $\pi=42.0649$ which comes very close...

On top of that, Prince Ali found a p-norm with $p<1$ for which $\pi=42$. Thank you very much!