All of these assume your background metric is Euclidean.
If your background 2D metric is a projection of a warped 3D space, you can make π as big as you want by tugging on the centre of the circle.
If your background 2D metric is a projection of a warped 3D space, you can make π as big as you want by tugging on the centre of the circle.
Any metric that "pulls on the origin" compared to Euclidean distance will have to do the mapping in a continuous way. This will basically result in both the radius and circumference being expanded in that metric.
Matter of fact, I linked an article that proves that for _all_ metrics, the value of π is always between 3 and 4 (inclusive). Unfortunately the article might have gotten the hug of death so here is an alternative link: https://www.researchgate.net/publication/353330827_Extremal_...
And I can think of a counterexample on a sphere, just using Euclidean distance on the surface. Consider a circle with centre at North Pole and radius being the distance from the North Pole to a point on the equator. For this circle it is easy to find out that pi=2
Your observation is correct and the surface of the sphere is a metric. The ratio of radius to circumference is not constant with that metric though so I feel like something should disqualify it. But I am not sure how.
So I think your observation shows that we need a stronger constraint than just being a metric. Other commenters have hinted that you need a normed vector space but I am not sure if that's sufficient.
Math is far more elegant than public school allows it to appear.
https://raypatrick.xyz/blog/2023/10/27/were-you-mathematical...