# The connection between diameter and pi

So I was having the worst headache and lying in bed thinking about simple geometry problems waiting for the pain killer to kick in. One of the things I did was compute the perimeter/circumference of the Reuleaux triangle. Okay so that one’s parameter is composed of three sixths of a circle with radius twice the widthof the triangle so if the width of the triangle is $D$ that’s $O = 3 \cdot 2\pi D / 6 = \pi D$. Similarly for all the other Reuleaux polygons the ($n = 2k+1$)-sided reuleaux polygon $O = n \cdot 2\pi/(2n) D = \pi D$. Huh… I thought. That’s kind of odd. All of these figures with constant widths have the same circumference formulas as the circle which of course has constant width called the diameter.

And looking up the wiki-article on shapes of constant widths and this is apparently a thing proven in 1860. All curves with constant width have a perimeters which is $\pi$ times it’s width. and it’s called Barbier’s theorem.

I also get the feeling that this should be possible to be reworked into a pro-pi argument in the tau-vs-pi debate. Of course by the law of the equivalence between pi and tau implies it could be made into a pro-tau-argument as well but since I’ve never seen it come up in one of those debates I would like to see what happens when a tauist brings up that defining the circle constant via the diameter of a circle was a mistake since the radius is a more fundamental property and then have the piist counter that at least there is a whole class of objects with diameters and they all have perimeters which are $\pi$ times the diameter : P