Areas of a series of annular sectors

[jfox]

Hi all

Hoping someone can figure out a problem. In the attached figure, A is the area of openings in a disc. Assume the segments 82, 84, 86, 88 are along a radius and are equal. Is it possible to prove that A4/A3 > A3/A2 > A2/A1?


In thanks I'll share this, hope it's not old news. Some wit wrote it on a bathroom wall when I was an undergrad, I still think it's funny.


$\sqrt{(Doing)^{2} + (Being)^{2}}$= $Doobee\,Doobee\,BingDing$

J Fox

 

[jfox]

Never mind. Trying an idealized example of r=R, .75R, .5R, .25R it comes down to

A4$\propto$ 1²-.75²
A3$\propto$ .75²-.5²
A2$\propto$ .5²-.25²

A4/A3 < A3/A2

But it was cool to find the math font/language!