# Areas of a series of annular sectors

by jfox
Tags: annular, areas, sectors, series
 P: 2 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 Attached Thumbnails
 P: 2 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!