# Areas of a series of annular sectors

1. Mar 15, 2014

### 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

#### Attached Files:

• ###### Series of Sectors.jpg
File size:
20.6 KB
Views:
96
2. Mar 15, 2014

### 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!

Last edited: Mar 15, 2014