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Areas of a series of annular sectors

by jfox
Tags: annular, areas, sectors, series
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jfox
#1
Mar15-14, 06:41 PM
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.


[itex]\sqrt{(Doing)^{2} + (Being)^{2}}[/itex]= [itex]Doobee\,Doobee\,BingDing[/itex]

J Fox
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Series of Sectors.jpg  
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jfox
#2
Mar15-14, 08:57 PM
P: 2
Never mind. Trying an idealized example of r=R, .75R, .5R, .25R it comes down to

A4[itex]\propto[/itex] 12-.752
A3[itex]\propto[/itex] .752-.52
A2[itex]\propto[/itex] .52-.252

A4/A3 < A3/A2

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


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