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Areas of a series of annular sectors 
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#1
Mar1514, 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 


#2
Mar1514, 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] 1^{2}.75^{2} A3[itex]\propto[/itex] .75^{2}.5^{2} A2[itex]\propto[/itex] .5^{2}.25^{2} A4/A3 < A3/A2 But it was cool to find the math font/language! 


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