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Areas of a series of annular sectors

  1. Mar 15, 2014 #1
    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

    Attached Files:

  2. jcsd
  3. Mar 15, 2014 #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!
    Last edited: Mar 15, 2014
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