- #1

jfox

- 2

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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

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