Areas of a series of annular sectors

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The discussion revolves around proving the inequality A4/A3 > A3/A2 > A2/A1 for the areas of annular sectors in a disc. The user initially seeks help with the problem, indicating that the segments along the radius are equal. After exploring an idealized example with specific radii, they conclude that A4/A3 is actually less than A3/A2. The conversation highlights the mathematical exploration of annular areas and the user's appreciation for discovering mathematical notation. The thread combines mathematical inquiry with a light-hearted reference to a humorous quote.
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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.


\sqrt{(Doing)^{2} + (Being)^{2}}= Doobee\,Doobee\,BingDing

J Fox
 

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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!
 
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