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A circle of area 1 is partitioned into square pieces in an iterative fashion, wherein each step of the iteration the largest square possible is cut out from all non-square pieces remaining from the previous step. Because squares do not tile a circle, there will be an infinite number of iterations producing an infinite number of ever-smaller squares.

As the number of iterations approaches infinity, what will the average area of the squares converge to? What will the square root of the average area$^2$ converge to?

Thanks in advance!

Xander Henderson
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TheNewGuy
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  • The area will converge to the area of the circle. Not sure what the second question is, though. Can you make it more precise? –  Dec 02 '17 at 23:35
  • Even the greatest square is less than the area of the whole circle, so the average of the squares could not possibly be the area of the circle. The second question has you averaging the area$^2$ of the squares, then taking the square root of that average. – TheNewGuy Dec 02 '17 at 23:44
  • I missed the word 'average' in the first question. Now I get what you're asking, hmm... –  Dec 02 '17 at 23:47
  • It is actually still unclear what you are asking. Do you want the average area of the sum of the squares or the average area of each individual square? – Isaac Browne Dec 03 '17 at 00:48
  • @TheNewGuy What have you tried? – Brethlosze Dec 03 '17 at 00:53
  • Clearly, both averages are zero (why?) – Brethlosze Dec 03 '17 at 00:55
  • @IsaacBrowne I am wanting the average area of the individual squares. – TheNewGuy Dec 03 '17 at 01:06
  • @hyprfrcb It is not clear to me. Could you post an answer explaining this? Thanks! – TheNewGuy Dec 03 '17 at 01:07

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Since we never would count an area greater than the area of a circle, the sum of the areas of the squares must be a finite value, as the area of the circle is finite. Thus as the number $n$ of squares approaches infinity, the value $A/n$ would approach $0$ as $A$ is finite, and $n$ is approaching infinity.

The average of the areas squared will also be $0$, as since the area $a_n$ of each square is less than $1$, the square of the area is less than the area itself. Thus, if S is the sum of all the squares of the areas, we have $S < A$, so $S$ is also finite. Thus the average (and the square root of the average) $S/n$, is 0.

Isaac Browne
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  • This is a little off-topic, but could you tell me the coordinates for the Mandelbrot copy in your profile picture? – Robert Soupe Jun 21 '18 at 02:30
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    I'm actually not sure. I lost my old phone where I had it saved as one of my "cool places". I think it is somewhere along the negative real line. If you do find it, please let me know – Isaac Browne Jun 21 '18 at 15:39