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A series of three concentric squares, with a circle inscribed in the smallest one. The circle is grey, labeled "City", and has r = 1000 A. The smallest square has a side length of 2000A. The second square has a side length of 4000 A. The remaining area of the smallest and middle squares are filled in with orange and labeled "Migrash." The outermost square has side length of 6000 A, and its remaining area is filled with green and labeled "Techum."

The Gemara Eiruvin 56b said that the Migrash (the orange area) is one quarter of the area of the sados (the green area). Let's look at the math. (This is how I understood the maskana)

  • greenArea = (2r+2000+2000)^2 - orangeArea - cityArea (since the length of the migrash is 1000 amos + the length of the techum is 1000)
  • orangeArea = (2r+1000+1000)^2 - cityArea
  • cityArea = 3r^2


  • greenArea = (2r+4000)^2 - (2r+2000)^2 - 3r^2
  • orangeArea = (2r+2000)^2 - 3r^2

greenArea/orangeArea = ((2r+4000)^2 - (2r+2000)^2 - 3r^2 + 3r^2)/((2r+2000)^2 - 3r^2)

greenArea/orangeArea = ((2r+4000)^2 - (2r+2000)^2)/((2r+2000)^2 - 3r^2)

According to Raba, this works if r = 1000 (the city is 2000 x 2000).

However, the result will be 20/13? Where did I go wrong?


I saw the Soncino Gemara and they explain that the Migrash is also a circle. In that case, we have the following picture:

A large square, containing a concentric circle, containing a concentric square, containing an inscribed circle. The smaller circle is filled with grey. The smaller square has a side length of 2000 A. The remaining area inside the smaller square and the larger circle is filled with orange. The larger circle has diameter 4000 A. The remaining area of the larger square is filled with green. The larger square has a side length of 6000 A.

  1. BigArea = (2(r+1000+1000))^2
  2. CompleteMigrashArea = 3(r+1000)^2
  3. City = 3r^2

  1. GreenArea = BigArea-CompleteMigrashArea
  2. migrashArea = CompleteMigrashArea-City
  3. GreenArea = BigArea-(MigrashArea+City)
  4. GreenArea = BigArea-MigrashArea-City
  5. GreenArea = (2(r+1000+1000))^2-3(r+1000)^2-3r^2

  1. MigrashArea = 3(r+1000)^2 - 3r^2

  1. GreenArea/MigrashArea = ((2(r+1000+1000))^2-3(r+1000)^2-3r^2)/(3(r+1000)^2 - 3r^2)

If r = 1000, then then the ratio is 7/4

Doesn't work either.


Abaye says the same method will work if the city is 1000 by 1000. (r=500).

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The 3r^2 cancels which calculating the green area. What is the goal here? What does r=1000 yield for abaye? – Double AA May 13 '13 at 22:40
@DoubleAA "The Gemara said that the Migrash (the orange area) is one quarter of the area of the sados " – Shmuel Brin May 13 '13 at 22:42
So you want green/orange == 4 – Double AA May 13 '13 at 22:43
@DoubleAA that's how I understood it. Maybe I'm wrong. – Shmuel Brin May 13 '13 at 22:44
I have a book called Rabbinical Mathematics and Astronomy he has 5 pages on this topic with diagrams and mathematical figures, I don't know if there's a way to upload these here but if you want I could send it privately but it will have to wait till after Yom Tov. – Meir Zirkind May 14 '13 at 5:21
up vote 8 down vote accepted

In Rava's scenario the entire square is the numerator, so it's not (green/orange), but ((green+orange+gray)/orange). So for your second diagram, the formula should be:

BigArea/MigrashArea = ((2(r+1000+1000))^2)/(3(r+1000)^2 - 3r^2)

or 3.6x10^7 / 9x10^6 = 4

Abaye, on the other hand, doesn't include the city, so for him it's ((green+orange)/orange). Since he takes r=500, then for him it's

BigArea/MigrashArea = (((2(r+1000+1000))^2)-3r^2)/(3(r+1000)^2 - 3r^2)

or (2.5x10^7-7.5x10^5)/6x10^6 = 4.0167

[Tosafos ד"ה אביי says that we actually subtract the city from the total area - though not from the migrash - as if it were squared off, so:

BigArea/MigrashArea = (((2(r+1000+1000))^2)-((2r)^2)/(3(r+1000)^2 - 3r^2)

or (2.5x10^7-1x10^6)/6x10^6 = 4

Gra, on the other hand, says that the figure "one-quarter" doesn't have to be precise.]

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