How big does a pentagonal Sukkah have to be?

Question

What is the minimum length requirement for each of the walls of a 5-sided sukkah?

What about 6 sides?

What about 7?

...

What about ∞ sides (i.e. a circle)?

Answer 1

Don't take my word on the translation, but Shulchan Aruch 634:2 says:

If it's round, it must contain within it a square of seven by seven t'fachim.

And MB adds that any other shape has the same rule and that one need not sit in the contained square.

You ask about wall length, though. For a circle, a contained square of 7×7 means, Baer Hetev and others there say, the wall length is 29.4 t'fachim (because 7×√2×π=7×1.4×3=29.4).

Other polygons' wall lengths can vary. The 'shortest' perimeter n-gon (n-sided polygon) containing a 7×7 square is one that approximates such a square, like this pentagon:

(Shortest is in scare quotes because there is no shortest really: it's always possible to get a slightly closer approximation of the square by moving the outlying vertices slightly closer to it.)

As for regular polygons, S. J. Dilworth and S. R. Mane, in a 2011 Journal of Geometry paper, say that if a regular pentagon of radius 1 (i.e. length 1 from the center to a corner) has n sides, then:

• If n is 5 or 9, then the largest square in the polygon has radius (cos(π/n)+sin(π/2n))/(cos(π/2n)+sin(π/2n)).
• If n>9 is 4k+1 for some integer k, then the largest square in the polygon has radius (1−sin(π/2n))(cos(π/2n)+sin(π/2n))/cos(π/4n).
• If n=4k−1 for an integer k, then the largest square in the polygon has radius (1+sin(π/2n))(cos(π/2n)−sin(π/2n))/cos(π/4n).
• If n is even but not divisible by 4, then the largest square has radius cos(π/n)/cos(π/2n).
• If n is divisible by 4, then the largest square has radius 1.

(A tip of the hat to sateesh mane for pointing me to that paper.)

In that case, we have that a 7×7 square (whose radius is 7/√2) sits inside a regular n-gon of radius:

• If n is 5 or 9: 7(cos(π/2n)+sin(π/2n))/(cos(π/n)+sin(π/2n))√2.
• If n>9 is 4k+1 for some integer k: 7cos(π/4n)/(1−sin(π/2n))(cos(π/2n)+sin(π/2n))√2.
• If n=4k−1 for an integer k: 7cos(π/4n)/(1+sin(π/2n))(cos(π/2n)−sin(π/2n))√2.
• If n is even but not divisible by 4: 7cos(π/2n)/cos(π/n)√2.
• If n is divisible by 4: 7/√2.

Now, the radius of a regular n-gon is s/2sin(π/n) where s is the length of a side of the n-gon. Thus, the side length of your suka would have to be:

• If n is 5 or 9: 7√2sin(π/n)(cos(π/2n)+sin(π/2n))/(cos(π/n)+sin(π/2n)).
• If n>9 is 4k+1 for some integer k: 7√2sin(π/n)cos(π/4n)/(1−sin(π/2n))(cos(π/2n)+sin(π/2n)).
• If n=4k−1 for an integer k: 7√2sin(π/n)cos(π/4n)/(1+sin(π/2n))(cos(π/2n)−sin(π/2n)).
• If n is even but not divisible by 4: 7√2sin(π/n)cos(π/2n)/cos(π/n).
• If n is divisible by 4: 7√2sin(π/n).

For example, a regular pentagon would need sides of length 7√2sin(π/5)(cos(π/10)+sin(π/10))/(cos(π/5)+sin(π/10)), which comes (using modern techniques, not those that yielded the 29.4 figure above) to a smidgen more than 6.3 t'fachim; a regular hexagon, a smidgen more than 5.52 t'fachim; a regular heptagon, a smidgen more than 4.64 t'fachim.

Caveat: Some of these calculations may be off. (In particular, I got a smaller sidelength earlier for the smallest pentagon (as you can see in an earlier revision of this answer), which shouldn't be possible, and I'm not sure whether that was mistaken or this is.) Also, I haven't read through the paper by Dilworth and Mane, and can't vouch for it.