- #1

- 52

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mprm86
- Start date

- #1

- 52

- 0

- #2

- 3

- 0

Note that the angle of a corner of a regular n-gon equals (n-2).Pi/n.

To fit the plane, k copies of a regular n-gon must be able to touch with their corners and therefore k angles should make up for a total arc of 2.Pi.

So, k should be chosen such that

k(n-2)Pi/n = 2Pi eq.

k(n-2) = 2n (*) eq.

k = 2n/(n-2).

So, we must choose n such that n-2 | 2n.

This looks already as if there are only few possibities. First note that n must be larger than 2. (Otherwise we don't even have a polygon).

n=3 gives 1|6 which is true, and k = 6/1 = 6

n=4 gives 2|8 which is true, and k = 8/2 = 4

n=5 gives 3|10 which is NOT true

n=6 gives 4|12 which is true, and k=12/4 = 3

How does it go on?

Well, the next quotient will be smaller than 3, so it must be 2, but then this would mean that just two corners n-gon fill an arc of 2Pi and this corner should be Pi, but this does not happen for a finite n.

Therefore, there are no other regular n-gons that tesselate the plane.

---

You can also look at solving the same equation (*) for n, we get:

n = 2k/(k-2)

So k should satisfy k-2 | 2k

So either

(i) k-2 = 1 or

(ii) k-2 = 2 or

(iii) k-2 is odd and k-2|k

(iv) k-2 is even and (k-2)/2 | k

Ad (i) k=3 and n=6

Ad (ii) k=4 and n=4

Ad (iii) k=2i+1 and 2i-1 | 2i+1. It is clear that if i>1 then (2i+1)/(2i-1) < 2, so this leaves no solutions

Ad (iv) k=2i and (i-1)|2i eq. i=2 OR i=3 only. i=2 gives k=4 and we had this already. i=3 gives k=6 and n=3.

This approach gives the same solutions.

QED

- #3

- 52

- 0

Thanks alot.

Share:

- Replies
- 1

- Views
- 17K