# Course:Harris, Fall 08: Diary Week 14

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- We explored the number of "nearest neighbors" to a point in the three geometries, using a regular tessellation by triangles.
- For the Euclidean plane, we chose a {3,6} tessellation. We picked a central point and counted:
- points 1 away: 6
- points 2 away: 12
- points 3 away: 18
- (guessing:) points <math>n</math> away: 6<math>n</math>

- For the sphere, we chose a {3,5} tessellation (the icosohedral tessellation):
- points 1 away: 5
- points 2 away: 5
- points 3 away: 1
- points <math>n</math> away for <math>n</math> > 3: 0

- For the hyperbolic plane, we chose a {3,7} tessellation
- We had to build this out of paper and equilateral triangles, striving to put 7 triangles together at each vertex.
- We ran out of tape, but not before getting enough done to make some counts:
- points 1 away: 7
- points 2 away: 21

- For the Euclidean plane, we chose a {3,6} tessellation. We picked a central point and counted:
- This was a digital approximation of the question, "How fast do circles grow in circumference, with increasing radius?"
- For the Euclidean plane, we know the answer:
- C = 2<math>\pi</math>R

- For the sphere:
- For circles of small radius, it looks pretty similar to the Euclidean plane (consider circles of high latitude near the North Pole).
- But for circles of larger radius, it's clear they're growing slower than the Euclidean formula (circles of latitude grow only slightly as they get into lower latitudes).
- And as the radius of a circle gets larger than distance from pole to equator, the circles actually start shrinking in circumference (in the sou thern hemisphere), eventually being just a point (the South Pole).
- For the hyperbolic plane, it's not so clear, and that's why we went to the digital approximation.

- For the Euclidean plane, we know the answer:

Wed, Fri: Thanksgiving break