# Course:Harris, Fall 08: Diary Week 14

Mon:

• 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 $n$ away: 6$n$
• 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 $n$ away for $n$ > 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
• 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$\pi$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.

Wed, Fri: Thanksgiving break