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Amanda Lutje
Math 329
September 21, 2015
Homework 4 – Spheres and Triangle Exploration
Part 1
1) a. The shortest path between Anchorage and Saint Petersburg is calculated in 1b, which is if
you head directly west on a great circle between these cities. This path was found to be
13343.391 kilometers. We can find the distance heading directly east by subtracting this
from the circumference of the great circle connecting these cities. If the radius is the 6371
km, then the circumference is 40030.17359 by formula. So we get 40030.17359 km 13343.391 km = 26686.78259 km.
b.
2)
Assuming the borders of Ideal City are as such, the different colors represent the different
districts. I calculated the distances between Roosevelt and Franklin, and Jefferson and
Franklin. Using the distance I halved it to get the midpoint, which was the radius for the
circles. I made boundaries cutting through where these circles meet to determine where
they are separated. They are not perfect squares, but I used the properties of circles to
visualize the areas that are actually closest to each school.
Part 2
1)
Equilateral Triangle: Three-sided polygon. All sides are the same length, and angles are
equal to 60 degrees.
Isosceles Triangle: Three-sided polygon. At least two sides and two angles are equal.
Scalene Triangle: Three-sided polygon where no sides or angles are equal.
Acute Triangle: Three-sided polygon where at least one angle is less than 90 degrees, all
angles greater than 0 degrees.
Right Triangle: Three-sided polygon where only one angle equal to 90 degrees.
Obtuse Triangle: Three-sided polygon where only one angle is greater than 90 degrees,
but less than 180.
2) On the sphere you can have a triangle with three right angles. But you can also have an
acute and obtuse triangle. So the sum of the interior angles in a triangle on a sphere can
be up to 300 degrees considering these triangles.
3) A sphere can be cut up into 8 different right-angles triangles, using perpendicular great
circles to separate them. They satisfy the definition of a triangle in that the 3 points are
connected by a line, but the angles are greater than the Euclidean space would allow.
4)
Above is what the “two-sided polygon” would look like. In my definition this would not
qualify as a triangle, because I explicitly define triangles as having three straight sides. I
would classify this as an irregular polygon though, because it is a closed figure with
vertices.
5) The Venn diagram I made would almost work for triangles on the sphere. You would
have to allow for equilateral triangles to be right triangles. But the rest should still hold.