Instructor:Spherical Geometry Exercises Solutions

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Spherical Geometry Exercises

General Questions

  1. Yes, every point on the sphere has exactly one antipodal point.
  2. There are really two valid choices here: 1) A is between B and C if A is on a geodesic segment joining B and C, or 2) A is between B and C if A is on the short geodesic segment joining B and C. In both cases, St. Louis is between the poles. In case 1, the north pole is between the south pole and St. Louis, but not in case 2.
  3. Draw the picture.
  4. Draw the picture.
  5. Draw the picture (it should look a lot like question 4's picture).
  6. 300°
  7. <math>360 - \frac{(n+2)180^\circ}{n}</math>
  8. Draw a geodesic segment connecting two corners of the quadrilateral. This splits the quadrilateral into two triangles. The sum of angles in the quadrilateral is the sum of the angles in the two triangles, which is larger than 180° + 180° = 360°.
  9. The north and south edges of Colorado are not geodesics - they are made from parallels. This means Colorado is not a quadrilateral, it has curved edges.
  10. draw some biangles
  11. 5- and 6 gons
  12. 2-gons with angle 90 degrees.
  13. Corner angles are 90°-120°-90°-120°.

Defects and Area

  1. defect = sum of angles - 180; to obtain the area we first find the fraction of the sphere: defect/720. Then we compute fraction * area of the whole sphere.
  2. defect square = sum of angles - 360. Fraction = defect / 720. Area = fraction*area of sphere.
  3. a. 90°; b. 1/8; c. 90°; d. 1/8; e. 36°; f. 1/20; g. 90°; h. 1/16; i. 90°; j. 180°
  4. 720°
  5. 720°
  6. A biangle with corner angle <math>\theta</math> covers <math>\theta/360^\circ</math> of the sphere.
  7. Corner angles are 90°-120°-90°-120°.

Spheres and Polyhedra

  1. No. Any three points on the sphere give a flat triangle, but it's possible to pick four points on a sphere that don't lie at the corners of a flat quadrilateral.
  2. Build the models!
  3. Tetrahedrons.
  4. Octahedrons.
  5. 12.
  6. 60.
  7. Draw the pictures.
  8. The regular tessellation by squares (it's self-dual).
  9. a. Tetrahedron; b. Octahedron; c. Cube; d. Icosahedron; e. Dodecahedron
  10. Tetrahedron-colored.svg
  11. Five triangles at each vertex gives 360°/5 = 72° corner angles, so each triangle has angle sum 72°+72°+72° = 216° for a defect of 36°, and an area fraction of 36°/720° = 1/20.
  12. Because there are two different corner angles for the rhombus.
  13. Defect is 60°. Covers 1/12 of the sphere.

Euler Characteristic

  1. 12 faces, 24 edges, 14 vertices. Yes, 12 - 24 + 14 = 2.
  2. a. 60 faces, 90 edges, 32 vertices; b. It's a soccer ball.
  3. 144 faces, 324 edges, 180 vertices. Euler characteristic is 0.
  4. 60 faces, 120 edges, 60 vertices. Euler characteristic is 0.

Escher and Spherical Geometry

  1. Although tessellations of the plane suggest infinity because they can be continued forever, Escher felt the necessity of an edge harmed the effect. Escher says that as you turn the ball, the neverending series of motifs suggests infinity. On the other hand, there are only finitely many motifs on the ball. Which is more compelling?
  2. a. 45°-90°-60°; b. 15°; c. 48; d. 48 faces, 72 edges, 26 vertices.


  1. He's relating duality of polyhedron to gender roles. In particular, he (arbitrarily) puts the "female" solids inside the "male" solids because they are subservient. He's also got to reach to include the tetrahedron, which is self-dual, and therefore a hermaphrodite (!??). Good thing Kepler lived around the turn of the 16th century, because this sort of gender stereotyping was typical then.
  2. a. They are close, since the diameter is 10 and circumference should be <math>10\pi</math>, about 31.4 cubits; b. On a sphere, the diameter of a circle with diameter <math>d</math> is less than <math>d\times\pi</math>, so it could possibly be built; c. A sphere of diameter about 19.1 cubits (a pretty small planet!).