Difference between revisions of "Instructor:Spherical Geometry Exercises Solutions"

From EscherMath
Jump to navigationJump to search
(Exercise renumbering to match the problem renumbering.)
Line 52: Line 52:
<li><ol style="list-style-type:lower-alpha">
<li><ol style="list-style-type:lower-alpha">
<li>[[Day and Night]] Night/day.  Black/white.
<li>[[Day and Night]] Night/day.  Black/white.
<li>[[Sun and Moon]] Sun/moon.  Night/day.
<li>[[Sun and Moon]] Sun/moon.  Night/day. Figure/ground.
<li>[[Order and Chaos]] Order/chaos.
<li>[[Order and Chaos]] Order/chaos.
<li>[[Sky and Water I]] Sky/water.  Black/white.
<li>[[Sky and Water I]] Sky/water.  Black/white.

Latest revision as of 22:27, 20 November 2013

Spherical Geometry Exercises

Geometry on the Sphere

  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. Drawing of a 90-90-90 triangle.
  4. Drawing of a triangle with one angle > 180.
  5. Drawing of three mutually perpendicular geodesics (it should look a lot like question 4's picture).
  6. Upper limit of corner of equilateral triangle is 300°
  7. Upper limit for corner of regular n-gon: <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 in the soccer ball
  12. Biangles with angle 90 degrees in the juggling balls.
  13. Corner angles are 90°-120°-90°-120° for the Ivory Ball Study.
  14. 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!).

Defect 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° is upper limit for defect of triangle
  5. 720° is upper limit for defect of any polygon
  6. A biangle with corner angle <math>\theta</math> covers <math>\theta/360^\circ</math> of the sphere.
  7. Ivory Ball Study: a) Defect is 60°. b) Covers 1/12 of the sphere.
  8. Concentric Rinds: a. 45°-90°-60°; b. 15° defect; c. 48 cover.


  1. No, spherical tessellations do not always give polyhedra. 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. Double Planetoid has tetrahedrons.
  4. Stars has octahedrons.
  5. 12 lizards.
  6. 60 seashells.
  7. Tetrahedron-colored.svg
  8. Icosahedron has five triangles at each vertex, giving 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.
  9. Ivory Ball Study is not regular, because there are two different corner angles for the rhombus.


  1. Draw pictures illustrating hexagon-triangle duality.
  2. The regular tessellation by squares (it's self-dual).
  3. Duals are: a. Tetrahedron; b. Octahedron; c. Cube; d. Icosahedron; e. Dodecahedron
    1. Day and Night Night/day. Black/white.
    2. Sun and Moon Sun/moon. Night/day. Figure/ground.
    3. Order and Chaos Order/chaos.
    4. Sky and Water I Sky/water. Black/white.
    5. Double Planetoid City/wilderness.
    6. Bond of Union Male/female
  4. 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.

Euler Characteristic

  1. Ivory Ball Study has 12 faces, 24 edges, 14 vertices. Yes, 12 - 24 + 14 = 2.
  2. Concentric Rinds has 48 faces, 72 edges, 26 vertices, <math>\chi = 2</math>.
  3. Pentakis dodecahedron has: a. 60 faces, 90 edges, 32 vertices; b. It's a soccer ball.
  4. Mobius strip II has 144 faces, 324 edges, 180 vertices. Euler characteristic is 0.
  5. The torus has: 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?