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

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<li>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°.</li>
 
<li>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°.</li>
 
<li>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.</li>
 
<li>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.</li>
<li>A biangle with corner angle <math>\theta</math> covers <math>\theta/360^\circ</math> of the sphere.</li>
+
<li> draw some examples</li>
 
<li> 5- and 6 gons </li>
 
<li> 5- and 6 gons </li>
 
<li> 2-gons with angle 90 degrees. </li>
 
<li> 2-gons with angle 90 degrees. </li>
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<li>720°</li>
 
<li>720°</li>
 
<li>720°</li>
 
<li>720°</li>
<li>Draw the picture.</li>
+
<li>A biangle with corner angle <math>\theta</math> covers <math>\theta/360^\circ</math> of the sphere.</li>
 +
<li>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.</li>
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</ol>
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 +
===Spheres and Polyhedra===
  
<li>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!).</li>
 
 
<li>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.</li>
 
<li>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.</li>
 
<li>Build the models!</li>
 
<li>Build the models!</li>
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<li>[[Image:tetrahedron-colored.svg]]</li>
 
<li>[[Image:tetrahedron-colored.svg]]</li>
 
<li>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?</li>
 
<li>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?</li>
<li>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.</li>
+
 
 
<li>Because there are two different corner angles for the rhombus.</li>
 
<li>Because there are two different corner angles for the rhombus.</li>
 
<li>Corner angles are 90°-120°-90°-120°.</li>
 
<li>Corner angles are 90°-120°-90°-120°.</li>
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<li>144 faces, 324 edges, 180 vertices.  Euler characteristic is 0.</li>
 
<li>144 faces, 324 edges, 180 vertices.  Euler characteristic is 0.</li>
 
<li>60 faces, 120 edges, 60 vertices.  Euler characteristic is 0.</li>
 
<li>60 faces, 120 edges, 60 vertices.  Euler characteristic is 0.</li>
 +
<li>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!).</li>
 
</ol>
 
</ol>

Revision as of 17:06, 24 October 2009


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 examples
  11. 5- and 6 gons
  12. 2-gons with angle 90 degrees.

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. 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.

Spheres and Polyhedra

  • 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.
  • Build the models!
  • Tetrahedrons.
  • Octahedrons.
  • 12.
  • 60.
  • Draw the pictures.
  • The regular tessellation by squares (it's self-dual).
  • a. Tetrahedron; b. Octahedron; c. Cube; d. Icosahedron; e. Dodecahedron
  • 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.
  • Tetrahedron-colored.svg
  • 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?
  • Because there are two different corner angles for the rhombus.
  • Corner angles are 90°-120°-90°-120°.
  • Defect is 60°. Covers 1/12 of the sphere.
  • 12 faces, 24 edges, 14 vertices. Yes, 12 - 24 + 14 = 2.
  • a. 45°-90°-60°; b. 15°; c. 48; d. 48 faces, 72 edges, 26 vertices.
  • a. 60 faces, 90 edges, 32 vertices; b. It's a soccer ball.
  • 144 faces, 324 edges, 180 vertices. Euler characteristic is 0.
  • 60 faces, 120 edges, 60 vertices. Euler characteristic is 0.
  • 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!).