Exercises of Chapter 1 (Part 6/6), Computational Geometry

Exercise 1.59 Find the dihedral angles of the tetrahedra in Figure 1.26(c) and (d).

Solution. Omitted. □


Exercise 1.60 Find the dihedral angles of a regular dodecahedron.


Solution. Due to the regularity and symmetry of the shape, the dihedral angle can be calculated on any of the two adjacent planes. Angles at vertices are all 3π/5, so if we pick one of the vertices of the dodecahedron and focus on it,  we'll find that a line segment starting from a point at a distance of r from this central vertex on the edge incidental to the vertex and perpendicular to one of the two edges sharing the central vertex with the one aforementioned measures r1 = r sin(3π/5) in length, and the distance between the above starting point and its counterpart on the edge remaining untouched of the three is
r2 = r√2-2cos(3π/5) . We can also create a perpendicular line segment from that point similarly. Switching to the triangle formed by the two perpendicular lines both r₁ in length and the line segment connecting the two starting points r2 in length as its edges, we can see the angle between the two perpendicular lines is the questioned angle, for which if denoted by θ it is easy to figure out that the following relationship holds,
cosθ =  -(cos2(2π/5) + cos(2π/5)) / sin2(2π/5). □

Exercise 1.65. Show that a regular tetrahedron cannot be scissors congruent with a cube.


Proof. There are only angles of rational multiples of π in a cube, to be precise, only angles of π/2 whether it be vertex based on dihedral. While regular tetrahedron has dihedral angles of irrational multiples of π. Note that all dihedral angles of a regular tetrahedron are the same also. According to Corollary 1.62, they are not scissors congruent. □


Exercise 1.66. Show that no Platonic solid is scissors congruent to any other Platonic solid.

Proof. We can choose d-function to be identity function on irrational multiples of π and any Dehn invariant that uses this d-function will show that it has different values between Platonic solids of different types (with it being zero for cube). □

Exercise 1.69. Let P be a 2×1×1 rectangular prism. Cut P into eight or fewer pieces and rearrange the pieces to form a cube.


Solution. Making use of the result by Exercise 1.52, we can first turn the prism into a ³√4  ×³√2 ×1 one, and then we cut the prism against the ³√4 long edge with a proportion of  ³√2 to ³√2(³√2 - 1). The larger piece can serve as a base for the ultimate cube, while the smaller can further easily be shaped to a top with dimensions of ³√2 ×³√2×(√2-1). □

References

[1] Dodecahedron, Wikipedia, http://en.wikipedia.org/wiki/Dodecahedron
[2] Platonic Solid, Wikipedia, http://en.wikipedia.org/wiki/Platonic_solid