Polyhedra Flashcards
what are polyhedra?
solid figures bound by plane polygons
what is a regular polyhedron?
all faces are congruent regular polygons with the same number of faces meeting at each vertex
what does is mean for a polyhedron to be convex
if P&Q are pts inside the polyhedron or on its boundary, then PQ lies inside the polyhedron or on its boundary
how may convex regular polyhedra are there?
5
Euclid “proves” this but he doesn’t state the assumptions of convexity and regularity
what is a tetrahedron
4 equilateral triangles, 3 meet at each vertex
what is a cube
6 equal squares, 3 meet at each vertex
what is an octahedron
8 equilateral triangles, 4 meet at each vertex
what is an icosahedron
20 equilateral triangles, 5 meet at each vertex
what is a dodecahedron
12 pentagons, 3 meet at each vertex
what is a face of a polyhedron
one of the plane polygons that contains it, when 2 of these meet they must have an entire edge in common
where is a dihedral angle formed
where 2 faces meet along an edge
what is a face angle of a polyhedron
the angle in any face passing through a vertex
prove that in a convex regular polyhedron, the only possible configurations at a vertex are 3, 4, or 5 triangles, 3 squares, or 3 pentagons
- there must be at least 3 faces meeting at each vertex (because otherwise the shape will not be closed)
- because of convexity at a vertex, the sum of the face angles at the vertex must be less than 2𝜋
- the 5 cases listed are the only possibilities
state Euler’s thm for polyhedra
if G is any connected plane figure with v vertices, e edges, and f faces, then v-e+f = 2
the proof follows by induction on e
how many vertices does a tetrahedron have
4 △s → f = 4
e = 4(3/2) = 6
(4 since we have 4△s, 3 since each △ has 3 sides, and /2 since each edge is part of 2 △s)
then v = 2 + e - f = 2 + 6 - 4 = 4 vertices
how many vertices does a cube have
6▢s → f = 6
e = 6(4/2) = 12
(6 since we have 6 ▢s, 4 since each ▢ has 4 sides, and /2 since each edge is part of 2 ▢s)
then v = 2 + e - f = 2 + 12 - 6 = 8 vertices
how many vertices does an octahedron have
8 △s → f = 8
e = 8(3/2) = 12
(8 since we have 8 △s, 3 since each △ has 3 sides, and /2 since each edge is part of 2 △s)
then v = 2 + e - f = 2 + 12 - 8 = 6 vertices
how many vertices does an icosahedron have
20 △s → f = 20
e = 20(3/2) = 30
(20 since we have 20 △s, 3 since each △ has 3 sides, and /2 since each edge is part of 2 △s)
then v = 2 + e - f = 2 + 30 - 20 = 12 vertices
how many vertices does a dodecahedron have
12 ⬠s → f = 12
e = 12(5/2) = 30
(12 since we have 12 ⬠s, 5 since each ⬠ has 5 sides, and /2 since each edge is part of 2 ⬠s)
then v = 2 + e - f = 2 + 30 - 12 = 20 vertices
what are two examples of dual pairs within the polyhedra? why?
cube & octahedron
if you choose the central pt of each face of the cube and join 2 such pts when the faces share an edge, you get the octahedron and vice versa
icosahedron & dodecahedron
if you choose the central pt of each face of the icosahedron and join 2 such pts when the faces share an edge, you get the dodecahedron and vice versa
what is the defect at a vertex
the defect at a vertex x is 2𝜋 minus the sum of the face angles at x, it is denoted 𝛅x
e.g. for tetrahedron: 2𝜋 - 3(𝜋/3) = 𝜋
what is Descartes corollary for convex polyhedra
in any convex polyhedron ∑𝛅x = 4𝜋
is it possible to construct a polyhedron with 4 equilateral △s and a ▢ that all have side lengths 1?
- sum of defects would be 4𝜋
- construct a perpendicular from the centre of the ▢abcx call the centre y
- then |xy| = (√2)/2, since all sides have length 1
- need h such that we have a right △ so h = (√2)/2
- let z be the point at the height h on the perpendicular
- construct za, zb, zc, zx, by design these all have length 1
we can do the same with a △ base and a ⬠ base
how can we construct the octahedron?
make 2 pyramids and “glue” them together at their bases
how can we construct the icosahedron?
make 2 pentagonal pyramids, the “glue” 5 triangles like teeth on the bottom of each pyramid
state Cauchy’s Rigidity thm
if P&Q are convex polyhedra and there exists a 1-1 correspondence ɸ between the vertices of P and the vertices of Q such that:
(1) XY is an edge of P iff ɸ(X)ɸ(Y) is an edge of Q
(2) corresponding faces are congruent,
then P&Q are congruent
