Polyhedra

Question 1

what are polyhedra?

Answer 1

solid figures bound by plane polygons

Question 2

what is a regular polyhedron?

Answer 2

all faces are congruent regular polygons with the same number of faces meeting at each vertex

Question 3

what does is mean for a polyhedron to be convex

Answer 3

if P&Q are pts inside the polyhedron or on its boundary, then PQ lies inside the polyhedron or on its boundary

Question 4

how may convex regular polyhedra are there?

Answer 4

5

Euclid “proves” this but he doesn’t state the assumptions of convexity and regularity

Question 5

what is a tetrahedron

Answer 5

4 equilateral triangles, 3 meet at each vertex

Question 6

what is a cube

Answer 6

6 equal squares, 3 meet at each vertex

Question 7

what is an octahedron

Answer 7

8 equilateral triangles, 4 meet at each vertex

Question 8

what is an icosahedron

Answer 8

20 equilateral triangles, 5 meet at each vertex

Question 9

what is a dodecahedron

Answer 9

12 pentagons, 3 meet at each vertex

Question 10

what is a face of a polyhedron

Answer 10

one of the plane polygons that contains it, when 2 of these meet they must have an entire edge in common

Question 11

where is a dihedral angle formed

Answer 11

where 2 faces meet along an edge

Question 12

what is a face angle of a polyhedron

Answer 12

the angle in any face passing through a vertex

Question 13

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

Answer 13

• 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

Question 14

state Euler’s thm for polyhedra

Answer 14

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

Question 15

how many vertices does a tetrahedron have

Answer 15

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

Question 16

how many vertices does a cube have

Answer 16

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

Question 17

how many vertices does an octahedron have

Answer 17

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

Question 18

how many vertices does an icosahedron have

Answer 18

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

Question 19

how many vertices does a dodecahedron have

Answer 19

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

Question 20

what are two examples of dual pairs within the polyhedra? why?

Answer 20

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

Question 21

what is the defect at a vertex

Answer 21

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

Question 22

what is Descartes corollary for convex polyhedra

Answer 22

in any convex polyhedron ∑𝛅x = 4𝜋

Question 23

is it possible to construct a polyhedron with 4 equilateral △s and a ▢ that all have side lengths 1?

Answer 23

• 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

Question 24

how can we construct the octahedron?

Answer 24

make 2 pyramids and “glue” them together at their bases

Question 25

how can we construct the icosahedron?

Answer 25

make 2 pentagonal pyramids, the “glue” 5 triangles like teeth on the bottom of each pyramid

Question 26

state Cauchy’s Rigidity thm

Answer 26

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