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Learning Math Home
Session 9: Solutions
 
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Solutions for Session 9, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6| A7 | A8


Problem A1

a. 

Three triangles around a vertex: You will get a figure with four triangular faces. It is called a tetrahedron.

b. 

Four triangles around a vertex: You will get a figure with eight triangular faces. It is called an octahedron.

Five triangles around a vertex: You will get a figure with 20 triangular faces. It is called an icosahedron.

Six triangles around a vertex: It lies flat, and you can't make a solid.

<< back to Problem A1


 

Problem A2

a. 

Follow the instructions. The construction works, and you can construct a cube. A cube uses a total of six squares in the construction.

b. 

It is impossible to make a solid if we try to connect four or more squares at a common vertex.

<< back to Problem A2


 

Problem A3

a. 

Follow the instructions. The construction works, and you can construct a dodecahedron. A dodecahedron has 12 pentagon faces.

b. 

It is impossible to make a solid if we try to connect four or more regular pentagons at a common vertex.

<< back to Problem A3


 

Problem A4

It is impossible to create a solid by connecting regular hexagons at a common vertex. If you have ever seen a soccer ball, you may have noticed that it is made up of hexagons and pentagons, but not of hexagons alone!

<< back to Problem A4


 

Problem A5

There are only five Platonic solids. The reason is that the sum of the interior angles of the regular polygons meeting at a vertex must not equal or surpass 360°. Otherwise, the figure will lie flat or even fold in on itself.

We need at least three faces to meet at a vertex, but after hexagons, the regular polygons all have interior angles that are more than 120°. So if three of them met at a vertex, there would be more than 360° there. We have already seen that this same angle restriction leads to just three possibilities for triangles (there must be at least three triangles but fewer than six), squares, and pentagons (in each case, there must be at least three but fewer than four).

This leaves a total of just five possible solids that fit the definition of "regular" or "Platonic" solid.

<< back to Problem A5


 

Problem A6

Solid

Vertices

Faces

Edges

Tetrahedron

4

4

6

Octahedron

6

8

12

Icosahedron

12

20

30

Cube

8

6

12

Dodecahedron

20

12

30

<< back to Problem A6


 

Problem A7

The pattern emerging from each row of the table is that, if f is the number of faces, e is the number of edges, and v is the number of vertices, we have v + f = e + 2. (Note that there are many equivalent ways of writing this relationship!)

<< back to Problem A7


 

Problem A8

It turns out that v + f = e + 2 holds for all convex polyhedra. This is known as Euler's formula, named for one of the most famous mathematicians of all time; he created the formula and proved that it always held.

<< back to Problem A8


 

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