Which solids can we make?
Problem
Which Solids Can We Make printable sheet
Which Solids Can We Make printable sheet - extension
This problem follows on from Semi-regular Tessellations.
Here are pictures of the five Platonic Solids - solids made from just one type of polygon, with the same number of polygons meeting at each vertex:
tetrahedron, cube, octahedron, dodecahedron, icosahedron
Can you convince yourself that there are no more?
The angle deficit at a vertex of a polyhedron is a measure of how far short each angle sum is from $360^{\circ}$.
For example, in a dodecahedron, three pentagons with interior angles of $108^{\circ}$ meet at each vertex, so the angle sum is $324^{\circ}$ and the angle deficit is $36^{\circ}$:
Archimedean Solids have two properties:
- They are formed by two or more types of regular polygons, each with the same side length
- Each vertex has the same pattern of polygons around it.
Here is a picture of an Archimedean Solid with 24 vertices. Its vertex form {3, 3, 3, 3, 4} is defined by the polygons that meet at each vertex: triangle, triangle, triangle, triangle, square.
Calculate the total angle deficit for this solid. Does it match your observations about the Platonic solids?
Try to suggest some other vertex forms which might give rise to Archimedean solids, assuming all solids share the property you have discovered. If you have access to construction sets such as Polydron, you could test out your ideas.
Below are some vertex forms you might like to try: some of them give rise to solids and some of them don't. Can you decide which will work before testing them out?
Getting Started
Sometimes it helps to think about two dimensions to get a better understanding of what happens in three dimensions:
In two dimensions, the interior angles of convex polygons are always less than $180^{\circ}$. Can you explain why?
In three dimensions, the sum of the interior angles at each vertex of convex polyhedra must be less than $360^{\circ}$. Can you explain why?
The exterior angles of polygons are a measure of how far short the angles are from $180^{\circ}$.
The angle deficit at a vertex of a polyhedron is a measure of how far short each angle sum is from $360^{\circ}$.
The sum of the exterior angles of a polygon is always $360^{\circ}$.
What do you notice about the total angle deficit of a solid?
This table might help:
| Angle Sum | Angle Deficit | Number of Vertices | Total Angle Deficit | |
| Cube | 270 | 90 | 8 | 720 |
| Tetrahedron | ||||
| Octahedron | ||||
| Icosahedron | ||||
| Dodecahedron |
Student Solutions
Thank you to Ella from Trinity C of E School for sending us Euclid's geometric proof that there are no more platonic solids.
Each vertex of any solid must coincide with more than two faces. At each vertex of the solid the angle must be less than $360^{\circ}$. Since the angles at all the vertices of all the faces of a Platonic solid are identical, each vertex of each face must be less than ${360^{\circ}\over{3}}=120^{\circ}$. Only regular polygons with less than 6 sides have angles less than $120^{\circ}$ so the common face must be a triangle, square or pentagon and only the listed solids can be made with these faces.
Camilla from St Bernard's School sent us her workings for the angle deficit.
Tetrahedron: 3 triangles. $60^{\circ}\times3=180^{\circ}$ and $360^{\circ}-180^{\circ}=180^{\circ}$
Cube: 3 squares. $90^{\circ}\times3=270^{\circ}$ and $360^{\circ}-270^{\circ}=90^{\circ}$
Octahedron: 4 triangles. $60^{\circ}\times4=240^{\circ}$ and $360^{\circ}-240^{\circ}=120^{\circ}$
Icosahedron: 5 triangles. $60^{\circ}\times5=300^{\circ}$ and $360^{\circ}-300^{\circ}=60^{\circ}$
She also worked out that the total angle deficit is no. of vertices $\times$ angle deficit. For Platonic solids this is always $720^{\circ}$. Here is her working for the Archimedean solid.
{3, 3, 3, 3, 4} is 4 triangles and 1 square. $60^{\circ}\times4+90^{\circ}\times1=330^{\circ}$ and $360^{\circ}-330^{\circ}=30^{\circ}$. Number of vertices is 24 so the total angle deficit is $30^{\circ}\times24=720^{\circ}$ which is the same as a Platonic solid.
You can use this method to test whether vertex forms will work. For example, {4, 5, 8} is 1 square, 1 pentagon and 1 octagon. $90^{\circ}+108^{\circ}+135^{\circ}=333^{\circ}$ and $360^{\circ}-333^{\circ}=27^{\circ}$ and ${720^{\circ}\over27^{\circ}}=26.7$ but you can't have a non-integer number of vertices so this doesn't work. Whereas, {3, 8, 8} is 1 triangle and 2 octagons. $60^{\circ}\times1+135^{\circ}\times2=330^{\circ}$ and $360^{\circ}-330^{\circ}=30^{\circ}$ and ${720^{\circ}\over30^{\circ}}=24$ which does work.
For more information about this topic, Camilla suggests reading this article which helped her with her solution.
Teachers' Resources
Why do this problem
This problem builds on work done in finding Semi-Regular Tessellations and offers a great opportunity to create the Archimedean solids using students' knowledge of interior angles of polygons and angles around a point.
Possible approach
These printable worksheets may be useful: Which Solids Can We Make?
Which Solids Can We Make? - Extension
The problem has been structured in such a way that a class can spend some time answering each of the questions in bold, with time set aside to discuss approaches, thoughts and solutions.
The problem builds up to the search for Archimedean solids using what students learn about the angle deficit of Platonic solids. One way of introducing the concept of angle deficit could be to make a solid out of Polydron and then unfold it into its net and look at the gap at a vertex.
| Vertex Form | Angle Sum | Angle Deficit | Number of Vertices | Total Angle Deficit | |
| Cube | $4, 4, 4$ | $270$ | $90$ | $8$ | $720$ |
| Tetrahedron | |||||
| Octahedron | |||||
| Icosahedron | |||||
| Dodecahedron |
Key questions
Why do the interior angles meeting at a vertex always add up to less than $360^{\circ}$?
(Hint: think about how "pointy" a tetrahedron is compared with a dodecahedron)
Possible support
Starting with some two-dimensional work on Semi-Regular Tessellations may help to prepare students for the three-dimensional thinking required in this task.
Possible extension
Once students have suggested a possible vertex form, challenge them to explain how they could work out how many of each type of polygon will be required to make the solid.