The Mathematics Education Department
EMT 725, Lisa F. Garey
This problem was posed in class: to create the simplest solid that has equal number of squares as triangles.
We know that this solid will have to satisfy Euler's formula V + F=E + 2.
Here is a proof of Euler's formula presented in class.
V=# vertices E=# edges F=# faces
The basic assumption is that no matter how the solid is transformed,
the angle sum of the faces remain the same. Take one face of the solid.
Each face is a polygon with a certain number of sides. There are F faces,
so there are S1, S2, S3, ...................., SF sides. We know that the
sum of interior angles of an n-gon is 
. Thus, the angle sum for
all F faces
Since each edge is made up of 2 sides, S1+S2+S3+.......+SF = 2E, so
----------------Eq.1
Now consider stretching one face very large. The solid then becomes flat with the enlarged face at the bottom, and the other faces collapsed on top of it.
Hence angle sum of solid = angle sum of base + angle sum of rim + angle sum of center
Let r = # vertices of enlarged face
Then angle sum of solid 
---------------- Eq.2
Therefore, Eq.1=Eq.2 yields 
There are 5 Platonic Solids, i.e polyhedrons with congruent dihedral angles and congruent polygons as faces. Take the one with 3 pentagons at each vertex as an example.
The 3 pentagonal faces at each vertex contribute 5 edges each and each edge is counted twice.
Also, at each vertex, 3 edges come together to form a vertex.
Thus, E=30, so F=12 and V=20 which gives us the dodecahedron
| At each vertex | E | F | V | |
| 3 triangles | 6 | 4 | 4 | Tetrahedron |
| 3 squares | 12 | 6 | 8 | Cube |
| 3 pentagons | 30 | 12 | 20 | Dodecahedron |
| 4 triangles | 12 | 8 | 6 | Octahedron |
| 5 triangles | 30 | 20 | 12 | Icosahedron |
Now let's return to the original problem
I began considering squares & triangles at each vertex. 1 square & 1 triangle at each vertex would not make a solid. 2 squares & 2 triangles at each vertex could make a solid. 3 squares & 3 triangles would not make a solid because the angles add up to 360. So I proceeded with my pattern of 2 squares & 2 triangles, but did not obtain a polyhedra. My erroneous assumption was that this solid would have a homogeneous patterning at each vertex.
Abandoning that idea, I just put the pieces together making sure that for every triangle, I attach a square. The figure below is the resulting polyhedra with 4 squares & 4 triangles. There are other polyhedra with more # squares & triangles, but this is the simplest.
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