Count the number of shapes in a polyhedron.

Question

So this is a question that was asked in the International Kangaroo Math Contest 2017. The question is:

The faces of the following polyhedron are either triangles or squares. Each triangle is surrounded by $3$ squares and each square is surrounded by $4$ triangles. If there are $6$ square faces, how many triangular faces are there?

What I did:

Each square shares each of its four neighboring triangles with two more squares. So we can say that for 6 squares we have $6\times4\ -\ 2 \times 6 = 12$ triangles. However, I still know that this calculation of mine is quite wrong and based on an awkward thinking. So, what is the correct answer and how?

Thanks for the attention.

Answer 1

Each edge of the polyhedron is shared between exactly one triangle and exactly one square, as can be inferred from the question statement. Thus, given six squares, there are 24 edges ($6×4$), and thus eight triangles ($24÷3$).

The polyhedron is called a cuboctahedron.

Answer 2

Each square is bordered by four triangles and $6\times4=24$. However every triangle is bordered by three different squares, so it was counted three times in the multiplication above. This means there are $24/3=8$ triangles.

Answer 3

This isn't rigorous, but if you don't have to write a proof, just get the right number, it's clear from the illustration. You can see "one hemisphere" of the polyhedron except for a triangle parallel with the line of sight, so the total number of sides of each type are just double what apppear in that hemisphere (i.e. what you see plus the hidden triangle).