Or search by topic
For the tetrahedron, I have:
I tested all Platonic solids: the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron.
For the tetrahedron, there was simple - there were only four points to study. I obtained more possible arrangements, corresponding to different solutions. The other, not shown is to go on the outside circuit up to the last-but-one point, than to go to the centre and finally at the starting point. There are different solutions only if the start - finish is represented on the figure, otherwise this solution is obtained from the first by a rotation.
For the cube, I worked on an applet:
The strategy I used is the following: I go on the outer lines, up to the last-but-one point. Then I go through the inner one. I can use symmetrical combinations, obtaining more than one path.
Looking at the octahedron, I take the same strategy as for the cube => there are many combinations.
For the dodecahedron, I worked on an applet, taking the same strategy as for the cube, the difference is that here there are more layers.
The last polyhedron I test is the icosahedron:
Here, I worked in the same manner as before, obtaining the figure shown. There are naturally more combinations.
Can you number the vertices, edges and faces of a tetrahedron so that the number on each edge is the mean of the numbers on the adjacent vertices and the mean of the numbers on the adjacent faces?
Each of these solids is made up with 3 squares and a triangle around each vertex. Each has a total of 18 square faces and 8 faces that are equilateral triangles. How many faces, edges and vertices does each solid have?
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?