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
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
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
For the cube, I worked on the applet on the Internet:
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 the applet on the internet,
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.