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.
Starting with four different triangles, imagine you have an
unlimited number of each type. How many different tetrahedra can
you make? Convince us you have found them all.
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.