We need to wrap up this cube-shaped present, remembering that we
can have no overlaps. What shapes can you find to use?
You want to make each of the 5 Platonic solids and colour the faces
so that, in every case, no two faces which meet along an edge have
the same colour.
In this problem you have to place four by four magic squares on the
faces of a cube so that along each edge of the cube the numbers
Correct solutions to the first part of the
question were received from: Alastair H (Forres Academy), Andrei L
(School 205, Bucharest). The solution below is based on Andrei's
submission. Well done to both of you.
First I created a small cube from the net in the figure and
observed the cubes from the problem:
The first cube cannot be formed from that net, because the
square marked with a red arrow should be in the position marked
with a blue arrow:
Looking at the cube made from the net: It is possible to see
that the second cube (horizontally) can be made from that net and
also the last three
In conclusion, the 4 cubes denominated before are the same one,
created from the net (B, D, E and F below).
The second part of the problem:
First shade the three faces of the view B of the cube which are
not visible. Then do the same with the other three views of the
same cube (D,E and F).
A B C D E F
You end up shading all the faces. This means that you can see
all the faces of the cube in the four views B, D, E and F so there
are no hidden faces where you can shade additional sections.