THE FLY ON THE OCTAHEDRON.
(
Unicursal and Route Problems)
"Look here," said the professor to his colleague, "I have been watching
that fly on the octahedron, and it confines its walks entirely to the
edges. What can be its reason for avoiding the sides?"
"Perhaps it is trying to solve some route problem," suggested the other.
"Supposing it to start from the top point, how many different routes are
there by which it may walk over all the edges, without ever going twice
along the same edge in any route?"
The problem was a harder one than they expected, and after working at it
during leisure moments for several days their results did not agree--in
fact, they were both wrong. If the reader is surprised at their failure,
let him attempt the little puzzle himself. I will just explain that the
octahedron is one of the five regular, or Platonic, bodies, and is
contained under eight equal and equilateral triangles. If you cut out
the two pieces of cardboard of the shape shown in the margin of the
illustration, cut half through along the dotted lines and then bend them
and put them together, you will have a perfect octahedron. In any route
over all the edges it will be found that the fly must end at the point
of departure at the top.
Answer:
Though we cannot really see all the sides of the octahedron at once, we
can make a projection of it that suits our purpose just as well. In the
diagram the six points represent the six angles of the octahedron, and
four lines proceed from every point under exactly the same conditions as
the twelve edges of the solid. Therefore if we start at the point A and
go over all the lines once, we must always end our route at A. And the
number of different routes is just 1,488, counting the reverse way of
any route as different. It would take too much space to show how I make
the count. It can be done in about five minutes, but an explanation of
the method is difficult. The reader is therefore asked to accept my
answer as correct.
