[Illustration]
There are thirty edges, of which eighteen were visible in the original
illustration, represented in the following diagram by the hexagon
NAESGD. By this projection of the solid we get an imaginary view of the
remaining twelve edges, and are able to see at once their direction and
the twelve points at which all the edges meet. The difference in the
length of the lines is of no importance; all we want is to present their
direction in a graphic manner. But in case the novice should be puzzled
at only finding nineteen triangles instead of the required twenty, I
will point out that the apparently missing triangle is the outline HIK.
In this case there are twelve odd nodes; therefore six distinct and
disconnected routes will be needful if we are not to go over any lines
twice. Let us therefore find the greatest distance that we may so travel
in one route.
It will be noticed that I have struck out with little cross strokes five
lines or edges in the diagram. These five lines may be struck out
anywhere so long as they do not join one another, and so long as one of
them does not connect with N, the North Pole, from which we are to
start. It will be seen that the result of striking out these five lines
is that all the nodes are now even except N and S. Consequently if we
begin at N and stop at S we may go over all the lines, except the five
crossed out, without traversing any line twice. There are many ways of
doing this. Here is one route: N to H, I, K, S, I, E, S, G, K, D, H, A,
N, B, A, E, F, B, C, G, D, N, C, F, S. By thus making five of the routes
as short as is possible--simply from one node to the next--we are able
to get the greatest possible length for our sixth line. A greater
distance in one route, without going over the same ground twice, it is
not possible to get.
It is now readily seen that those five erased lines must be gone over
twice, and they may be "picked up," so to speak, at any points of our
route. Thus, whenever the traveller happens to be at I he can run up to
A and back before proceeding on his route, or he may wait until he is at
A and then run down to I and back to A. And so with the other lines that
have to be traced twice. It is, therefore, clear that he can go over 25
of the lines once only (25 × 10,000 miles = 250,000 miles) and 5 of the
lines twice (5 × 20,000 miles = 100,000 miles), the total, 350,000 miles,
being the length of his travels and the shortest distance that is
possible in visiting the whole body.
It will be noticed that I have made him end his travels at S, the South
Pole, but this is not imperative. I might have made him finish at any of
the other nodes, except the one from which he started. Suppose it had
been required to bring him home again to N at the end of his travels.
Then instead of suppressing the line AI we might leave that open and
close IS. This would enable him to complete his 350,000 miles tour at A,
and another 10,000 miles would take him to his own fireside. There are a
great many different routes, but as the lengths of the edges are all
alike, one course is as good as another. To make the complete 350,000
miles tour from N to S absolutely clear to everybody, I will give it
entire: N to H, I, A, I, K, H, K, S, I, E, S, G, F, G, K, D, C, D, H, A,
N, B, E, B, A, E, F, B, C, G, D, N, C, F, S--that is, thirty-five lines
of 10,000 miles each.