THE GARDEN WALLS.
(
Patchwork Puzzles)
A speculative country builder has a circular field, on which he has
erected four cottages, as shown in the illustration. The field is
surrounded by a brick wall, and the owner undertook to put up three
other brick walls, so that the neighbours should not be overlooked by
each other, but the four tenants insist that there shall be no
favouritism, and that each shall have exactly the same length of wall
space for his wall fruit trees. The puzzle is to show how the three
walls may be built so that each tenant shall have the same area of
ground, and precisely the same length of wall.
Of course, each garden must be entirely enclosed by its walls, and it
must be possible to prove that each garden has exactly the same length
of wall. If the puzzle is properly solved no figures are necessary.
Answer:
The puzzle was to divide the circular field into four equal parts by
three walls, each wall being of exactly the same length. There are two
essential difficulties in this problem. These are: (1) the thickness of
the walls, and (2) the condition that these walls are three in number.
As to the first point, since we are told that the walls are brick walls,
we clearly cannot ignore their thickness, while we have to find a
solution that will equally work, whether the walls be of a thickness of
one, two, three, or more bricks.
The second point requires a little more consideration. How are we to
distinguish between a wall and walls? A straight wall without any bend
in it, no matter how long, cannot ever become "walls," if it is neither
broken nor intersected in any way. Also our circular field is clearly
enclosed by one wall. But if it had happened to be a square or a
triangular enclosure, would there be respectively four and three walls
or only one enclosing wall in each case? It is true that we speak of
"the four walls" of a square building or garden, but this is only a
conventional way of saying "the four sides." If you were speaking of the
actual brickwork, you would say, "I am going to enclose this square
garden with a wall." Angles clearly do not affect the question, for we
may have a zigzag wall just as well as a straight one, and the Great
Wall of China is a good example of a wall with plenty of angles. Now, if
you look at Diagrams 1, 2, and 3, you may be puzzled to declare whether
there are in each case two or four new walls; but you cannot call them
three, as required in our puzzle. The intersection either affects the
question or it does not affect it.
If you tie two pieces of string firmly together, or splice them in a
nautical manner, they become "one piece of string." If you simply let
them lie across one another or overlap, they remain "two pieces of
string." It is all a question of joining and welding. It may similarly
be held that if two walls be built into one another--I might almost say,
if they be made homogeneous--they become one wall, in which case
Diagrams 1, 2, and 3 might each be said to show one wall or two, if it
be indicated that the four ends only touch, and are not really built
into, the outer circular wall.
The objection to Diagram 4 is that although it shows the three required
walls (assuming the ends are not built into the outer circular wall),
yet it is only absolutely correct when we assume the walls to have no
thickness. A brick has thickness, and therefore the fact throws the
whole method out and renders it only approximately correct.
Diagram 5 shows, perhaps, the only correct and perfectly satisfactory
solution. It will be noticed that, in addition to the circular wall,
there are three new walls, which touch (and so enclose) but are not
built into one another. This solution may be adapted to any desired
thickness of wall, and its correctness as to area and length of wall
space is so obvious that it is unnecessary to explain it. I will,
however, just say that the semicircular piece of ground that each tenant
gives to his neighbour is exactly equal to the semicircular piece that
his neighbour gives to him, while any section of wall space found in one
garden is precisely repeated in all the others. Of course there is an
infinite number of ways in which this solution may be correctly varied.
