"But, for my part, it was Greek to me."
_Julius Cæsar_, i. 2.
Many people are accustomed to consider the cross as a wholly Christian
symbol. This is erroneous: it is of very great antiquity. The ancient
Egyptians employed it as a sacred symbol, and on Greek sculptures we
find representations of a cake (the supposed real origin of our hot
cross buns) bearing a cross. Two such cakes were discovered at
Herculaneum. Cecrops offered to Jupiter Olympus a sacred cake or _boun_
of this kind. The cross and ball, so frequently found on Egyptian
figures, is a circle and the _tau_ cross. The circle signified the
eternal preserver of the world, and the T, named from the Greek letter
_tau_, is the monogram of Thoth, the Egyptian Mercury, meaning wisdom.
This _tau_ cross is also called by Christians the cross of St. Anthony,
and is borne on a badge in the bishop's palace at Exeter. As for the
Greek or mundane cross, the cross with four equal arms, we are told by
competent antiquaries that it was regarded by ancient occultists for
thousands of years as a sign of the dual forces of Nature--the male and
female spirit of everything that was everlasting.
[Illustration: Fig. 5.]
The Greek cross, as shown in Fig. 5, is formed by the assembling
together of five equal squares. We will start with what is known as the
Hindu problem, supposed to be upwards of three thousand years old. It
appears in the seal of Harvard College, and is often given in old works
as symbolical of mathematical science and exactitude. Cut the cross into
five pieces to form a square. Figs. 6 and 7 show how this is done. It
was not until the middle of the nineteenth century that we found that
the cross might be transformed into a square in only four pieces. Figs.
8 and 9 will show how to do it, if we further require the four pieces to
be all of the same size and shape. This Fig. 9 is remarkable because,
according to Dr. Le Plongeon and others, as expounded in a work by
Professor Wilson of the Smithsonian Institute, here we have the great
Swastika, or sign, of "good luck to you "--the most ancient symbol of
the human race of which there is any record. Professor Wilson's work
gives some four hundred illustrations of this curious sign as found in
the Aztec mounds of Mexico, the pyramids of Egypt, the ruins of Troy,
and the ancient lore of India and China. One might almost say there is a
curious affinity between the Greek cross and Swastika! If, however, we
require that the four pieces shall be produced by only two clips of the
scissors (assuming the puzzle is in paper form), then we must cut as in
Fig. 10 to form Fig. 11, the first clip of the scissors being from a
to b. Of course folding the paper, or holding the pieces together
after the first cut, would not in this case be allowed. But there is an
infinite number of different ways of making the cuts to solve the puzzle
in four pieces. To this point I propose to return.
[Illustration: Fig. 6]
[Illustration: Fig. 7]
[Illustration: Fig. 8]
[Illustration: Fig. 9]
[Illustration: Fig. 10]
[Illustration: Fig. 11]
It will be seen that every one of these puzzles has its reverse
puzzle--to cut a square into pieces to form a Greek cross. But as a
square has not so many angles as the cross, it is not always equally
easy to discover the true directions of the cuts. Yet in the case of the
examples given, I will leave the reader to determine their direction for
himself, as they are rather obvious from the diagrams.
Cut a square into five pieces that will form two separate Greek crosses
of _different sizes_. This is quite an easy puzzle. As will be seen in
Fig. 12, we have only to divide our square into 25 little squares and
then cut as shown. The cross A is cut out entire, and the pieces B, C,
D, and E form the larger cross in Fig. 13. The reader may here like to
cut the single piece, B, into four pieces all similar in shape to
itself, and form a cross with them in the manner shown in Fig. 13. I
hardly need give the solution.
[Illustration: FIG. 12.]
[Illustration: FIG. 13.]
Cut a square into five pieces that will form two separate Greek crosses
of exactly the _same size_. This is more difficult. We make the cuts as
in Fig. 14, where the cross A comes out entire and the other four pieces
form the cross in Fig. 15. The direction of the cuts is pretty obvious.
It will be seen that the sides of the square in Fig. 14 are marked off
into six equal parts. The sides of the cross are found by ruling lines
from certain of these points to others.
[Illustration: FIG. 14.]
[Illustration: FIG. 15.]
I will now explain, as I promised, why a Greek cross may be cut into
four pieces in an infinite number of different ways to make a square.
Draw a cross, as in Fig. 16. Then draw on transparent paper the square
shown in Fig. 17, taking care that the distance c to d is exactly
the same as the distance a to b in the cross. Now place the
transparent paper over the cross and slide it about into different
positions, only be very careful always to keep the square at the same
angle to the cross as shown, where a b is parallel to c d. If
you place the point c exactly over a the lines will indicate the
solution (Figs. 10 and 11). If you place c in the very centre of the
dotted square, it will give the solution in Figs. 8 and 9. You will now
see that by sliding the square about so that the point c is always
within the dotted square you may get as many different solutions as you
like; because, since an infinite number of different points may
theoretically be placed within this square, there must be an infinite
number of different solutions. But the point c need not necessarily be
placed within the dotted square. It may be placed, for example, at point
e to give a solution in four pieces. Here the joins at a and f may
be as slender as you like. Yet if you once get over the edge at a or
f you no longer have a solution in four pieces. This proof will be
found both entertaining and instructive. If you do not happen to have
any transparent paper at hand, any thin paper will of course do if you
hold the two sheets against a pane of glass in the window.
[Illustration: FIG. 16.]
[Illustration: FIG. 17.]
It may have been noticed from the solutions of the puzzles that I have
given that the side of the square formed from the cross is always equal
to the distance a to b in Fig. 16. This must necessarily be so, and
I will presently try to make the point quite clear.
We will now go one step further. I have already said that the ideal
solution to a cutting-out puzzle is always that which requires the
fewest possible pieces. We have just seen that two crosses of the same
size may be cut out of a square in five pieces. The reader who
succeeded in solving this perhaps asked himself: "Can it be done in
fewer pieces?" This is just the sort of question that the true puzzle
lover is always asking, and it is the right attitude for him to adopt.
The answer to the question is that the puzzle may be solved in four
pieces--the fewest possible. This, then, is a new puzzle. Cut a square
into four pieces that will form two Greek crosses of the same size.
[Illustration: FIG. 18.]
[Illustration: FIG. 19.]
[Illustration: FIG. 20.]
The solution is very beautiful. If you divide by points the sides of the
square into three equal parts, the directions of the lines in Fig. 18
will be quite obvious. If you cut along these lines, the pieces A and B
will form the cross in Fig. 19 and the pieces C and D the similar cross
in Fig. 20. In this square we have another form of Swastika.
The reader will here appreciate the truth of my remark to the effect
that it is easier to find the directions of the cuts when transforming a
cross to a square than when converting a square into a cross. Thus, in
Figs. 6, 8, and 10 the directions of the cuts are more obvious than in
Fig. 14, where we had first to divide the sides of the square into six
equal parts, and in Fig. 18, where we divide them into three equal
parts. Then, supposing you were required to cut two equal Greek crosses,
each into two pieces, to form a square, a glance at Figs. 19 and 20 will
show how absurdly more easy this is than the reverse puzzle of cutting
the square to make two crosses.
Referring to my remarks on "fallacies," I will now give a little example
of these "solutions" that are not solutions. Some years ago a young
correspondent sent me what he evidently thought was a brilliant new
discovery--the transforming of a square into a Greek cross in four
pieces by cuts all parallel to the sides of the square. I give his
attempt in Figs. 21 and 22, where it will be seen that the four pieces
do not form a symmetrical Greek cross, because the four arms are not
really squares but oblongs. To make it a true Greek cross we should
require the additions that I have indicated with dotted lines. Of course
his solution produces a cross, but it is not the symmetrical Greek
variety required by the conditions of the puzzle. My young friend
thought his attempt was "near enough" to be correct; but if he bought a
penny apple with a sixpence he probably would not have thought it "near
enough" if he had been given only fourpence change. As the reader
advances he will realize the importance of this question of exactitude.
[Illustration: FIG. 21.]
[Illustration: FIG. 22.]
In these cutting-out puzzles it is necessary not only to get the
directions of the cutting lines as correct as possible, but to remember
that these lines have no width. If after cutting up one of the crosses
in a manner indicated in these articles you find that the pieces do not
exactly fit to form a square, you may be certain that the fault is
entirely your own. Either your cross was not exactly drawn, or your cuts
were not made quite in the right directions, or (if you used wood and a
fret-saw) your saw was not sufficiently fine. If you cut out the puzzles
in paper with scissors, or in cardboard with a penknife, no material is
lost; but with a saw, however fine, there is a certain loss. In the case
of most puzzles this slight loss is not sufficient to be appreciable,
if the puzzle is cut out on a large scale, but there have been
instances where I have found it desirable to draw and cut out each part
separately--not from one diagram--in order to produce a perfect result.
[Illustration: FIG. 23.]
[Illustration: FIG. 24.]
Now for another puzzle. If you have cut out the five pieces indicated in
Fig. 14, you will find that these can be put together so as to form the
curious cross shown in Fig. 23. So if I asked you to cut Fig. 24 into
five pieces to form either a square or two equal Greek crosses you would
know how to do it. You would make the cuts as in Fig. 23, and place them
together as in Figs. 14 and 15. But I want something better than that,
and it is this. Cut Fig. 24 into only four pieces that will fit together
and form a square.
[Illustration: FIG. 25.]
[Illustration: FIG. 26.]
The solution to the puzzle is shown in Figs. 25 and 26. The direction of
the cut dividing A and C in the first diagram is very obvious, and the
second cut is made at right angles to it. That the four pieces should
fit together and form a square will surprise the novice, who will do
well to study the puzzle with some care, as it is most instructive.
I will now explain the beautiful rule by which we determine the size of
a square that shall have the same area as a Greek cross, for it is
applicable, and necessary, to the solution of almost every dissection
puzzle that we meet with. It was first discovered by the philosopher
Pythagoras, who died 500 B.C., and is the 47th proposition of Euclid.
The young reader who knows nothing of the elements of geometry will get
some idea of the fascinating character of that science. The triangle ABC
in Fig. 27 is what we call a right-angled triangle, because the side BC
is at right angles to the side AB. Now if we build up a square on each
side of the triangle, the squares on AB and BC will together be exactly
equal to the square on the long side AC, which we call the hypotenuse.
This is proved in the case I have given by subdividing the three squares
into cells of equal dimensions.
[Illustration: FIG. 27.]
[Illustration: FIG. 28.]
It will be seen that 9 added to 16 equals 25, the number of cells in the
large square. If you make triangles with the sides 5, 12 and 13, or with
8, 15 and 17, you will get similar arithmetical proofs, for these are
all "rational" right-angled triangles, but the law is equally true for
all cases. Supposing we cut off the lower arm of a Greek cross and place
it to the left of the upper arm, as in Fig. 28, then the square on EF
added to the square on DE exactly equals a square on DF. Therefore we
know that the square of DF will contain the same area as the cross. This
fact we have proved practically by the solutions of the earlier puzzles
of this series. But whatever length we give to DE and EF, we can never
give the exact length of DF in numbers, because the triangle is not a
"rational" one. But the law is none the less geometrically true.
[Illustration: FIG. 29.]
[Illustration: FIG. 30.]
Now look at Fig. 29, and you will see an elegant method for cutting a
piece of wood of the shape of two squares (of any relative dimensions)
into three pieces that will fit together and form a single square. If
you mark off the distance _ab_ equal to the side _cd_ the directions of
the cuts are very evident. From what we have just been considering, you
will at once see why _bc_ must be the length of the side of the new
square. Make the experiment as often as you like, taking different
relative proportions for the two squares, and you will find the rule
always come true. If you make the two squares of exactly the same size,
you will see that the diagonal of any square is always the side of a
square that is twice the size. All this, which is so simple that anybody
can understand it, is very essential to the solving of cutting-out
puzzles. It is in fact the key to most of them. And it is all so
beautiful that it seems a pity that it should not be familiar to
everybody.
We will now go one step further and deal with the half-square. Take a
square and cut it in half diagonally. Now try to discover how to cut
this triangle into four pieces that will form a Greek cross. The
solution is shown in Figs. 31 and 32. In this case it will be seen that
we divide two of the sides of the triangle into three equal parts and
the long side into four equal parts. Then the direction of the cuts will
be easily found. It is a pretty puzzle, and a little more difficult than
some of the others that I have given. It should be noted again that it
would have been much easier to locate the cuts in the reverse puzzle of
cutting the cross to form a half-square triangle.
[Illustration: FIG. 31.]
[Illustration: FIG. 32.]
[Illustration: FIG. 33.]
[Illustration: FIG. 34.]
Another ideal that the puzzle maker always keeps in mind is to contrive
that there shall, if possible, be only one correct solution. Thus, in
the case of the first puzzle, if we only require that a Greek cross
shall be cut into four pieces to form a square, there is, as I have
shown, an infinite number of different solutions. It makes a better
puzzle to add the condition that all the four pieces shall be of the
same size and shape, because it can then be solved in only one way, as
in Figs. 8 and 9. In this way, too, a puzzle that is too easy to be
interesting may be improved by such an addition. Let us take an example.
We have seen in Fig. 28 that Fig. 33 can be cut into two pieces to form
a Greek cross. I suppose an intelligent child would do it in five
minutes. But suppose we say that the puzzle has to be solved with a
piece of wood that has a bad knot in the position shown in Fig. 33--a
knot that we must not attempt to cut through--then a solution in two
pieces is barred out, and it becomes a more interesting puzzle to solve
it in three pieces. I have shown in Figs. 33 and 34 one way of doing
this, and it will be found entertaining to discover other ways of doing
it. Of course I could bar out all these other ways by introducing more
knots, and so reduce the puzzle to a single solution, but it would then
be overloaded with conditions.
And this brings us to another point in seeking the ideal. Do not
overload your conditions, or you will make your puzzle too complex to be
interesting. The simpler the conditions of a puzzle are, the better. The
solution may be as complex and difficult as you like, or as happens, but
the conditions ought to be easily understood, or people will not attempt
a solution.
If the reader were now asked "to cut a half-square into as few pieces as
possible to form a Greek cross," he would probably produce our solution,
Figs. 31-32, and confidently claim that he had solved the puzzle
correctly. In this way he would be wrong, because it is not now stated
that the square is to be divided diagonally. Although we should always
observe the exact conditions of a puzzle we must not read into it
conditions that are not there. Many puzzles are based entirely on the
tendency that people have to do this.
The very first essential in solving a puzzle is to be sure that you
understand the exact conditions. Now, if you divided your square in half
so as to produce Fig. 35 it is possible to cut it into as few as three
pieces to form a Greek cross. We thus save a piece.
I give another puzzle in Fig. 36. The dotted lines are added merely to
show the correct proportions of the figure--a square of 25 cells with
the four corner cells cut out. The puzzle is to cut this figure into
five pieces that will form a Greek cross (entire) and a square.
[Illustration: FIG. 35.]
[Illustration: FIG. 36.]
The solution to the first of the two puzzles last given--to cut a
rectangle of the shape of a half-square into three pieces that will form
a Greek cross--is shown in Figs. 37 and 38. It will be seen that we
divide the long sides of the oblong into six equal parts and the short
sides into three equal parts, in order to get the points that will
indicate the direction of the cuts. The reader should compare this
solution with some of the previous illustrations. He will see, for
example, that if we continue the cut that divides B and C in the cross,
we get Fig. 15.
[Illustration: FIG. 37.]
[Illustration: FIG. 38.]
The other puzzle, like the one illustrated in Figs. 12 and 13, will show
how useful a little arithmetic may sometimes prove to be in the solution
of dissection puzzles. There are twenty-one of those little square cells
into which our figure is subdivided, from which we have to form both a
square and a Greek cross. Now, as the cross is built up of five squares,
and 5 from 21 leaves 16--a square number--we ought easily to be led to
the solution shown in Fig. 39. It will be seen that the cross is cut out
entire, while the four remaining pieces form the square in Fig. 40.
[Illustration: FIG. 39]
[Illustration: FIG. 40]
Of course a half-square rectangle is the same as a double square, or two
equal squares joined together. Therefore, if you want to solve the
puzzle of cutting a Greek cross into four pieces to form two separate
squares of the same size, all you have to do is to continue the short
cut in Fig. 38 right across the cross, and you will have four pieces of
the same size and shape. Now divide Fig. 37 into two equal squares by a
horizontal cut midway and you will see the four pieces forming the two
squares.
[Illustration: FIG. 41]
Cut a Greek cross into five pieces that will form two separate squares,
one of which shall contain half the area of one of the arms of the
cross. In further illustration of what I have already written, if the
two squares of the same size A B C D and B C F E, in Fig. 41, are cut in
the manner indicated by the dotted lines, the four pieces will form the
large square A G E C. We thus see that the diagonal A C is the side of a
square twice the size of A B C D. It is also clear that half the
diagonal of any square is equal to the side of a square of half the
area. Therefore, if the large square in the diagram is one of the arms
of your cross, the small square is the size of one of the squares
required in the puzzle.
The solution is shown in Figs. 42 and 43. It will be seen that the small
square is cut out whole and the large square composed of the four pieces
B, C, D, and E. After what I have written, the reader will have no
difficulty in seeing that the square A is half the size of one of the
arms of the cross, because the length of the diagonal of the former is
clearly the same as the side of the latter. The thing is now
self-evident. I have thus tried to show that some of these puzzles that
many people are apt to regard as quite wonderful and bewildering, are
really not difficult if only we use a little thought and judgment. In
conclusion of this particular subject I will give four Greek cross
puzzles, with detached solutions.