I have found that a large number of people imagine that the following is
a correct solution of the problem. Using the letters in the diagram
below, they argue that if you make the distance BA one-third of BC, and
therefore the area of the rectangle ABE equal to that of the triangular
remainder, the card must hang with the long side horizontal. Readers
will remember the jest of Charles II., who induced the Royal Society to
meet and discuss the reason why the water in a vessel will not rise if
you put a live fish in it; but in the middle of the proceedings one of
the least distinguished among them quietly slipped out and made the
experiment, when he found that the water _did_ rise! If my
correspondents had similarly made the experiment with a piece of
cardboard, they would have found at once their error. Area is one thing,
but gravitation is quite another. The fact of that triangle sticking its
leg out to D has to be compensated for by additional area in the
rectangle. As a matter of fact, the ratio of BA to AC is as 1 is to the
square root of 3, which latter cannot be given in an exact numerical
measure, but is approximately 1.732. Now let us look at the correct
general solution. There are many ways of arriving at the desired result,
but the one I give is, I think, the simplest for beginners.
[Illustration]
Fix your card on a piece of paper and draw the equilateral triangle BCF,
BF and CF being equal to BC. Also mark off the point G so that DG shall
equal DC. Draw the line CG and produce it until it cuts the line BF in
H. If we now make HA parallel to BE, then A is the point from which our
cut must be made to the corner D, as indicated by the dotted line.
A curious point in connection with this problem is the fact that the
position of the point A is independent of the side CD. The reason for
this is more obvious in the solution I have given than in any other
method that I have seen, and (although the problem may be solved with
all the working on the cardboard) that is partly why I have preferred
it. It will be seen at once that however much you may reduce the width
of the card by bringing E nearer to B and D nearer to C, the line CG,
being the diagonal of a square, will always lie in the same direction,
and will cut BF in H. Finally, if you wish to get an approximate measure
for the distance BA, all you have to do is to multiply the length of the
card by the decimal .366. Thus, if the card were 7 inches long, we get 7
× .366 = 2.562, or a little more than 2½ inches, for the distance from B
to A.
But the real joke of the puzzle is this: We have seen that the position
of the point A is independent of the width of the card, and depends
entirely on the length. Now, in the illustration it will be found that
both cards have the same length; consequently all the little maid had to
do was to lay the clipped card on top of the other one and mark off the
point A at precisely the same distance from the top left-hand corner!
So, after all, Pappus' puzzle, as he presented it to his little maid,
was quite an infantile problem, when he was able to show her how to
perform the feat without first introducing her to the elements of
statics and geometry.