Answer :
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.

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.

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!