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 x .366 = 2.562, or a little more than 21/2 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.

200.—­A KITE-FLYING PUZZLE.

Solvers of this little puzzle, I have generally found, may be roughly divided into two classes:  those who get within a mile of the correct answer by means of more or less complex calculations, involving “pi,” and those whose arithmetical kites fly hundreds and thousands of miles away from the truth.  The comparatively easy method that I shall show does not involve any consideration of the ratio that the diameter of a circle bears to its circumference.  I call it the “hat-box method.”

[Illustration]

Supposing we place our ball of wire, A, in a cylindrical hat-box, B, that exactly fits it, so that it touches the side all round and exactly touches the top and bottom, as shown in the illustration.  Then, by an invariable law that should be known by everybody, that box contains exactly half as much again as the ball.  Therefore, as the ball is 24 in. in diameter, a hat-box of the same circumference but two-thirds of the height (that is, 16 in. high) will have exactly the same contents as the ball.

Now let us consider that this reduced hat-box is a cylinder of metal made up of an immense number of little wire cylinders close together like the hairs in a painter’s brush.  By the conditions of the puzzle we are allowed to consider that there are no spaces between the wires.  How many of these cylinders one one-hundredth of an inch thick are equal to the large cylinder, which is 24 in. thick?