(related to Problem: Papa's Puzzle)

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.

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 $0.366.$ Thus, if the card were $7$ inches long, we get $7 \times 0.366 = 2.562,$ or a little more than $2\frac 12$ 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.

**Dudeney, H. E.**: "Amusements in Mathematics", The Authors' Club, 1917

This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook.