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Papa

 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!

Math Genius