I am aware that this is not the simplest possible solution.

198.—­THE EIGHT STICKS.

The first diagram is the answer that nearly every one will give to this puzzle, and at first sight it seems quite satisfactory.  But consider the conditions.  We have to lay “every one of the sticks on the table.”  Now, if a ladder be placed against a wall with only one end on the ground, it can hardly be said that it is “laid on the ground.”  And if we place the sticks in the above manner, it is only possible to make one end of two of them touch the table:  to say that every one lies on the table would not be correct.  To obtain a solution it is only necessary to have our sticks of proper dimensions.  Say the long sticks are each 2 ft. in length and the short ones 1 ft.  Then the sticks must be 3 in. thick, when the three equal squares may be enclosed, as shown in the second diagram.  If I had said “matches” instead of “sticks,” the puzzle would be impossible, because an ordinary match is about twenty-one times as long as it is broad, and the enclosed rectangles would not be squares.

[Illustration]

199.—­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.

[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.