Part 84

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."

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? Circles are to one another as the squares of their diameters. The square of 1/100 is 1/100000, and the square of 24 is 576; therefore the large cylinder contains 5,760,000 of the little wire cylinders. But we have seen that each of these wires is 16 in. long; hence 16 x 5,760,000 = 92,160,000 inches as the complete length of the wire. Reduce this to miles, and we get 1,454 miles 2,880 ft. as the length of the wire attached to the professor's kite.

Whether a kite would fly at such a height, or support such a weight, are questions that do not enter into the problem.

201.--HOW TO MAKE CISTERNS.

Here is a general formula for solving this problem. Call the two sides of the rectangle a and b. Then

a + b - (a squared + b squared - ab)^1/2 --------------------------- 6

equals the side of the little square pieces to cut away. The measurements given were 8 ft. by 3 ft., and the above rule gives 8 in. as the side of the square pieces that have to be cut away. Of course it will not always come out exact, as in this case (on account of that square root), but you can get as near as you like with decimals.

202.--THE CONE PUZZLE.

The simple rule is that the cone must be cut at one-third of its altitude.

203.--CONCERNING WHEELS.

If you mark a point A on the circumference of a wheel that runs on the surface of a level road, like an ordinary cart-wheel, the curve described by that point will be a common cycloid, as in Fig. 1. But if you mark a point B on the circumference of the flange of a locomotive-wheel, the curve will be a curtate cycloid, as in Fig. 2, terminating in nodes. Now, if we consider one of these nodes or loops, we shall see that "at any given moment" certain points at the bottom of the loop must be moving in the opposite direction to the train. As there is an infinite number of such points on the flange's circumference, there must be an infinite number of these loops being described while the train is in motion. In fact, at any given moment certain points on the flanges are always moving in a direction opposite to that in which the train is going.

In the case of the two wheels, the wheel that runs round the stationary one makes two revolutions round its own centre. As both wheels are of the same size, it is obvious that if at the start we mark a point on the circumference of the upper wheel, at the very top, this point will be in contact with the lower wheel at its lowest part when half the journey has been made. Therefore this point is again at the top of the moving wheel, and one revolution has been made. Consequently there are two such revolutions in the complete journey.

204.--A NEW MATCH PUZZLE.

1. The easiest way is to arrange the eighteen matches as in Diagrams 1 and 2, making the length of the perpendicular AB equal to a match and a half. Then, if the matches are an inch in length, Fig. 1 contains two square inches and Fig. 2 contains six square inches--4 x 11/2. The second case (2) is a little more difficult to solve. The solution is given in Figs. 3 and 4. For the purpose of construction, place matches temporarily on the dotted lines. Then it will be seen that as 3 contains five equal equilateral triangles and 4 contains fifteen similar triangles, one figure is three times as large as the other, and exactly eighteen matches are used.

205.--THE SIX SHEEP-PENS.

206.--THE KING AND THE CASTLES.

There are various ways of building the ten castles so that they shall form five rows with four castles in every row, but the arrangement in the next column is the only one that also provides that two castles (the greatest number possible) shall not be approachable from the outside. It will be seen that you must cross the walls to reach these two.

207.--CHERRIES AND PLUMS.

There are several ways in which this problem might be solved were it not for the condition that as few cherries and plums as possible shall be planted on the north and east sides of the orchard. The best possible arrangement is that shown in the diagram, where the cherries, plums, and apples are indicated respectively by the letters C, P, and A. The dotted lines connect the cherries, and the other lines the plums. It will be seen that the ten cherry trees and the ten plum trees are so planted that each fruit forms five lines with four trees of its kind in line. This is the only arrangement that allows of so few as two cherries or plums being planted on the north and east outside rows.

208.--A PLANTATION PUZZLE.

The illustration shows the ten trees that must be left to form five rows with four trees in every row. The dots represent the positions of the trees that have been cut down.

209.--THE TWENTY-ONE TREES.

I give two pleasing arrangements of the trees. In each case there are twelve straight rows with five trees in every row.

210.--THE TEN COINS.

The answer is that there are just 2,400 different ways. Any three coins may be taken from one side to combine with one coin taken from the other side. I give four examples on this and the next page. We may thus select three from the top in ten ways and one from the bottom in five ways, making fifty. But we may also select three from the bottom and one from the top in fifty ways. We may thus select the four coins in one hundred ways, and the four removed may be arranged by permutation in twenty-four ways. Thus there are 24 x 100 = 2,400 different solutions.