Page:Amusements in mathematics.djvu/195

Rh by the other. That is, if the two heights are a and b respectively, then $$\frac{ab}{a+b}$$ will give the height of the intersection. In the particular case of our puzzle, the intersection was therefore 2 ft. 11 in. from the ground. The distance that the poles are apart does not affect the answer. The reader who may have imagined that this was an accidental omission will perhaps be interested in discovering the reason why the distance between the poles may be ignored.

187.—THE MILKMAID PUZZLE.

a straight line, as shown in the diagram, from the milking-stool perpendicular to the near bank of the river, and continue it to the point A, which is the same distance from that bank as the stool. If you now draw the straight line from A to the door of the dairy, it will cut the river at B. Then the shortest route will be from the stool to B and thence to the door. Obviously the shortest distance from A to the door is the straight line, and as the distance from the stool to any point of the river is the same as from A to that point, the correctness of the solution will probably appeal to every reader without any acquaintance with geometry.

188.—THE BALL PROBLEM.

a round ball is placed on the level ground, six similar balls may be placed round it (all on the ground), so that they shall all touch the central ball.

As for the second question, the ratio of the diameter of a circle to its circumference we call pi; and though we cannot express this ratio in exact numbers, we can get sufficiently near to it for all practical purposes. However, in this case it is not necessary to know the value of pi at all. Because, to find the area of the surface of a sphere we multiply the square of the diameter by pi; to find the volume of a sphere we multiply the cube of the diameter by onesixth of pi. Therefore we may ignore pi, and have merely to seek a number whose square shall equal one-sixth of its cube. This number is obviously 6. Therefore the ball was 6 ft. in diameter, for the area of its surface will be 36 times pi in square feet, and its volume also 36 times pi in cubic feet.

189.—THE YORKSHIRE ESTATES.

triangular piece of land that was not for sale contains exactly eleven acres. Of course it is not difficult to find the answer if we follow the eccentric and tricky tracks of intricate trigonometry; or I might say that the application of a. well-known formula reduces the problem to finding one-quarter of the square root of (4×370×116)-(370+116—74)$2$—that is a quarter of the square root of 1936, which is onequarter of 44, or II acres. But all that the reader really requires to know is the Pythagorean law on which many puzzles have been built, that in any right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. I shall dispense with all "surds" and similar absurdities, notwithstanding the fact that the sides of our triangle are clearly incommensurate, since we cannot exactly extract the square roots of the three square areas.

In the above diagram ABC represents our triangle. A D B is a right-angled triangle, A D measuring 9 and B D measuring 17, because the square of 9 added to the square of 17 equals 370, the known area of the square on A B. Also A E C is a right-angled triangle, and the square of 5 added to the square of 7 equals 74, the square estate on A C. Similarly, C F B is a right-angled triangle, for the square of 4 added to the square of 10 equals 116, the square estate on B C. Now, although the sides of our triangular estate are incommensurate, we have in this diagram all the exact figures that we need to discover the area with precision.

The area of our triangle A D B is clearly half of 9×17, or 76½ acres. The area of A E C is half of 5×7, or 17½ acres; the area of C F B is half of 4 X 10, or 20 acres; and the area of the oblong E D F C is obviously 4×7, or 28 acres. Now, if we add together 17½, 20, and