by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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Make a fold in the paper, as shown by the dotted line in the illustration. Then, taking any two points, as A and B, describe semicircles on the line alternately from the centres B and A, being careful to make the ends join, and the thing is done. Of course this is not a true spiral, but the puzzle was to produce the particular spiral that was shown, and that was drawn in this simple manner.
184.?HOW TO DRAW AN OVAL.?solution
If you place your sheet of paper round the surface of a cylindrical bottle or canister, the oval can be drawn with one sweep of the compasses.
185.?ST. GEORGE'S BANNER.? solution
As the flag measures 4 ft. by 3 ft., the length of the diagonal (from corner to corner) is 5 ft. All you need do is to deduct half the length of this diagonal (2Y 2 ft.) from a quarter of the distance all round the edge of the flag (3/4 ft.)?a quarter of 14 ft. The difference (1 ft.) is the required width of the arm of the red cross. The area of the cross will then be the same as that of the white ground.
186.?THE CLOTHES LINE PUZZLE.? solution
Multiply together, and also add together, the heights of the two poles and divide one result by the other. That p 9 183 is, if the two heights are a and b respectively, then ab /( a + ^ 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.? solution
RIVER
Door
S TOOL
Draw 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.? solution
If 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 ofp/atall. Because, to find the area of the surface of a sphere we multiply the square of the diameter byp/; to find the volume of a sphere we multiply the cube of the diameter by one-sixth 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.? solution
The 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 x 370 x 116) - (370 + 116 - 74) 2 ?that is a quarter of the square root of 1936, which is one-quarter of 44, or 11 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.
