your compasses at A and with the distance A D describe the arc cutting H B at E. Then place the point of your compasses at D and with the distance D E describe the arc cutting the circumference at F. Now, D F is one of the sides of your pentagon, and you have simply to mark off the other sides round the circle. Quite simple when you know how, but otherwise somewhat of a poser.
Having formed your pentagon, the puzzle is to cut it into the fewest (3/4)
round the circumference. But a pentagon is quite another matter. So, as my puzzle has to do with the cutting up of a regular pentagon, it will perhaps be well if I first show my less experienced readers how this figure is to be correctly drawn. Describe a circle and draw the two lines H B and D G, in the diagram, through the centre at right angles. Now find the point A, midway between C and B. Next place the point of (2/4)
155. THE PENTAGON AND SQUARE. I wonder how many of my readers, amongst those who have not given any close attention to the elements of geometry, could draw a regular pentagon, or five-sided figure, if they suddenly required to do so. A regular hexagon, or six-sided figure, is easy enough, for everybody knows that all you have to do is to describe a circle and then, taking the radius as the length of one of the sides, mark off the six points (1/4)
SOLUTION TO 347. COUNTING THE RECTANGLES. (1/2)
There are 1,296 different rectangles in all, 204 of which are squares,
counting the square board itself as one, and 1,092 rectangles that are not squares. The general formula is that a board of n² squares contains ((n² + n)²)/4 rectangles, of which (2n³ + 3n² + n)/6 are squares and (3n^4 + 2n³ - 3n² - 2n)/12 are rectangles that are not squares. It is curious and interesting that the total number of
SOLUTION TO 156. THE DISSECTED TRIANGLE. (2/2)
Now, if we want two triangles, 1 will be one of them, and 2, 3, 4, and 5 will fit together, as in B, to form the other. If we want three equilateral triangles, 1 will be one, 4 and 5 will form the second, as in C, and 2 and 3 will form the third, as in D. In B and C the piece 5 is turned over; but there can be no objection to this, as it is not forbidden, and is in no way opposed to the nature of the puzzle.
SOLUTION TO 156. THE DISSECTED TRIANGLE. (1/2)
Diagram A is our original triangle. We will say it measures 5 inches (or 5 feet) on each side. If we take off a slice at the bottom of any equilateral triangle by a cut parallel with the base, the portion that remains will always be an equilateral triangle; so we first cut off piece 1 and get a triangle 3 inches on every side. The manner of finding directions of the other cuts in A is obvious from the diagram.
156. THE DISSECTED TRIANGLE. A good puzzle is that which the gentleman in the illustration is showing to his friends. He has simply cut out of paper an equilateral triangle--that is, a triangle with all its three sides of the same length. He proposes that it shall be cut into five pieces in such a way that they will fit together and form either two or three smaller equilateral triangles, using all the material in each case. Can you discover how the cuts should be made?
321. THE ROOK'S JOURNEY. This puzzle I call "The Rook's Journey," because the word "tour"
(derived from a turner's wheel) implies that we return to the point from which we set out, and we do not do this in the present case. We should not be satisfied with a personally conducted holiday tour that ended by leaving us, say, in the middle of the Sahara. The rook here makes twenty-one moves, in the course of which journey it visits every square (1/2)
SOLUTION TO 403. THE SPANISH DUNGEON. (6/6)
extended calling out the moves. He and No. 10 did most of the work, each changing his cell five times. No. 12, the man with the crooked leg, was lame, and therefore fortunately had only to pass from his cell into the next one when his time came round.
SOLUTION TO 403. THE SPANISH DUNGEON. (5/6)
cells. This position may be reached in as few as thirty-seven moves. Here are the moves: 15, 14, 10, 6, 7, 3, 2, 7, 6, 11, 3, 2, 7, 6, 11,
10, 14, 3, 2, 11, 10, 9, 5, 1, 6, 10, 9, 5, 1, 6, 10, 9, 5, 2, 12, 15,
3. This short solution will probably surprise many readers who may not find a way under from sixty to a hundred moves. The clever prisoner was No. 6, who in the original illustration will be seen with his arms
SOLUTION TO 403. THE SPANISH DUNGEON. (4/6)
which position can be reached in the fewest moves. I am afraid, however,
it is only after considerable study and experience that the solver is able to get such a grasp of the various "areas of disturbance" and methods of circulation that his judgment is of much value to him.
The second diagram is a most favourable magic square position. It will be seen that prisoners 4, 8, 13, and 14 are left in their original
SOLUTION TO 403. THE SPANISH DUNGEON. (3/6)
This can best be solved by working backwards--that is to say, you must first catch your square, and then work back to the original position. We must first construct those squares which are found to require the least amount of readjustment of the numbers. Many of these we know cannot possibly be reached. When we have before us the most favourable possible arrangements, it then becomes a question of careful analysis to discover
SOLUTION TO 403. THE SPANISH DUNGEON. (1/6)
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| 1 | 2 | 3 | 4 | | 10 | 9 | 7 | 4 |
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| 5 | 6 | 7 | 8 | | 6 | 5 | 11 | 8 |
Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"
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