The Canterbury Puzzles Part 27

66.--_The Buried Treasure._

The field must have contained between 179 and 180 acres--to be more exact, 179.37254 acres. Had the measurements been 3, 2, and 4 furlongs respectively from successive corners, then the field would have been 209.70537 acres in area.

One method of solving this problem is as follows. Find the area of triangle APB in terms of _x_, the side of the square. Double the result=_xy_. Divide by _x_ and then square, and we have the value of _y_^{2} in terms of _x_. Similarly find value of _z_^{2} in terms of _x_; then solve the equation _y_^{2}+_z_^{2}=3^{2}, which will come out in the form _x_^{4}-20_x_^{2}=-37. Therefore _x_^{2}=10+(sqrt{63})=17.937254 square furlongs, very nearly, and as there are ten acres in one square furlong, this equals 179.37254 acres. If we take the negative root of the equation, we get the area of the field as 20.62746 acres, in which case the treasure would have been buried outside the field, as in Diagram 2.

But this solution is excluded by the condition that the treasure was buried in the field. The words were, "The doc.u.ment ... states clearly that the field is square, and that the treasure is buried in it."

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THE PROFESSOR'S PUZZLES

67.--_The Coinage Puzzle._

The point of this puzzle turns on the fact that if the magic square were to be composed of whole numbers adding up 15 in all ways, the two must be placed in one of the corners. Otherwise fractions must be used, and these are supplied in the puzzle by the employment of sixpences and half-crowns. I give the arrangement requiring the fewest possible current English coins--fifteen. It will be seen that the amount in each corner is a fractional one, the sum required in the total being a whole number of shillings.

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68.--_The Postage Stamps Puzzles._

The first of these puzzles is based on a similar principle, though it is really much easier, because the condition that nine of the stamps must be of different values makes their selection a simple matter, though how they are to be placed requires a little thought or trial until one knows the rule respecting putting the fractions in the corners. I give the solution.

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[1/2 d]

[4-1/2 d] [1 d] [3 d]

[2 d] [3 d] [4 d]

[2-1/2 d] [5 d] [1-1/2 d] ]

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[4 d] [1/2 d]

[3 d] [1-1/2 d]

[9 d]

[10 d] [6 d] [2 d]

[1 d] [1 s.] [5 d] ]

I also show the solution to the second stamp puzzle. All the columns, rows, and diagonals add up 1_s._ 6_d._ There is no stamp on one square, and the conditions did not forbid this omission. The stamps at present in circulation are these:--1/2_d._, 1_d._, 1-1/2_d._, 2_d._, 2-1/2_d._, 3_d._, 4_d._, 5_d._, 6_d._, 9_d._, 10_d._, 1_s._, 2_s._ 6_d._, 5_s._, 10_s._, 1, and 5. In the first solution the numbers are in arithmetical progression--1, 1-1/2, 2, 2-1/2, 3, 3-1/2, 4, 4-1/2, 5. But any nine numbers will form a magic square if we can write them thus:--

1 2 3 7 8 9 13 14 15

where the horizontal differences are all alike and the vertical differences all alike, but not necessarily the same as the horizontal.

This happens in the case of the second solution, the numbers of which may be written:--

0 1 2 5 6 7 10 11 12

Also in the case of the solution to No. 67, the Coinage Puzzle, the numbers are, in shillings:--

2 2-1/2 3 4-1/2 5 5-1/2 7 7-1/2 8

If there are to be nine _different_ numbers, 0 may occur once (as in the solution to No. 22). Yet one might construct squares with negative numbers, as follows:--

-2 -1 0 5 6 7 12 13 14

69.--_The Frogs and Tumblers._

It is perfectly true, as the Professor said, that there is only one solution (not counting a reversal) to this puzzle. The frogs that jump are George in the third horizontal row; Chang, the artful-looking batrachian at the end of the fourth row; and Wilhelmina, the fair creature in the seventh row. George jumps downwards to the second tumbler in the seventh row; Chang, who can only leap short distances in consequence of chronic rheumatism, removes somewhat unwillingly to the gla.s.s just above him--the eighth in the third row; while Wilhelmina, with all the sprightliness of her youth and s.e.x, performs the very creditable saltatory feat of leaping to the fourth tumbler in the fourth row. In their new positions, as shown in the accompanying diagram, it will be found that of the eight frogs no two are in line vertically, horizontally, or diagonally.

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70.--_Romeo and Juliet._

This is rather a difficult puzzle, though, as the Professor remarked when Hawkhurst hit on the solution, it is "just one of those puzzles that a person might solve at a glance" by pure luck. Yet when the solution, with its pretty, symmetrical arrangement, is seen, it looks ridiculously simple.

It will be found that Romeo reaches Juliet's balcony after visiting every house once and only once, and making fourteen turnings, not counting the turn he makes at starting. These are the fewest turnings possible, and the problem can only be solved by the route shown or its reversal.

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71.--_Romeo's Second Journey._

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In order to take his trip through all the white squares only with the fewest possible turnings, Romeo would do well to adopt the route I have shown, by means of which only sixteen turnings are required to perform the feat. The Professor informs me that the Helix Aspersa, or common or garden snail, has a peculiar aversion to making turnings--so much so that one specimen with which he made experiments went off in a straight line one night and has never come back since.

72.--_The Frogs who would a-wooing go._