Chapter 14

Triangles

In this nest of triangles there are no less than six hundred and fifty-three distinct triangles of various shapes and sizes.

In a chamber of the Great Pyramid an ancient Egyptian jar was found, marked with the device now known as Pharaoh’s seal.

Vase

Can you count the number of triangles or pyramids, of many sizes, but all of similar shape that are expressed on it? Solvers should draw the figure on a larger scale.

Solution

It is interesting to note that the repeated addition of odd numbers to one another can be so arranged as to produce cube numbers in due sequence.Thus:—

and so on, to any extent.

In a large old-fashioned garden walks were arranged round a central fountain in the shape of a Maltese cross.

Garden

If four persons started at noon from the fountain, walking round the four paths at two, three, four and five miles an hour respectively, at what time would they meet for the third time at their starting-point, if the distance on each track was one-third of a mile?

Solution

When the tens of two numbers are the same, and their units added together make ten, multiply the units together, increase one of the tens by unity, and multiply it by the other ten. The result is the product of the two original numbers, if the first result follows the other.Thus:—

43 × 47 = 2021.

Can you cut Fig. A into two parts, and so rearrange these that they form either Fig. B or Fig. C?

Puzzle

The two parts of A must not beturned roundto form B or C, but must retain their original direction.

Solution

Coal may fail us, but we can never run short of material for “words that burn.” It has been calculated that if a man could read 100,000 words in an hour, and there were 4,650,000 men available, they could not pronounce the possible variations which could be formed from the alphabet in 70,000 years!

It is possible, in a sense, by the following neat method, to take 45 from 45, and find that 45remains:—

Cut out in stiff cardboard four pieces shaped as Fig. 1, four as Fig. 2, and four as Fig. 3, taking care that they are all exactly true to pattern in shape and proportion to one another.

Pieces

Now see whether you can put the twelve pieces together so as to form a perfect octagon.

Solution

Here is a proof that 7, if it cannot rival the mystic 9, has quaint properties of itsown:—

Here is an arithmeticalcuriosity:—

A farmer’s wife kept a pure strain of Aylesbury ducks for market on a square pond, with a duck-house at each corner. As trade grew brisk she found that she must enlarge her pond. An ingenious neighbour undertook to arrange this without altering the shape of the pond, and without disturbing the duck-houses. What was his plan?

Duck pond

Solution

It would seem impossible to subtract 69 from 55, but it can be arranged thus, with six as aremainder:—

Cut out in cardboard twenty triangular pieces exactly the size and shape of this one, and try to place them together so that they form a perfect square.

Triangle

Solution

The decimal equivalent of1⁄13is .076923. This (omitting the point), multiplied by 1, 3, 4, 9, 10, or 12, yields results in which the same figures appear in varied order, but similar sequence, and multiplied by 2, 5, 6, 7, 8, or 11, it yields a different series, with similar characteristics.Thus:—

A kaleidoscope cylinder contains twenty small pieces of coloured glass. As we turn it round, or shake it, so as to make ten changes of pattern every minute, it will take the inconceivable space of time of 462,880,899,576 years and 360 days to exhaust all the possible symmetrical variations. (The 360 days is good!)

Here is an amusing little exercise for the ingenuity of our solvers.

Grid

Take six sharp pins, and puzzle out how to stick them into six of the black dots, so that no two pins, are on the same line, in any direction, vertical, horizontal, or diagonal.

Solution

The middle of a large playground was paved with sixty-four square flagstones of equal size, which are numbered on this diagram from one to sixty-four.

One of the schoolmasters, who had a head for puzzles, took his stand upon the square here numbered 19, and offered a prize to any boy who, starting from the square numbered 46, could make his way to him, passing through every square once, and only once. It was after many vain attempts that the course was at last discovered. Can you work it out?

Solution

Place twelve draughtsmen, or buttons, in a square, so that you count four along each side of it,thus:—

Buttons

Now take the same men or buttons, and arrange them so that they form another square, and you can count five along each side of it.

Solution

Here is a good specimen of the eccentricities and powers ofnumbers:—

She was quite an old maid, and her age was a most absolute secret. Determined to discover it, her scapegrace nephew, on Christmas Eve, produced these tables, and asked her with well simulated innocence on which of them she could see the number of her age.

From her answer he was able to calculate that the old lady was fifty-five.

The tell-tale tables disclosed her age thus:—As it appeared in tables A, B, C, E, and F, he added together the numbers at the top left-hand corners, and found the total to be fifty-five. This rule applies in all cases.

Of the many paper-cutting tricks which appeal to us none is more simple and attractive thanthis:—

Altar

Take a piece of paper, say 5 inches by 3 inches, but any oblong shape and size will do, and after folding it four times cut it lengthways up the centre. Unfold the pieces, and to your surprise you will find a perfect cross and other pieces in pairs of the shapes shown above. The puzzle is how to fold the paper.

Rectangle

The paper must be folded first so that B comes upon C, then so that A comes upon D, then from D to C, and lastly from E to C. If it is now cut lengthways exactly along the centre the figures shown on the original diagram will be formed, which resemble a cross and lighted candles on an altar.

Take a thin board, about eight inches square, and mark it out into thirty-six equal parts; bore a hole in the centre of each part, and then fit in a small wooden peg, leaving about a quarter inch above the surface, as is shown in Fig. 1, the section below the diagram.

Puzzle

Prepare thirty-six pieces of white or coloured cardboard of the length A to B, and place them over the pegs in any direction in which they will fit so as to form some such symmetrical pattern as is given on the second diagram, putting two holes only on each peg. Chess-players will see that this is the regular knight’s move.

Puzzle

Quite a number of beautifuldesigns can be thus formed, and those who have not the means at hand for making a complete set can enjoy the puzzle by merely marking out thirty-six squares, and drawing lines from centre to centre of the exact length from A to B, with black or coloured pencils.

Mitre

Divide this figure into four similar and equal parts.

Solution

The solution of the pretty little problem: place three twos in three different groups, so that twice the first group, or half the third group equals the second group, isthis:—

22 + 2=122 -22= 12 + 22= 2

The following figure, which represents part of a brick wall, cannot be marked out along all the edges of the bricks in less than six continuous lines without going more than once over the sameline:—

Brick wall

Here, in strong contrast to the simple figure given above, which could not be traced without lifting the pen six times from the paper, is an intricate design, the lines of which, on the upper or on the lower half, can be traced without any break at all.

Brick wall

The general rule that governs such cases is, that where an uneven number of lines meet a fresh start has to be made. In the diagram now given the only such points are at the extremities of the upper and lower halves of the figure at A and X. At all other points two, or four, or six lines converge, and there is no break of continuity in a tracing of the figure.

Can you suggest quite a simple and practical way to fix the points on the sides of a square which will be at the angles of an octagon formed by cutting off equal corners of the square, as shown below?

Square

Solution

Very interesting and curious are the properties of the figures 142857, used in varied order but always in similar sequence, in connection with 7 and9:—

The subjoined diagram shows how a square with sides that measure each 12 yards can be divided into five triangles, no two of which are of equal area, and of which the sides and areas can be expressed in yards by wholenumbers:—

Triangles

The areas of these triangles are 6, 12, 24, 48, and 54 square yards respectively, and the sum of these, 144 square yards, is the area of the square.

Our readers may remember the remarkable fact that the figures of the sum, £12, 12s. 8d., when written thus, 12,128, exactly represent the number of farthings it contains. Now this, so far as we know, is the only instance of the peculiarity, but there are at least five other cases which come curiously near to it. They arethese:—

If a ladder, with rungs 1 foot apart, rests against a wall, and its thirteenth rung is 12 feet above the ground, the foot of the ladder is 25 feet from the wall.

Ladder

Proof.—Drop a perpendicular from A to B. Then, as A B C is a right angle, and the squares on A C, A B, are 169 feet and 144 feet, the square on C B must be 25 feet, and the length of C B is 5 feet. We thus move 5 feet towards the wall in going 13 feet up the ladder, and in mounting 65 feet (five times as far) we must cover 25 feet.

A prettily ingenious method of dividing the area of a circle into quarters, each of them a perfect curve, with perimeter (or enclosing line) equal to the circumference of the circle, and with which four circles can be formed, is clearly shown by the subjoineddiagrams:—

Circles

The host of a large hotel at Cairo noticed that his Visitors’ Book contained the names of an Austrian, a Brazilian, a Chinaman, a Dane, an Englishman, a Frenchman, a German, and a Hungarian. Moved by this curious alphabetical list, he offered them all free quarters and the best of everything if they could arrange themselves at a round dining-table so that not one of them should have the same two neighbours on any two occasions for 21 successive days.

The following is one of many ways in which this arrangement can be made, and it seems to be the simplest of them all.

Number the persons 1 to 8; and for our first day set them down in numerical orderexcept that the two centre ones (4 and 5) change places:

Keep the 1 and the 7 unaltered but double each of the other numbers. When the product is greater than 8, divide by 7, and only set down the remainder. Thus we get:

(Here the fourth figure 3 is 5 × 2 ÷ 7, givingremainder3, and so on.)

Repeat this operation once more:

To fill in the intermediate days we have only to keep 1 unchanged and let the remaining numbers run downwards insimple numerical order, following 8 with 2, 2 with 3, and so on.Thus:—

This completes the schedule. It will be found on examination that every number is between every pair of the other numbers once, and once only.

In order to reduce our first-day ring to exact numerical order we have only to interchange the numbers 4 and 5 throughout. The first three lines for example would then become:

or, by putting letters for figures,

An arrangement of the guests is thus arrived at for twenty-one successive days, so that not one of them has the same two neighbours on any two occasions.

Can you apply the two oblongs drawn below to the two concentric squares, so as to produce thirty-one perfect squares?

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