Amusements in Mathematics Part 23

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Another correspondent made the side of his square 1 of the side of the pentagon. As a matter of fact, the ratio is irrational. I calculate that if the side of the pentagon is 1--inch, foot, or anything else--the side of the square of equal area is 1.3117 nearly, or say roughly 1+3/10. So we can only hope to solve the puzzle by geometrical methods.

156.--THE DISSECTED TRIANGLE.

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.

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.

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157.--THE TABLE-TOP AND STOOLS.

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One object that I had in view when presenting this little puzzle was to point out the uncertainty of the meaning conveyed by the word "oval." Though originally derived from the Latin word ovum, an egg, yet what we understand as the egg-shape (with one end smaller than the other) is only one of many forms of the oval; while some eggs are spherical in shape, and a sphere or circle is most certainly not an oval. If we speak of an ellipse--a conical ellipse--we are on safer ground, but here we must be careful of error. I recollect a Liverpool town councillor, many years ago, whose ignorance of the poultry-yard led him to subst.i.tute the word "hen" for "fowl," remarking, "We must remember, gentlemen, that although every c.o.c.k is a hen, every hen is not a c.o.c.k!" Similarly, we must always note that although every ellipse is an oval, every oval is not an ellipse. It is correct to say that an oval is an oblong curvilinear figure, having two unequal diameters, and bounded by a curve line returning into itself; and this includes the ellipse, but all other figures which in any way approach towards the form of an oval without necessarily having the properties above described are included in the term "oval." Thus the following solution that I give to our puzzle involves the pointed "oval," known among architects as the "vesica piscis."

[Ill.u.s.tration: THE TWO STOOLS.]

The dotted lines in the table are given for greater clearness, the cuts being made along the other lines. It will be seen that the eight pieces form two stools of exactly the same size and shape with similar hand-holes. These holes are a trifle longer than those in the schoolmaster's stools, but they are much narrower and of considerably smaller area. Of course 5 and 6 can be cut out in one piece--also 7 and 8--making only six pieces in all. But I wished to keep the same number as in the original story.

When I first gave the above puzzle in a London newspaper, in compet.i.tion, no correct solution was received, but an ingenious and neatly executed attempt by a man lying in a London infirmary was accompanied by the following note: "Having no compa.s.ses here, I was compelled to improvise a pair with the aid of a small penknife, a bit of firewood from a bundle, a piece of tin from a toy engine, a tin tack, and two portions of a hairpin, for points. They are a fairly serviceable pair of compa.s.ses, and I shall keep them as a memento of your puzzle."

158.--THE GREAT MONAD.

The areas of circles are to each other as the squares of their diameters. If you have a circle 2 in. in diameter and another 4 in. in diameter, then one circle will be four times as great in area as the other, because the square of 4 is four times as great as the square of 2. Now, if we refer to Diagram 1, we see how two equal squares may be cut into four pieces that will form one larger square; from which it is self-evident that any square has just half the area of the square of its diagonal. In Diagram 2 I have introduced a square as it often occurs in ancient drawings of the Monad; which was my reason for believing that the symbol had mathematical meanings, since it will be found to demonstrate the fact that the area of the outer ring or annulus is exactly equal to the area of the inner circle. Compare Diagram 2 with Diagram 1, and you will see that as the square of the diameter CD is double the square of the diameter of the inner circle, or CE, therefore the area of the larger circle is double the area of the smaller one, and consequently the area of the annulus is exactly equal to that of the inner circle. This answers our first question.

[Ill.u.s.tration: 1 2 3 4]

In Diagram 3 I show the simple solution to the second question. It is obviously correct, and may be proved by the cutting and superposition of parts. The dotted lines will also serve to make it evident. The third question is solved by the cut CD in Diagram 2, but it remains to be proved that the piece F is really one-half of the Yin or the Yan. This we will do in Diagram 4. The circle K has one-quarter the area of the circle containing Yin and Yan, because its diameter is just one-half the length. Also L in Diagram 3 is, we know, one-quarter the area. It is therefore evident that G is exactly equal to H, and therefore half G is equal to half H. So that what F loses from L it gains from K, and F must be half of Yin or Yan.

159.--THE SQUARE OF VENEER.

[Ill.u.s.tration: +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | | | | :| | |: | | :| | |: | || | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | | | | :| | |: | | :| | |: | || | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | |_ _|___|__:|___|___|:__|___|__:|___|___|:__|__||___| +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | | | | :| | |: | | :| | |: | || | +---+---+---+---+---+---+---+---+---+---+===+===+---+ | | | :| | |: | | :| | ||: | | | | | | :| | |: | | :| | ||: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ |||:|||:||:||||:||| | | | :| | |: | | :| | ||: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | ||: | | | | | | :| | |: | B | :| | ||: | C | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | ||: | | | |_ _|___|__:|___|___|:__|___|__:|___|__||:__|___|___| +---+---+---+---+---+---+---+---+===+===+===+===+===+ | | | :| | |: | | :|| | |: | | | | | | :| | |: | | :|| | |: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :|| | |: | | | | | | :| | |: | | :|| | |: | A | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ |||:|||:||:||||:||| | | | :| | |: | | :|| | |: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :|| | |: | | | | | | :| | |: | | :|| | |: | | | +===+===+===+===+===+===+===+===+---+---+---+---+---+ | | | :| | |: | | :|| | |: | | | | | | :| | D |: | | :|| | |: | | | +---+---+---+---+---+---+---+---+===+===+===+===+===+ ]

Any square number may be expressed as the sum of two squares in an infinite number of different ways. The solution of the present puzzle forms a simple demonstration of this rule. It is a condition that we give actual dimensions.

In this puzzle I ignore the known dimensions of our square and work on the a.s.sumption that it is 13n by 13n. The value of n we can afterwards determine. Divide the square as shown (where the dotted lines indicate the original markings) into 169 squares. As 169 is the sum of the two squares 144 and 25, we will proceed to divide the veneer into two squares, measuring respectively 12 12 and 5 5; and as we know that two squares may be formed from one square by dissection in four pieces, we seek a solution in this number. The dark lines in the diagram show where the cuts are to be made. The square 5 5 is cut out whole, and the larger square is formed from the remaining three pieces, B, C, and D, which the reader can easily fit together.

Now, n is clearly 5/13 of an inch. Consequently our larger square must be 60/13 in. 60/13 in., and our smaller square 25/13 in. 25/13 in. The square of 60/13 added to the square of 25/13 is 25. The square is thus divided into as few as four pieces that form two squares of known dimensions, and all the sixteen nails are avoided.

Here is a general formula for finding two squares whose sum shall equal a given square, say a. In the case of the solution of our puzzle p = 3, q = 2, and a = 5.

________________________ 2pqa / a( p + q) - (2pqa) --------- = x; --------------------------- = y p + q p + q Here x + y = a.

160.--THE TWO HORSESHOES.

The puzzle was to cut the two shoes (including the hoof contained within the outlines) into four pieces, two pieces each, that would fit together and form a perfect circle. It was also stipulated that all four pieces should be different in shape. As a matter of fact, it is a puzzle based on the principle contained in that curious Chinese symbol the Monad. (See No. 158.) [Ill.u.s.tration]

The above diagrams give the correct solution to the problem. It will be noticed that 1 and 2 are cut into the required four pieces, all different in shape, that fit together and form the perfect circle shown in Diagram 3. It will further be observed that the two pieces A and B of one shoe and the two pieces C and D of the other form two exactly similar halves of the circle--the Yin and the Yan of the great Monad. It will be seen that the shape of the horseshoe is more easily determined from the circle than the dimensions of the circle from the horseshoe, though the latter presents no difficulty when you know that the curve of the long side of the shoe is part of the circ.u.mference of your circle. The difference between B and D is instructive, and the idea is useful in all such cases where it is a condition that the pieces must be different in shape. In forming D we simply add on a symmetrical piece, a curvilinear square, to the piece B. Therefore, in giving either B or D a quarter turn before placing in the new position, a precisely similar effect must be produced.

161.--THE BETSY ROSS PUZZLE.

Fold the circular piece of paper in half along the dotted line shown in Fig. 1, and divide the upper half into five equal parts as indicated. Now fold the paper along the lines, and it will have the appearance shown in Fig. 2. If you want a star like Fig. 3, cut from A to B; if you wish one like Fig. 4, cut from A to C. Thus, the nearer you cut to the point at the bottom the longer will be the points of the star, and the farther off from the point that you cut the shorter will be the points of the star.

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162.--THE CARDBOARD CHAIN.

The reader will probably feel rewarded for any care and patience that he may bestow on cutting out the cardboard chain. We will suppose that he has a piece of cardboard measuring 8 in. by 2 in., though the dimensions are of no importance. Yet if you want a long chain you must, of course, take a long strip of cardboard. First rule pencil lines B B and C C, half an inch from the edges, and also the short perpendicular lines half an inch apart. (See next page.) Rule lines on the other side in just the same way, and in order that they shall coincide it is well to p.r.i.c.k through the card with a needle the points where the short lines end. Now take your penknife and split the card from A A down to B B, and from D D up to C C. Then cut right through the card along all the short perpendicular lines, and half through the card along the short portions of B B and C C that are not dotted. Next turn the card over and cut half through along the short lines on B B and C C at the places that are immediately beneath the dotted lines on the upper side. With a little careful separation of the parts with the penknife, the cardboard may now be divided into two interlacing ladder-like portions, as shown in Fig. 2; and if you cut away all the shaded parts you will get the chain, cut solidly out of the cardboard, without any join, as shown in the ill.u.s.trations on page 40.

It is an interesting variant of the puzzle to cut out two keys on a ring--in the same manner without join.

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164.--THE POTATO PUZZLE.

As many as twenty-two pieces may be obtained by the six cuts. The ill.u.s.tration shows a pretty symmetrical solution. The rule in such cases is that every cut shall intersect every other cut and no two intersections coincide; that is to say, every line pa.s.ses through every other line, but more than two lines do not cross at the same point anywhere. There are other ways of making the cuts, but this rule must always be observed if we are to get the full number of pieces.

The general formula is that with n cuts we can always produce (n(n + 1) + 1)/2 . One of the problems proposed by the late Sam Loyd was to produce the maximum number of pieces by n straight cuts through a solid cheese. Of course, again, the pieces cut off may not be moved or piled. Here we have to deal with the intersection of planes (instead of lines), and the general formula is that with n cuts we may produce ((n - 1)n(n + 1))/6 + n + 1 pieces. It is extremely difficult to "see" the direction and effects of the successive cuts for more than a few of the lowest values of n.

165.--THE SEVEN PIGS.

The ill.u.s.tration shows the direction for placing the three fences so as to enclose every pig in a separate sty. The greatest number of s.p.a.ces that can be enclosed with three straight lines in a square is seven, as shown in the last puzzle. Bearing this fact in mind, the puzzle must be solved by trial.

[Ill.u.s.tration: THE SEVEN PIGS.]

166.--THE LANDOWNER'S FENCES.

Four fences only are necessary, as follows:-- [Ill.u.s.tration]

167.--THE WIZARD'S CATS.

The ill.u.s.tration requires no explanation. It shows clearly how the three circles may be drawn so that every cat has a separate enclosure, and cannot approach another cat without crossing a line.

[Ill.u.s.tration: THE WIZARDS' CATS.]

168.--THE CHRISTMAS PUDDING.

The ill.u.s.tration shows how the pudding may be cut into two parts of exactly the same size and shape. The lines must necessarily pa.s.s through the points A, B, C, D, and E. But, subject to this condition, they may be varied in an infinite number of ways. For example, at a point midway between A and the edge, the line may be completed in an unlimited number of ways (straight or crooked), provided it be exactly reflected from E to the opposite edge. And similar variations may be introduced at other places.

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169.--A TANGRAM PARADOX.

The diagrams will show how the figures are constructed--each with the seven Tangrams. It will be noticed that in both cases the head, hat, and arm are precisely alike, and the width at the base of the body the same. But this body contains four pieces in the first case, and in the second design only three. The first is larger than the second by exactly that narrow strip indicated by the dotted line between A and B. This strip is therefore exactly equal in area to the piece forming the foot in the other design, though when thus distributed along the side of the body the increased dimension is not easily apparent to the eye.

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170.--THE CUs.h.i.+ON COVERS.

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The two pieces of brocade marked A will fit together and form one perfect square cus.h.i.+on top, and the two pieces marked B will form the other.

171.--THE BANNER PUZZLE.

The ill.u.s.tration explains itself. Divide the bunting into 25 squares (because this number is the sum of two other squares--16 and 9), and then cut along the thick lines. The two pieces marked A form one square, and the two pieces marked B form the other.

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172.--MRS. SMILEY'S CHRISTMAS PRESENT.

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The first step is to find six different square numbers that sum to 196. For example, 1 + 4 + 25 + 36 + 49 + 81 = 196; 1 + 4 + 9 + 25 + 36 + 121 = 196; 1 + 9 + 16 + 25 + 64 + 81 = 196. The rest calls for individual judgment and ingenuity, and no definite rules can be given for procedure. The annexed diagrams will show solutions for the first two cases stated. Of course the three pieces marked A and those marked B will fit together and form a square in each case. The a.s.sembling of the parts may be slightly varied, and the reader may be interested in finding a solution for the third set of squares I have given.

173.--MRS. PERKINS'S QUILT.

The following diagram shows how the quilt should be constructed.

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There is, I believe, practically only one solution to this puzzle. The fewest separate squares must be eleven. The portions must be of the sizes given, the three largest pieces must be arranged as shown, and the remaining group of eight squares may be "reflected," but cannot be differently arranged.

174.--THE SQUARES OF BROCADE.

[Ill.u.s.tration: Diagram 1]

So far as I have been able to discover, there is only one possible solution to fulfil the conditions. The pieces fit together as in Diagram 1, Diagrams 2 and 3 showing how the two original squares are to be cut. It will be seen that the pieces A and C have each twenty chequers, and are therefore of equal area. Diagram 4 (built up with the dissected square No. 5) solves the puzzle, except for the small condition contained in the words, "I cut the two squares in the manner desired." In this case the smaller square is preserved intact. Still I give it as an ill.u.s.tration of a feature of the puzzle. It is impossible in a problem of this kind to give a _quarter-turn_ to any of the pieces if the pattern is to properly match, but (as in the case of F, in Diagram 4) we may give a symmetrical piece a _half-turn_--that is, turn it upside down. Whether or not a piece may be given a quarter-turn, a half-turn, or no turn at all in these chequered problems, depends on the character of the design, on the material employed, and also on the form of the piece itself.

[Ill.u.s.tration: Diagram 2]

[Ill.u.s.tration: Diagram 3]

[Ill.u.s.tration: Diagram 4]

[Ill.u.s.tration: Diagram 5]

175.--ANOTHER PATCHWORK PUZZLE.

The lady need only unpick the st.i.tches along the dark lines in the larger portion of patchwork, when the four pieces will fit together and form a square, as shown in our ill.u.s.tration.

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176.--LINOLEUM CUTTING.

There is only one solution that will enable us to retain the larger of the two pieces with as little as possible cut from it. Fig. 1 in the following diagram shows how the smaller piece is to be cut, and Fig. 2 how we should dissect the larger piece, while in Fig. 3 we have the new square 10 10 formed by the four pieces with all the chequers properly matched. It will be seen that the piece D contains fifty-two chequers, and this is the largest piece that it is possible to preserve under the conditions.

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177.--ANOTHER LINOLEUM PUZZLE.

Cut along the thick lines, and the four pieces will fit together and form a perfect square in the manner shown in the smaller diagram.

[Ill.u.s.tration: ANOTHER LINOLEUM PUZZLE.]

178.--THE CARDBOARD BOX.

The areas of the top and side multiplied together and divided by the area of the end give the square of the length. Similarly, the product of top and end divided by side gives the square of the breadth; and the product of side and end divided by the top gives the square of the depth. But we only need one of these operations. Let us take the first. Thus, 120 96 divided by 80 equals 144, the square of 12. Therefore the length is 12 inches, from which we can, of course, at once get the breadth and depth--10 in. and 8 in. respectively.

179.--STEALING THE BELL-ROPES.

Whenever we have one side (a) of a right-angled triangle, and know the difference between the second side and the hypotenuse (which difference we will call b), then the length of the hypotenuse will be a b --- + -. 2b 2 In the case of our puzzle this will be 48 48 ------- + 1 in. = 32 ft. 1 in., 6 which is the length of the rope.

180-- THE FOUR SONS.

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The diagram shows the most equitable division of the land possible, "so that each son shall receive land of exactly the same area and exactly similar in shape," and so that each shall have access to the well in the centre without trespa.s.s on another's land. The conditions do not require that each son's land shall be in one piece, but it is necessary that the two portions a.s.signed to an individual should be kept apart, or two adjoining portions might be held to be one piece, in which case the condition as to shape would have to be broken. At present there is only one shape for each piece of land--half a square divided diagonally. And A, B, C, and D can each reach their land from the outside, and have each equal access to the well in the centre.

181.--THE THREE RAILWAY STATIONS.

The three stations form a triangle, with sides 13, 14, and 15 miles. Make the 14 side the base; then the height of the triangle is 12 and the area 84. Multiply the three sides together and divide by four times the area. The result is eight miles and one-eighth, the distance required.

182.--THE GARDEN PUZZLE.

Half the sum of the four sides is 144. From this deduct in turn the four sides, and we get 64, 99, 44, and 81. Multiply these together, and we have as the result the square of 4,752. Therefore the garden contained 4,752 square yards. Of course the tree being equidistant from the four corners shows that the garden is a quadrilateral that may be inscribed in a circle.

183.--DRAWING A SPIRAL.

Make a fold in the paper, as shown by the dotted line in the ill.u.s.tration. 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.

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184.--HOW TO DRAW AN OVAL.

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 compa.s.ses.

185.--ST. GEORGE'S BANNER.

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 (2 ft.) from a quarter of the distance all round the edge of the flag (3 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.

Multiply together, and also add together, the heights of the two poles and divide one result by the other. That is, if the two heights are a and b respectively, then 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.