Dissection Puzzles Special Issue || Fraudulent Dissection Puzzles: A Tour of the Mathematics of...

Fraudulent Dissection Puzzles: A Tour of the Mathematics of BamboozlementAuthor(s): John SharpSource: Mathematics in School, Vol. 31, No. 4, Dissection Puzzles Special Issue (Sep., 2002), pp.7-13Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30212198 .

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Frandulent dissection pussles - a tour of the mathematics of bambooslement

by John Sharp

Bamboozlement and hornswoggling are two wonderful sounding words which might have come out of Harry Potter, but they are terms applied to a type of dissection puzzle which is presented in the form of a paradox. They have been given to this type of puzzle by Greg Frederickson in his gathering of dissection puzzles Dissections Plane and Fancy (Frederickson, 1997). Such puzzles have much to offer at all levels of mathematics with the added benefit of adding a bit of fun to mathematics. In this article, I want to take you on a tour of such puzzles, historically and mathematically, and to show that there are still uncharted waters waiting to be explored. The mathematics will take us through geometry, number sequences with suggestions for investigations and historically we will meet some mathematicians known and not so well known, as well as an architect or two.

Fig. 1

The inspiration for the tour comes from the well-known dissection puzzle shown in Figure 1. This puzzle does not seem to have a name. It is usually described under a heading of 'a geometrical paradox'. John Bradshaw (page 3) has written about how he has used it in the classroom. I want to delve deeper into its mathematics and suggest how it might have been created or discovered.

It offers potential for a variety of mathematics at many levels. It is normally given with a square of side 8 and a rectangle of sides 5 and 13. Anyone who has been exposed to the Fibonacci numbers will recognize them immediately and will probably know that for any two Fibonacci numbers, ifF, = 1 and F2 = 1, :

Fn2 = Fn_1 Fn+1 - (-1)n (1)

Since F6 = 8, then the area of the square is less than the rectangle by one unit.

Lewis Carroll's Notes on the Puzzle

The puzzle seems to have appeared in this form about the middle of the 19th century. I have a theory as to how it arose which I will describe in a moment. It was a favourite of Lewis Carroll and among the papers unpublished in his lifetime are notes on a generalization of the problem, where he was the first to ask what other values can the sides of the square and rectangle have. The notes are incomplete and vague, but a possible reconstruction of his argument was published by Warren Weaver (Weaver, 1938). Using Carroll's method for seeing how the puzzle works, apart from providing some useful ideas for investigations, gives some mathematical clues to how the puzzle might have arisen.

n-a a

n-a n n-a

n-a n

a n n-a

Fig. 2

Using the notation as in Figure 2, since the rectangle has an area one unit more than the square:

(2n - a)(n - a) - n2 = 1 (2)

giving

a2 -3na + n2 - 1 = 0 (3)

Note that this equation is symmetrical in n and a and if we solve for a we get

3n + 5n2 + 4 a

2

Now we know we want a to be an integer, so there are lots of questions we can ask for an investigation.

Investigation Find values of n for which a is an integer. 1. The expression above the line needs to be even to

make a an integer. Does it make a difference ifn is odd or even?

2. What values of n will result in an integer when the square root is taken?

3. Does the + sign matter, or should only one of them be taken?

In a class everyone could be given a value of n to calculate or, alternatively, you could set up a spreadsheet.

Table 1 shows integer results which answer these questions.

Table 1

n evaluation using + sign evaluation using - sign

1 3 0 3 8 1 8 21 3

21 55 8

Using Lewis Carroll's method, we can thus see that there are an infinite number of solutions. You can then use these

Mathematics in School, September 2002 The MA web site www.m-a.org.uk 7



values to calculate the size of the rectangle. But note, because of the symmetry of Equation (3), that values of a take the same values as those for n, though of course not at the same time. Although the values form a series, they are not the Fibonacci series, but alternate terms of it. They are part of another recurrence sequence, however.

Suppose the rth value of this series is Ur. From the above table, note how, if a value of a is Ur, then the value of n is Ur+ 1. Now since a and n obey Equation (3) we can write:

Ur+12 - 3Ur+1Ur + U,2 - 1 = 0 (4)

Ur2 - 3UrUr- + Ur12 - 1 = 0 (5)

subtracting and factorizing gives

(Ur-1 -

Ur+1) (Ur-1 + Ur+1 - 3Ur) = 0 (6)

which means

Ur +1 = 3 Ur - Ur-1 (7)

Other Versions of the Puzzle

Figure 1 is a puzzle where the rectangle has an area one unit larger than the square. Equation (1) says that for the rectangle to be one unit more in area than the square, then the square must have the side of an even Fibonacci number, but it also says that if an odd one is used, then the rectangle has an area of one less than the area of the square.

This means that Equation (3) becomes

a2 -3na + n2 +1 = 0 (8)

If we apply the same reasoning as above, then we get another recurrence sequence in the equivalent of Table 1 (i.e. 1, 2, 5, 13, 34 ...) with the same relationship as Equation (7) but with different starting values. This can be used to extend the investigation.

There are other similar puzzles based on Fibonacci dissections. The pieces of Figure 1 can be rearranged as shown in Figure 3, so that using the results of Figure 1, it is possible to demonstrate that 63 = 64 = 65! This variation is sometimes attributed to the famous American puzzler Sam Loyd but, prolific though he was, he was a bamboozler too and often claimed other people's work as his own. The true inventor seems to have been Walter Dexter who published it in the Boy's Own Paper in 1901.

Fig. 3

Figure 4 is a variation I found in one of my notebooks, butt I do not know where it came from and is not mentioned in Frederickson (1997). It shows that 132 - 8 x 21.

Fig. 4

Investigation

Can you find any more variations on the puzzle of Figure 1 using the same pieces?

What apparent area do you get in Figure 3 if you use the variation where the rectangle of Figure 1 has an area of one less than the area of the square?

Investigate the mathematics of Figure 4 using Lewis Carroll's methods.

How the Puzzle Might have Been Discovered

Rouse Ball (1892) gives a reference to the puzzle of Figure 1 which goes back to a German journal in 1868 but I have not been able to trace a copy. However, there is a similar puzzle which goes back to the 18th century and this may have an origin in a mistake of a 16th century architect, so let's start there and then see how the puzzle might have arisen from variations on this.

Sebastian Serlio and the table to door problem

Serlio (1475-1554) was an Italian architect who published a five volume work in 1551 which was the encyclopaedia of its day on architecture. It was translated into many languages and the English version of 1611 is still in print (Serlio, 1982). The first volume is 'entreating of geometrie'. Folio 13 shows the diagram shown in Figure 5.

8 Mathematics in School, September 2002 The MA web site www.m-a.org.uk



A E ~La~ C A

W

T C

cl~lC~ F

The book is printed in a typeface that is hard to read, with the added complication of different spellings and somewhat quaint English. He describes the problem as follows:

It may also fall out, that a man should find a table of ten foote long and three foote broade: with this Table a man would make a door of seven foote high and four foote wyde. Now to doe it, a man would saw the Table longwise in two parts and, setting them one under another, and so they would be but five foote high, and it should bee seven: and againe, if they would cut it three foote shorter, and so make it four foot broade, then the one side shall be too much pieced. Therefore he must do it in this sort: Take the Table of ten foote long and three foote broad, and marke it with A. B. C. D. then sawe it Diagonall wise, that is, from the corner C to B, with two equall parts, then draw one piece thereof three foote backwards towards the corner B, then the line A.E shall be foure foote broad and so shall the line E.D also hold foure foote broad: by this meanes you shall have your doore A.E.ED. seven foote long, and foure foote broade, and you shall yet have the three cornerd pieces marked E.B.G. and C.E and C left for some other use.

If we draw this and create the door from the table, then we have a set of diagrams like Figure 6.

a

6=

l~c~l ~=-.;~

Fig. 6

The action would be to cut the table of Figure 6a and then slide it along so that you end up after cutting the triangular pieces off with two pieces like Figure 6c. Figure 6a has an area of 30 units, but Figure 6c has a combined area of 31 (28 + 3) units. Where has the extra area come from? Well, if you are exact in the way you get from 6a to 6c, then you see that there is a gap between the two triangles which is used to make the extra unit of length. So a true version of figure 6c would have gaps along the diagonal also.

I am pretty sure that Serlio did not know there was a problem. He was a practical man finding a solution to a need. It took another architect, Pietro Cataneo (Cataneo, 1567), to point out the error and show that there was a paradox.

Investigation

Cataneo came up with an alternative dissection, using Serlio's method, of a 12 by 4 rectangle into a pair of rectangles of sides 9 by 5 and 3 by 1. Now, you will notice that there is no bamboozlement with these. Investigate what are the conditions for being able to get a perfect dissection or a paradoxical one.

Ozanam and Hooper's Puzzles

Mathematics books come in a wide range of complexity. There are very advanced textbooks on mathematics, but there are also recreational mathematics books by authors like Martin Gardner and Ian Stewart which aim to amuse by means of mathematical games and puzzles. This is nothing new. One of the most famous recreational mathematics books is by the French mathematician Jacques Ozanam. Another is William Hooper's Rational Recreations of 1794. Such geometrical puzzle books are a source of some very interesting mathematics and went through many editions. Many such recreations were copied, so there may have been a number of variations to avoid outright plagiarism, although I have a book of puzzles and pastimes dated 1852, which copies Hooper's version with the exact same words. Both Ozanam and Hooper have a puzzle which was obviously derived from Serlio's. Hooper's puzzle (shown in Figure 7) he called 'The Geometrical Money'. There is a facsimile of the pages from Hooper's book in Slocum (1994). Figures 7 and 8 show Hooper's original diagrams.

( - -1 E

G B

Fig. 7

Hooper says take a piece of pasteboard and mark squares on it as in Figure 7. Draw diagonal AD and cut along the diagonal and also along the lines EF and GH to give two triangles and two four-sided irregular figures. Then he says place them together in their original positions and "in each square you are to draw the figure of a piece of money; observing to make those in the squares, through which the line AD passes, something imperfect". As they stand he says you will now count thirty pieces of money. This is the origin of the name he gives it "The Geometrical Money".

Then place them together as in Figure 8. Then you see that there are now thirty-two pieces. (Note that he is very bad at labelling since the letters do not correspond to their place on the first diagram.)

A. B

C D Fig. 8




Ozanam's version is slightly different in that he starts from an 11 by 3 grid, instead of Hooper's 10 by 3. When Charles Hutton translated Ozanam in 1803, he said "This deception, it must be allowed, is very puerile in the eyes of a geometrician". The Ozanam version is a better puzzle and he says so "because it is harder to view the deficiency". Looking at the difference between Hooper's and Ozanam's figures is a good exercise in observation.

Fig. 9

Ozanam's diagram in Hutton's translation (reconstructed in Figure 9) creates two points where the diagonal intersects the horizontal line of the grid. Figure 10 is the equivalent of Hooper's diagrams of Figure 7, but with a slight modification. It is actually better to work with vertical lines on the grid, from the fourth in from each end as shown at the top of Figure 10. Ozanam shows that 33 equals 34.

Fig. 10

Drawing the reassembled pieces show the discrepancies more than if you made the puzzle out of card. But if you compare the version of Ozanam's puzzle (Figure 11).

Fig. 11

with the equivalent one for Hooper's (Figure 12)

Fig. 12

then it is obvious that the squares of the grid would look better with Ozanam's version.

Placing the pieces of the Ozanam puzzle on the square grid as in figure 11, instead of abutting them, shows where the extra area comes from.

Fig. 13

Using Carroll's Method for Analysing these Puzzles

With Figures 12 and 13, we have a puzzle which is not quite the same as Figure 1. If we combine the two rectangles of these figures we do not get a square. Applying the logic that Lewis Carroll used on the puzzle of Figure 1, gives some clues as to how these puzzles might have evolved into the one we know today. However, looking at the general case of the Hooper/Ozanam puzzle using Lewis Carroll's method is not as easy because there are more variables. Since Ozanam has a difference of one, like the puzzle of Figure 1, I am going to use his puzzle as the basis of the following. The equivalent marked up version of Figure 2 is shown in Figure 14.

b c

n-a a

a

cb n-a

cc b

a

b

Fig. 14

Since the combined area after dissection is one unit greater than the original.

nb + ac = (n - a) (b + c) + 1 (9)

giving

2ac = nc - ab + 1 (10)

This has four variables, which we can simplify further by saying that a = 2, if we want an Ozanam type solution, but it is still not easy to solve. Since we are looking for solutions for which all values are integers, it is known as a Diophantine equation after the Greek mathematician who first studied problems having integer solutions. Some such




equations can be solved by using special tricks, but nowadays the easiest way is to use computers to try a range of numbers and checking if the equation holds.

The following table shows some possibilities for matching the two sides of the equation.

Table 2

a b c n b + c n-a

2 2 5 5 7 3 2 3 7 5 10 3 2 4 7 5 11 3 2 4 9 5 13 3 2 5 11 5 16 3 2 6 13 5 19 3 2 7 5 7 12 5 2 10 7 7 17 5 2 12 5 9 17 7 3 2 7 7 9 4 3 3 5 8 8 5 3 3 10 7 13 4 3 4 13 7 17 4 3 5 8 8 13 5 3 7 11 8 18 5 3 8 5 11 13 8 3 9 7 10 16 7 4 2 9 9 11 5 4 3 13 9 16 5 4 5 7 11 12 7 5 2 11 11 13 6 5 7 9 14 16 9 5 7 12 13 19 8 6 4 5 17 9 11 6 8 7 19 15 13 6 9 11 17 20 11 7 2 5 17 7 10 7 3 11 16 14 9 7 5 6 20 11 13 8 4 11 19 15 11

Note that lines two and three are marked in bold. The first starts with a rectangle the same as Hooper's, but divides it up in a different way. The second is Ozanam's. Also, unlike the Lewis Carroll result, there is no obvious pattern to the numbers, because there is so much more freedom to choose numbers when there are four variables.

Hooper had his areas differing by two. They could differ by any number, even zero; but then this is not a puzzle any more, or at least a different one. These are a few examples.

Table 3

a b c n area difference

2 3 6 5 0 2 4 8 5 0 2 5 5 6 0 2 5 10 5 0 2 5 13 5 3 5 9 12 14 3 3 12 8 11 4 4 12 13 12 4 4 7 11 11 5 5 12 13 15 5

There is also the possibility, if n and c are equal, of rotating the long rectangle by 90' and dissecting one rectangle into another, but with the areas being different. So for example, one case from Table 2, where a = 2, b = 10, c = 7 and n = 7 allows you to 'prove' that 84 equals 85:

10 7

5 22

10 2

Fig. 15

Having done this, it is natural to ask if, instead of making a rectangle in this manner, then what about a square. This requires not only that n and c are equal, but also that they must equal the sum b + a. When a = 3, b = 5, c = 8 and n = 8, these conditions are satisfied, and it's not surprising that they are Fibonacci numbers and we are back where we started with the original puzzle. Seen in the Ozanam orientation the pieces are like this:

5

5

3

3

8 5

Fig. 16

How and when the puzzle changed to a square and who made the discovery is still unknown, but the circumstantial evidence for its origins is fairly strong. Since the original puzzle has variations, with differences before and after dissection of values other than one, there are other puzzles waiting to be discovered. The mathematics is relatively easy, with the hardest part finding integer values to fit Equation (10).

Inventing New Puzzles

As we have seen above there is a whole family of puzzles here, not just the simple one of Figure 1. You could investigate rectangle puzzles like those in Figure 15, but what




about square puzzles? Are there any other possible variations on the above puzzles, like Figure 4? What about other number sequences? The following are a few suggestions and some solutions to start you thinking. This is unexplored territory as far as I can see. Frederickson (1997) has been far- ranging on his survey of the field, but I believe the following are unpublished.

The Fibonacci sequence is the first of a family of recurrence sequences of the general type:

Un+1 = allUn + bUn-_1 (11)

where a and b are integers and both equal to 1 in the case of the Fibonacci series. The corresponding general equation to (1) is:

Un2 = Unl . Un+1 - (-b)n (12)

which surprisingly does not contain a. There are many possibilities to explore, especially if you use different pairs of numbers to start the sequence, for example beginning with 1, 3 and having the same recurrence relationship as the Fibonacci sequence (a and b are both equal to 1) gives the Lucas numbers 1, 3, 4, 7, 11, 18, 29 and so on. You could make a puzzle like Figure 1 using these numbers and try it on someone who knows all about Fibonacci numbers and the Figure 1 puzzle. Will it puzzle them if they see it with unfamiliar numbers?

The following examples show puzzles based on the sequence with a = 2 and b = 1, which gives the sequence 1, 2, 5, 12, 29, 70 and so on. The numbers increase at a faster rate than the Fibonacci series, so it is harder to see where the missing square went when you cut out the pieces to make the puzzles. They all show how 122 = 5 x 29.

Figure 17 I call Pavitt's puzzle, since it was invented by a friend of mine (Pavitt, 1974).

Fig. 17

It is not quite as simple as the Fibonacci puzzle of Figure 1, and contains edges of length 4 and 7 which are not part of the sequence. Figure 18 is a simpler version of the same dissection which also has a length of side 3 which is not in the series.

Fig. 18

Fig. 19




Figure 19 is even simpler, and now all sides have values which are part of the series.

What new puzzles can you invent?

More Pieces to the Puzzle

Since writing this, I filled in a gap on page 8 and found the first known reference to the puzzle (Schl6milch, 1868). Unfortunately, it is little more than a description of the puzzle which seems to have been used to fill a gap in the page. The solution is not given because it is seen as a good topic to be given to students to solve. About 10 years later, it seems to be well known in England because Charles Darwin's son George (Darwin, 1877) when writing about it is able to say that "the following puzzle was shown to me by a friend" again without mentioning the origin.

References Cataneo, Pietro 1567 L'Archittura di Pietro Cataneo Senese, Venice. Darwin, G. H. 1877 'A Geometrical Puzzle', The Messenger of Mathematics,

VI, p.87. Frederickson, Greg N. 1997 Dissections Plane and Fancy, Cambridge

University Press. Pavitt, Ken 1974 Personal communication. Rouse Ball, W. W. 1892 Mathematical Recreations and Essays, Dover. Schl6milch, O. 1868 'Ein geometrisches Paradoxon', Zeitschriftfiir Mathematik

und Physik, 13, p.162. Serlio, Sebastian 1611, 1982 The Five Books ofArchitecture, Dover. Slocum, Jerry 1994 'Puzzles Old and New: Some Historical Notes'. In Guy,

Richard and Woodrow, Robert (Eds) The Lighter Side of Mathematics, Mathematical Association of America.

Weaver, Warren 1938 'Lewis Carroll and a Geometrical Paradox', American Mathematical Monthly, April, p.234-5.

Keywords: Dissection puzzles; Fibonacci puzzle; Lewis Carroll.

Author John Sharp, 20 The Glebe, Watford, Hertfordshire WD25 OLR. e-mail: [email protected]

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