Geometrical Paradox

Bamboozlement is a term suggested by Greg Frederickson to describe plane dissections, followed by rearrangment of pieces that result in a figure of supposedly different area.

Curry's Paradox

From where comes the hole?
This is Martin Gardner's modified version of Curry's paradox. An extensive history and explanations are given in [Martin Gardner, 1956]. See also [Frederickson, 1997]. The picture made Daniel Takacs.

A Chessboard Paradox

Sam Loyd presented this fallacy one the first American Chess congress 1858. The first time this was seen in print was (1868), a "Kurze Mitteilung" of the Editor O. Schlömilch. Eleven years later (1879) Schlegel published the Fibonacci generalization.
A fake dissection
Alex Bogomolny made an applet Fibonacci Bamboozlement where you can shift the pieces for squares of Fibonacci number size.

Rectangular Transformation Paradox (4 Pieces)

Given three numbers x0, x1, x2 in almost geometric progression which means that x1 x1 - x2 x0 = µ is small. The following figure shows half of a x1 by x2 + x1 rectangle.
   +  ,
   |     '  ,
   |           '  ,
   |                 '  ,
   |                    |  '  ,
   x1                   |        '  ,
   |                    |              '  ,
   |                    x0                   '  ,
   |                    |                          '  ,
   |                    |                                '  ,
   +------ x2-x0 -------+---------- x1+x0 ---------------------+
Rearranging the four pieces gives a x0 + x1 by x2 rectangle.
The area difference between the first and the second rectangle is µ.

The classical parameters given by Sam Loyd are (x0, x1, x2, µ) = (3, 5, 8, 1), three consecutive Fibonacci numbers, in which case one rectangle is a square as x0 + x1 = x2. Other transformations are given in the table below.
x0 x1 x2  µ   x1 × x1+x2   x0+x1 × x2   area1  area2   reference

 4  5  6  1    5 ×  11       9   ×  6     55     54    [Walter Lietzmann]
 3  5  8  1    5 ×  13       8   ×  8     65     64    [Sam Loyd]
 5  6  7  1    6 ×  13      11   ×  7     78     77
 4  7 12  1    7 ×  19      11   × 12    133    132
 5  8 13 -1    8 ×  21      13   × 13    168    169
 2  7 24  1    7 ×  31       9   × 24    217    216    [Hermann Schubert]
 4  9 16  1    9 ×  25      14   × 16    225    224
 5 11 24  1   11 ×  35      16   × 24    385    384    [Torsten Sillke]
Typically you find the note that this transformation can be made for arbitrary rectangles A*D and B*C if |A*D - B*C| = 1. But this is not true. [Lietzmann 1972] and [Devendran 1985] correctly say that we must have A+B=D in addition. I think my parametrization is more suitable for this dissection.

Rectangular Transformation Paradox (6 Pieces)

Given three numbers x0, x1, x2 in almost geometric progression which means that x1 x1 - x2 x0 = µ is small. The following figure shows half of a x1 by x2 + 2x1 rectangle.
   +   ,
   |       '   ,
   |               |   ,
   |               |       '   ,
   |               |               '   ,
   x1              |                       '   ,
   |               |                               '   ,
   |             2*x0                                  |   '   ,
   |               |                                   |           '   ,
   |               |                                   x0                  '   ,
   |               |                                   |                           '   ,
   +--- x2-4*x0 ---+------------- x1+2*x0 -------------+------------- x1+2*x0 -------------+
Rearranging the six pieces gives a 2x0 + x1 by x2 rectangle.
The area difference between the first and the second rectangle is .
x0 x1 x2   µ  x1 × 2x1+x2  2x0+x1 × x2   area1  area2   reference

 2  5 12  1   5 ×   22       9   × 12    110    108    [Torsten Sillke]
 3  8 21  1   8 ×   37      14   × 21    296    294    [Torsten Sillke]
This construction can be generalized. So we have an 8 piece transformation of a x1 by x2 + 3x1 rectangle into a 3x0 + x1 by x2 rectangle where the difference of their areas is .

Cassini's Fibonacci identity

Many Bamboozlements use Fibonacci numbers. One of the oldest theorems about Fibonacci numbers, due to the French astronomer Jean-Dominique Cassini in 1680, is the identity Fn+1 × Fn-1 - Fn × Fn = (-1)n. The shortest proof of this identity is via the matrix identity which is another way of writing the recurrence formular. Then use the theorem that the determinant is multiplicative.
            n
     [ 1 1 ]       [ Fn+1 Fn   ]           n     | Fn+1 Fn   |
     [ 1 0 ]   =   [ Fn   Fn-1 ]   =>  (-1)   =  | Fn   Fn-1 |

There is a geometric reasoning if we compare the two rectangles Fn+1 × Fn-1 and Fn × Fn. They both contain the rectangle Fn × Fn-1. This means

Showing the relation for n=1 with F2 = F1 = 1 and F0 = 0 completes the proof.

A combinatorial interpretation of the relation of Cassini was published by N. Werman and D. Zeilberger.

Supplement Rectangles for Curry's Paradox

For a Curry triangle we can make a two piece zig-zag dissection of the supplement rectangle if the width of the rectangles differ by one. A first sequence of pairs of rectangles is
  x x     x x x x x       x x x x x x x x
  + x     + + x x x       + + + x x x x x
  + x     + + + x x       + + + + + x x x
  + +     + + + + +       + + + + + + + +

  + x x   + + x x x x x   + + + x x x x x x x x
  + . x   + + + . x x x   + + + + + . x x x x x
  + + x   + + + + + x x   + + + + + + + + x x x
A second sequence of pairs of rectangles is
  x   x x x x x     x x x x x x x x x
  x   + x x x x     + + x x x x x x x
  .   + + . x x     + + + + . x x x x
  +   + + + + x     + + + + + + + x x
  +   + + + + +     + + + + + + + + +

  x   + x x x x x   + + x x x x x x x x x
  x   + + x x x x   + + + + x x x x x x x
  +   + + + + x x   + + + + + + + x x x x
  +   + + + + + x   + + + + + + + + + x x
The hole need not be in the middle but this looks nicer.

Euclid's Supplement Parallelograms

A parallelogram is dissected by a diagonal into to congruent triangles.
        C-----------F-------------------I
       /           /               ,   /
      /           /           ,       /
     /           /       ,           /
    /           /   ,               /
   B-----------E-------------------H
  /       ,   /                   /
 /   ,       /                   /
A-----------D-------------------G
Select a point E on the diagonal. Draw lines through E which are parallel to the sides of the parallelogram. As a result we get four small parallelograms. Now Euclid says: The two supplement parallelograms DGHE and BEFC have the same area.

If E is not on the diagonal then we have two quadrangles AGIE and ACIE. And area(AGIE) - area(ACIE) = area(DGHE) - area(BEFC) but this is not null. This explains the paradoxes. The 'traingles' are quadrangles.

Area of a Parallelogram

What is the area of a parallelogram generated by two vectors X=(a,b) and Y=(c,d)? The corners of the parallelogram are 0, X, Y, and X+Y. Now assume that a, b, c, and d are positive. Our 'paradox' shows that the area is |a*d - b*c| the determinant of the vectors X and Y. The angle t between X and Y is t = arctan(b/a) - arctan(d/c). Using the scalar product we get the equations |X|*|Y|*sin(t) = Area and |X|*|Y|*cos(t) = X*Y. Division gives tan(t) = Area/(X*Y). Substitute the parameter of the chess board dissection gives X=(8,3), Y=(5,2). Then the area=1 and X*Y=46. So t = arctan(1/46) which is nearly 1/46. This follows from the alternating series arctan(x) = x - x3/3 + x5/5 - ... . Converting Rad to Degree gives 1.245 which is rather small.

References

Hardware


Torsten Sillke

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Last Update: 2004-10-13