1. Introduction
The purpose of this paper is to demonstrate the dynamic geometric
dissections. With the help of Cabri Geometry II, a tool for geometric
constructions, we can explore these geometric dissections discovered by
great mathematicians, examine their correctness and display their
animations from figures to figures.
The dissections explored here are masterpieces that have been
discovered by many people. However, most of them are displayed as
plane figures. It’s thus difficult for beginners to feel how charming an
unmoved dissection has. This paper attempts to provide the readers with a
different aspect of viewing the magic and beauty of the geometric
dissections.
In the beginning of this paper, some geometric figures that will
appear in later constructions are introduced. Dissecting techniques that
have been widely used in every construction of figures are then discussed
in the second part of this thesis. The main topics of this paper are
illustrated by regular polygons, star polygons, dissected curves and other
figures. Following this are special types of hinged dissections such as
swing-hinged dissections, flip-hinged dissections and twist-hinged
dissections. Finally, I will demonstrate how the dissection goes forward to
three dimensions such as dissecting a solid figure or dissecting the
surface of a solid. The figures identified here should help the reader better
explore the beauty of geometric dissections.
1
2. Geometric Figures
Geometric figures are shown here with names, basic definitions and
areas to illustrate the relationships between two figures.
2.1 Two-dimensional Figures
 Regular Polygon: {n}
An n-sided regular polygon is a polygon in which the length of each
side is the same and every side is placed symmetrically around one
common center. (i.e., the polygon is both equiangular and equilateral),
denoted by {n}.
Let s be the side length, and R the circumradius of a regular
polygon.
Figure 1: Regular polygon for n = 5
1
2
The area of an n–side regular polygon {n} is: A{n}  nR 2 sin(
2
)
n
This formula will be frequently used in calculating two different figures
with equal size of area.
 Star Polygon: {p/q}
Figure 2: Star polygon: {5/2}
2
A star polygon {p/q}, with p, q positive integers, is a figure formed
by connecting every q-th point from p regularly placed points on a
circumference. The number q represents the density of the star polygon.
Without loss of generality, we may take q< p/2
The usual definition of a star polygon {p/q} requires that p and q be
relatively primes (Coxeter 1969). However, the star polygon can also be
generalized to become the star figure (or "improper" star polygon) when
p and q share a common divisor (Savio and Suryanaroyan 1993). For
such a figure, it is necessary to start the procedure with the first
unconnected point and repeat it afterwards. Repeat the whole procedure
until all points are connected.
For (p,q)≠ 1, the {p/q} symbol can be factored as {p/q}= {p’/q’},
where p’= p/n and q’= q/n to give n {p’/q’} figures, each rotated by 2π/p
radius, or 360°/p.
To calculate the area of a star polygon {p/q}, we can cut the star
polygon into p parts, each part has the shape of a kite, and its area is equal
to a larger triangle subtract a smaller triangle. Then the area of a star
polygon {p/q} is:
A{ p / q}  pR 2 sin(



)[cos( )  sin( ) tan(
p
p
p
(q  1)
)]
p
 Greek Cross
Figure 3: The Greek Cross, Latin Cross and Cross of Lorraine
These are the crosses with the simple shape and are composed of
identical squares. From the left of Figure3: the first one is the Greek
Cross which is denoted by {GC} with area equal to five squares; then
comes the Latin Cross which is denoted by {LC} with area equal to six
squares; the last one is the Cross of Lorraine which is denoted by {L’}
with area equal to thirteen squares (Frederickson, 1997).
3
2.2 Three-dimensional Figures
 Cube
A cube is a widely seen figure with six
squares and a volume equal to the third power
of its side length.
 Tetrahedron
A tetrahedron has four faces of the
equilateral triangles. It can be constructed from
joining four vertexes of a cube. If a cube
centers at (0,0,0) with radius of 3 unit, connect
the vertexes of (1,1,1), (1,-1,-1), (-1,1,-1) and
(-1,-1,1), a tetrahedron with volume equal to
4/3 units can be found.
 Octahedron
An octahedron has eight faces of the
equilateral triangles. During the construction,
we can place the center at (0,0,0), with a radius
equal to 1 unit, and with each vertex laying on
the intersection of a unit sphere and three axes.
The volume of such an octahedron is thus equal
to 4/3 units.
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 Truncated octahedron
A truncated octahedron has eight
hexagonal faces and six square faces. It can be
constructed from cutting every corner of an
octahedron until the faces of a triangle in
octahedron become a hexagon. Suppose the
radius of an octahedron is equal to 3 units, and
then the volume of a truncated octahedron is
equal to 32 units.
 Cuboctahedron
A cuboctahedron is a solid between a cube
and an octahedron. It has six square faces and
eight faces of the equilateral triangles. If we
start from cutting every corner of an octahedron
having a radius of 3 units until each face of the
equilateral triangles becomes a quarter the area
of the original equilateral triangle. We then get
a cuboctahedron with volume equal to 27 units.
 Hexagonal prism
A hexagonal prism has two hexagonal
faces and six rectangular faces. If the radius of
the hexagon is R and the height of prism is h,
then the volume of a hexagonal prism is equal
to
3
3R 2 h .
2
5
3. Definition of Hinged Dissections
In order to distinguish all kinds of hinged dissections in the contents,
different types of hinged dissections are illustrated in chapter 3.1, while
additional details of swing-hinged dissections are presented in chapter
3.2.
3.1 Types of hinged dissections:
 Swing-hinged dissection:
For swing-hinged dissection of two
pieces, we mean that the pieces stay in the
plane as we swing them around their hinges
(hinged points). Thus we turn over no pieces
when we use swing hinges (Frederickson,
1997). As shown in figure 4, the dots indicate
the hinges.
Figure 4:
Swing-hinged pieces
 Flip-hinged dissection:
A flip hinge is the rotation of a piece
through 180° in the third dimension along a
vertex that connects the two pieces
(Frederickson, 1997). Figure 5 shows the flip
hinge where the star “*” indicates the turned
pieces and the small circle “。” indicates the
rotated vertex.
Figure 5: Flip-hinged
pieces
 Twist-hinged dissection:
A twist hinge is the rotation of a piece
through 180° in the third dimension along an
axis of rotation that is perpendicular and
interior to a shared edge (Frederickson, 1997).
As shown in Figure 6, the star “*” indicates
the turning pieces while the small circle “。” on
the edges indicates where the rotation axis is
perpendicular to.
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Figure 6: Twist-hinged
pieces
3.2 Hinged patterns in swing-hinged dissections:
In swing-hinged dissections, we present more details on how many
pieces are hinging in the dissections or what kinds of swing-hinged
patterns we referred to.
 Partially hinged:
For some dissections we can hinge most, rather than all, of the
pieces. We call such a dissection a partially hinged dissection
(Frederickson, 1997). Examples include the dissection of a pentagon to a
triangle in Figure 21, or the dissection of an octagon to a square in Figure
48.
 Fully hinged:
A dissection is fully hinged if all pieces are connected with hinges
into a chain so that rotating the pieces around one way assembles them
into one of the figures of the dissection, while rotating the pieces the
other way assembles them into the other figures of the dissection
(Frederickson, 1997). Examples include the transformation of a square to
a triangle is illustrated in Figure 19, or a dodecagon to a square is in
Figure 170.
 Variously hinged:
A dissection is variously hinged if we can hinge it with the largest
possible number of hinges in several different ways (Frederickson, 1997).
Examples indlude the transformation of a square to a triangle in Figure 19,
or a Greek Cross to a square in Figure 156.
 Cyclicly hinged:
If we can remove a set of hinged pieces without disconnecting them,
we call this dissection a cyclicly hinged dissection (Frederickson, 1997).
Examples include the transformation of a Greek Cross to a hexagon in
Figure 173, or the transformation of two hexagons to a hexagon in Figure
162.
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4. Dissecting Technique:
To dissect one figure into another, there are some techniques for
general purposes that could help us to deal with two or more figures. The
techniques discussed here are tessellation techniques, plain-strip
techniques, twin-strip techniques, quadrilateral slides and trapezoid
slides.
4.1 Tessellation Technique
A tessellation is the tilling of an n-dimensional figure in
n-dimensional space. We focus on two dimensions. A tessellation element
is such a figure that tiles the plane. Generally, a tessellation element does
not need to be that of a regular polygon, it may be that of any figure that
tiles the plane, even for combinations of line segments and curves.
A regular tessellation is composed of regular polygons
symmetrically tile on the plane. There are exactly three regular
tessellations, which are composed of triangles, squares and hexagons.
Figure 7: Tessellation: triangles
Figure 8: Tessellation: squares
A semiregular tessellation is composed of two or more convex
regular polygons such that the same polygons surround each polygonal
vertex in the same order. There are eight such tessellations in the plane.
Figure 9: Octagon and squares
Figure 10: Dodecagon and triangles
A different form of tessellation is the non-edge-to-edge tessellation.
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One polygon slides on another polygon to produce the tessellation.
Sometimes it is not required to add an extra polygon as in Figure 11, and
sometimes it may need to add an extra polygon to tile the plane as in
Figure 12.
Figure 11: Non-edge-to-edge squares
Figure 12: Unequal squares
Examples:
 Just tessellation:
If two polygons can tile the plane by their original figures, or by the
transformed elements, and their repeating patterns are the same. Then, we
can find an economical dissection from superposing two tessellations.
Examples include the dissection of a Greek Cross to a square (Figure
154), and a dodecagon to a square (Figure 44).
 Complete the Tessellation:
When a polygon cannot tiles the plane by it original figure, or it
cannot be transformed into an element that tiles the plane, we may add
another polygon to tile the plane. If the added polygon can also be added
to the target polygon to form the tessellation and their repeating patterns
were the same, then they can be dissected from superposing two
tessellations. Harry Lindgren called it complete the tessellation
(Frederickson, 1997). Examples include the dissection of an octagon to
square (Figure 48), and a hexagram to a hexagon (Figure 160).
 Tile the surface of a polyhedron:
Sometime the element of a polygon cannot tile the plane, but it still
worked well if it can tile the surfaces of a 3-dimensional polyhedron
whose faces are of the target polygons. Examples include the dissection
of a hexagram to a triangle (Figure 122), and a {10/2} to a heptagon
(Figure 129).
4.2 Plain-Strip Technique
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Figure 13: Crossposition for Plain-strips of {5} and {4}
If one figure can be transformed into an element that tiles a plain
strip with its repeating element, and the other figure can also tiles the
plain strip, we can superpose these two strips such that the width of one
strip is coincident with the width of the element of another strip. This
technique is called the “plain-strip” or “P-strip”, technique (Frederickson,
1997).
In fact, such technique is a variation of tessellation technique for the
strips can repeat non-edge-to-edge to tile the plane. One example shown
in Figure 10 illustrates the crossposition for strips of pentagons and
squares. The strip of pentagons is composed of filling a parallelogram in
a plain strip, and the strip of squares is composed of repeating squares.
We can superpose these strips such that the width of strip of squares is
coincident with the width of the element of pentagon.
The limitation of the plain strip technique occurs when the width of
the first strip is greater than the width of the element in the second strip.
Since the areas of these two figures are the same, the width of the second
strip will also be greater than the width of the first element in its strip. It
will not be possible to make a crossposition for these strips. However, it
doesn’t include a special case that the two strips have the same width. In
such case, we may cover one strip paralleled with the other strip.
Examples:





Parallelogram strips: a hexagon to a square (Figure 31)
Non-parallelogram strips: a hexagram to a square (Figure 119)
Optimized strips: a octagon to a hexagon (Figure 55)
Customized strips: a decagon to a square (Figure 78)
Bumpy plain strips: a star polygon of {12/2} to a square (Figure 136)
4.3 T-strip technique
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Figure 14: Crossposition for Twin-strips of {4} and {3}
The T-strip technique is a variation of the P-strip technique called
the “twin-strip” or “T-strip” technique (Frederickson, 1997). In P-strips,
we cut a figure into pieces that rearranged to form a repeating element
that fills out a strip. But in T-strip, there are two equally elements placed
symmetrically about a center point and they fill out a plain strip by the
pairs of twined elements. This center point usually lies on the middle of
their boundary. Frederickson (1997) called such a point an anchor point.
Such a strip also can be used to derive a dissection, but we cannot
translate the strip superposed on it as freely as on a P-strip. We have two
ways to crosspose these strips. One of them is to crosspose the edge of
one strip passing through the anchor point of second strip. The other way
is to crosspose the second T-strip over the first strip with their anchor
points coincident with each other.
When a P-strip and a T-strip are used simultaneously, Lindgren
called this a PT dissection (Lindgren, 1972). When two strips were
T-strips, he called it a “TT dissection”. Since the twined strip is composed
of two equally elements with each rotated 180 degrees around the anchor
point to match with the other one, such elements can be hinged during the
transformation.
Examples:
 TT dissection: a square to a triangle (Figure 20)
 PT dissection: a pentagon to a triangle (Figure 23)
4.4 Quadrilateral Slide
11
Figure 15: Lindgren’s Q-slide
Harry Lindgren introduced his quadrilateral slide, or Q-slide
(Lindgren 1956). Such slide converts one quadrilateral to another
quadrilateral with the same angles. As shown in Figure15, there is a taller
quadrilateral on the left and a shorter quadrilateral on the right.
Lindgren first located the point E on side AB in order that segment
BE is the desired side length of the edge on the right. Then he locates the
point F such that DF is the desired side length of the edge on the left. Let
the point G be the midpoint of AE, and let the point H be the midpoint of
CF. Make a cut from GH, and a second cut from E parallel to segment AD,
a third cut from F parallel to segment BC. Note that once the point E is
located, the location of point F is thus determined.
This Q-slide can be hinged cyclicly form one quadrilateral to another
as in Figure 16.
Figure 16: Cyclicly hinged: one quadrilateral to another
4.5 Trapezoid Slide
12
Figure 17: Hanegraaf’s T-slide
A dissection shown in Figure 17 is that of Hanegraff, a dissection
which converts a trapezoid to a parallelogram. This technique is similar to
that of Lindgren’s Q-slide so that Frederickson has called it a T-slide
technique (Frederickson, 1997).
Starting with the trapezoid ABCD on the left, he first located the
point E on AB such that the length of BE is equal to three times of
AB/CD subtract 1. Then the length of AE will be the side length of the
new parallelogram. Next he located point F on CD such that the sum of
the lengths of BE and DF equals the length of AE. Finally, he let point G
be the midpoint of CF, and let the point H be the midpoint of BC. He cut
from F to H, and then he made two cuts from G and E, both parallel to
AD.
This T-slide can be cyclicly hinged as shown on the left of Figure 14.
A general case of this T-slide is that the point E is located freely on
AB. In this case the resulting figure will not be a parallelogram, instead it
will be composed of two parallelograms such that the upper one has the
width equal to length of AE and the lower one has the width equal the
sum of length BE and CG.
5. Dissecting Plane Figures
In dissecting plane figures, we may start with regular polygons, star
13
polygons, curved figures and other figures such as Greek Crosses. We
then go forward to some special types of dissections such as fully
swing-hinged dissections, flipped-hinged dissections and twist-hinged
dissections.
5.1 Regular Polygons
The notation {n} will be used to denote a regular n-polygon, while
the notation of (m) will denote the number of pieces needed in the
dissection.
 {4} to {3} (4)
Figure18: Dudeney’s triangle to square
This is a beautiful dissection of a triangle to a square shown in
Figure 18. It was Henry Ernest Dudeney who first posed this puzzle on
the newspaper and gave this solution (Frederickson, 1997).
Figure 19: Crossposition of triangles and square
In reproducing this dissection, we can use the T-strip technique. We
may simply take the triangle as the element of a T-strip. Rotating the
triangle 180 degrees with the midpoint of one side of it to make a T-strip
14
element. The strip for square is also formed by repeating the squares.
Crossposing these two strips as in Figure 19 (Frederickson, 1997). The
relation between each figure is 2s2{4}=3sin120oR2{3}, while the width of
strip for triangles is equal to 3R{3}/2. The dots indicate the anchor points
of both strips.
Figure 20: Hinged pieces
When moving the pieces from triangle to square, we can fix one of
the pieces of the triangle. We then rotate the other pieces 180 degrees
around the anchor points. These pieces are fully hinged, connected
linearly by the anchor points and can be variously hinged. If we move
away one piece from the right of a chain in Figure 20, connecting it with
the leftmost piece by anchor point, we then get a new type of hinged
animation.
 {5} to {3} (6)
Figure 21: Goldberg’s pentagon to triangle
In 1952, Michael Goldberg gave a 6-piece dissection of a pentagon
to a triangle, shown in Figure 21. Lindgren (1964b) showed how to derive
this dissection by crossposing the P-strip element for pentagons and the
T-strip element for triangles, as shown in Figure 23 (Frederickson, 1997).
15
The element of strip for pentagons can be derived from cutting along a
diagonal that gives an isosceles triangle and a trapezoid. This isosceles
triangle can be cut into two pieces that fit the left side of the trapezoid as
a parallelogram, as shown in Figure 22. The relation between each figure
is 5sin72oR2{5}=3sin120oR2{3}, while the width of strip for pentagons is
2sin36ocos18oR{5}. Since the strip for triangles is a T-strip, we can
partially hinge it, forming two chains. One chain connects the top two
pieces of pentagon in Figure 21 and the remaining four pieces can be
fully and variously hinged (ibib).
Figure 22: Element of {5}
Figure 23: Crossposition for {5}, {3}
 {5} to {4} (6)
Figure 24: Brodie’s pentagon to square
We start from Busschop’s dissection of a pentagon to a square in
terms of P-strips. He picked up the element for pentagon in Figure 22 and
formed a P-strip. Then he crossposed it with the strip of squares. The
16
relation between each figure is 5sin72oR2{5}=2s2{4}. But unfortunately,
Busschop's dissection was not minimal, just because he chose the wrong
points from each element to coincide (Frederickson, 1997). Using the
same elements, the Scotsman Robert Brodie (1891) gave a 6-piece
dissection, which can be derived by using the same two strips as shown in
Figure 25 (Frederickson, 1997).
Figure 25: Crossposition of pentagons and squares
In fact, there are many dissections that take only 6 pieces. One of
them is to remove the vertex of square in Figure 25 to the adjacent point
of the right side. Then infinitely many dissections will form by sliding the
strip of squares such that two smaller pieces of pentagon lay inside the
squares.
 {6} to {3} (5)
Figure 26: Lindgren’s hexagon to triangle
In 1940, Michael Goldberg gave a 6-piece dissection of a hexagon to
a triangle as a problem in the American Mathematical Monthly. And in
17
1951, Lindgren found such a 5-piece dissection shown in Figure 26
(Frederickson, 1997). The T-strip element of hexagon is in Figure27, and
the T-strip element of triangle simply takes its original shape. Then he
crossposed two T-strips with their anchor point coincide with each other
as shown in Figure 28 (Frederickson, 1997). The relation between the
radius each figure is 2sin60oR2{6}= sin120oR2{3}, while the width of
strip for triangle is 3R{3}/2.
Figure 27: Hexagon element
Figure 28: Crossposed {6}, {3}
Since it uses the T-strips, we can partially and variously hinge four
pieces of this dissection and translate the small triangle to the desired
place.
 {6} to {4} (5)
Figure 29: Busschop’s hexagon to square
This is Busschop’s dissection of a hexagon to a square. He first cut
the hexagon along a diagonal into two equal trapezoids that rearranged
18
into a parallelogram as in Figure 30. Then, he crossposed this strip with
the strip of squares as in Figure 31 (Frederickson, 1997). The relation
between each figure is 3sin60oR2{6}= s2{4}.
Figure 30: {6}’s element
Figure 31: Crossposition: {6}, {4}
To minimize the number of pieces, Busschop had a vertex of the
square coincide with a vertex of the hexagon. The resulting 5-piece
dissection is shown in Figure 29 (Frederickson, 1997). In the animation,
there are two pairs of pieces sliding one another to the desired position.
 {6} to {5} (7)
Figure 32: Lindgren’s hexagon to pentagon
Lindgren (1964b) gave this 7-piece dissection of a hexagon to a
pentagon. He chose the element of pentagon in Figure 22 and the element
of hexagon in Figure 33, crossposing these strips in the minimal position
as Figure 34 (Frederickson, 1997). The relation between the radius of
each figure is 6sin60oR2{6}=5sin72oR2{5}, while the width of strip for
19
pentagons is 3R{6}/2. There are infinitely many dissections by sliding the
strip of hexagons horizontally such that the isosceles triangle always lies
inside the largest piece of strip of pentagons.
Figure 33: {6}’s element
Figure 34: Crossposition : {6}, {5}
 {7} to {3} {8}
Figure 35: Theobald’s heptagon to triangle
For dissecting a heptagon into other figures, Gavin Theobald has
produced an ingenious element of heptagon that only takes four pieces.
As shown in Figure 36 (Frederickson, 1997), Theobald first cut an
isosceles triangle from a diagonal that fit the button of the heptagon.
Them he cut a long-thin trapezoid and a small isosceles triangle and
rearranged them into a P-strip element as shown on the right of Figure 36.
20
Figure 36: Theobald’s partition and element of heptagon
He made a cut from the midpoints of two sides of a triangle and
rearranged them into a parallelogram that is a P-strip element. He
crossposed these strips as Figure 37 in order that both the long-thin
isosceles triangle within the heptagon and the equilateral triangle within
the big triangle be prevented from cutting any part of the figure. Infinitely
many dissections are possible while two strips follow these restrictions.
The relation between each figure is 7sin(2π/7)R2{7}=3sin120oR2{3},
while the width of strip for triangles is 3R{3}/4. The resulting 8-piece
dissection of a heptagon to a triangle shown in Figure 35 is due to
Theobald (2001).
Figure 37: Crossposition for strips of {7} and {3}
21
 {7} to {4} (7)
Figure 38: Theobald’s heptagon to square
In 1927, Henry Dudeney posed the puzzle of dissecting a heptagon
to a square and George Wotherspoon gave a 10-piece solution. In 1964,
Lindgren found three different 9-piece dissections of a heptagon to a
square. Then, Anton Hanegraaf founded an 8-piece dissection that uses
the technique of his T-slide. Finally, Theobald found a 7-piece dissection
as shown in Figure (Theobald, 2001). And it was Theobald (1997) who
reproduced it and gave the details of its construction.
Figure 39: Desired side length
Figure 40: Parhexagon
It is impossible to crosspose the strip of heptagons over the strip of
squares since the width of strip of squares is greater than the width of the
element of heptagon. We cannot crosspose them to derive a dissection.
Theobald found another way to deal with this problem. He found a line
segment that perpendicular to the edge of the largest piece and having
length equal to the desired side length as shown in Figure 39 with long
dashed lines. The relation between each figure is 7sin(2 π
/7)R2{7}=2s2{4}. Then he shifted the three pieces from the right side to
22
the left, producing another strip element with a right angle in it as in
Figure 40. Finally, he made two cuts along the dashed lines of Figure 40
and rearranged them into a square as in Figure 38.
 {7} to {5} (9)
Figure 41: Theobald’s heptagon to pentagon
Choosing Theobald’s heptagon element in Figure 36 and the
pentagon element in Figure 22. We can reproduce his 9-piece dissection
of a heptagon to a pentagon with the method P-strips. The relation
between each figure is 7sin(2π/7)R2{7}=5sin72oR2{5}, while the width
of strip for pentagons is 2sin36o cos18 o R{5}. We crosspose these strips as
shown in Figure 42 so that the long-thin isosceles triangle of heptagon
and two smaller pieces of pentagon are prevented from cutting any part of
the figure. Under these restrictions, infinitely many dissections are
possible. We can hinge two pieces of this dissection.
Figure 42: Crossposition for strips of {7} and {5}
23
 {7} to {6} (8)
Figure 43: Theobald’s heptagon to hexagon
By using Theobald’s heptagon element shown in Figure 36 and the
element of hexagon shown in Figure 33, we can crosspose them with
plain strips as in Figure 44 so that every smaller pieces is prevented from
cutting. The resulting 8-piece dissection is shown in Figure 43 (Theobald,
2001). The relation between the radius of each figure is
7sin(2π/7)R2{7}=6sin60oR2{6}, while the width of strip for hexagon is
3R{6}/2.
Figure 44: Crossposition for strips of {7} and {6}
24
 {8} to {3} (7)
Figure 45: Theobald’s Octagon to triangle
Theobald produced an impressive T-strip element for the octagon,
whose partition and rearrangement are shown in Figure 46
(Frederickson,1997). He cuts two small right triangles and a trapezoid,
one small triangle fits atop and the other one combines with the trapezoid
to a bigger right triangle fits adown. Crossposing these T-strips as in
Figure 47, the relation between the radius of octagon and triangle is
8sin45 o R2{8}=3sin120 o R2{3}, while the width of strip for triangles is
3R{3}/2. This 7-piece dissection of an octagon to a triangle improved
upon Lindgren’s 1964 dissection by one piece.
Figure 46: Octagon partition and T-strip element
25
Figure 47: Crossposition for octagon and triangle
 {8} to {4} (5)
Figure 48: Bennett’s octagon to square
In the May 1926 issue, Dudeney had solved a difficult puzzle, that of
cutting a regular octagon into a square in only seven pieces.
Dudeney transformed the octagon into a rectangle in four piece and
then converted the rectangle to a square with a P-slide, introducing three
more pieces. But Bennett found a new type of dissection, using a novel
technique that Harry Lindgren has called completing the tessellation
(Frederickson,1997)
26
Figure 49: Tessellation: {8}, {4}
Adding a square to an octagon with equal sides forms a tessellation
element with repeating pattern of squares. Any two squares form a
tessellation element with repeating pattern that is a square, too. Since
their repeating pattern were the same, we can take them into
superposition to form the dissection. First, taking the larger square in the
second tessellation to be of area equal to the octagon and then taking the
remaining square in that tessellation to be identical to the square in the
first tessellation gives two tessellations that we can superpose as in Figure
49 (Frederickson, 1997). The relation between octagon and square is
4sin45oR2{8}= s2{4}. Since the repetition of octagons is side by side, we
superpose the square by the width of octagon and then adding the small
square equal to that of octagon’s to form a tessellation.
In 1951, Lindgren first published such superposition. The 4-fold
rotational symmetry of each tessellation leads to 4-fold rotational
symmetry in the dissection. Also, these four pieces and be fully and
variously hinged.
 {8} to {5} (9)
Figure 50: Theobald’s octagon to pentagon
27
Theobald first cut the octagon with two isosceles to form a P-strip
element as shown in Figure 51. Then he combine this strip with
pentagon’s P-strip as its element shown in Figure 22. Their crossposition
is shown in Figure 52 and resulting a 9-piece dissection is shown in
Figure 50. The relation between the radius of octagon and pentagon is
8sin45o R2{8}=5sin72 o R2{5}, while the width of strip for pentagon
equals 2sin36o cos18 o R{5}.
Figure 51: Element of {8}
Figure 52: Crossposition : {8}, {5}
 {8} to {6} (8)
Figure 53: Theobald’s octagon to hexagon
In 1964, Lindgren gave a 9-piece dissection of an octagon to a
hexagon, but Theobald has improved one piece by using the optimized
strip approach (Frederickson, 1997).
28
Figure 54: Sample cuts
Figure 55: Crossposing {8}, {6}
Theobald started with Lindgren's tessellation element for an octagon
(Figure 54 in solid lines). There are an infinite number of possible cuts
that will transform it into a strip element, with two examples shown with
the long dashed line and with small dashed line (Frederickson, 1997).
Theobald crossposed a hexagon strip based on Figure 30 to avoid cutting
the octagon's rhombus. In order to have the boundaries of the hexagon
strip cross the octagon strip at intersection points of line segments, he
used the dashed cut in Figure 54 to produce the octagon strip
(Frederickson, 1997). The relation between the radius of octagon and
hexagon is 4sin45 o R2{8}=3sin60 o R2{6}, while the width of strip for
hexagon is sin60 o R{6}. The crossposition of strips is shown in Figure55,
and the resulting 8-piece dissection is shown in Figure 53.
 {8} to {7} (11)
Figure 56: Theobald’s octagon to heptagon
Except for producing a T-strip element of octagon (Figure 46),
29
Theobald has produced a P-strip element of octagon that uses the same
cuts but different arrangement as shown in Figure 57. And for heptagon,
he used his P-strip element shown in Figure 36. He crossposed these
strips to prevent some small pieces from cutting as shown in Figure 58.
The relation between the radius of octagon and heptagon is
8sin45 o R2{8}=7sin(2π/7) R2{7}, while the width of strip for octagon is
2s{8}=4sin22.5 o R{8}. The 11-piece dissection of octagon to heptagon is
shown in Figure 56 (Theobald, 2001).
Figure 57: {8}’s P-strip element
Figure 58: Crossposition: {8}, {7}
 {9} to {3} (8)
Figure 59: Theobald’s enneagon to triangle
In 1964, Lindgren reported that Ernest Irving Freese had found a
9-piece dissection of an enneagon to triangle. After that, Rober Reid
found an elegant 8-piece dissection that uses the Q-slide. However, he
needed to turn over one of the pieces. Finally, Gavin Theobald found a
different 8-piece dissection, and with no pieces turned over
(Frederickson,1997).
30
Figure 60: Partition: {9} Figure 61: Crossposing for trapezoids
Theobald started with Freese’s basic idea, of cutting three outer
trapezoids off the enneagon to leave a revealed triangle. Freese used a
T-strip to convert three trapezoids to a larger trapezoid that fit below of
the revealed triangle to form a triangle. Theobald cut a fourth trapezoid
away form the revealed triangle and combine the new trapezoid with the
one beneath it. His partition of enneagon is shown in Figure 60. The
nibbled triangle plus a trapezoid form a triangle. Then Theobald
converted the rest two pieces to a larger trapezoid by T-strips. His
crossposition is shown in Figure 61 (Frederickson,1997). The width of
strip for the larger trapezoid is 3(R{3}-R{9})/2 , while the relation
between the radius of each figure is 3sin40o R2{9}=sin120o R2{3}.
 {9} to {4} (9)
Figure 62: Theobald’s enneagon to square
In 1964, Lindgren gave a 12-piece dissection of an enneagon to a
square. By designing a clever P-strip element, Hanegraaf found a
31
10-piece dissection. His P-strip element is derived from converting a
trapezoid to a parallelogram that uses the T-slide technique which has
mentioned before. In 1995, David Paterson also gave a 10-piece
dissection. But these records have been broken by Gavin Theobald for
producing a strip element that gave better results.
Figure 63: Partition of {9}
Figure 64: Theobald’s element of {9}
Theobald partitioned the enneagon into six pieces and rearranged
them to a P-strip element as shown in Figure 63 and Figure 64. But in the
rearrangement, the vertices at the lower right of the largest piece and the
long piece do not exactly coincide. This is because the length of the long
piece is 1+2cos40°-1/(4cos40°)≒2.206 (where the side of the enneagon is
assumed to be 1), while the length of the two edges is
2sin10°+2cos20° ≒ 2.227. Theobald then converted his tessellation
element to a strip element (Figure 64) by cutting along the long dashed
line, which spans from the rightmost vertex of the largest piece to
approximately 2.227~2.206 below the leftmost vertex of the largest piece
(Frederickson,1997).
Figure 65: Theobald’s {9} and {4} crossposition
32
When crossposing strips of enneagon and square in Figure 65, the
relation between each figure is 9sin40oR2{9}=2s2{4}, while the resulting
9-piece dissection is shown in Figure 62.
 {9} to {5} (12)
Figure 66: Theobald’s enneagon to pentagon
For dissecting an enneagon to another figure, Theobald has found a
different element that used 5 pieces. He first cut the isosceles triangle
atop from enneagon, next he cut additional triangle off the trapezoid to fit
the right top corner, and finally he cut a small triangle off the remaining
trapezoid to be combined with it to a parallelogram. His partition and the
rearrangement shown in Figure 67 is a new P-strip element for the
enneagon.
Figure 67: Theobald’s partition and element for an enneagon
The P-strip element for pentagon is shown in Figure 22. In order to
33
prevent the isosceles triangle from partitioning, Theobald crossposed
these two strips as in Figure 68. The relation between the radius of
enneagon and pentagon is 9sin40oR2{9}=5sin72oR2{5}, while the width
of strip for pentagon is 2sin36o cos18 o R{5}. This gave a 12-piece
dissection of an enneagon to a pentagon as shown in Figure 66.
Figure 68: Theobald’s croospotition of {9} and {5}
 {9} to {6} (11)
Figure 69: Theobald’s enneagon to hexagon
For dissecting an enneagon to a hexagon, Theobald used his
enneagon element shown in Figure 67 and the element of hexagon shown
in Figure 30. The croosposition for strips is shown in Figure 70. The
relation between the radius of enneagon and hexagon is
3sin40oR2{9}=3sin60oR2{6}, while the width of strip for hexagon is
sin60oR{6}. Theobald’s 11-piece dissection of enneagon to hexagon is
shown in Figure 69 (Theobald, 2001).
34
Figure 70: Crossporition of {9} and {6}
 {9} to {7} (14)
Figure 71: Theobald’s enneagon to heptagon
Theobald based on his ingenious strip element of heptagon and
enneagon shown in Figure 36 and 64 respectively. He produced a
14-piece dissection by crossposing these strips as shown in Figure 72.
The relation between the radius of enneagon and heptagon is
9sin40 o R2{9}=7sin(2π/7) R2{7}, while the width of heptagon’s strip
element is 2sin(2π/7)R{7}. The resulting 14-piece dissection of an
enneagon to a heptagon is shown in Figure 71 (Theobald, 2001).
35
Figure 72: Crosspotition for element of {9} and {7}
 {10} to {3} (7)
Figure 73: Theobald’s decagon to triangle
This is Gavin Theobald’s 7-piece dissection of a decagon to a
triangle. It uses a customized strip based on the general-purpose decagon
strip suggested in Figure 76, and its crossposition shown in Figure 74
(Frederickson,1997). The relation between the radius of decagon and
triangle is 10sin36oR2{10}=3sin120oR2{3}, while the width of strip for
triangle is 3R{6}/2.
Figure 74: Crossposition of strips for {10} and {3}
36
 {10} to {4} (7)
Figure 75: Theobald’s decagon to square
An exciting feature of Gavin Theobald's work is his invention of a
technique which is derived from a crossposition for a specific dissection,
with the strip altered so that certain of its line segments coincide with line
segments in the other strip. Frederickson has called it the customized
strips technique (Frederickson,1997).
Figure 76: Partition of {10}
Figure 77: Element of {10}
First, Theobald partitioned a decagon into four pieces that forms a
general purposed decagon strip element, as shown with solid and dotted
lines in Figures 76 and 77. Second, he crossposed this strip with a strip of
squares. The doted lines in Figure 76 and 77 shows the original partitions
and the dashed lines indicate the new partitions been altered. The final
corssposition is shown in Figure 78. The relation between the radius of
decagon and square is 5sin36 o R2{10}= s2{4}. The resulting 7-piece
dissection of a decagon to a square is shown in Figure 75 improved that
of Lindgren’s in 1964 by one piece.
37
Figure 78: Theobald’s crossposition for decagon and square
 {10} to {7} (11)
Figure 79: Theobald’s decagon to heptagon
For dissecting a decagon to a heptagon, Theobald again used the
method of customized strip approach but on a different strip element for
decagon. First, he cut the decagon into four pieces and rearranged them
into P-strip element as shown in Figure 80.
Figure 80: First partition and element of decagon
38
The P-strip element for heptagon is shown in Figure 36. He then
crossposed these two strips. In order that the triangle inside the heptagon
be prevented from partitioning, he made a second cut as in Figure 81,
again producing a second element.
Figure 81: Second partition and element of decagon
Finally, he crossposed these two strips as in Figure 82. The relation
between
the
radius
of
decagon
and
heptagon
is
o 2
2
10sin36 R {10}=7sin(2/7)R {7}, while the width of heptagon’s strip
element is 2sin(2 π /7)R{7}. The resulting 11-piece dissection of a
decagon to a heptagon is shown in Figure 79 (Theobald, 2001).
Figure 82: Crossposition of decagon and heptagon
39
 {10} to {8} (10)
Figure 83: Theobald’s decagon to octagon
This 10-piece dissection of a decagon to an octagon was found by
Theobald. He used customized strips approach for two figures, both for
decagon’s strip element and octagon’s strip element. The original
partition of the octagon is shown in Figure 54, while the original partition
of the decagon is shown in Figure 77. And their customized strip
elements are shown in Figure 84.
Figure 84: Customized strip element for {8}’s and {10}’s
The crossposition of strips for decagon and octagon is shown in
Figure 85. The relation between the radius of decagon and octagon is
5sin36o R2{10}=4sin45o R2{8}, while the width of octagon’s strip
element is 2sin45o R{8}. The resulting 10-piece dissection of a decagon
to a octagon is shown in Figure 79 (Theobald, 2001).
40
Figure 85: Crossposition for octagon and decagon
 {10} to {9} (13)
Figure 86: Theobald’s decagon to enneagon
As the element of decagon shown in Figure 67, Theobald found a
similar way of cutting the enneagon into 5 pieces to form a strip element
as shown in Figure 87. Instead of cutting the isosceles triangle from the
top of enneagon, he cut it away from the bottom of the largest piece. Then
he picked the decagon's strip element as shown in Figure 80. The
crossposition for strips of decagon and enneagon is shown in Figure 88.
The relation between the radius of decagon and enneagon is
10sin36oR2{10}=9sin40oR2{9}, while the width of strip of enneagon is
2sin40oR{9}. Theobald's 13-piece dissection of a decagon to an enneagon
is shown in Figure 86.
41
Figure 87: Element of {9}
Figure 88: Crossposition for {10} and {9}
 {12} to {4} (6)
Figure 89: Lindgren’s dodecagon to square
It was Lindgren (1951) who found an impressive dissection of a
dodecagon to a square that used a technique of superposing tessellations.
Lindgren first cut the dodecagon into four pieces that rearrange to give
the tessellation element in Figure 90. He then superposed this tessellation
over a tessellation of squares (Figure 91) to give a 6-piece dissection
shown in Figure 89 (Frederickson, 1997). In this dissection, we can
partially hinged four equal pieces and translate the rest pieces.
Figure 90: {12} element
Figure 91: Tessellations : {12}, {4}
42
 {12} to {5} (10)
Figure 92: Dodecagon to pentagon
This time, Theobald used his technique of customized strip approach
on the strip of pentagon. He first partitioned the decagon into four pieces
and rearranged them into a strip element as shown in Figure 93. He then
crossposed it with the strip of pentagon whose element is shown in Figure
22. Crossposing two strips would produce two small triangles inside the
pentagon’s element as the left of Figure 94.
Figure 93: Partition and element of {12}
Figure 94: First and second element of pentagon
43
For minimizing pieces, he altered the left two pieces to the dashed
line and cut a small triangle off the right-down side to rotate it 180
degrees to the right-up side and made a new element that is a
parallelogram. He crossposed these strips as in Figure 95. The relation
between the radius of each figure is 12sin30oR2{12}=5sin72oR2{5}, while
the width of strip for pentagon is 2sin36ocos18oR{5}. The resulting
10-piece dissection of a dodecagon to a pentagon is shown in Figure 92
(Theobald, 2001).
Figure 95: Crossposition for dodecagon and pentagon
 {12} to {6} (6)
Figure 96: Freese’s dodecagon to hexagon
We construct this dissection by the method of completing the
tessellations. First, adding two equilateral triangles to the dodecagon
forms a tessellation element with repeating pattern that is a hexagon. Also,
a hexagon and two identical equilateral triangles form the tessellation
44
element of its repeating pattern that is a hexagon. In both cases the
repeating pattern is the same, thus we can superpose them over to
produce the dissection. If the tessellation of dodecagon and hexagon are
superposed when the areas of the dodecagon and hexagon are equal and
the triangles are all identical, then Figure 97 results. The relation between
the radius of each figure is 12sin30oR2{12}=6sin60oR2{6}, while the
width of hexagon is 2cos30oR{6}. We can construct the position for
hexagons by the repeating patterns or by constructing the width of
hexagon from vertex of dodecagon to the fourth adjacent vertex. The
associated 6-piece dissection is shown in Figure 96 Lindgren (1964b)
credits this dissection to Ernest Irving Freese (Frederickson, 1997).
Figure 97: Tessellation: {12}, {6}
Observed that this dissection has 2-fold rotational symmetry and can
be partially hinged. The four pieces other than the two small triangles can
be hinged in two ways.
 {12} to {7} (11)
Figure 98: Theobald’s dodecagon to heptagon
45
Choosing the element of dodecagon in Figure 99, and Theobald’s
heptagon element in Figure 36, arranging them in crossposition as Figure
100. The relationship between the radius of each figure is
12sin30o R2{12}=7sin(2π/7)o R2{8}, while the width heptagon’s element
is 2sin(2π/7)R{7}. The resulting 11-piece dissection of a dodecagon to a
heptagon is shown in Figure 98.
Figure 99: Partition and element of {12}
Figure 100: Crossposition for {12} and {7}
 {12} to {8} (10)
Figure 101: Theobald’s dodecagon to octagon
46
This 10-piece dissection still uses the technique of customized strip.
Since octagon has lots of partition according in Figure 54. We choose the
element of octagon together with the dodecagon’s element in Figure 99.
Rearrange them in crossposition and then cut the octagon element to the
desired customized strip. Their crossposition is shown in Figure 102. The
relationship
between
the
radius
of
each
figure
is
o 2
o 2
3sin30 R {12}=sin45 R {8}, while the width of the element of octagon is
2sin45o R{8}. The resulting 10-piece dissection is shown in Figure 101.
Figure 102: Crossposition for strips of {12} and {8}
 {12} to {9} (14)
Figure 103: Theobale’s dodecagon to enneagon
Theobald begin with his enneagon element in Figure 67, together
with dodecagon’s element in Figure 99. The relationship between the
radius of each figure is 12sin30oR2{12}=9sin40oR2{9}, while the width of
octagon’s strip element is 2sin45oR{8}. He crossposed them as Figure
104. This results in a 14-piece dissection of dodecagon to enneagon
47
shown in Figure 103.
Figure 104: Crosspotition for {12} and (9)
 {12} to {10} (12)
Figure 105: Theobald’s dodecagon to decagon
Theobald first partitioned a decagon into four pieces and found the
general-purpose decagon strip element, as shown is Figure 77. Then he
chose the element of dodecagon in Figure 99, crossposing it with
decagon’s element as in Figure 107, then alter decagon’s element as a
customized strip element as Figure 106. In decagon’s element, the
original cuts are shown in dashed lines and final cuts in real lines. This
results in a 12-piece dissection as shown in Figure 105.
48
Figure 106: element {10}
Figure 107: Crossposition : {12}, {10}
Actually, Theobald has found an 11-piece dissection of a dodecagon
to a decagon in which two piece have to turn over.
49
5.2 Star Polygon
 {5/2} to {4} (7)
Figure 108: Tilson’s pentagram to square
Lindgren (1958) gave an 8-piece dissection of dissecting a
pentagram to a square. But Philip Grahom Tilson found a better result
that use only 7 pieces (Frederickson, 1997).
Figure 109: {5/2} element
Figure 110: Crossposition of {5/2}, {4}
Tilson found a way to cut the pentagram into four pieces that form a
P-strip element, as shown in Figure 109. He crosspossed this strip with
the strip of square as shown in Figure 110. This gives a 7-piece dissection
shown in Figure 108.
When we crosspose these strips, the area of a pentagram can be
compute from its P-strip element. And the square root of its area is the
side length of the desired square.
50
 {5/2} to {10} (6)
Figure 111: Pentagram to decagon
Lindgren (1964b) shown a 6-piece dissection of a pentagram to a
decagon as in Figure 111. He found that a decagon consists of five 72 o
-rhombuses and five 36 o –rhombuses. And he also found that the
relationship between the side lengths of the pentagram and the decagon is
s{5/2}=2cos(π/10)s{10} (Frederickson, 1997). Based on these ideas, he
found such a dissection.
When rearranging the pieces from a pentagram to a decagon, there
are two pieces that can be hinged.
 {6/2} to {3} (5)
Figure 112: Mott-Smith’s {6/2} to triangle
Geoffrey Mott-Smith (1964) found several 5-piece dissections of a
hexagram to a triangle (Frederickson, 1997). This dissection shown in
Figure 112 is one of them that can be fully and variously hinged.
51
Figure 113: {6/2} element
Figure114: Tessellation for {6/2}
Mott-Smith’s first partitioned the hexagram into two pieces that
form an element as shown in Figure 113. This element has a spectacular
characterization that it can tile the surface of an octahedron.
Figure 114 shows the portion of how to rearrange the hexagram’s
element to tile the surface of an octahedron, the dashed lines indicates the
triangular faces of an octahedron. We can fold the eight hexagram
elements on an octahedron along the dashed lines to derive the surface as
in Figure 115.
With the help of Maple 8, we can construct a 3D version of the
tilling of the hexagram element on an octahedron. We first form an
octahedron centers at (0,0,0) with radius of 1 unit. Then, we construct the
patterns on one of its surface by the command of “segment”. Once a face
is constructed, the other faces can be formed by rotation and reflection.
The source code is given afterwards.
Figure 115: On an octahedron
52
Construct polyhedral tessellation with Maple 8:
> restart;
> with(plots):
> with(geom3d):
>point(A,0,0,1);point(B,0,1,0);point(C,1,0,0);point(D,1/2,1/2,0);point(E,1
/2,0,1/2);point(F,0,1/2,1/2);point(O,1/3,1/3,1/3);point(G,1/6,1/6,2/3);poi
nt(H,1/6,2/3,1/6);point(J,2/3,1/6,1/6);
>segment(AG,[A,G]);segment(GE,[G,E]);segment(EJ,[E,J]);segment(EO,
[E,O]);segment(JC,[J,C]);segment(OD,[O,D]);segment(DH,[D,H]);seg
ment(HB,[H,B]);
>p1:=draw(AG):p2:=draw(GE):p3:=draw(EJ):p4:=draw(EO):p5:=draw(J
C):p6:=draw(OD):p7:=draw(DH):p8:=draw(HB):
> q1:=display(p1,p2,p3,p4,p5,p6,p7,p8):
> with(plottools):
> q2:=rotate(q1,0,0,Pi/2):
> q3:=rotate(q1,0,0,Pi):
> q4:=rotate(q1,0,0,3*Pi/2):
> q5:=rotate(q1,0,0,2*Pi):
> r1:=display(q2,q3,q4,q5,scaling=constrained):
> s1:=reflect(q1,[[0,0,0],[0,0,1],[0,1,0]]):
> s2:=reflect(s1,[[0,0,0],[1,0,0],[0,1,0]]):
> q6:=rotate(s2,0,0,Pi/2):
> q7:=rotate(s2,0,0,Pi):
> q8:=rotate(s2,0,0,3*Pi/2):
> q9:=rotate(s2,0,0,2*Pi):
>t1:=display(r1,q6,q7,q8,q9,color=black,scaling=constrained):
> with(geom3d):
> octahedron(k,point(a,[0,0,0]),1);
> t2:=draw(k,style=patchnogrid):
> display(t1,t2);
53
Another way of cutting {6/2} to triangle (5)
Figure 116: Hexagram to triangle
This 5-piece dissection of hexagram to triangle can be fully and
variously hinged as shown in Figure 116.
 {6/2} to {4} (5)
Figure 117: Bradley’s hexagram to square
In the United States, Sam Loyd (Eliz.J., 1908a) had given a 7-piece
dissection of a hexagram to a square. But remarkably, Bradley found a
dissection that only needs 5-pieces as shown in Figure 117 (Frederickson,
1997).
Bradley first cut two equilateral triangles off a hexagram and
rearranged them as a P-strip element as shown in Figure 118. Then he
crossposed this strip with the strip of squares as in Figure 119. The area
of the hexagram can be compute from its P-strip element.
54
Figure 118: {6/2} element
Figure 119: Crossposition: {6/2}, {4}
 {6/2} to {6} (6)
Figure 120: Frederickson’s hexagram to hexagon
Lindgren (1972) found a 7-piece dissection of a hexagram to a
hexagon that used the PP dissection. Frederickson (1997) got one piece
better by using the technique of tessellation. He found a 6-piece
dissection in which two pieces have to turn over.
Except for dissecting a hexagram into a strip element, a hexagram
can form a tessellation by adding two equilateral triangles to each
hexagram. Fortunately, a hexagon forms a non-edge-to-edge tessellation
by additional two equilateral triangles to each hexagon. Since both of the
tessellation elements have the same repeating pattern that is a hexagon,
we can superpose these tessellations by letting the area of hexagon equal
to the area of hexagram and letting the additional equilateral triangles of
hexagon equal to the added triangles of hexagram.
55
Figure 121: Tessellations: {6/2}, {6}
We superpose them as Figure 121, and let one of the added
equilateral triangles placed inside the hexagram. It results in an added
triangle of the hexagram been cut into three pieces. In the superposition
of two tessellations, if both of the added figures are placed inside the
target figure, we don’t need to do any changes. But if the added figure
does not lie wholly inside the target figure, we need to do some change.
As this equilateral triangle been cut into three pieces as in Figure 121, we
need to turn over two pieces which is used and leave the third pieces
unused in the dissection.
 {8/2} to {4} (7)
Figure 122: Frederickson’s {8/2} to square
Lindgren (1964b) discovered how to cut the {8/2} into five pieces
that form a tessellation element, which let to an 8-piece dissection of an
{8/2} to a square. However, Frederickson improved this to seven pieces
in (Frederickson 1972d) and give a variation of it in Figure 122. In Figure
56
123, we cut the star into four pieces to form a tessellation element and
superpose its tessellation with one for squares in Figure 124.
Figure 123: {8/2} element
Figure 124: Overlay {8/2}, square
The relationship between each figure is 4cos22.5os2{8/2}= s2{4}.
Another way of cutting {8/2} to {4} (5)
Figure 125: {8/2} to square
Figure 125 is another way of cutting {8/2} to a square in which one
piece has one full side of the square plus a half of another side
(Frederickson, 1997).
57
 {8/3} to {4} (8)
Figure 126: Frederickson’s {8/3} to square
When completing the tessellation with the stars, sometimes adding
regular polygons is not possible to tile the plane. We will then add some
irregular figures, like this {8/3} to a square.
Figure 127: Tessellation: {8/3}, {4}
By adding a 4-armed twisted cross to each square, we can form a
tessellation and its repeating pattern is a square. Similarity, we can form a
tessellation by adding a small square to a large square. If the large square
has equal area with the {8/3} and the added small square has equal area
with the 4-armed twisted cross, we can superpose these strips as Figure
127. The relationship between the {8/3} and the square is
8sin22.5o(cos22.5o-sin22.5o)R2{8/3}= s2{4}.
The remaining problem is to cut a 4-armed twisted cross to a square.
Lucky, this 4-armed twisted cross can form a tessellation by itself. We can
superpose the tessellation of 4-armed crosses with the tessellation of
58
squares over it as shown in Figure 128. The relationship between the
twisted cross and the small square is (5-4cos45o)s2{8/3}=x2, where x is
the side length of the small square.
Figure 128: Tessellation: 4-armed twisted crosses and squares
This dissection is found by Frederickson (1997). It has the same
number of pieces as that of by Lindgren. But this dissection behaved
greater symmetry. The four pieces of the star are translational and can be
hinged. And the other four pieces are tranglational.
 {10/2} to {5} (7)
Figure 129: Frederickson’s {10/2} to pentagon
Lindgren (1964) found an 8-pieces dissection of a {10/2} to a
pentagon by using trial and errors. Frederickson (1974) found this 7-piece
dissection that use the technique of tessellation on regular polyhedrons.
59
Figure 130: {10/2} element
Frederickson first partitioned a {10/2} and rearranged it as a
tessellation element with its repeating pattern that is a pentagon as shown
in Figure 130. The dashed lines indicate the desired pentagon and the
arrows indicate how the tessellation is formed. He tessellates these
elements on the dodecahedral. Its tessellation is shown in Figure 131 with
twelve {10/2} elements. We can fold this tessellation on a dodecahedral
as in Figure 132.
Figure 131: Dodecahedral tessellation for {10/2}
We can construct the tessellation for {10/2} on a dodecahedral with
maple 8. The relationship between the {10/2} and the pentagon is
4cos18ocos36os2{10/2}= R2{5}. We thus let s{10/2} equal to 1 during the
construction. We construct one face of a dodecahedron with the command
of “segment”. The rest faces of the dodecahedron can be formed by
reflection and rotation. The resulting 3D version is in Figure 132.
60
Figure 132: On a dodecahedron
Construct the tessellation on a dodecahedron with Maple 8:
> restart;
> with(plots):
> with(geom3d):
> ang:=Pi/10;
> r:=2*sqrt(cot(ang)*cos(2*ang));
> tau:=(sqrt(5)+1)/2;
> k:=tau*r*sin(2*ang);
> h:=k*tau^2/sqrt(2*tau^2+1/tau^2-2);
>point(A,r*cos(ang),r*sin(ang),h):point(B,r*cos(5*ang),r*sin(5*ang),h):
point(C,r*cos(9*ang),r*sin(9*ang),h):point(D,r*cos(13*ang),r*sin(13*
ang),h):point(E,r*cos(17*ang),r*sin(17*ang),h):point(F,r*cos(17*ang),r
*sin(17*ang)+1,h):point(G,(r*cos(13*ang)+r*cos(17*ang))/2,(r*sin(13
*ang)+r*sin(17*ang))/2,h):point(H,(r*cos(13*ang)+r*cos(17*ang))/2,(r
*sin(13*ang)+r*sin(17*ang))/2+1,h):point(J,r*cos(13*ang),r*sin(13*an
g)+1,h):point(K,(r*cos(13*ang)+r*cos(17*ang))/2+cos(ang),(r*sin(13*
ang)+r*sin(17*ang))/2+1-sin(ang),h):point(L,(r*cos(13*ang)+r*cos(17
*ang))/2+cos(ang),(r*sin(13*ang)+r*sin(17*ang))/2+sin(ang),h):point(
M,r*cos(13*ang)+cos(ang),r*sin(13*ang)+1-sin(ang),h):point(N,(r*cos
(13*ang)+r*cos(17*ang))/2+cos(ang)+cos(ang*3),(r*sin(13*ang)+r*sin
(17*ang))/2+1-sin(ang)+sin(ang*3),h):point(P,(r*cos(13*ang)+r*cos(17
*ang))/2+cos(ang)+cos(ang*3),(r*sin(13*ang)+r*sin(17*ang))/2+2-sin(
ang)+sin(ang*3),h):point(Q,(r*cos(13*ang)+r*cos(17*ang))/2+2*cos(a
ng)+cos(ang*3),(r*sin(13*ang)+r*sin(17*ang))/2+2+sin(ang*3),h):poin
t(R,r*cos(13*ang)+cos(ang)-cos(3*ang),r*sin(13*ang)+1-sin(ang)+sin(
3*ang),h):point(S,r*cos(13*ang)+2*cos(ang)-cos(3*ang),r*sin(13*ang)
61
+1+sin(3*ang),h):point(T,r*cos(13*ang)+cos(ang)-cos(3*ang),r*sin(13
*ang)+2-sin(ang)+sin(3*ang),h):point(U,r*cos(13*ang)+2*cos(ang)-co
s(3*ang),r*sin(13*ang)+2+sin(3*ang),h):point(V,r*cos(13*ang)-cos(3*
ang),r*sin(13*ang)+2+sin(3*ang),h):
>segment(AB,[A,B]):segment(BC,[B,C]):segment(CD,[C,D]):segment(D
E,[D,E]):segment(AE,[A,E]):segment(EF,[E,F]):segment(GH,[G,H]):se
gment(DJ,[D,J]):segment(KF,[K,F]):segment(KH,[K,H]):segment(LE,[
L,E]):segment(LG,[L,G]):segment(MH,[M,H]):segment(MJ,[M,J]):seg
ment(KN,[K,N]):segment(NP,[N,P]):segment(PQ,[P,Q]):segment(QA,[
Q,A]):segment(MR,[M,R]):segment(RS,[R,S]):segment(RT,[R,T]):seg
ment(US,[U,S]):segment(UT,[U,T]):segment(VT,[V,T]):segment(VC,[V
,C]):
>p1:=draw(AB):p2:=draw(BC):p3:=draw(CD):p4:=draw(DE):p5:=draw(
AE):p6:=draw(EF):p7:=draw(GH):p8:=draw(DJ):p9:=draw(KF):p10:=
draw(KH):p11:=draw(LE):p12:=draw(LG):p13:=draw(MH):p14:=draw
(MJ):p15:=draw(KN):p16:=draw(NP):p17:=draw(PQ):p18:=draw(QA):
p19:=draw(MR):p20:=draw(RS):p21:=draw(RT):p22:=draw(US):p23:=
draw(UT):p24:=draw(VT):p25:=draw(VC):
>11:=display(p1,p2,p3,p4,p5,p6,p7,p8,p9,p10,p11,p12,p13,p14,p15,p16,p
17,p18,p19,p20,p21,p22,p23,p24,p25,color=black,scaling=constrained):
> odecahedron(dod,point(aaa,[0,0,0]),k*sqrt(tau^2+1/tau^2));
> d1:=draw(dod):
> with(plottools):
> ang1:=arccos(tau/sqrt(2*tau^2+(1/tau^2)-2)):
> d2:=rotate(d1,0,ang1,-Pi/10):
> q1:=rotate(q11,0,0,0):
> q21:=reflect(q1,[[0,0,0],[0,1,-tau],[1,1,-tau]]):
> q2:=reflect(q21,[[0,0,0],[0,1,0],[0,0,1]]):
> q31:=rotate(q1,0,0,4*Pi/5):
> q32:=reflect(q31,[[0,0,0],[0,1,-tau],[1,1,-tau]]):
> q33:=reflect(q32,[[0,0,0],[0,1,0],[0,0,1]]):
> q3:=rotate(q33,0,0,-2*Pi/5):
> q41:=rotate(q1,0,0,2*Pi/5):
> q42:=reflect(q41,[[0,0,0],[0,1,-tau],[1,1,-tau]]):
> q43:=reflect(q42,[[0,0,0],[0,1,0],[0,0,1]]):
> q4:=rotate(q43,0,0,-4*Pi/5):
> q51:=rotate(q1,0,0,-2*Pi/5):
> q52:=reflect(q51,[[0,0,0],[0,1,-tau],[1,1,-tau]]):
62
> q53:=reflect(q52,[[0,0,0],[0,1,0],[0,0,1]]):
> q5:=rotate(q53,0,0,-6*Pi/5):
> q61:=rotate(q1,0,0,6*Pi/5):
> q62:=reflect(q61,[[0,0,0],[0,1,-tau],[1,1,-tau]]):
> q63:=reflect(q62,[[0,0,0],[0,1,0],[0,0,1]]):
> q6:=rotate(q63,0,0,2*Pi/5):
> r1:=display(q1,q2,q3,q4,q5,q6):
> r2:=rotate(r1,Pi,0,0):
> display(r2,r1,d2);
 {10/3} to two {5/2}s (10)
Figure 133: Lindgren’s {10/3} to two {5/2}s
Lindgren (1972) discovered this masterful dissection of a {10/3} to
two {5/2}s as shown in Figure 133. He found that the relationship
between the {10/3} and the pentagram is R2{10/3}=2s2{5/2}.
This dissection possesses 10-fold rotational symmetry. We can cut a
{10/3} into two parts and each part consists of five identical pieces that
can be hinged to form a pentagram.
63
 {12/2} to {3} (6)
Figure 134: Lindgren’s {12/2} to triangle
It is Lindgren's 6-piece dissection of a {12/2} to a triangle. The
relation between each figure is s{3}=8sin60o(sin60o+sin30o)s{12/2}. And
its cut is only along the sides or diagonals of the constituent rhombuses
(Frederickson, 1997). This dissection is beautiful for its reflection
symmetry on the pieces.
 {12/2} to {4} (8)
Figure 135: Theobald’s {12/2} to square
The strips mentioned above are of fixed width. This time we
introduce Robert Reid’s (1987) bumpy strip.
Reid partitioned the {12/2} into five pieces and rearranged them into
a bumpy P-strip as shown in Figure 136. The plain strip of {12/2} is
composed of the dashed lines and solid lines. That is, the outward bumps
matches the inward bumps. But Reid did not get the minimal dissection.
Theobald modified Reid’s dissection and gave a crossposition in Figure
136. This improved that of Reid’s by two pieces.
64
Figure 136: Bumpy crossposition of {12/2}s and squares
Since this dissection has two fold of rotational symmetry, we can
move the piece elegantly. We first move the four outer small pieces from
the {12/2} and rearrange them into the desired place. Second, we
swing-hinge the inner four pieces to get a square.
 {12/2} to {6} (8)
Figure 137: Frederickson’s {12/2} to hexagon
For dissecting a {12/2} to a hexagon, Lindgren (1964) found a
10-piece dissection. But Frederickson (1972) found better result that used
only 8 pieces.
65
Figure 138: {12/2} element
Figure 139: Overlay {12/2}, hexagon
Frederickson cut a {12/2} star into five pieces that rearrange to a
tessellation element as shown in Figure 138. Its repetition pattern is a
hexagon. He then superposed this tessellation with the tessellation of
hexagons as in Figure 139. This gave an 8-piece dissection. The relation
between each figure is 2tan30oR{12/2}= R{6}. During the animation, we
first move the smaller four pieces to the desired place. We then
swing-hinge the bottom three pieces to the top of the {12/2}.
 {12/3} to {4} (9)
Figure 140: Frederickson’s {12/3} to square
Frederickson (1972) found a 10-piece dissection of a {12/3} to a
square. Again, he found a 9-piece dissection in which two pieces have to
turn over (Frederickson, 1997).
66
Figure 141: First {12/3} element
Figure 142: Second {12/3} element
He first cut away four outside pieces of the {12/3} and leaves a
square as shown in Figure 141. The four outside pieces form a square.
The first partition and rearrangement is shown in Figure 141. Since the
two squares forms a tessellation of squares, he combined the two pairs of
pieces and rearranged them to the tessellation element of a square as
shown in Figure 142. He superposed this tessellation with the one of
square to derive this 9-piece dissection as shown in Figure 140. Two
pieces denoted by “*” have to turn over.
 {12/4} to {4} (11)
Figure 143: Frederickson’s {12/4} to square
67
Frederickson (1997) found this 11-piece dissection of a {12/4} to a
square by using the tessellation technique. The {12/4} forms a
tessellation element with an additional ragged polygon as shown in
Figure 144. Its repetition pattern is a square. He thus superposed this
tessellation with the non-edge-to-edge tessellation of squares. The area of
the larger square is equal to the area of a {12/4} and the area of the
smaller square is equal to that of the ragged polygon. The relation
between
the
{12/4}
and
the
larger
square
is
12sin15o(cos15o-sin15o)R2{12/4}= s2{4}. The relation between the {12/4}
and the smaller square is 5s2{12/4}= x2, where x denotes the side length
of the smaller square. Since the ragged polygon can be cut into five
pieces that rearranged to form two squares, two tessellations were not
placed at center. The extra two right triangles shown in Figure 143 make
the five pieces prevent from partitioning. The resulting 11-piece
dissection is shown in Figure 143.
Figure 144: Tessellations: {12/4}, {4}
68
5.3 Dissected Curves
 Disk to oval seat tops (8)
Figure 146: Jackson’s disk to oval seat tops
For dissecting a disk to two oval seat tops with a handhold in each
oval, it becomes a competition. John Jackson (1821) gave an 8-piece
dissection as shown in Figure 146, which may be the earliest curved
dissection. He cut the disk with a smaller circle and two perpendicular
lines. The smaller circle has half the radius of the original one. Since this
dissection possesses four folds of rotational symmetry, we can translate
all the pieces during the arrangement.
 Disk to oval seat tops (6)
Figure 145: Sam Loyd’s disk to oval seat tops
Jackson gave an 8-pieces dissection, but Sam Loyd (1901) found a
better dissection that used only 6 pieces. His dissection is based on the
69
Chinese tai chi.pattern (Yin and Yang symbol). Loyd first partitioned the
disk into two parts that formed the shape of Yin and Yang. He then cut the
tails of these pieces in order to fit the shape of the ovals. And finally, he
cut two handholds inside the Yin and Yang pieces that fit the top and
down of Yin and Yang symbols. The resulting 6-piece dissection is shown
in Figure 145. It has two folds of rotational symmetry.
 Disk to oval seat tops (6)
Figure 147: Frederickson’s disk to oval seat tops
Except for Loyd’s dissection that used 6 pieces, Frederickson (1997)
also found a 6-piece dissection that is a variation of Jackson’s dissection.
Since the outer pieces of the disk attached the inner pieces when
rearranging to two ovals, Frederickson combined these two pairs of
pieces that reduced the pieces to six. This dissection has two fold of
symmetry and all the pieces are translational.
 Disk to different oval seat tops (4)
Figure 148: Loyd’s disk to different oval seat tops
70
Instead of placing the handholds paralleled with the ovals, Loyd
(1927) rotated the handholds 90 degrees to give this dissection as shown
in Figure 148 that only needs four pieces. This dissection has four folds
of rotational symmetry and all the pieces are translational.
 Disk to ovals with thinner holes (6)
Figure 149: Dudeney’s disk to ovals with thinner holes
This is Dudeney’s (1902) 6-piece dissection of a disk to two ovals
with thinner holes as shown in Figure 149. Dudeney pointed out the
uncertainty of the meaning conveyed by the word “oval.” “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 with out necessarily having the
properties above described are included in the term “oval.”
Figure 150: In the construction
71
In the construction, as shown in Figure 150, we let arc AD having
1/8 of total disk length, and let segment AB equal to BO, arc BC has the
same radius with arc AC. Follow the same way, we then get this
dissection.
 Disk to horseshoes (4)
Figure 151: Dudeney’s one disk to two horseshoes
Dudeney (1958) found this four-piece dissection of a disk to two
horseshoes such that all the pieces are different in shape. As a matter of
fact, it is based on the principle contained in that curious Chinese symbol
the Monad – the Yin and the Yang of great Monad. The above figure
gives the correct solution to this problem. When constructing it, we first
make a Chinese Yin and Yang symbol inside the disk. The rest short arcs
in the disk are quarter of the circumference of small Yin-and-Yang circle.
72
5.4 Dissecting other Figures
There are some figures that are not hard to dissect. We focus here on
some types of Greek Crosses that have the simplest shapes.
 {GC} to {3} (5)
Figure 152: Lindgren’s Greek Cross to triangle
Depending on the T-strip technique, Dudeney (1902) found a 6-piece
dissection of a Greek Cross to a triangle that can be cyclicly hinged.
Lindgren (1961) improved one piece by using the same technique. But he
used the different T-strip element for the Greek Cross.
Figure 153: Crossposition of Greek Cross and triangle
Lindgren cut a small square from the Greek Cross and rearranged
them to a T-strip element. He crossposed the T-strip of Greek Cross over
the T-strip for triangles as in Figure 153, making their anchor points
coincide with each other. The relation between each figure is
73
10s2{GC}=3sin60oR2{3}. This gave a 5-piece dissection as shown in
Figure 152. During the rearrangement, we can partially and variously
hinge all pieces except for the small square. The small square can slide on
its adjacent piece.
 {GC} to {4} (4)
Figure 154: Greek Cross to square
The earliest dissection of a Greek Cross to a square is discovered by
Lemon (1890) as shown in Figure 154. This dissection can be derived
from superposing two tessellations from each figure. The Greek Cross is
itself a tessellation element and its repetition pattern is a square. We thus
superpose this tessellation with one of squares as shown in Figure 155. In
fact, there are infinitely many ways to superpose these two tessellations.
Figure 155 has just shown one of them. The resulting 4-piece dissection
of a Greed to a square is shown in Figure 154.
Figure 155: Overlaying Greek Cross, square
74
During the rearrangement, we can fully and variously hinge these
four pieces. We swing-hinge these pieces in one way to get a Greek Cross
and we swing-hinge in the other way to get a square.
Another {GC} to {4}
Figure 156: Greek Cross to square
This is a symmetrical dissection of a Greek to a square that takes
four pieces. This dissection is derived from superposing two tessellations
such that the corners of the squares are positioned at the center of the
Greek Cross. This dissection can be fully and variously hinged.
 {L’} to {4} (7)
Figure 157 : Szeps’s {L’} to square
There is another type of the Greed Cross called the Cross of Lorraine.
It consists of 13 identical squares and is denoted by {L’} (Frederickson,
1997).
75
Bernard Lemaire (1974) discovered an 8-piece dissection of a {L’}
to a square by finding a tessellation element of {L’} for the plane. But Mr.
Szeps then found a 7-piece dissection by finding a tessellation element of
{L’} for the surface of a polyhedron in which the faces is the desired
figure. That is, he found a tessellation element of {L’} for the surface of a
cube.
Figure 157: {L’} element
Szeps partitioned the {L’} into three pieces and rearranged them into
a tessellation element as shown in Figure 158. The real lines indicate the
element while the dashed lines indicate the desired shape of a square.
This element cannot tile the plane, but it can tile the surface of a cube. We
pick six elements and arrange them as the expansion of the surface of a
cube as shown in Figure 159. We fold this expansion along the dashed
lines in Figure 159. Then, we have a tessellation on the surface of a cube.
Figure 159: Tessellation for the Cross of Lorraine
76
We can construct a 3D version of the tilling of the {L’} element on a
cube with the help of Maple 8. We construct one face of the cube, and the
other faces can be formed by the command of “rotation” and “reflection”.
First, we construct one face of a square as the right of Figure 158. We let
the side length of an identical square of {L’} equals to 1, and the large
square is centered at y-axis. We construct this face with the command of
“point” and “segment”. Then, we rotate this square tan -1(2/3) around the
y-axis. Second, we use the command of “rotation” and “reflection” to
rotate this face to the rest faces of the cube. And finally, we construct a
cube having radius of 39 and centered at (0,0,0). The resulting 3D
version is in Figure 160.
Figure 160: Cross of Lorraine tessellation superposed on the cube
Construct polyhedral tessellation with Maple 8:
> restart;
> with(plots):
> with(geom3d):
> a:=sqrt(13);
>point(A,-1,a,5);point(B,-5,a,-1);point(C,1,a,-5);point(D,5,a,1);point(E,3,
a,7/3);point(F,3,a,-2);point(G,-11/3,a,1);point(H,-2,a,-3);point(J,-1,a,3);
point(K,1,a,3);point(L,1,a,1);point(M,-3,a,1);point(N,-3,a,-1);point(O,-1
,a,-1);point(P,1,a,-1);point(Q,1,a,-3);point(R,-1,a,-3);
>segment(AJ,[A,J]);segment(JK,[J,K]);segment(KL,[K,L]);segment(GL,[
G,L]);segment(MN,[M,N]);segment(BP,[B,P]);segment(PC,[P,C]);segm
ent(QH,[Q,H]);segment(OR,[O,R]);segment(EF,[E,F]);
>p1:=draw(AJ):p2:=draw(JK):p3:=draw(KL):p4:=draw(GL):p5:=draw(M
N):p6:=draw(BP):p7:=draw(PC):p8:=draw(QH):p9:=draw(OR):p10:=d
raw(EF):
77
> q1:=display(p1,p2,p3,p4,p5,p6,p7,p8,p9,p10):
> with(plottools):
> q2:=rotate(q1,0,arctan(2/3),0):
> r1:=display(q2):
> r11:=rotate(r1,0,Pi,0):
> r2:=rotate(r11,0,0,Pi/2):
> r22:=rotate(r2,Pi,0,0):
> r3:=rotate(r22,0,Pi/2,0):
> r33:=rotate(r3,0,0,Pi):
> r4:=rotate(r33,-Pi/2,0,0):
> r44:=rotate(r1,0,Pi,0):
> r5:=rotate(r44,Pi/2,0,0):
> r55:=rotate(r5,0,0,Pi):
> r6:=rotate(r55,0,-Pi/2,0):
>s1:=display(r1,r2,r3,r4,r5,r6,scaling=constrained,color=black):
> with(geom3d):
> cube(c,point(cc,[0,0,0]),sqrt(39));
> s2:=draw(c,style=patchnogrid):
> display(s1,s2);
78
5.5 Swing-hinged Dissections
 Hinged dissection of two hexagons to one with areas 1, 12 and 13
(7)
Figure 161: Frederickson’s cyclicly hinged dissection of hexagons with
areas 1, 12 and 13
For two hexagons to one, Frederickson (1997) has found a method to
generate various kinds of these dissections with different area relations
that only need 7 pieces. One of them which we have seen in Figure 161 is
of areas 1+12=13. Besides having 3-fold rotational symmetry and
translation with no rotation, the pieces from the second hexagon can be
cyclicly hinged, as shown in Figure 162.
Figure 162: Cyclicly hinged: two hexagons to one
During the animation of this dissection, the locus of a hinge is an arc.
That is, the hinge moves on a circle.
79
 Hinged {8} to {4} (7)
Figure 163: Frederickson’s hingeable dissection of octagon to a square
Based on the 5-piece unhingeable dissection of an octagon to a
square, Frederickson have found a 7-piece hingeable dissection
(Frederickson, 2002). In order to have the center square attach with the
other pieces, he cut two identical triangles from one of the four pieces,
and combine it with the square to a trapezoid. The original superposition
for two tessellations is shown in Figure 165 and the cutting of one piece
in shown in Figure 164. These two identical make a fully hinged
dissection possible.
Figure 164: Equal triangles
Figure 165: Tessellation of {8}, {4}
Figure 166: Hinged pieces for an octagon to a square
80
We can fix the rightmost triangle and swing hinge the rest pieces
linearly as shown in Figure 166. Each piece rotate around its adjacent
hinge 180 degrees off one side to make one figure, and rotate around the
other side makes the other figure.
 Hinged {12} to {4} (8)
Figure 167: Frederickson’s hingeable dissection of a dodecagon to a
square
Harry Lindgren (1951) gave a beautiful 6-piece unhingeable
dissection of a dodecagon to a square. In July 2001, Frederickson has
found an 8-piece hingeable dissection based on Lindgren’s 6-piece
dissection. Frederickson first found a hingeable tessellation element as in
Figure 168 which can be formed by cutting and hinging. This element has
same shape as that of Lindgren’s. In Figure 168, the dots indicate the
points of rotational symmetry. Rotating the element around each point as
in Figure 169 thus forms the tessellation.
168: element for {12}
169: Tessellations for {12} and {4}
81
When we superposed the tessellations of dodecagons and squares we
would get such 8-piece hinged dissection as in Figure 167. Frederickson
also show the hinges for a dodecagon to a square as Figure 170
(Frederickson, 2002). In constructing the animation of this dissection, we
may use two steps that one splits the pieces and one closes the pieces.
Figure 170: Hinges for a dodecagon to a square
 Hinged {GC} to {6} (8)
Figure 171: Frederickson’s Greek Cross to Hexagon
For dissection of Greek Crosses to hexagon, Lindgren (1964b) gave
several 7-piece unhingeable dissections, Frederickson found an 8-piece
hingeable dissection as shown in Figure 171 (Frederickson, 2002). He
created a hingeable strip element for Greek Cross and a hingeable strip
element for hexagon. Then he crossposed these two strips as a TT
dissection as in Figure 172. This makes the fully hinged possible. But
surprisingly, it not only can fully hinge all pieces but also can cyclicly
hinge. It is because the sum of the rotating angle from each piece is zero,
thus the linearly hinged pieces can hinge cyclicly.
82
Figure 172:Crossposition: {GC}, {6} Figure 173: Hinges {GC}, {6}
Since the total sum of rotating angle is zero, we can fix one piece
and rotate the rest pieces as what we usually do. This would already make
the pieces behave cyclicly. If we prefer a better animation, we only need
to place the center of a cycle at the original center. Figure 173 shows the
cyclicly hinged pieces.
83
5.6 Flip-hinged Dissection
 Cyclicly flip-hinged {4} to {3} (4)
Figure 174: Cyclicly flip-hinged dissection of square to triangle
As the dissection of square to triangle, we not only can fully hinge
four pieces but also can use another kind of hinges named flip-hinged. A
flip hinge is of a piece rotating 180° through the third dimension along a
vertex that connect with two pieces. For this dissection, we even can flip
all the pieces cyclicly. As Figure 175 shown, the dashed lines indicates
the flipping axis and the dot indicates hinges.
Figure 175: Cyclicly flip-hinged: {4} to {3}
Note that during the construction, when we flip over four pieces
along the dashed lines, we do not get exactly the figure we want. We need
to rotate four pieces a small angle to make them coincide to a straight line
and connect hinges to form a cycle.
84
 Partially cyclicly flip-hinged {8} to {4} (5)
Figure 176: Partially cyclicly flip-hinged dissection of octagon to square
In the 5-piece dissection of a octagon to a square, since the outer
four pieces can fully hinged and their pieces were connected end to end,
we can flip these four pieces along four dashed segments of joining their
hinges in Figure 176. Then rotate the square around the center of octagon
about 20.5302 degree. Figure 177 shows the motion of flipping from
octagon to square.
Figure 177: Cyclicly flip-hinged: octagon to square
85
 Partially Cyclicly flip-hinged {6} to {3} (5)
Figure 178: Partially and cyclicly flip-hinged dissection of a hexagon to
a triangle
As we have seen a 5-piece dissection of hexagon to triangle in
Figure 178. It has four pieces swing-hinged and connected end to end.
Thus we can flip these four pieces cyclicly along the segments connecting
their hinges and move the small equilateral triangle to the desired place.
Note that after flipping four points along these segments we should move
some of the pieces to make them into a cycle.
Figure 179: Cyclicly flip-hinged of outer four pieces
86
 Fully Cyclicly flip-hinged {6/2} to {3} (5)
Figure 180: Fully and cyclicly flip-hinged dissection of hexagram to
triangle
Geoffrey Mott-Smith (1964) found several 5-piece dissections of
hexagram to a triangle as we have seen one of them in Figure 112, a
dissection of hexagram to triangle that can fully and variously hinged.
Although these pieces were connected end to end, but they are not rotated
180 degree during the swing. So we choose another dissection of him
shown in Figure 116 which can be fully swing-hinged, pieces connect end
to end and each piece rotate 180 degree. We make it to flip-hinge
dissection. As Figure 181 shown, we flip four pieces over the segments
which joining their hinges, then a equilateral triangle appears in front of
us. No extra movement were needed.
Figure 181: Cyclicly flip-hinged for {6/2} to {3}
87
5.7 Twist-hinged Dissections
 Twist-hinged dissection of {4} to {3} (8)
Figure 182: Frederickson’s twist-hinged dissection of square to triangle
For swing-hinged dissections, Frederickson has discovered general
techniques for converting many swing-hinged dissection to twist-hinged
ones. His technique were to make the new piece be the union of two
isosceles triangles, one carved from each piece attached to the swing
hinge (Frederickson, 2002). A simply way is to draw a small circle center
at the hinges, this would easily get such isosceles triangles and the new
triangle is a right triangle. This construction used eight pieces, which is
differed from that of Frederickson’s 7-pieces dissection for we can start to
twist-hinge this dissection from any pieces with star notations.
In Figure 182, the star “*” indicates the pieces has to turn over for
odd times and the small circle indicates the twisting center.
Figure 183: Twist-hinged pieces
During the animation, we first twist the rightmost piece of a square,
and twist the following pieces turn by turn until a triangle is appeared.
88
 Twist-hinged dissection of {12} to {6} (9)
Figure 184: Twist-hinged dissection of dodecagon to hexagon
This twist-hinged dissection as shown in Figure 184 is a
modification from that of Frederickson’s that used the same pieces with
him (Frederickson, 2002). We begin from superposing the tessellations of
dodecagon and hexagon as in Figure 185 such that the added small
triangle is placed at the center of dodecagon. We add two isosceles
triangles to each small triangles, this would produce three triangles that
we can twist hinge it. We then cut a trapezoid between two equilateral
triangles such that we can twist them to from each other. And finally, we
attach an isosceles triangle to the middle triangle to adjust the position of
triangle. The resulting 9-pieces dissection is shown in Figure 184.
Figure 185: Tessellation for dodecagons and hexagons
89
The animation of this dissection can be constructed from fixing one
of the pieces that can connect to the rest pieces linearly. And
twist-hinging the rest pieces step by step. Figure 186 shows how the
animation of twist-hinge is formed.
Figure 186: Twist-hinged pieces
 Twist-hinged dissection of three {6}s to one
Figure 187: Frederickson’s twist-hinged dissection of three hexagons to
one
For hexagrams having areas with 12 + 22 + 22 = 32 , there is a simple
6-piece unhinged dissection. Robert Reid found a nifty 7-piece hinged
dissection. Frederickson converted Robert’s hinged dissection to get a
nifty 8-piece twist-hinged dissection shown in Figure 187 (Frederickson,
2002).
90
Figure 188: Twist-hinged pieces
For constructing the animation of these figures, we should find the
untwisted pieces first. Then step by step flip the rest pieces linearly.
Figure 188 shows the how to start the flip.
 Twist-hinged dissection of two pentagon plus two pentagram to a
decagon (8)
Figure 189: Frederickson’s twist-hinged dissection for two {5}s plus two
{5/2}s to one {10}
Vlfred Varsady found an unhinged 6-piece dissection of two
pentagons plus two pentagrams to a decagon having the equal side length.
Frederickson found a 12-piece swing-hinged dissection. It is interesting
that not after long Frederickson found a twist-hinged dissection that only
needs 8 pieces as shown in Figure 189 (Frederickson, 2002).
91
Figure 190: Twist-hinged pieces
We start flipping from the pieces that can be connected to the rest
pieces linearly. Then step by step twist the following pieces. Figure 190
shows the animation during the flipping.
92
6. Dissecting Three-dimensional
Figures
Theorems have proved that any two-dimensional rectilinear figures
with equal area can be dissected into a finite number of pieces to form
each other. But what has happened when we shift two-dimensional to
three-dimensional figures? David Hibert (1900) conjectured in his famous
23 problems that there are pairs of polyhedra with equal area that can’t be
dissected one into another in finite number of pieces. And his student
Max Dehn (1900) proved it. But there are still some solids that we can
dissect it to another figure easily.
In order to simulate a three-dimensional figure on a plane screen, we
should construct a frame in three-dimensional space and realize how it
works when demonstrating three-dimensional animations.
6.1 Frame in Three-dimensional Space
First, we construct the rotation matrix as following:
Equation
Condition
Rotation Matrix
Rotate around x-axis
counterclockwise with
degree of θ.
w1  x

w2  y cos  z sin 
 w  y sin   z cos
 3
0
1
0 cos

0 sin 
Rotate around y-axis
counterclockwise with
degree of θ.
w1  x cos  z sin 

w2  y
 w   x sin   z cos
 3
 cos 
 0

 sin 
Rotate around z-axis
counterclockwise with
degree of θ.
 w1  x cos  y sin 

w2  x sin   y cos
w  z
 3
cos 
 sin 

 0
93
0 
 sin  
cos 
0 sin  
1
0 
0 cos  
 sin 
cos 
0
0
0
1
If we rotate a vector around x-axes with x degree, around y-axes
with y degree, and around z-axes with z degree, we can get a rotated
matrix M from multiply each rotation matrixes:
cos( z )  sin( z ) 0  cos( y ) 0 sin( y ) 
M =  sin( z ) cos( z ) 0  0
1
0 
 0
0
1  sin( y ) 0 cos( y )
0
0 
1
0 cos( x)  sin( x)


0 sin( x) cos( x) 
Considering three basic vectors in 3-dimensional space:
v1 = (1, 0, 0)
v2 = (0, 1, 0)
v3 = (0, 0, 1)
Multiplying each vector with rotation matrix M, we have :
M v1 = ( cos(z)cos(y), sin(z)cos(y), -sin(y) )
M v2 = ( -sin(z)cos(x)-cos(z)sin(y)sin(x), cos(z)cos(x)-sin(z)sin(y)sin(x),
-cos(y)sin(x) )
M v3 = ( -sin(z)sin(x)+cos(z)sin(y)cos(x), cos(z)sin(x)+sin(z)sin(y)cos(x),
cos(y)cos(x) )
If we want to model a 3-dimensional figure in computer screen, we
just need to project these rotated vectors onto x-z plane since y-direction
towards us. Then new frame could be constructed from above vectors
having their y-component to disappear. Whatever constructing them by
calculating or by vector performing, a new frame thus formed.
In my constructions, in order to simplify some complicated
condition I restricted each rotated angles from zero to half of pi.
94
6.2 Skills in Constructing
Three-dimensional Animations
Part 1: When the direction point x rotates around y-direction, its locus
is an ellipse on the x-z plane, where
1. the major axis of this ellipse is
perpendicular to the y-direction.
2. the base circle of this ellipsehave the
same radius with the unit circle.
3. this ellipse passing through another
direction point z.
Thus this, ellipse can be constructed
with compressed method.
Figure 191
Part2: When a point rotates uniformly, its motion follows equal angle
motion respect to the point on base circle.
1. That is, if point A→A’ and B→B’ rotate
the same angle in their plane, then the
points on base circle respect to these
points rotate the same angle.
2. We can say that the rotated point on base
circle respect to the point of ellipse
follows uniform angle motion.
The construction thus needs to find the
point on base circle respect to the point of
Figure 192
ellipse, and make uniform angle rotation on
base circle, then finally project it onto the
ellipse.
95
Given a point P on ellipse, we first find
its corresponding point P’ on base circle,
make an uniform angle motion on base circle
to the point Q’, and finally project it to the
point Q of ellipse. This may let the rotation
behaves as the real rotation in
three-dimensions.
Figure 193
Part3: To Decide the rotated direction: clockwise or counterclockwise!
1. As Figure 194 shown, if we want to rotate
the x-direction to the direction of z, we
need to rotate counterclockwise.
2. But when base axes change, the
x-direction and y-direction may change,
too. As Figure 195 shown, if we still want
to rotate from x-direction to z-direction,
we need to rotate clockwise now.
Figure 194
Thus, we may bisect the ellipse into two
parts with major axis. By observing the
positions of x-direction and z-direction, we
can find one direction passes through the
major axis whenever the rotated direction
changes.
Base on this direction we can make a
Figure 195
criterion resulting in {1} or {-1} while the
rotated direction behaves clockwise or
counterclockwise respectively. This criterion
prevents us from constructing while the
rotated direction changes.
96
6.3 Polyhedral Dissections
We now shift our attention from two dimensions to three dimensions.
In 1990, David Hibert conjectured in the third problem of his famous list
of 23 problems that there exist a pairs of polyhedrons that we couldn’t
dissect one into another in a finite number of pieces. And his student Max
Dehn (1900) proved it. But there are still many solid dissections that have
beautiful results. Thus in this section, we will demonstrate the animation
of them.
 Hinged dissection of two cubes to a rhombic dodecahedron (7)
By observing the structure of a rhombic dodecahedron in Figure 196,
we can find that it consists of one cube plus six equal pyramidal pieces on
each face of this cube. In 1953, H. Martyn Cundy and A. P. Rollett
described this simple hinged dissection. Arthur Loeb (1976) and Henk
Mulder (1977) also described this dissection (Frederickson, 1997). We
dissect one cube into six equal pyramidal pieces and cover each face of
another cube a pyramidal piece. The resulting 7-piece dissection is shown
in Figure 196. The animation let the six pyramidal pieces expand like
unfolding the surface of a cube, and then wrap the other cube in the same
way.
Figure 196: Hinged dissection of two cubes to a rhombic dodecahedron
97
 Two truncated octahedrons to cube (6)
David Paterson (1988) and Anton Hanegraaf both found a simple
way to dissect two identical truncated octahedrons to a cube
(Frederickson, 1997). They first cut a truncated octahedral into four
identical pieces by a horizontal slice and a vertical slice. Then they cut
another truncated octahedral with one slice perpendicular to the former
two slices.
In the animations of this dissection, we can hinge the four equal
pieces and translate the other two equal pieces to a cube. The resulting
6-piece dissection is shown in Figure 197.
Figure 197: Two truncated octahedrons to a cube
98
 Hinged dissection of two truncated octahedrons to a cube (8)
Except for unhinged dissection explained by Paterson and Hanefraaf,
Jan Slothouber (1973) gave a 9-piece hinged dissection. But Frederickson
found a better result in only 8 pieces (Frederickson, 2002). He cut both
truncated octahedrons into four equal parts as shown in Figure 198. One
chain consists of four pieces that can be cyclicly hinged and the other
chain is just linearly hinged. The animation of the cyclicly chain can be
demonstrated from swing four pieces before connecting them into a cycle.
Then swing hinge the linearly pieces to wrap around the cyclicly pieces.
Figure 198: Frederickson’s hinged dissection of two truncated
octahedrons to a cube
99
 Hinged dissection of a truncated octahedron to a hexagonal
prism (7)
This hingeable 7-piece dissection of a truncated octahedron to a
hexagonal prism is found by Anto Hanegraaf as shown in Figure 199
(Frederickson, 2002). Hanegraaf cut a truncated octahedron into three
levels, and the thickness of each level is equal to the thickness of the
desired hexagonal prism. The truncated octahedron can be constructed
from cutting every corner of an octahedron with radius of 3 units until
every face of a square became a regular hexagon. Its volume then equal to
32, and the thickness of the desired prism is thus equal to 2 2 / 3 . In the
animation, the middle level stay unmoved to be the center of a prism, the
top level is cut into three equal pieces and swing down to the middle level,
and the bottom level is also cut into three equal pieces and swing up to
the middle level.
Figure 199: Hanegraaf’s hinged dissection of a truncated octahedron to a
hexagonal prism
100
 A cuboctahedron plus an octahedron to two truncated
octahedrons (10)
Anto Hanegraaf found this 10-piece dissection of one cuboctahedron
plus one octahedron to two truncated octahedrons as shown in Figure 200
(Frederickson, 1997). We can construct a truncated octahedron from
cutting the faces of a cuboctahedron until every face of an equilateral
triangle becomes a regular hexagon. Hanegraaf first cut 8 identical pieces
off a cuboctahedron in ordered to leave an uncut truncated octahedron
inside. Then he swing hinged these 8 pieces three times to wrap around
an octahedron, and resulting in a new truncated octahedron.
Figure 200: Hanefraaf’s dissection of one cuboctahedron plus one
octahedron to two truncated octahedrons
In constructing a cuboctahedron, we first construct a truncated
octahedron with its volume equal to 32 units as the former construction.
Thus it has a volume of 32 units. We then expand every square from
radius of 1 unit to 2 units on the plane of its face. The new figure is the
desired cuboctahedron having volume equal to 32×5/3. Finally, we
101
construct an octahedron with radius of 2 units, thus the volume of it is
32/3. The animation of this dissection takes three steps of swing hinge to
convert these 8 pieces from wrapped of a cuboctahedron to wrap around
an octahedron.
 Hinged dissection of a (2×1×1)-rectangular block to a cube (7)
For dissecting a (2×1×1)-rectangular block to a cube, Willian Fitch
Cheney (1933) gave an 8-piece unhingeable dissection. Albert Wheeler
Cheney (1933), a math teacher discovered a 7-piece unhingeable
dissection. Michael Goldberg (1966) claimed a 10-piece hinged
dissection. And finally, in 1969, Anto Hanegraaf found a 7-piece cyclicly
hinged dissection. And in 1989, he found a 6-piece unhingeable
dissection (Frederickson, 2002)
In Hanegraaf’s 7-piece hingeable dissection (Figure 201). He first
converted a (2×1×1)-rectangular block to a (r×r2×1)-rectangular block
with a T-slide where r= 3 2 . Then, he converted the second rectangular to
a cube of side length of r with a T-slide. In the animation, we can cyclicly
hinge these pieces twice to form a cube.
Figure 201: Hanefraaf’s hinged dissection of a (2×1×1)-rectangular
block to a cube
102
 Hinged dissection of a truncated octahedron to a cube (14)
In Hanegraaf’s 14-piece hingeable dissection of a truncated
octahedron to a cube which is shown in Figure 202. He began with
Theobald’s approach of cutting a truncated octahedron into six pieces
plus a rectangular block such that the six pieces can be swing hinge to a
new rectangular block whose length and width equal to the desired side
length.
In the construction, we first construct a truncated octahedron with
volume of 32 units, then the desired cube has side length of
s= 3 32  3.1744. The new block which consists of six pieces is thus a
(2×s×s)-rectangular block, and the remaining block is a
(2×a×a)-rectangular block, where a= ( s 3  2s 2 ) / 2  2.4336. We then
convert a (2×a×a)-rectangular block to a thinker but higher
((2-s)×(a+b)×a)-rectangular block, where b=(4-s)a/(s-2)  1.7108. And
then, we convert a ((2-s)×(a+b)×a)-rectangular block to a
((2-s)×s×s)-rectangular block. Finally, we combine this block with the
block beneath it and we get the cube with volume equal to 32. In the
animation, the last 2-th and 4-th swing is of Hanefraaf’s T-swing.
103
Figure 202: Hanefraaf’s truncated octahedron to a cube
104
6.4 Dissecting the Surface of Polyhedrons
In our above discussion, we have discussed some solid dissections in
three dimensions. In this section, we will change our dissection model
from the topic of dissections between solids to the dissections between
surfaces of solids.
 Surface of cube to surface of tetrahedron (2)
Figure 203: Theobald’s surface of cube to surface of tetrahedron
Theobald’s found this 2-piece dissection of dissecting the surface of
a cube to the surface of a tetrahedron as shown in Figure 203
(Frederickson, 1997). We illustrate here again with the help of Cabri
Geometry II to demonstrate its animation in three-dimensional space.
First, we construct the expansion of the surface of a cube as a zigzag
fashion, which will be a P-strip element. Then, we construct the
expansion of the surface of a tetrahedron as a parallelogram. We
crosspose these two strips as in Figure 204. We have the dissection
105
between two surfaces that need only two pieces. We can fold these two
pieces to the expansion of a cube or the expansion of a tetrahedron.
Figure 204: Theobald’s crossposition of cube and tetrahedron surfaces
In the animation, we first construct the expansions of each figures at
the middle of one round, then fold the surface of a cube along its folding
lines as time goes backwards, and fold the surface of a tetrahedron along
its folding lines as time goes upwards. The animation of dissecting and
folding between these surfaces is thus complete.
106
Results
This paper has given the dynamic demonstration of 70 plane
dissections and 8 three-dimensional dissections. These plane dissections
include dissecting regular polygons, star polygons, Greek Cross and other
curved figures such as disk and ovals. Special types of plane dissections
include the swing-hinged dissections, the flip-hinged dissections and the
twist-hinged dissections. These illustrations are just tips of an iceberg.
There are still many plane figures waiting for us to explore, such as the
Maltese Cross, the swastika, the golden rectangle, the dissection of family
of regular polygons to one, and even the three-figure dissections.
The three-dimensional dissections referred here include the
dissections between solids and the dissections between the surfaces of
polyhedrons. This paper provides a model for performing
three-dimensional animations with Cabri Geometry II. However, this
model is not a suitable circumstance for 3D animations. It should have
some professional 3D soft ware to deal with such problems. My effort,
however, was to demonstrate that the dynamic of the three-dimensional
dissections can be illustrated by Cabri Geometry II, a simple geometric
construction tool.
107
Index of Dissections
In two-dimensional figures, a regular n-polygon is represented by
{n}, a star polygon is represented by {p/q}, Greek Cross by {GC} and
Cross of Lorrain by {L’}. But in three-dimensional figures, no
representation was used. And the parentheses behind the figures are the
number of pieces used in the dissection.
{12} to {6} (6)
{12} to {7} (11)
{12} to {8} (10)
{12} to {9} (14)
{12} to {10} (12)
Regular Polygons
{4} to {3} (4)
{5} to {3} (6)
{5} to {4} (6)
{6} to {3} (5)
{6} to {4} (5)
{6} to {5} (7)
{7} to {3} {8}
{7} to {4} (7)
{7} to {5} (9)
{7} to {6} (8)
{8} to {3} (7)
{8} to {4} (5)
{8} to {5} (9)
{8} to {6} (8)
{8} to {7} (11)
{9} to {3} (8)
{9} to {4} (9)
{9} to {5} (12)
{9} to {6} (11)
{9} to {7} (14)
{10} to {3} (7)
{10} to {4} (7)
{10} to {7} (11)
{10} to {8} (10)
{10} to {9} (13)
{12} to {4} (6)
{12} to {5} (10)
14
15
16
17
18
19
20
22
23
24
25
26
27
28
29
30
31
33
34
35
36
37
38
40
41
42
43
44
45
46
47
48
Star Polygons
{5/2} to {4} (7)
{5/2} to {10} (6)
{6/2} to {3} (5)
Another way of {6/2} to
{3} (5)
{6/2} to {4} (5)
{6/2} to {6} (6)
{8/2} to {4} (7)
Another way of {8/2} to
{4} (7)
{8/3} to {4} (8)
{10/2} to {5} (7)
{10/3} to two {5/2}s (10)
{12/2} to {3} (6)
{12/2} to {4} (8)
{12/2} to {6} (8)
{12/3} to {4} (9)
{12/4} to {4} (11)
108
50
51
51
54
54
55
56
57
58
59
63
64
64
65
66
67
Dissected Curves
Disk to oval seat tops (6)
Disk to oval seat tops (8)
Disk to oval seat tops (6)
Disk to different oval seat
tops (4)
Disk to ovals with thinner
holes (6)
Disk to horseshoes (4)
Twist-hinged Dissections
69
69
70
70
{4} to {3} (7)
{12} to {6} (9)
three {6}s to one (8)
two {5}s plus two {5/2} to
one {10} (8)
71
72
Polyhedral Dissections
Two cubes to a rhombic
dodecahedron (7)
Two truncated octahedral
to cube (6)
Two truncated octahedral
to a cube (8)
A truncated octahedron to a
hexagonal prism (7)
Two truncated octahedral
to a cuboctahedron and an
octahedron (10)
A (2×1×1)-rectangular
block to a cube (7)
A truncated octahedron to a
cube (14)
Dissecting other Figures
{GC} to {3} (5)
{GC} to {4} (4)
Another {GC} to {4}
{L’} to {4} (7)
73
74
75
75
Swing-hinged Dissections
two {6}s to one {6} (7)
{8} to {4} (7)
{12} to {4} (8)
{GC} to {6} (8)
79
80
81
82
88
89
90
91
97
98
99
100
101
102
103
Flip-hinged Dissections
{4} to {3} (4)
{8} to {4} (5)
{6} to {3} (5)
{6/2} to {3} (5)
Dissected surfaces
84
84
86
87
Surface of cube to surface
of tetrahedron (2)
109
105
Index of Elements
Figure 22
{5} to {3} (6)
{5} to {4} (6)
{5} to {6} (7)
{5} to {7} (9)
{5} to {8} (9)
{5} to {9} (12)
{5} to {12} (10)
{6} to {4} (5)
{6} to {8} (8)
{6} to {9} (11)
{6} to {3} (5)
Figure 27
{6} to {5} (7)
{6} to {7} (8)
Figure 30
Figure 36
Figure 33
{7} to {3} {8}
{7} to {4} (7)
{7} to {5} (9)
{7} to {6} (8)
{7} to {8} (11)
{7} to {9} (14)
{7} to {10} (11)
{7} to {12} (11)
{8} to {5} (9)
{8} to {3} (7)
Figure 46
{8} to {6} (8)
{8} to {10} (10)
{8} to {12} (10)
Figure 51
Figure 54
{8} to {7} (11)
{9} to {4} (9)
{9} to {7} (14)
Figure 57
Figure 64
110
{9} to {5} (12)
{9} to {6} (11)
{9} to {12} (14)
{10} to {3} (7)
{10} to {4} (7)
{10} to {8} (10)
{10} to {12} (12)
Figure 67
Figure 77
{10} to {7} (11)
{10} to {9} (13)
{9} to {10} (13)
Figure 80
Figure 87
{12} to {5} (10)
{12} to {7} (11)
{12} to {8} (10)
{12} to {9} (14)
{12} to {10} (12)
Figure 93
Figure 99
{5/2} to {4} (7)
{6/2} to {3} (5)
Figure 109
Figure 113
{6/2} to {4} (5)
{8/2} to {4} (7)
Figure 118
Figure 123
111
{10/2} to {5} (7)
Figure 130
{12/2} to {6} (8)
Figure 138
{L’} to {4} (7)
{12/3} to {4} (9)
Figure 142
Figure 157
Hinged {12} to
{4} (8)
Figure 168
112
Reference
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