The New Version of the Tangram Puzzle
Lecturer：Wen-Hsien SUN
President of Chiu Chang Mathematics Education Foundation
Classic Chinese Tangram
Tangram is a kind of jigsaw puzzle: assemble seven
geometric pieces, without overlap, to form charming,
elegant, sophisticated, and sometimes paradoxical figures.
We can obtain a set of the Tangram by dissecting a
square into seven pieces as the figure shown below.
Classic Chinese Tangram
The seven pieces can be rearranged to form thousands
of different figures of geometric shapes (triangles,
parallelograms, polygons and etc.), people in motion,
animals(cats, dogs,pigs, horses and etc.), and bridges,
houses, pagodas, letters of the alphabet and some
Chinese characters.
Arrange the following shapes
using Tangram:
Early Tangram
The 14 pieces puzzle on the
right can be seen as an early
version of Tangram. It was
first referred in Archimedes’
work in 3rd century BC as
“Stomachion,” meaning
‘problem that makes one
insane’. A discovery in
1906 in Saint Sabba of Jerusalem shows the puzzle was
dissected from a big square not the previously known two
smaller squares.
蝶几圖
The book, “蝶几圖” written in 1617 by Chinese
author 戈汕, included the following puzzle. The set
contains 13 pieces, with extra piece for the two *
marked pieces below.
Pythagorean Theorem
Prove the Pythagorean Theorem using two sets of
Tangram:
Tangram Table
The Tangram table (rosewood) below was made in 1840,
Guangdong, China. There is another similar Tangram
table with detail carvings on mahogany wood in Taiwan.
Ivory Tangram
There are also ivory made Tangram for the royals.
Ivory Tangram
This is an ivory Tangram made in 1802 exporting to
France.
Ivory Tangram
The ivory Tangram
on the right was
made in England,
with gold leaf
plating on the
surface.
Japanese Tangram: 智慧板
The Japanese Tangram published in Japan, 1796:
Puzzle from China
“The Fashionable Chinese Puzzle,” published in 1817
across Europe and America.
Napoleon and Tangram
The Swiss. And Dutch books about Tangram reported
that Napoleon also likes Tangram.
Tangram Stamps
Left: Dutch Stamps, 1997,
with Chinese zodiac made by
Tangram.
Right: Finnish stamp set,
May 2000.
Tangram Stamps
Hong Kong stamp set, May 2009. The
stamp sheetlet uses the tangram to
demonstrate different actions, such as
diving, playing football and practicing
martial arts, hence showing the versatility
and endless possibilities of the game.
Tangram Crafts
Ceramic containers in the shape of Tangram.
Tangram in France
Book of Tangram
Chinese book: 七巧圖解, published in 1815.
The Basic Construction of Tangram
Chinese Tangram is an intelligent puzzle based on
isosceles right triangles. There is only one monotan (the
unit triangles), but there are 3 ditans (formed of 2 unit
triangles), 4 tritans (formed of 3 unit triangles) and 14
tetratans (formed of 4 unit triangles).
Japanese Tangram
Shape by Shape
The Pieces
Shape by Shape
Shape by Shape
Strategy of Solving the Problem
Never satisfied with only one solution.
Always try all the possibilities and find all
the solutions.
The difference from mathematics to games
is that games need one solution only while
mathematics requires finding all possible
solutions and proving there are no other
possible answers.
Puzzles of the Square
Use pieces (a), (b), (c), (d) and (e) to form a
bigger square.
Use pieces (b), (c), (d) and (e) to form a
square.
b
a
c
d
e
Puzzles of the Square
Suppose that each symmetric triangle has area of
1: square piece (a) and triangle (b) has area of 2;
triangle (c) has area of 4; piece (d) and (e) has
area of 5.
2
2
b
a
c
d
e
Puzzles of the Square
If there are 2 copies of each of the 5 pieces (a, b, c, d, e),
how to assemble all the pieces to form a square?
b
a
c
b
a
c
e
d
d
e
The Method of Analyzing the Area
a+b+c+d+e=2+2+4+5+5=18 units. If the first
question is solvable, the length of the sides of such
3 2
a square is supposed to be
2
b
a
c
d
e
The Method of Analyzing the Area
b+c+d+e=16 units. If the second question is
solvable, the length of the sides of the square
should be 2.
b
c
d
e
The Method of Analyzing the Area
2a+2b+2c+2d+2e=36 units. If the last question is
solvable, the length of the sides of the square is
supposed to be 3.
b
a
c
b
a
c
e
d
d
e
Solutions
The square of 18 units.
The square of 16 units.
Solutions
The following figure shows one of the solutions to the
square of 36 unit isosceles right triangles.
Arnold transformation
Arnold transformation
Tangram
Tangram
Tangram
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=31.72
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=30.93
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=32.11
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=20.17
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=19.65
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=33.74
Public image
Secret image
Generated secret image
Compress ration=3.2 , PSNR=34.76
Public image
Generated public image
PSNR=37.96
Secret image
Generated secret image
PSNR=35.47
Public image
Secret image
Generated public image
Generated secret image
PSNR=40.72
PSNR=38.43
Public image
Secret image
Generated public image
Generated secret image
PSNR=29.17
PSNR=31.77
Public image
Secret image
Generated public image
Generated secret image
PSNR=31.24
PSNR=32.14
Public image
Secret image
Generated public image
Generated secret image
PSNR=30.13
PSNR=29.81
Public image
Generated public image
PSNR=33.46
Secret image
Generated secret image
PSNR=27.90
Public image
Secret image
Generated public image
Generated secret image
PSNR=34.73
PSNR=28.76
Public image
Secret image
Generated public image
Generated secret image
PSNR=37.41
PSNR=30.19
2B or Not
2B
A
E
D
F
C
G
B
N
2B
H
L
I
J
K
M
To be or not to be –
that is the question.
---------From Hamlet
A
E
D
F
C
G
B
H
N
2B
L
I
J
K
M
Shapes of Tetratans
A
B
H
I
J
C
K
D
L
E
M
F
G
N
Shapes of Tetratans
There are up to 14 shapes of tetratans.
B
G
K
I
E
F
M
L
N
H
C
D
A
J
Super-Tangram
2A+2B+2C+2D+E+F+
G+H+I+J+the 3
ditans+ 2 monotan
Super-Tangram
M+N+K+L+2 of each of the
tritans+the 3 ditans+4 monotan
The classical 2B Question─Tangram
唐圖─Tangram,shown as the figure below, consists
of square of 2 copies of the monotan, 1 copy of each
of the 3 ditans and 2 identical copies of the Btetratans. The area of the square is 16 units.
The pieces of Tangram
a
b
c
d
= 2a+b+c+d+ 2B
2B
Question?
There are 2 identical copies of the B-tetratans in the
classical Tangram. We wonder why this particular
tetratan was chosen. Is there any other possibility such
as choosing 2A、2C、2D…. or 2M 、 AB 、 AC...？
A
C
M
D
Proof
There are 105 combinations.
How can we find and prove
which combinations are
indeed infeasible?
Consider the central region of
the square depicted in the
figure below. We may take its
area to be 8 units. What is the
minimum overlap in area of A
～M with this region?
8
Proof
The following chart shows the minimum overlap
in area of A~N：
2 units
A
1 unit
C
Proof
A
E
F
D
G
B
C
The minimum
overlap in area
Proof
K
H
M
L
I
J
The minimum
overlap in area
Possible Combinations
2
2
1
2
3
3
2
3
3
3
4
4
2
X
A
B
C
D
E
F
G
H
I
J
K
L
M
N
X
X
N
X
X
X
X
X
X
X
X
X
X
X
X
X
2
M
4
4
3
4
5
5
4
5
5
5
6
6
4
4
L
6
6
5
6
7
7
6
7
7
7
8
8
4
K
6
6
5
6
7
7
6
7
7
7
8
3
J
5
5
4
5
6
6
5
6
6
6
3
I
5
5
4
5
6
6
5
6
6
3
H
5
5
4
5
6
6
5
6
2
G
4
4
3
4
5
5
4
3
F
5
5
4
5
6
6
3
E
5
5
4
5
6
2
D
4
4
3
4
1
C
3
3
2
2
B
4
4
2
A
4
Proof
b
c
d
The minimum
overlap in area
b
c
d
2
1
0
Proof
Now the square-ditan (b) must be entirely within this
region, so must half of parallelogram-ditan (c), for a
total overlap of area at least 3 units. This consideration
rules out the following combinations which exceed 5
units for a total overlap of area:(These are marked pink
in the chart followed)
2E，2F，2H，2I，2J，2K，2L，AK，AL，BK，
BL，DK，DL，EF，EH，EI，EJ，EK，EL，FH，
FI，FJ，FK，FL，GK，GL，HI，HJ，HK，HL，
IJ，IK，IL，JK，JL，KL，KM，LM。
Proof
Those involving N are all impossible because it does not
even fit inside our square . (These are marked blue in
the chart in p.40)
N
Possible Combination
2
2
1
2
3
3
2
3
3
3
4
4
2
X
A
B
C
D
E
F
G
H
I
J
K
L
M
N
X
X
N
X
X
X
X
X
X
X
X
X
X
X
X
X
2
M
4
4
3
4
5
5
4
5
5
5
6
6
4
4
L
6
6
5
6
7
7
6
7
7
7
8
8
4
K
6
6
5
6
7
7
6
7
7
7
8
3
J
5
5
4
5
6
6
5
6
6
6
3
I
5
5
4
5
6
6
5
6
6
3
H
5
5
4
5
6
6
5
6
2
G
4
4
3
4
5
5
4
3
F
5
5
4
5
6
6
3
E
5
5
4
5
6
2
D
4
4
3
4
1
C
3
3
2
2
B
4
4
2
A
4
Case: 2M
?
Proof
Of the other combinations, after placing the 2 tetratans in
2M,AJ,BJ,DJ,EG,FG,FM,GJ,GM,HM and IM, the piece b
will not fit. The figure below shows possible placements
of the 2 tetratans, but they are inconsequential . (These are
marked Green in the chart followed)
Chart of Feasible
Combination
2
2
1
2
3
3
2
3
3
3
4
4
2
X
A
B
C
D
E
F
G
H
I
J
K
L
M
N
X
X
N
X
X
X
X
X
X
X
X
X
X
X
X
X
2
M
4
4
3
4
5
5
4
5
5
5
6
6
4
4
L
6
6
5
6
7
7
6
7
7
7
8
8
4
K
6
6
5
6
7
7
6
7
7
7
8
3
J
5
5
4
5
6
6
5
6
6
6
3
I
5
5
4
5
6
6
5
6
6
3
H
5
5
4
5
6
6
5
6
2
G
4
4
3
4
5
5
4
3
F
5
5
4
5
6
6
3
E
5
5
4
5
6
2
D
4
4
3
4
1
C
3
3
2
2
B
4
4
2
A
4
Case: 2G
?
Proof
In the remaining
combinations, namely, 2G,
AG, AH, AI, AM, BG, BH,
BI, BM, CK, CL, CM, DE,
DF, DM, EM, GH, GI, JM
After placing d and b, the
piece c will not fit. The
figure at right shows possible
placements of the 2 tetratans,
but they are inconsequential .
(These are marked Purple in
the chart followed)
Chart of Feasible Combination
2
2
1
2
3
3
2
3
3
3
4
4
2
X
A
B
C
D
E
F
G
H
I
J
K
L
M
N
X
X
N
X
X
X
X
X
X
X
X
X
X
X
X
X
2
M
4
4
3
4
5
5
4
5
5
5
6
6
4
4
L
6
6
5
6
7
7
6
7
7
7
8
8
4
K
6
6
5
6
7
7
6
7
7
7
8
3
J
5
5
4
5
6
6
5
6
6
6
3
I
5
5
4
5
6
6
5
6
6
3
H
5
5
4
5
6
6
5
6
2
G
4
4
3
4
5
5
4
3
F
5
5
4
5
6
6
3
E
5
5
4
5
6
2
D
4
4
3
4
1
C
3
3
2
2
B
4
4
2
A
4
Case: 2D
Conclusion─create the new 22
sets of the Tangram
Among so many constructions of the tangram,there are
only the following 23 combinations which substitute Btetratans without any change of the pieces a, b, c and d.
Tetratans of symmetric
triangles
Six of the 14 tetratans are convex quadrangles, which
are B、F、G、I、K and N. However, only be 2B, BF
could be used to form Tangram sets.
B
G
E
F
M
L
K
I
N
H
C
D
A
J
Combinations of Convex
Tetratans ─ 2B and BF
At most how many edges of tangram
are there?
1. Improper tangram
七巧圖中的各組塊只通
過頂點與頂點或頂點與
邊相連接起來的，稱為
非常規的七巧圖。
There are at most 23 edges.
At most how many edges of tangram
are there?
2. Proper tangram
七巧圖中的各組塊是由
邊與邊相連接起來的，
稱為常規的七巧圖。
There are at most 23 edges.
At most how many edges of tangram
are there?
3. Regular tangram
七巧圖中的各組塊是經
由有理邊與有理邊相連
接、無理邊與無理邊相
起來，且每個邊的邊段
都要吻合相接，稱為正
規的七巧圖。
There are at
most 18 edges.
What does this
mean？
At most how many edges of tangram
are there?
因七巧板各組塊的邊長為可分為1的倍
數與無理數 2 的倍數兩種，故這裡的
邊段可分為兩類：
1. 有理邊段：一個有理邊段就是一個單位
長的線段。
2. 無理邊段：一個無理邊段就是一個長為
2 的線段。
4個有理邊段
2個有理邊段，
1個無理邊段，
共3個邊段
2個有理邊段，
2個無理邊段，
共4個邊段
2個有理邊段，
2個無理邊段，
共4個邊段
4個有理邊段，
2個無理邊段，
共6個邊段
At most how many edges of tangram
are there?
整組七巧版共有30個邊段
（其中有理邊段20個，無
理邊段10個）。
七片組合在一起，至少有
6個邊段要相貼合在一起，
每一個貼合處就至少減了
2個邊段，因此正規七巧
圖至多有30-12=18條邊。
Convex polygons made of Tangram
For any pair of points inside a convex polygon,
any point on the straight line that joints the pair
would also be inside the polygon. Else it would
be a concave polygon.
convex polygon
concave polygon
Convex polygons made of Tangram
Suppose Tangram pieces can only be connected
along its sides, then we have the following:
1. All internal angles of a convex polygon made of
Tangram must be multiples of 45°, (i.e. 45°，
90°，135°).
45°
90°
135°
Convex polygons made of Tangram
Suppose Tangram pieces can only be connected
along its sides, then we have the following:
2. A convex polygon made of Tangram has at most
8 sides.
Since the sum of external angles of any convex
polygon equals 360°. Convex polygons made of
Tangram has minimal external angles of 45°,
therefore the number of sides cannot exceed
360÷45＝8.
Convex polygons made of Tangram
Convex polygons that satisfies the conditions
are:
1. Octagon：
All internal angles equal 135°
135°
Convex polygons made of Tangram
2. Heptagon：
Six of the internal angles equals 135°, last
one equals 90°.
135°
Convex polygons made of Tangram
3.Hexagon：
Ⅰ Four 135° angles,
Two 90° angles;
135°
Ⅱ Five 135° angles,
One 45° angle
135°
135°
45°
135°
Convex polygons made of Tangram
4. Pentagon：
Ⅰ Two 135°, three 90° angles；
Ⅱ Three 135°, one 90°, one 45° angle.
45°
135°
135°
135°
135°
45°
Convex polygons made of Tangram
5. Tetragon：
ⅠTwo 135°,
two 45° angles;
ⅡOne 135°,
Two 90°,
One 45° angle;
ⅢAll internal angles
equal 90°.
135°
45°
135°
45°
45°
135°
135°
45°
Convex polygons made of Tangram
6. Triangle：
One internal angle equals 90°, two equal
45°.
45°
Convex polygons made of Tangram
By observing the above convex polygons, we
notice that:
1. 4 pairs of parallel
sides in Octagon.
135°
2. 3 pairs of parallel
sides in Heptagon.
135°
Convex polygons made of Tangram
3. At least 2 pairs of parallel sides in Hexagon.
135°
135°
135°
135°
45°
Convex polygons made of Tangram
4. At least 1 pair of parallel sides in Pentagon.
135°
135°
45°
45°
135°
135°
Convex polygons made of Tangram
5. Other than the one on the bottom right corner,
the tetragons has at least 1 pair of parallel
sides.
135°
45°
45°
135°
45°
135°
45°
135°
Convex polygons made of Tangram
Lemma 1：
The side length of a convex polygon made of
Tangram is either all made of rational side of unit
triangles, or all made of irrational side of the unit
triangles.
1
2
45°
45°
45°
67.5°
Convex polygons made of Tangram
Lemma 2：
The rectangle inscribed by a convex polygon made
of Tangram has side lengths either all irrational
numbers, or all rational numbers. (i.e. it is either
integers, or integer multiples of 2 .)
(a + b 2)(c + d 2 )= 8，a、b、c、d are integers.
 ac + (bc + ad) 2 + 2bd = 8
 b = 0，d = 0 （all lengths are integers.）
or c = 0，a = 0 （all lengths are multiples of
2）
A tangram must be a square if the
lengthes of the edges of the tangram are
all irrational
【PROOF】
七巧圖的邊若全為無理邊，則其邊必與坐標線
相交45°，所有拐角必定是90°或270°，每條邊
必須是 的倍數，因七巧圖的面積是8，它必
2
定可分割成四個面積為2的正方形。而四個正
方形的組合即四方塊，只能是以下5種：
A tangram must be a square if the
lengthes of the edges of the tangram are
all irrational
前4種四方塊中：
第一個圖，小正方形不論放在那個位置，
二個大三角形都無法放入。
在第二、三、四圖裡，放入小正方形，無
法同時放入二個大三角形及平行四邊形，
故只有第五圖可以拼出。
A tangram must be a square if the
lengthes of the edges of the tangram are
all irrational
由以上證明可知七巧圖拼出的凸多邊形，
除了正方形外，其他的都必須有有裡邊。
我們把有理邊放置在水平位置，則此七巧
圖一定可被一有理邊長的矩形緊密地包圍
起來，即七巧圖內接於此矩形。
Convex polygons made of Tangram
P
a
If the octagon ABCDEFGH H
B
A
is made of Tangram, and is
inscribing rectangle PQRS.
Then assume length of PQ G
equals x, QR equals y.
d
S
Q
b
C
D
c
F
E
R
Convex polygons made of Tangram
Since the area of the octagon made of Tangram equals 8,
we get:
1
1
1
1
xy - a 2 - b 2 - c 2 - d2 = 8 P
a
2
2
2
2
H
B
A
that is:
2 xy - 16 = a 2 + b 2 + c 2 + d 2
where a+b≦x
c+d≦x
a+d≦y
and b+c≦y
C
G
d
S
Q
b
D
c
F
E
R
Convex polygons made of Tangram
The system 2xy - 16 = a2 + b2 + c 2 + d2 has the
following 20 positive integer solutions:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
x
3
4
2
4
2
2
2
3
3
2
y
5
4
4
4
6
5
6
3
4
5
a
3
2
0
4
2
2
2
1
2
1
b
0
2
0
0
0
0
0
0
0
1
c
1
2
0
0
2
0
0
0
0
1
d
2
2
0
0
0
0
2
1
2
1
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
x
3
3
3
4
8
5
5
1
1
1
y
5
3
4
6
9
5
5
8
9
9
a
3
1
2
4
8
4
5
0
1
1
b
0
0
0
0
0
1
0
0
0
0
c
2
1
2
4
8
4
3
0
1
0
d
1
0
0
0
0
1
0
0
0
1
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(2)
(1)
x
y
a
b
c
d
x
y
a
b
c
d
3
5
3
0
1
2
4
4
2
2
2
2
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(4)
(3)
x
y
a
b
c
d
x
y
a
b
c
d
2
4
0
0
0
0
4
4
4
0
0
0
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(6)
(5)
x
y
a
b
c
d
x
y
a
b
c
d
2
6
2
0
2
0
2
5
2
0
0
0
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(8)
(7)
x
y
a
b
c
d
x
y
a
b
c
d
2
6
2
0
0
2
3
3
1
0
0
1
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(10)
(9)
x
y
a
b
c
d
x
y
a
b
c
d
3
4
2
0
0
2
2
5
1
1
1
1
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(12)
(11)
x
y
a
b
c
d
x
y
a
b
c
d
3
5
3
0
2
1
3
3
1
0
1
0
Convex polygons made of Tangram
The first 1~13 sets can be shown as the configurations
below:
(13)
x
y
a
b
c
d
3
4
2
0
2
0
Convex polygons made of Tangram
Set 14~20 cannot be made from Tangram:
(15)
(14)
x
y
a
b
c
d
x
y
a
b
c
d
4
6
4
0
4
0
8
9
8
0
8
0
Convex polygons made of Tangram
Set 14~20 cannot be made from Tangram:
(16)
x
y
a
b
c
d
5
5
4
1
4
1
Convex polygons made of Tangram
Set 14~20 cannot be made from Tangram:
(18)
(17)
x
y
a
b
c
d
x
y
a
b
c
d
5
5
5
0
3
0
1
8
0
0
0
0
Convex polygons made of Tangram
Set 14~20 cannot be made from Tangram:
(20)
(19)
x
y
a
b
c
d
x
y
a
b
c
d
1
9
1
0
1
0
1
9
1
0
0
1
Convex polygons made of Tangram
Therefore there can only be 13 convex polygons made
of Tangram.
Do It Yourself～
x
y
a
b
c
d
2
5
1
1
1
1
Do It Yourself～
x
y
a
b
c
d
3
3
1
0
1
0
Do It Yourself～
x
y
a
b
c
d
3
5
3
0
1
2
Do It Yourself～
x
y
a
b
c
d
4
4
2
2
2
2
Do It Yourself～
x
y
a
b
c
d
2
4
0
0
0
0
Do It Yourself～
x
y
a
b
c
d
4
4
4
0
0
0
Do It Yourself～
x
y
a
b
c
d
2
6
2
0
2
0
Do It Yourself～
x
y
a
b
c
d
2
5
2
0
0
0
Do It Yourself～
x
y
a
b
c
d
2
6
2
0
0
2
Do It Yourself～
x
y
a
b
c
d
3
4
2
0
0
2
Do It Yourself～
x
y
a
b
c
d
3
5
3
0
2
1
Do It Yourself～
x
y
a
b
c
d
3
3
1
0
0
1
Do It Yourself～
x
y
a
b
c
d
3
4
2
0
2
0
Further thoughts～
This set cannot be made from Tangram. Why?
(16)
x
y
a
b
c
d
5
5
4
1
4
1
Further thoughts～
As shown, there are only three possible locations for
the square:
Further thoughts～
Location 1：
Further thoughts～
Location 2：
Further thoughts～
Location 3：
Further thoughts～
When the square was put in one of its possible locations,
there is only one way to insert the 2 triangles.
However, it is then impossible to put in the last
parallelogram!!
How many distinct pentagons can
Tangram form?
Doesn’t matter it is a concave or convex
pentagon, its internal angles MUST BE multiples
of 45°. In other words, it has to be n×45°,
where n equals 1，2，3，4，5，6，7.
Sum of all pentagon’s internal angles equals
540°, so we have:
(n1 + n 2 + n3 + n 4 + n5 )× 45°
= 540°
n1 + n 2 + n 3 + n 4 + n 5 = 12
How many distinct pentagons can
Tangram form?
We record the angles as n1、n2、n3、n4、n5 either
counter-clockwise or clockwise, to easily classify
each pentagons.
We assume the Mirror Images are identical. For
example, 61122 and 62211 are the following:
45°
270° 45°
45°
45° 270°
How many distinct pentagons can
Tangram form?
Therefore there can only be the following 20
possible combinations:
72111，71211，63111，61311，
62211，62121，62112，61221，
53211，53121，53112，52311，
52131，51321，52221，52122，
33321，33231，32322，33222。
How many distinct pentagons can
Tangram form?
Considering the cases
individually. For example,
pentagon 72111 might
looks like this:
By observation
8>z>x, z>y, y>x.
z－GE=GF+x+y
GE = GF =
z - (x + y)
2
z + (x + y)
AG =
2
A
z
D
G
E
F
x
C y B
How many distinct pentagons can
Tangram form?
The area of the pentagon
equals 8, that is:
△AGB+△GEF-△CFD＝8
1  z + (x + y)  + 1  z - (x + y)  - 1 x 2
=8


 
2
 2
2
2
 2
2
A
z
D
2
z 2 + 2 xy + y 2 - x 2 = 32
G
E
F
x
C y B
How many distinct pentagons can
Tangram form?
1. If x, y, z are positive integers, and 8>z>x,
z>y, y>x, solving the system we get:
z=5，y=2，x=1
The pentagon would look like this:
How many distinct pentagons can
Tangram form?
2. If x, y, z are integer multiples of 2 , and
8>z>x, z>y, y>x.
Let x= 2 x’, y= 2y’, z= 2z’, where x’, y’, and
z’ are positive integers, substitute into the
equation
We have:
z 2 + 2 xy + y 2 - x 2 = 32
2
2
2
z’ + 2 x’y’+ y’ - x’ = 16
How many distinct pentagons can
Tangram form?
Solving the equation z2 +2xy + y2 - x2 =16
we have:
z’=3，y’=2，x’=1
or
z=3 2，y=2 2 ，x= 2
The pentagon would look like this:
How many distinct pentagons can
Tangram form?
Applying the similar process to the 20 cases above,
some have multiple solutions where some have
none!
The following 18 pentagons are the result after
examining all 20 possible cases, including 2
convex pentagons.
How many distinct pentagons can
Tangram form?
Which two
pentagons
are convex?
How many distinct pentagons can
Tangram form?
A1. 72111
Regular tangram
A2. 72111
Regular tangram
A3. 72111
Proper tangram
How many distinct pentagons can
Tangram form?
B1. 71211
B3. 71211
Regular tangram
Proper tangram
B2. 71211
Proper tangram
How many distinct pentagons can
Tangram form?
C1. 63111
Regular tangram
C2. 63111
Proper tangram
How many distinct pentagons can
Tangram form?
C3. 63111
Proper tangram
C4. 63111
Proper tangram
How many distinct pentagons can
Tangram form?
D1. 61311
Regular tangram
D3. 61311
Regular tangram
D2. 61311
Regular tangram
How many distinct pentagons can
Tangram form?
D4. 61311
Proper tangram
D5. 61311
Proper tangram
How many distinct pentagons can
Tangram form?
D6. 61311
Proper tangram
D7. 61311
Proper tangram
How many distinct pentagons can
Tangram form?
E1. 62211
Regular tangram
E2. 62211
Proper tangram
E3. 62211
Proper tangram
How many distinct pentagons can
Tangram form?
F1. 62121
Regular tangram
F2. 62121
Proper tangram
How many distinct pentagons can
Tangram form?
F3. 62121
Proper tangram
F4. 62121
Proper tangram
How many distinct pentagons can
Tangram form?
G1. 62112
Proper tangram
How many distinct pentagons can
Tangram form?
H1. 61221
Regular tangram
H2. 61221
Proper tangram
How many distinct pentagons can
Tangram form?
H3. 61221
Proper tangram
H4. 61221
Proper tangram
H5. 61221
Proper tangram
How many distinct pentagons can
Tangram form?
J1. 53121
Regular tangram
J2. 53121
Proper tangram
J3. 53121
Proper tangram
How many distinct pentagons can
Tangram form?
K1. 53112
Regular tangram
K2. 53112
Proper tangram
How many distinct pentagons can
Tangram form?
L1. 52311
Regular tangram
L3. 52311
Proper tangram
L2. 52311
Proper tangram
How many distinct pentagons can
Tangram form?
M1. 52131
Regular tangram
M3. 52131
Proper tangram
M2. 52131
Regular tangram
How many distinct pentagons can
Tangram form?
N1. 51321
Regular tangram
N2. 51321
Regular tangram
How many distinct pentagons can
Tangram form?
N3. 51321
Regular tangram
N4. 51321
Regular tangram
How many distinct pentagons can
Tangram form?
N5. 51321
Proper tangram
N6. 51321
Proper tangram
N7. 51321
Proper tangram
How many distinct pentagons can
Tangram form?
O1. 52221
Regular tangram
O2. 52221
Proper tangram
O3. 52221
Proper tangram
How many distinct pentagons can
Tangram form?
P1. 33231
Regular tangram
Q1. 32232
Regular tangram
These are the 2 convex pentagons
which can be formed by tangram.
How many distinct pentagons can
Tangram form?
Same method can be applied to hexagons!
It is known that there are 4 convex hexagons
No one has solved the cases for concave
hexagons yet!!
a) 如何用七巧板圍出一個內部有一個面積
為4的正方形空洞，且此正方形空洞的
頂點不在此七巧圖的邊界上？
b) 內部有二個面積為1的正方形空洞 ？
c) 內部有面積為1的正方形空洞和三角形
空洞各一個？
(b)
(a)
(c)
如何用七巧板圍出內部有空洞的七巧圖，使
得此空洞面積愈大愈好，且此空洞的所有頂
點都不在此七巧圖的邊界上？
目前已知的最大紀錄為10.985，您能打破
此紀錄嗎？
The pieces
A
E
C
G
H
F
B
D
J
I
L
M
c
a
b
a
d
K
N