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Use the following slider to specify the number of kites to use in a dissection of the disk and obtain a corresponding lower bound on π:

Out[]=

​
n 14 π ≈ 3.115

Computational Explanation

Liu Hui approximated the area of a disk by dividing it into congruent kite-shaped regions and making use of clever recurrence formulas derived from the Pythagorean theorem to iteratively obtain the area obtained by doubling the number of kites. To see how the process works, consider an inscribed 12-sided regular polygon consisting of six kites:

Out[]=

Call the radius of the circle

R

and denote the chordal diagonal of the kite by

a

n

. Now denote the apothem (distance from the center of the circle to the crossing point of the two kite diagonals) by

r

n

. As shown in the following diagram, the sagitta (distance from the crossing point of the kite's diagonals to its less-pointed apex) is then simply

R-

r

n

:

Out[]=

The lengths

r

n

,

R

and

a

n

are now related via the Pythagorean theorem using a triangle from the lower portion of the kite:

r

n



2

R

-

2

a

n

2

More interestingly, if we double the number of kites from

n

to

2n

, the chordal diagonal length for the new kite

a

2n

is also given by the Pythagorean theorem using a triangle from the upper portion of the kite:

a

2n



2

a

n

2

+

2

(R-

r

n

)

.

Therefore, if we know

a

n

for some value of

n

, we can immediately compute

a

2n

using the previous formulas:

In[]:=

r[n_]:=Simplify@Sqrt[R^2-(a[n]/2)^2]​​a[n_]:=Simplify@Sqrt[(a[n/2]/2)^2+(R-r[n/2])^2]

To begin the recursion, note that the edge length of a regular hexagon inscribed in a circle of radius

R

is

a

6

R

since the regular hexagon consists of six equilateral triangles arranged around the origin, so:

In[]:=

a[6]=R;

While not strictly necessary, we can simplify the form of computations involving symbolic radii by assuming

R

is positive:

In[]:=

$Assumptions=R>0;

Knowing

a

n

is sufficient to compute the area of a regular polygon using the area of a kite with diagonal lengths

p

and

q

as given by

A

kite



1

2

pq

. For the kites considered here and shown in the diagram above, one diagonal has length

a

n

and the other length

R

. Taking

n

such kites therefore gives the area of a polygon with

2n

sides as:

A

2n



1

2

npq

1

2

n

a

n

R

Writing in the Wolfram Language:

In[]:=

A[n_]:=

1

2

n

2

a

n

2

R

Liu considered a circle of radius 1 chi  10 zuen and expressed the areas obtained by doublings of the hexagon up to

A

192

as fractions over 625. These results are summarized in the following table:

Out[]=

A 6	150.	150
A 12	300.	300
A 24	310.583	310+ 364 625
A 48	313.263	313+ 164 625
A 96	313.935	313+ 584 625
A 192	314.103	314+ 64 625

Liu then observed that the area of the circle can be completely covered by replacing the upper portion of the kites with rectangular extensions tangent to the circle as shown in the following sequence of diagrams:

In[]:=

diagram[poly_,text_,ang_:2π/8]:=Module[{x=Sin[ang],y=Cos[ang]},Graphics[{FaceForm[FaceForm[ColorData[97][1]]],poly/.{"x"x,"y"y},​​text,​​Circle[],Line[{{0,0},{-x,y},{0,1},{x,y},{0,0},{0,1}}],Line[{{-x,y},{-x,1},{x,1},{x,y},{-x,y}}]},BaseStyle12]]​​Row[{​​diagram[Polygon[{{0,0},{-"x","y"},{"x","y"}}],Text[Subscript["S",96],{.2,.45}]],​​diagram[Polygon[{{0,0},{-"x","y"},{0,1},{"x","y"}}],Text[Subscript["S",192],{.2,.45}]],​​diagram[Polygon[{{-"x","y"},{0,1},{"x","y"}}],Text[Subscript["S",192]-Subscript["S",96],{0,1.3}]],​​diagram[Polygon[{{-"x","y"},{-"x",1},{"x",1},{"x","y"}}],Text[2(Subscript["S",192]-Subscript["S",96]),{0,1.3}]],​​diagram[Polygon[{{0,0},{-"x","y"},{-"x",1},{"x",1},{"x","y"}}],Text[Subscript["S",96]+2(Subscript["S",192]-Subscript["S",96]),{0,1.3}]]​​}]

Out[]=

Liu then used the identity:

A

192

<A<

A

96

+2(

A

192

-

A

96

)

to obtain bounds on the value of the polygon area and hence π:

In[]:=

Block[{R=1.},A[192]<A<A[96]+2(A[192]-A[96])]

Out[]=

3.14103<A<3.14271

Plugging in verifies Liu's bounds indeed bracket π:

In[]:=

3.14103<π<3.14271

Out[]=

True

Plotting values for larger

n

(though Liu only computed up to

n5

) shows rapid convergence to π:

In[]:=

ListPlot[Block[{R=1.},Table[A[6

n

2

],{n,10}]],FillingAxis,PlotRangeAll]

Out[]=

The convergence is even more evident on a logarithmic plot of the difference

π-

A

6

n

2

:

In[]:=

ListLogPlot[Block[{R=1.},Table[π-A[6

n

2

],{n,10}]]]

Out[]=

By extrapolating the values he calculated to larger

n

, Liu obtained a very good estimate of π:

In[]:=

3927

1250

//N

Out[]=

3.1416

There is more to be said, however. Since as

n

goes to infinity, the polygon approximates a disk of radius

R

and area π

2

R

more and more closely, the value of π is given by:

π

lim

n∞

A(6

n

2

)

2

R

Using the recurrence with symbolic instead of numeric values shows a beautiful pattern for successive approximations:

In[]:=

Block{R=1},Table

A[6

n

2

]

2

R

,{n,7}

Out[]=

3,6

2-

3

,12

2-

2+

3

,24

2-

2+

2+

3

,48

2-

2+

2+

2+

3

,96

2-

2+

2+

2+

2+

3

,192

2-

2+

2+

2+

2+

2+

3



The

n

th iteration can therefore be computed directly in the Wolfram Language as:

In[]:=

NestedRadicalA[1]:=3​​NestedRadicalA[n_]:=6

n-2

2

Sqrt[2-Nest[Function[Sqrt[2+#]],

3

,n-2]]

In[]:=

Table[NestedRadicalA[n],{n,8}]

Out[]=

3,6

2-

3

,12

2-

2+

3

,24

2-

2+

2+

3

,48

2-

2+

2+

2+

3

,96

2-

2+

2+

2+

2+

3

,192

2-

2+

2+

2+

2+

2+

3

,384

2-

2+

2+

2+

2+

2+

2+

3



These values have a direct connection with trigonometry by noting that the area of a regular

n

-gon inscribed in a circle of radius

R

is given by:

A

n



1

2

n

2

R

sin

2π

n

As a result, the above nested radical expressions may be immediately seen to correspond to:

In[]:=

Table2Sin

2Pi

6

n

2

,{n,4}//FunctionExpand//FullSimplify

Out[]=

1,

2-

3

,

2-

2+

3

,

2-

2+

2+

3



with a numerical prefactor of

6

n-2

2

. Interestingly, these nested radicals are very similar to expressions for

sin(π/

n

2

)

and

cos(π/

n

2

)

such as:

cos

π

32



1

2

2+

2+

2+

2

sin

π

32



1

2

2+

2-

2+

2

,

but with

2+

3

at their innermost level instead of a

2+

2

. This suggests such values might result from Liu's construction starting with a square (for which

a

4



2

R

) instead of a 6-gon, and indeed this is the case:

In[]:=

Table2Sin

2Pi

4

n

2

,{n,4}//FunctionExpand//FullSimplify

Out[]=



2

,

2-

2

,

2-

2+

2

,

2-

2+

2+

2



Other Resources

Additional Reading

• Dauben, J. W. The Chinese "Euclid," Liu Hui. Ch.3, § V in The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook (Ed. V. Katz.). Princeton, NJ: Princeton University Press, pp. 226–240, 2007.
• Joseph, G. G. The Crest of the Peacock: Non-European Roots of Mathematics, 3rd ed. Princeton, NJ: Princeton University Press, pp. 193–198 and 266–270, 2011.
• Liu, H. Jiu zhang suan shu. Various editions. 1896, 1899, 1975 photo-reprint, etc.
• Mankiewicz, R. The Story of Mathematics. Princeton, NJ: Princeton University Press, p. 37, 2004.
• Martzloff, J.-C. A History of Chinese Mathematics. Berlin: Springer-Verlag, pp. 278–282, 1987.
• Shen, K.; Crossley, J. N.; and Lun, A. W.-C. The Nine Chapters on the Mathematical Art: Companion and Commentary. Oxford and Beijing: Oxford University Press, 1999.

Additional Links

• Hathi Trust Digital Library: Jiu zhang suan shu xi cao tu shuo (1896)
• Hathi Trust Digital Library: Jiu zhang suan shu. v.1-3 (1899)
• Mathematical Treasures - Jiuzhang suanshu
• Wikipedia: Liu Hui
• Wikipedia: The Nine Chapters on the Mathematical Art