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Algorithm

Zhao Youqin started with an inscribed square in a circle with radius r.¹

If ℓ {\displaystyle \ell } denotes the length of a side of the square, draw a perpendicular line d from the center of the circle to side l. Let e denotes r − d. Then from the diagram:

d = r 2 − ( ℓ 2 ) 2 {\displaystyle d={\sqrt {r^{2}-\left({\frac {\ell }{2}}\right)^{2}}}} e = r − d = r − r 2 − ( ℓ 2 ) 2 . {\displaystyle e=r-d=r-{\sqrt {r^{2}-\left({\frac {\ell }{2}}\right)^{2}}}.}

Extend the perpendicular line d to dissect the circle into an octagon; ℓ 2 {\displaystyle \ell _{2}} denotes the length of one side of octagon.

ℓ 2 = ( ℓ 2 ) 2 + e 2 {\displaystyle \ell _{2}={\sqrt {\left({\frac {\ell }{2}}\right)^{2}+e^{2}}}} ℓ 2 = 1 2 ℓ 2 + 4 ( r − 1 2 4 r 2 − ℓ 2 ) 2 {\displaystyle \ell _{2}={\frac {1}{2}}{\sqrt {\ell ^{2}+4\left(r-{\frac {1}{2}}{\sqrt {4r^{2}-\ell ^{2}}}\right)^{2}}}}

Let l 3 {\displaystyle l_{3}} denotes the length of a side of hexadecagon

ℓ 3 = 1 2 ℓ 2 2 + 4 ( r − 1 2 4 r 2 − ℓ 2 2 ) 2 {\displaystyle \ell _{3}={\frac {1}{2}}{\sqrt {\ell _{2}^{2}+4\left(r-{\frac {1}{2}}{\sqrt {4r^{2}-\ell _{2}^{2}}}\right)^{2}}}}

similarly

ℓ n + 1 = 1 2 ℓ n 2 + 4 ( r − 1 2 4 r 2 − ℓ n 2 ) 2 {\displaystyle \ell _{n+1}={\frac {1}{2}}{\sqrt {\ell _{n}^{2}+4\left(r-{\frac {1}{2}}{\sqrt {4r^{2}-\ell _{n}^{2}}}\right)^{2}}}}

Proceeding in this way, he at last calculated the side of a 16384-gon, multiplying it by 16384 to obtain 3141.592 for a circle with diameter = 1000 units, or

π = 3.141592. {\displaystyle \pi =3.141592.\,}

He multiplied this number by 113 and obtained 355. From this he deduced that of the traditional values of π, that is 3, 3.14, ⁠22/7⁠ and ⁠355/113⁠, the last is the most exact.²

See also

• Liu Hui's π algorithm

References

1. Yoshio Mikami, Development of Mathematics in China and Japan, Chapter 20, The Studies about the Value of π etc., pp 135–138 /wiki/Yoshio_Mikami ↩

2. Yoshio Mikami, p136 ↩