# A geometric challenge from ancient China

Donald B. Wagner

17 February 2022

The famous mathematical classic Jiuzhang suanshu 九章算術 (Arithmetic in nine chapters, ca. 1st century CE) has this formula for the diameter of a circle inscribed in a right triangle:

The commentator Liu Hui 劉徽 in the 3rd century CE gave a proof by area dissection. His diagram has long been lost, but a plausible reconstruction is this:

The area of the triangle can be represented in two ways:

so that

Liu Hui then added that an alternate formula is

d = a + b - c

which follows from the fact that

b – ½d = c – (a – ½d)

In the 13th century Li Ye 李冶 in his book Ceyuan haijing 測圓海鏡 (Sea mirror of circle measurements, 1248) gives the interesting diagram seen here on the right and (among other things) gives formulas for the diameter of the circle given the dimensions of various right triangles within the large triangle. I give these in modern form directly below. He gives no proofs.

Your challenge is to prove each of these formulas. It turns out that all can be proved by Liu Hui’s area dissection method. When you give up, click on the button to see the proof.

In each of the formulas, a, b, and c are respectively the shorter and longer legs and the hypotenuse of the triangle in question; d is the diameter of the circle.

The sum of the blue and red areas in the upper rectangle is

½d (a + b + c) = ab + ad

so that

2ab = d (b + ca)

Further

b + ½d = c + (a – ½d)

d = ab + c

The sum of the blue and red areas in the upper rectangle is

½d(a + b + c = ab + bd

so that

2ab = d (ab + c)

Further,

a + ½d = c +(b – ½d)

d = b + ca

The sum of the blue and red areas in the upper rectangle is

½d (a + b + c) = ab + dc

so that

2ab = d (a + b - c)

Further,

c = (½da) + (½db)

d = a + b + c

The area of the triangle is

½ab = ¼cd

so that

2ab = cd

Twice the area of the triangle AQK is

ab = ½d (b+c)

so that

2ab = d (b + c)

The area of the triangle is

so that

2ab = d(a + c)

The area of the blue and red triangle is

so that

2ab = (ca)d

The area of the blue and red triangle is

¼(b + c)d = ½ab + ½bd

so that

2ab = (cb)d

Further formulas were added by Li Shanlan 李善蘭 (1810–1882) around 1876 in an unpublished commentary on Ceyuan haijing. He drew the line RT parallel to AB, passing through the point O, and stated the following equalities. Two of these are trivial. The other two can be proved using the same approach as above.

Obviously

a = ½d

d = 2a = 2ab/b

Obviously

b = ½d

d = 2b = 2ab/a

The area of the triangle is

½ab = ¼d2 + ¼d(a – ½d) + ¼d(b – ½d)

so that

2ab = d2 + ad – ½d2 + bd – ½d2

= d(a + b)

The area of the blue and red triangle is

½ab + ¼ad + ⅛d2 = ¼d(b + ½d)

so that

2ab + ad + ½d2 = bd + ½d2

and

2ab = d(ba)

### References

Most of the information above, except the proofs, is taken from

Qian Baocong 钱宝琮, Zhongguo shuxue shihua 中国数学史话, Beijing: Zhongguo Qingnian Chubanshe, 1958, pp. 75–78.

I have not seen the proofs published anywhere.

There are several useful translations of the Jiuzhang suanshu and Liu Hui’s commentary.

Karine Chemla and Guo Shuchun, Les neuf chapitres: Le classique mathématique de la Chine ancienne et ses commentaires, Paris, Dunod, 2004.

Guo Shuchun, Joseph W. Dauben, and Xu Yiba , Nine chapters on the art of mathematics, Shenyang, Liaoning: Liaoning Education Press, 2013.

Shen Kangshen, John N. Crossley, and Anthony W.–C. Lun, The nine chapters on the mathematical art: Companion and commentary, Oxford: University Press and Beijing: Science Press.

There has not been much published in Western languages on Li Ye and his Ceyuan haijing. The basic facts are given by

Jean-Claude Martzloff, A history of Chinese mathematics, Berlin: Springer, 1997, pp. 143–149.

On Li Shanlan see Martzloff, pp. 173–176, and

Arthur W. Hummel (ed.), Eminent Chinese of the Ch’ing period, Washington, D.C.: U.S. Government Printing Office, 1943, pp. 479–482.

Much on all of these subjects can also be found in Wikipedia and elsewhere on the Web.