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 + c – a)
Further
b + ½d = c + (a – ½d)
d = a – b + c
The sum of the blue and red areas in the upper rectangle is
½d(a + b + c) = ab + bd
so that
2ab = d (a – b + c)
Further,
a + ½d = c +(b – ½d)
d = b + c – a
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 = (½d – a) + (½d – b)
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
½ab = ¼ad + ¼cd
so that
2ab = d(a + c)
The area of the blue and red triangle is
¼(c + a)d = ½ab +½ad
so that
2ab = (c – a)d
The area of the blue and red triangle is
¼(b + c)d = ½ab + ½bd
so that
2ab = (c – b)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Further formulas were added by Li Shanlan 李善蘭 (1810–1882) around 1876 in an unpublished commentary on Ceyuan haijing.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(b – a)
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.