A History of Mathematics From Mesopotamia to Modernity

100 A History ofMathematics

This (ifzis the yield of low-grade) gives us 36z=99 from the first column, soz=^9936 = (^234)
as required. We now use the second column to findy, the medium-grade yield( 5 y+z= 24 ),
and so on.

5. (See Fig. 5.) The equation on the St Andrew’s website actually is not ‘the answer’, since it
   computes the radius (correctly, as 120). If we wantxto be the diameter, as the question asks,
   then we must have radius=x 2. We use the similar right-angled triangles ABC, ADE. We have
   AC=x+150, BC=208, and DE=x/2 easily. AD= 135 +x/2, so by Pythagoras’s theorem,
   AB=

√

2082 +(x+ 150 )^2. Since

AD

DE

=

AB

BC

AD·BC=AB·DE, and( 270 +x)· 208 =(

√

2082 +(x+ 150 )^2 )·x(doubling to remove the

halves). Now get rid of the square root by squaring everything; we get

2082 ( 270 +x)^2 =( 2082 +(x+ 150 )^2 )x^2

or

x^4 + 300 x^3 +22,500x^2 −23,362,560x−3,153,945,600= 0

I leave it to you to check thatx=240 is a solution. It would seem likely that Li knew the

answer in the first place not from solving the equation but (like most textbook writers) because

he had chosen the numbers to come out exactly; in this case so that the sides of the triangles

are in the ratio 8 : 15 : 17.