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A History of Mathematics
3 This (if z is the yield of low-grade) gives us 36z = 99 from the first column, so z = 99 36 = 2 4 as required. We now use the second column to find y, the medium-grade yield (5y + 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 want x to be the diameter, as the question asks, then we must have radius = 2x . 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
AB AD = DE 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 )x2 or x4 + 300x3 + 22,500x2 − 23,362,560x − 3,153,945,600 = 0 I leave it to you to check that x = 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.