Solutions to Problems; Chapter (^8 393)
- Let a, b, c be the lengths of the sides. Then u+b+c = 98, and, by Heron’s
formula for the area, 49(49 - a)(49 - b)(49 - c) = 4202. Let u = 49 - a,
v = 49 - b, w = 49 - c. Then u + v + w = 49 and uvw = 3600, so that U,
v, w are zeros of a polynomial of the form j(x) = x3 - 49x2 + rx - 3600.
Prom the given example, we know that (u, v, w) = (9,20,20) works and
we can write
f(x) = (x - 9)(x - 20)2 - sx.
Now, j(x) will have three real zeros _ the line y = sx intersects the
cubic curve y = (x - 9)(x - 20)2 ’ m three points (counting multiplicity).
This will occur for 0 5 s 5 k, where k is that positive value of s for which
f(x) has a double zero, i.e. j(x) = (x - 9)(x - 20)2 - sx and f’(x) =
(x - 2O)[(x - 20) + 2(x - 9)] - s have a zero in common.
But any common zero of f(x) and f’(x) is a zero of
f(x) - xf’(x) = -(x - 20)(x - 12)(2x + 15).
Hence, when s = k, f(x) has 12 as a double zero. When 0 5 s 5 k, f(x)
has a zero between 9 and 12 inclusive. Since such a zero must divide 3600,
the possibilities are 9, 10 and 12. Checking these out yields the solutions
(u,v, w) = (9,20,20), (10, 15, 24), (12, 12, 25) or (u,b,c) = (40,29,29),
(39, 34, 25), (37, 37, 24).- (x - y)2 + (y - %)2 + (% - x)?
- Multiply the difference of the two sides by the product of the denomi-
nators to obtain
x2y2r2(x + y + %) - 2xyz(xy + yz + ZX) + (“Y2 + j/z2 + %X2)
+ xY+/2 + yz2 + %X2) - 2334x + y + %) + (xy + y.z + ZX)
= xy(y + 1)(%X - 1)2 + yz(z + l)(xy - 1)2 + %X(X + l)(yz - 1)” 2 0.- Factoring the denominator, first as a difference of squares, and then
completely yields (u - x)(b - x)(u + x)(b + x). The numerator is equal to
[u(b - x) + x(u - x)12 - [x(u - x) + x(b - x)][u(b - x) + b(u - x)]
= (u” - ux)(b - x)~ + (x2 - bx)(u - x)” + (ax - bx)(u - x)(b - x)
= (u - x)(b - x)[u(b - x) - x(u - x) + (a - b)x] = (u - x)“(b - x)“.
The answer is (u - x)(b - x)/(u + x)(b + x).