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318 Answers to Exercises and Solutions to Problems

sphere and at the center of the circle of intersection, it must intersect this

circle in exactly two points, (xi, yi, zi) and (x2, ~2, ~2). It is easily verified

that (16 - xl, 9 - yi, 8 - zi) satisfies the three equations, and so this point

can be none other than (x2, ~2, z2).

8.29. Suppose if possible that at least two of x, y, z are equal. Then

x = y = z = t where t3 + (u - 1)t + b = 0. By Exercise 1.4.4, there is a

unique real value oft which satisfies this equation.

If x, y, z are all distinct, then from

z - y = (y - x)(y2 + y-z + x2 + a)

x - t = (z - y)(z2 + zy + y2 + a)

y - x = (x - %)(x2 + xz + z2 + a),

we have that

1 = (y2 + yx + x2 + a)(z2 + zy + y2 + a)(~” + xy + y2 + a) > u3,

which contradicts the condition on a.

8.30. Since a.z + cx = bz + cy = ay + bx, it follows that

x:y:z=a(b+ c - u) : b(c + a - b) : c(a + b - c).

Taking a suitable constant of proportionality, values of x, y, z satisfying

the given equations can be found.

8.31.

%2 - abz + (u” + b2 - 4) = ,z2 - (xy + xy-’ + x-‘y + x-‘y-‘)z

+ (x2 + x-2 + y2 + y-2) = [% - (“y-l + x-‘y)][z - (xy + x-‘y-l)].

8.32. Let t = ux/a” = vy/b2 = wzjc”. Then x/u = t(a/u), etc. so that
t2(u2/u2 + b2/v2 + c2/w2) = 1. Hence

(x/u + y/v + z/w)~ = (tu2/u2 + tb2/v2 + tc2/w2)2

= u2/u2 + b2/v2 + c2/w2.

8.33. A complete set of solutions of the equation fi + G = 1
can be described geometrically as follows. Let BC be a segment of length

1. Each solution (y, z) of the equation corresponds to a point P on the
   segment of length 1 parallel to and distant 4 from BC.

/qiy

BdF-= &Ti C
For the system of three equations, form a triangle ABC whose sides
are three segments of length 1. The solution of the system is given by the