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Solutions to Problems; Chapter 6 349

4.5. k = 86.

4.6. Let u = tan A, v = tan B, w = tanC, and let the fourth root be y.
Wehavep-y=u+v-tw=uvw=s/y,uv+uw+vw+y(u+v+w)=
q and (uv + uw + vw)y + uvw = r. Hence uv + uw + VW = q - s, so
(q - s)y + (p - y) = r. Hence (q - s - 1)y = r -p.
Suppose, if possible, that q = s + 1. Then uv + uw + VW = 1. Together
with u + v + w = uvw, this implies that

o= l-(u+v)w- uv = (1+ w”)(uv - 1)

so uv = 1. Similarly uw = VW = 1 and we obtain a contradiction. Hence
q - s - 1 # 0 and y = (r - p)/(q - s - 1).

Remark. It can also be seen that y2 - py + s = 0. Eliminating y from
the two equations gives a necessary constraint on the coefficients, which
reflects the fact that not every set of three numbers can be represented as
the tangents of the angles of a triangle.

4.7. x + y = 5 - z and

(x + y)z + xy = 3 * xy = z2 - 52. + 3 3 0 < (x - y)2 = (x + y)2 - 42y

= -(3z2 - 1oz - 13) = -(z + 1)(3z - 13),

from which the result follows,

4.8. Suppose a is as required. Then u + v + w = 6

ea (u-l)+(v-2)=-(w-3)

+=9 (u - 1)s + (v - 2)3 + 3(u - l)(v - 2)(u + v - 3)

= -(w - 3)3

w (u - l)(v - 2)(u + v- 3) = 0.

Similarly

and

(u - l)(w - 3)(u + w - 4) = 0

(v - 2)(w - 3)(v + w - 5) = 0.
Case 1. 21 = 1, v = 2, w = 3 are the zeros of the polynomial x3 - 6x2 +
11x - 6, which is not of the required form.
Case2.u=l,v+w=5,sothata=u(v+w)+vw=5-a+o=55/2.
In this case, v and w are the zeros of x2 - 5x - 512 and the cubic is

(x - 1)(x2 - 5x - 5/2) = x2 - 6z2 + 5x/2 + 5/2.

Case3.v=2,u+w=4,sothata=8-a/2ja=16/3.Thecubicis

(x - 2)(x2 - 4x - 8/3) = x3 - 6x2 + 16x/3 + 16/3.