380 Answers to Exercises and Solutions to Problemsasinu+bsinv+csinw=O.
Shifting two terms to the right side in each equation, squaring and adding
yields
a2 = b2 + c2 + Pbccos(v - w)
b2 = a2 + c2 + Paccos(u - 20)
c2 = a2 + b2 + 2ubcos(u - v).(*IOn the other hand, suppose u, v, w can be chosen to satisfy (*). Then= a2+b2+c2+2ab(cosucosv+sinusinv)+..,
= a2 + b2 + c2 + [c2 - u2 - b2] +... = 0.
Hence, the given system is solvable iff the system (*) is solvable.
We show that (*) is solvable iff Ial 5 lb/ + ICI, Ibl 5 Ial + ]c] and ]c] 2
IQ] + 161. If (*) is solvable, then a2 < b2 + c2 + 2lbl ICI = (lb1 + Ic])~, etc. On
the other hand, suppose the inequality conditions hold. Then we can form
a triangle with sides ]a], lbl, ]c] opposite angles A, B, C respectively. We
can select 21, 21, w to satisfylbcl cos A = -bccos(v -w)lacl cos B = -ac cos(u - w)
Iubl cos C = -ub cos(u - v).
For example, if bc > 0, ub < 0, UC < 0, we can choose u = B, v = B + C =
x - A, w = 0. [Solution by Gary Baumgartner.]
Second solution. Assume that ubc # 0, and let there be a solution
(x, y, z) = (cosu,cos v,cos w), where 0 < u,v, w 5 ?r. Then, for each 8,
we have that
u cos(u + e) + ~CO~(V + e) + CCOS(W + e)= cose(ucosu+bcosv+ccosw)
+ sinB(usinu+bsinv+csinw) =O.Taking 0 = -u yields a = -b cos(v -u)-ccos(w-u), whence Ial 5 Ibl+IcI.
Similarly, Ibl 5 IQ]+ ]c] and ]c] 5 ]a]+ lb/. The rest can be treated as before.
- Yes. Since there are five odd powers of x, the second player can guar-
antee that after the first player has completed four moves, there is a coeffi-
cient of one of these powers which has not been assigned. When he comes
to make his fourth move, he finds that the polynomial f(x) has the form
g(x) + axp + bxd where g(x) is determined, r is odd and a, b have yet to be
selected. Ifs is even, he notes that, since f(1) + f(-1) = g(1) +g(-1) + 2b
does not depend on a, he can choose b in such a way that f(l)+ f (-1) = 0.