Set 5 
 25.
 Let c be a positive integer and let f_{0} = 1, f_{1} = c
and f_{n} = 2f_{n1}  f_{n2} + 2 for n ³ 1. Prove that for
each k ³ 0, there exists m ³ 0 for which
f_{k} f_{k+1} = f_{m}.
 26.
 Let p(x) be a polynomial with integer coefficients for
which p(0) = p(1) = 1. Let a_{0} be any nonzero integer and define
a_{n+1} = p(a_{n}) for n ³ 0. Prove that, for distinct
nonnegative integers i and j, the greatest common divisor of
a_{i} and a_{j} is equal to 1.
 27.
 (a) Let n be a positive integer and let f(x) be a quadratic polynomial with real coefficients and real roots that differ by at least n. Prove that the polynomial g(x) = f(x) + f(x + 1) + f(x + 2) + ¼+ f(x + n) also has real roots.
 (b) Suppose that the hypothesis in (a) is replaced by
positing that f(x) is any polynomial for which the difference
between any pair of roots is at least n. Does the result
still hold?
 28.
 Find the locus of points P lying inside an equilateral
triangle for which ÐPAB + ÐPBC + ÐPCA = 90^{°}.
 29.
 (a) Prove that for an arbitrary plane convex quadrilateral, the ratio of the largest to smallest distance between pairs of points is at least Ö2.
 (b) Prove that for any six distinct points in the plane, the ratio of the largest to smallest distances between pairs of them is at least Ö3.
 30.
 Let a, b, c be positive real numbers. Prove that
 (a) 4(a^{3} + b^{3}) ³ (a + b)^{3};
 (b) 9(a^{3} + b^{3} + c^{3}) ³ (a + b + c)^{3},
 (c) More generally, establish that for each integer n and n
nonnegative reals that
n^{2} (a_{1}^{3} + a_{2}^{3} + ¼+ a_{n}^{3}) ³ (a_{1} + a_{2} + ¼+ a_{n})^{3} .
 25.
 First solution. A straightforward induction argument establishes that f_{n} = nc + (n1)^{2}, and it can be checked algebraically that

 25.
 Comment. This is basically a problem in experimentation and pattern recognition. Once the result is conjectured, the algebraic manipulations required are straightforward. One way to approach the problem is to consider f_{k} f_{k+1} = m^{2} + (c  2)m + 1 as a quadratic equation to be solved for m. Note that when c = 4, it turns out that f_{n} = (n+1)^{2}. If we let f(c, n) = nc + (n1)^{2}, then m = f(c+1, k). More generally, we have for every monic quadratic polynomial p with integer coefficients, p(k)p(k+1) = p(m) for some integer m.
 26.
 First solution. By the factor theorem, we have that p(x)  1 = x(x  1)q(x) for some polynomial q(x) with integer coefficients. Hence

 26.
 Second solution. Let p(x) = c_{d} x^{d} + ¼+ c_{1} x + c_{0}. The conditions imply that c_{0} = c_{d} + ¼+ c_{1} + c_{0} = 1. Let a be any nonzero integer. Then


 27.
 (a) First solution. By making a substitution of the form x¢ = x  k, we can dispose of the linear term in the quadratic. Thus we may assume that f(x) = x^{2}  r^{2} where 2r ³ n. Then


 27.
 (a) Second solution. [Z. AmirKhosravi] Let u and v be the two roots of f(x) with u + n £ v, with strict inequality when n = 1. Wolog, suppose that f(x) < 0 for x < u and x > v, while f(x) < 0 for u < x < v. Then g(x) is a quadratic polynomial for which

 When n = 1 and v = u + 1, then g(u) = 0. Since g(x) is a real quadratic with at least one real root, both roots are real.
 27.
 (b) First solution. Suppose that f(x), and hence g(x), has degree d, and let the roots of f(x) be a_{1}, a_{2}, ¼, a_{d} where a_{i+1} ³ a_{i} + n. Since f(x) has no multiple roots, it alternates in sign as x passes through the roots a_{i}.
 To begin with, let n ³ 2, and let a_{i} be one of the roots. Suppose, wolog, that f(x) < 0 for a_{i}  n < x < a_{i} and f(x) > 0 for a_{i} < x < a_{i} + n. Then


 Now suppose that n = 1. If the intervals [a_{i}  1, a_{i}] are all disjoint, then we can use the foregoing argument since g(a_{i}  1) = f(a_{i}  1) and g(a_{i}) = f(a_{i} + 1) are nonzero with opposite signs. Suppose, on the other hand, there are s roots of f(x), namely a_{r}, a_{r+1}, ¼, a_{r+s1} spaced unit distance apart (a_{r+i} = a_{r+i1} + 1 for 1 £ i £ s1) such that a_{r}  1 and a_{r+s1} + 1 are not roots of f(x). We will discuss the case that f(x) < 0 for a_{r1} £ x < a_{r} and f(x) > 0 for a_{r+s1} < x £ a_{r+s1} + 1; the other three cases can be handled analogously.
 In this case, all the roots of f(x) are simple, s is odd, and



 We can apply this analysis to the other ``blocks'' of s roots of f(x) spaced at unit distance to obtain disjoint intervals with at least s roots of g(x). From this, we can deduce that all the roots of g(x) must be real.
 28.
 First solution. Let ÐPAB = a, ÐPAC = a¢, ÐPBC = b, ÐPBA = b¢, ÐPCA = g and ÐPCB = g¢, so that a+ a¢ = b+ b¢ = g+ g¢ = 60^{°}. Let AP  = u, BP  = v and CP  = w. By the Law of Sines,




 Since also a¢+ b¢+ g¢ = 90^{°}, a similar identity holds for a¢, b¢ and g¢. Hence

 Now sinl+ sinm = 2sin^{1}/_{2}(l+m)cos^{1}/_{2}(l m) = 2 sin[(n)/2]cos[(lm)/2] and sinn = 2 sin[(n)/2]cos[(n)/2], whence 0 = 2 sin[(n)/2] [ cos[(n)/2]  cos[(l m)/2]]. Hence n = 0, n = lm or n = m l. If n = 0, then g = g¢ and P lies on the bisector of angle C. If l = m+ n, then l = 0 and P lies on the bisector of angle A. If m = l+ n, then m = 0 and P must lie on the angle bisector of the triangle ABC.
 Conversely, it is easily checked that any point P on these bisectors satisfies the given condition.
 28.
 Second solution. [Z. AmirKhosravi] See Figure 28.2. Let the vertices of the triangle in the complex plane be 1, 1 and Ö3i, and let z be a point on the locus. Then, with the angles as indicated in the diagram and cis q = cosq+ isinq, we have that





 29.
 First solution. (a) One of the angles q of the quadrilateral must be between 90^{°} and 180^{°} inclusive so that cosq £ 0. Let a and b be the lengths of the sides bounding this angle and let u be the length of the diagonal joining the endpoints of these sides (opposite q). Wolog, suppose that a £ b. Then

 (b) If ABC is a triangle for which a £ b £ c and C ³ 120^{°}, then 1 £ cosC £ 1/2 and

 Consider the smallest closed convex set containing the six points. If all six points are on the boundary, then they are the vertices of a (possibly degenerate) convex hexagon. Since the average interior angle of such a hexagon is 120^{°}, one of the angles is at least 120^{°} and we get the desired result.
 If the convex set has fewer than six points on the boundary, the remaining point(s) lie in its interior. Triangulate the convex set using only the boundary points; one of the interior points lies inside one of the triangles (possibly on a side), and one of the sides of the triangle must subtend at this point an angle of at least 120^{°}. Again, the desired result holds.
 29.
 Second solution. [L. Lessard] Suppose ABC is a triangle with a £ b £ c and 90^{°} £ C £ 180^{°}. Then 2A £ A + B = 180^{°}  C, so that by the Law of Sines,

 30.
 First solution. (a) Taking the difference of the two sides of the inequality yields 3 times

 (b) Using the result in (a), we obtain that

 30.
 Second solution. (b) Since the desired inequality is completely symmetrical in the variables, wolog we may suppose that a ³ b ³ c. Then





 30.
 First solution. The Chebyshev Inequality asserts that if x_{1} ³ x_{2} ³ ¼ ³ x_{n} ³ 0 and y_{1} ³ y_{2} ³ ¼ ³ y_{n} ³ 0, then

 Using this twice yields

 30.
 Second solution. The power mean inequality asserts that, for k > 1,

 30.
 Third solution. By the CauchySchwarz Inequality, we obtain

 30.
 Comment. L. Lessard began his solution by in effect replacing the original set { a_{i} } be a set whose elemenets are a little closer to the mean of these numbers.The result holds for n = 1. We assume as an induction hypothesis that it holds for n = K, viz. that


