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1998-1999 Olympiad Correspondence Problems

Set 5

25.
Let c be a positive integer and let f0 = 1, f1 = c and fn = 2fn-1 - fn-2 + 2 for n 1. Prove that for each k 0, there exists m 0 for which fk fk+1 = fm.
View solution
26.
Let p(x) be a polynomial with integer coefficients for which p(0) = p(1) = 1. Let a0 be any nonzero integer and define an+1 = p(an) for n 0. Prove that, for distinct nonnegative integers i and j, the greatest common divisor of ai and aj is equal to 1.
View solution
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) + + 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?
View solution
28.
Find the locus of points P lying inside an equilateral triangle for which PAB + PBC + PCA = 90.
View solution
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 2.
(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 3.
View solution
30.
Let a, b, c be positive real numbers. Prove that
(a) 4(a3 + b3) (a + b)3;
(b) 9(a3 + b3 + c3) (a + b + c)3,
(c) More generally, establish that for each integer n and n nonnegative reals that


n2 (a13 + a23 + + an3) (a1 + a2 + + an)3  .
View solution

Problem 25.

25.
First solution. A straightforward induction argument establishes that fn = nc + (n-1)2, and it can be checked algebraically that
fk fk+1 = fm
where m = kc + (k2 - k + 1) = fk + k.

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 fk fk+1 = m2 + (c - 2)m + 1 as a quadratic equation to be solved for m. Note that when c = 4, it turns out that fn = (n+1)2. If we let f(c, n) = nc + (n-1)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.

Problem 26.

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
an+1 - 1
= an (an - 1)q(an)
= = an an-1 a1 (a1 - 1) n

k = 1 
q(ak)
for each positive integer n. (Note that an+1 cannot be zero, as this would force the impossible an (an - 1)q(an) = -1.) It follows that am 1 (mod ar) whenever m > r and the result follows.

26.
Second solution. Let p(x) = cd xd + + c1 x + c0. The conditions imply that c0 = cd + + c1 + c0 = 1. Let a be any nonzero integer. Then
p(a) = cd ad + + c1 a + c0 c0 = 1   (mod  a) .
Suppose as an induction hypothesis that the ith iterate pi (a) 1 (mod a). Then
pi+1 (a) = cd pi (a)d + + c1 pi (a) + c0 cd + + c1 + c0 = 1     (mod  a) .
In particular, we find that an+r 1 (mod an) for each pair of positive integers n and r from which the result follows.


Problem 27.

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) = x2 - r2 where 2r n. Then
g(x)
= (n+1) [x2 - nx + n(2n+1)
6
- r2]
= (n+1)



x + n
2


2

 
+

n2 + 2n
12
- r2



 .
Since
r2 - n2 + 2n
12
n2
4
- n2 + 2n
12
= n(n-1)
6
  ,
g(x) 0 when x = -n/2 and so has real roots. These roots are coincident if and only if n = 1 and r = 1/2, i.e., when f(x) = x2 - 1/4.

27.
(a) Second solution. [Z. Amir-Khosravi] 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
g(u - n)
= f(u - n) + + f(u) > 0
g(u) = f(u) + + f(u + n) < 0
g(v) = g(v) + + g(v + n) > 0  .
By the Intermediate Value Theorem, g(x) has two roots.

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 a1, a2, , ad where ai+1 ai + n. Since f(x) has no multiple roots, it alternates in sign as x passes through the roots ai.
To begin with, let n 2, and let ai be one of the roots. Suppose, wolog, that f(x) < 0 for ai - n < x < ai and f(x) > 0 for ai < x < ai + n. Then
g(ai - n) = f(ai - n) + + f(ai -1) + f(ai) < 0
and
g(ai) = f(ai) + f(ai + 1) + + f(ai + n) > 0 .
By the Intermediate Value Theorem, g(x) has a root bi between ai - n and ai. Since the open intervals (ai - n) are disjoint, the bi are distinct. Since we have identified d (the degree of g(x)) roots of g(x), all of the roots of g(x) are real.

Now suppose that n = 1. If the intervals [ai - 1, ai] are all disjoint, then we can use the foregoing argument since g(ai - 1) = f(ai - 1) and g(ai) = f(ai + 1) are nonzero with opposite signs. Suppose, on the other hand, there are s roots of f(x), namely ar, ar+1, , ar+s-1 spaced unit distance apart (ar+i = ar+i-1 + 1 for 1 i s-1) such that ar - 1 and ar+s-1 + 1 are not roots of f(x). We will discuss the case that f(x) < 0 for ar-1 x < ar and f(x) > 0 for ar+s-1 < x ar+s-1 + 1; the other three cases can be handled analogously.
In this case, all the roots of f(x) are simple, s is odd, and
g(ar - 1) = f(ar - 1) + f(ar) = f(ar - 1) < 0
g(ar+i) = f(ar+i) + f(ar+i+1) = 0      for  0 i s-2
and
g(ar+s-1) = f(ar+s-1) + f(ar+s-1 + 1) = f(ar+s-1 + 1) > 0 .
Thus, g(x) must have an odd number of roots in the interval (ar - 1, ar+s-1) counting multiplicity. We can account for s - 1 of them, { ar, ar+1, , ar+s-2 }, so, since s-1 is even, there must be at least s roots in this interval.

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.

Problem 28.

28.
First solution. Let PAB = a, PAC = a, PBC = b, PBA = b, PCA = g and PCB = g, so that a+ a = b+ b = g+ g = 60. Let |AP | = u, |BP | = v and |CP | = w. By the Law of Sines,
sina
sinb
· sinb
sing
· sing
sina
= v
u
· w
v
· u
w
= 1
whence
sinasinbsing = sinasinbsing .
Now
sinasinbsing
= 1
2
[ cos(a- b) - cos(a+ b)]sing
= 1
4
[sin(g+ a- b) +sin(b+ g- a) + sin(a+ b- g) - sin(a+ b+ g)]  .
Suppose that a+ b+ g = 90. Then
sinasinbsing
= 1
4
[sin(90 - 2 a) + sin(90- 2 b) + sin(90 - 2g) - 1]
= 1
4
[cos2a+ cos2b+ cos2g- 1] .

Since also a+ b+ g = 90, a similar identity holds for a, b and g. Hence
0
= (cos2a- cos2a)+ (cos2b- cos2b) + (cos2g- cos2g)
= 2sin60[sinl+ sinm+ sinn]
where l = a- a, m = b- b and n = g- g. Note that |l|,|m|, |n| < 60 and that l+ m+ n = 0.

Now sinl+ sinm = 2sin1/2(l+m)cos1/2(l- m) = -2 sin[(n)/2]cos[(l-m)/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 = l-m 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. Amir-Khosravi] See Figure 28.2. Let the vertices of the triangle in the complex plane be 1, -1 and 3i, 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
1 - 3i
2
= z - 3i
|z - 3i |
cis  a
1 = 1 - z
|1 - z |
cis b
and
3i + 1
2
= z + 1
|z + 1 |
cis  g .
Since cis a·cis b·cis g = cis [(p)/2] = i,
1 = i (z - 3i)(1 - z2)
|z - 3i ||1 - z2 |
 .
Hence the real part of (z - 3i)(1 - z2) is 0, which implies that the real part of z - z3 + 3iz2 is 0. Setting z = x + yi, we find that
0
= x - x3 + 3xy2 - 23xy
= x[(1 - 3y)2 - x2]
= x(1 - 3y + x)(1 - 3y - x) .
Hence (x, y) lies on one of the lines x = 0, y = (1/3)(1 + x), y = (1/3)(1 - x), the three bisectors of the angles. Conversely, one can show that any point on the bisectors lies on the lines.


Problem 29.

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
u2 = a2 + b2 - 2abcosq a2 + b2 2a2
whence u2/a2 2. Since the largest distance is at least u and the smallest at most a, the result follows.

(b) If ABC is a triangle for which a b c and C 120, then -1 cosC -1/2 and
c2 = a2 + b2 - 2ab cos120 a2 + b2 + ab 3a2
so that c 3a. Thus to prove the result, we need to show that among the six given points, we can find three that are vertices of a (possibly degenerate with three vertices in a line) triangle with one angle at least 120, or in other words, a point rays from which to two other points make an angle of at least 120.

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,
c
a
= sinC
sinA
sinC
sin(90 - C/2)
= 2sin(C/2)cos(C/2)
cos(C/2)
= 2sin(C/2) .
If C = 90, then c 2a and if C 120, then c 3a. The solution can be completed as before.


Problem 30

The specific results

30.
First solution. (a) Taking the difference of the two sides of the inequality yields 3 times
a3 - a2 b - ab2 + b3 = (a2 - b2)(a + b) = (a + b)(a - b)2 0 .

(b) Using the result in (a), we obtain that
9 (a3 + b3 + c3)
= 4(a3 + b3) +4(a3 + c3) + 4(b3 + c3) + (a3 + b3 + c3)
(a + b)3 + (a + c)3 + (b + c)3 + (a3 + b3 + c3)
= (a + b + c)3 + 2(a3 + b3 + c3) - 6abc
(a + b + c)3 + 2(a3 + b3 + c3) - 2(a3 + b3 + c3)
= (a + b + c)3  ,
with the last inequality due to that of the arithmetic and geometric means.

30.
Second solution. (b) Since the desired inequality is completely symmetrical in the variables, wolog we may suppose that a b c. Then
a2 (a - b) + b2 (b - c) + c2 (c - a) = (a2 - c2)(a - b) + (b2 - c2)(b - c) 0
and
a2 (a - c) + b2 (b - a) + c2 (c - b) = (a2 - b2)(a - c) + (b2 - c2)(b - c) 0 ,
so that
a3 + b3 + c3 a2 b + b2 c + c2 a
and
a3 + b3 + c3 ab2 + bc2 + ca2 .
Hence
(a + b + c)3
= a3 + b3 + c3+ 3(a2 b + b2 c + c2 a) + 3(ab2 + bc2 + ca2) + 6abc
a3 + b3 + c3 + 3(a3 + b3 + c3) + 3(a3 + b3 + c3)+ 2(a3 + b3 + c3)
from which the result follows.

The general result

30.
First solution. The Chebyshev Inequality asserts that if x1 x2 xn 0 and y1 y2 yn 0, then
n(x1 y1 + x2 y2 + + xn yn ) (x1 + + xn)(y1 + + yn) ,
This can be seen by taking the difference of the two sides which turns out to be the sum of all nonnegative products of the form (xi - xj)(yi - yj) with i < j.

Using this twice yields
n2 (a13 + + an3)
n (a12 + + an2)(a1 + + an)
(a1 + + an)2 (a1 + + an) = (a1 + + an)3 ,

30.
Second solution. The power mean inequality asserts that, for k > 1,
a1 + + an
n


a1k + + ank
n


1/n

 
 .
Applying this to k = 3 yields the result.

30.
Third solution. By the Cauchy-Schwarz Inequality, we obtain
n2

n

i = 1 
ai3



n

i = 0 
ai

n2

n

i = 0 
ai2

1/2

 




n

i = 0 
1

·

n

i = 0 
ai2



2

 


n

i = 1 
ai

4

 
 .
The result follows.

30.
Comment. L. Lessard began his solution by in effect replacing the original set { ai } 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
k2 (a13 + a23 + + ak3) (a1 + a2 + + ak)3
for all suitable ai. Apply this to each k-tuple among { a1, a2, , ak+1 } and add the results to obtain
k3 (a13 + a23 + + ak+13) (S - a1)3 + (S - a2)3 + + (S - ak+1)3
where S = a1 + a2 + + ak+1. Let bi = (S - ai)/k. Then S = b1 + b2 + + bk and so
a13 + a23 + + ak+13 b13 + b23 + + bk+13 .
This process can be iterated. The idea is that the right side tends under the iterations to a (k+1)-fold sum each terms of which is S/(k+1). But this requires an additional step to show that the maximum of |S - ai | does indeed decrease sufficiently as to tend to zero as we keep iterating.


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