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4.36 Shortlisted Problems 1995 601
28. Let F(x) = f(x)−95 for x ≥ 1. Writing k for m+95, the given condition becomes
F(k +F(n)) = F(k) +n, k ≥ 96,n ≥ 1. (1)
Thus for x,z ≥ 96 and an arbitrary y we have F(x + y) + z = F(x + y + F(z)) = F(x + F(F(y) + z)) = F(x) + F(y) + z, and consequently F(x + y) = F(x) + F(y) whenever x ≥ 96. Moreover, since then F(x + y) + F(96) = F(x + y + 96) = F(x) + F(y +96) = F(x) +F(y) + F(96) for any x,y, we obtain
F(x+y) = F(x) +F(y), x,y ∈ N. (2)
It follows by induction that F(n) = nc for all n, where F(1) = c. Equation (1) becomes ck + c2n = ck + n, and yields c = 1. Hence F(n) = n and f(n) = n+95 for all n.
Finally, k=1 f(k) = 96+97+···+114 = 1995.
Second solution. First we show that f(n) > 95 for all n. If to the contrary f(n) ≤ 95, we have f(m) = n + f(m + 95 − f(n)), so by induction f(m) = kn+f(m+k(95−f(n))) ≥ kn for all k, which is impossible. Now for m > 95 we have f(m+f(n)−95) = n+f(m), and again by induction f(m + k(f(n) − 95)) = kn + f(m) for all m,n,k. It follows that with n
fixed,
f(m+k(f(n) −95)) n
k→∞ m+k(f(n) −95) f(n)−95
hence f(s) n s→∞ s f(n) −95
Hence f(n)−95 does not depend on n, i.e., f(n) ≡ cn+95 for some constant c. It is easily checked that only c = 1 is possible.
602 4 Solutions
4.37 Solutions to the Shortlisted Problems of IMO 1996
1. We have a5 + b5 − a2b2(a + b) = (a3 − b3)(a2 − b2) ≥ 0, i.e. a5 + b5 ≥ a2b2(a +b). Hence
ab ab abc2 c
a5 +b5 +ab a2b2(a+b) +ab a2b2c2(a +b)+abc2 a+b+c
Now, the left side of the inequality to be proved does not exceed a+b+c + a+b+c + a+b+c = 1. Equality holds if and only if a = b = c.
2. Clearly a1 > 0, and if p = a1, we must have an < 0, |an| > |a1|, and p = −an. But then for sufficiently large odd k, −ak = |an|k > (n−1)|a1|k, so that ak + ··· + ak ≤ (n − 1)|a1|k − |an|k < 0, a contradiction. Hence p = a1.
Now let x > a1. From a1 + ··· + an ≥ 0 we deduce j=2(x − aj) ≤
(n −1) x+ n−1 , so by the AM–GM inequality,
n−1
(x−a2)···(x−an) ≤ x+ n−1 ≤ xn−1+xn−2a1+···+an−1. (1) The last inequality holds because n−1 ≤ (n − 1)r for all r ≥ 0. Multi-plying (1) by (x−a1) yields the desired inequality.
3. Since a1 > 2, it can be written asa1 = b+b−1 for someb > 0. Furthermore, a2 −2 = b2 +b−2 and hence a = (b2 +b−2)(b +b−1). We prove that
an = b+b−1b2 +b−2b4 +b−4···b2n−1 +b−2n−1
by induction. Indeed, ann1 = an−1 2 − 2 = b2n−1 +b−2n−1 2 − 2 = b2n +b−2n.
Now we have
n 3
ai = 1+ b2 +1 + (b2 +1)(b4 +1) +··· 2n−1
··· + (b2 +1)(b4 +1)...(b2n +1).
Note that 1(a+2−√a2 −4) = 1+ 1; hence we must prove that the right side in (1) is less than b. This follows from the fact that
b2k
(b2 +1)(b4 +1)···(b2k +1)
= (b2 +1)(b4 +1)···(b2k−1 +1) − (b2 +1)(b4 +1)···(b2k +1) ;
hence the right side in (1) equals 1 1− (b2+1)(b4+1)...(b2n+1) , and this is
clearly less than 1/b.
4.37 Shortlisted Problems 1996 603
4. Consider the function
f(x) = x + x2 +···+ xn .
Since f is strictly decreasing from +∞ to 0 on the interval (0,+∞), there exists exactly one R > 0 for which f(R) = 1. This R is also the only positive real root of the given polynomial.
Since lnx is a concave function on (0,+∞), Jensen’s inequality gives us
n aj ln j ≤ ln⎛ n aj · j ⎞ = lnf(R) = 0. j=1 j=1
Therefore j=1 aj(lnA − j lnR) ≤ 0, which is equivalent to AlnA ≤ BlnR, i.e., AA ≤ RB.
5. Considering the polynomials ±P(±x) we may assume w.l.o.g. that a,b ≥ 0. We have four cases:
(1) c ≥ 0,d ≥ 0. Then |a| +|b|+|c|+ |d| = a+b+c+d = P(1) ≤ 1.
(2) c ≥ 0,d < 0. Then |a|+|b|+|c|+|d| = a+b+c−d = P(1)−2P(0) ≤ 3. (3) c < 0,d ≥ 0. Then
|a|+|b| +|c|+|d| = a+b−c +d
= 3P(1) − 3P(−1)− 3P(1/2)+ 3P(−1/2) ≤ 7.
(4) c < 0,d < 0. Then
|a|+|b| +|c|+|d| = a+b−c −d
= 3P(1) −4P(1/2)+ 3P(−1/2) ≤ 7.
Remark. It can be shown that the maximum of 7 is attained only for P(x) = ±(4x3 −3x).
6. Let f(x),g(x) be polynomials with integer coefficients such that
f(x)(x +1)n + g(x)(xn + 1) = k0. (∗)
Write n = 2rm for m odd and note that xn + 1 = (x2r + 1)B(x), where B(x) = x2r(m−1) −x2r(m−2) +···−x2r +1. Moreover, B(−1) = 1; hence
B(x) −1 = (x+1)c(x) and thus
R(x)B(x) +1 = (B(x) −1)n = (x+1)nc(x)n (1)
for some polynomials c(x) and R(x).
The zeros of the polynomial x2r + 1 are ωj, with ω1 = cos π + isin π , and ωj = ω2j−1 for 1 ≤ j ≤ 2r. We have
604 4 Solutions
(ω1 +1)(ω2 +1)···(ω2r+1 +1) = 2. (2)
From (∗) we also get f(ωj)(ωj + 1)n = k0 for j = 1,2,...,2r. Since A = f(ω1)f(ω2)···f(ω2r) is a symmetric polynomial in ω1,...,ω2r with integer coefficients, A is an integer. Consequently, taking the product over j = 1,2,...,2r and using (2) we deduce that 2nA = k2r is divisible by 2n = 22rm. Hence 2m | k0.
Furthermore, since ωj + 1 = (ω1 + 1)pj(ω1) for some polynomial pj with integer coefficients, (2) gives (ω1 + 1)2 p(ω1) = 2, where p(x) = p2(x)···p2r(x) has integer coefficients. But then the polynomial (x + 1)2 p(x) − 2 has a zero x = ω1, so it is divisible by its minimal poly-nomial x2 +1. Therefore
(x+1)2r p(x) = 2+(x2r +1)q(x) (3)
for some polynomial q(x). Raising (3) to the mth power we get (x + 1)np(x)n = 2m + (x2r + 1)Q(x) for some polynomial Q(x) with integer
coefficients. Now using (1) we obtain
(x+1)nc(x)n(x2r + 1)Q(x) = (x2r +1)Q(x) + (x2r +1)Q(x)B(x)R(x) = (x +1)np(x)n −2m +(xn +1)Q(X)R(x).
Therefore(x+1)nf(x)+(xn+1)g(x) = 2m for some polynomialsf(x),g(x) with integer coefficients, and k0 = 2m.
7. We are given that f(x+a+b)−f(x+a) = f(x+b)−f(x), where a = 1/6 and b = 1/7. Summing up these equations for x,x+b,...,x+6b we obtain f(x+a+1)−f(x+a)= f(x+1)−f(x). Summing up the new equations for x,x+ a,...,x+5a we obtain that
f(x+2)−f(x+1) = f(x+1) −f(x).
It follows by induction that f(x + n) − f(x) = n[f(x + 1) − f(x)]. If f(x + 1) = f(x), then f(x + n) − f(x) will exceed in absolute value an arbitrarily large number for a sufficiently large n, contradicting the assumption that f is bounded. Hence f(x+ 1) = f(x) for all x.
8. Putting m = n = 0 we obtain f(0) = 0 and consequently f(f(n)) = f(n) for all n. Thus the given functional equation is equivalent to
f(m+ f(n)) = f(m) +f(n), f(0) = 0.
Clearly one solution is (∀x) f(x) = 0. Suppose f is not the zero function. We observe that f has nonzero fixed points (for example, any f(n) is a fixed point). Let a be the smallest nonzero fixed point of f. By induction, each ka (k ∈ N) is a fixed point too. We claim that all fixed points of f are of this form. Indeed, suppose that b = ka + i is a fixed point, where i < a. Then
4.37 Shortlisted Problems 1996 605
b = f(b) = f(ka+i) = f(i+f(ka)) = f(i)+f(ka) = f(i) +ka;
hence f(i) = i. Hence i = 0.
Since the set of values of f is a set of its fixed points, it follows that for i = 0,1,...,a−1, f(i) = ani for some integers ni ≥ 0 with n0 = 0.
Let n = ka+i be any positive integer, 0 ≤ i < a. As before, the functional equation gives us
f(n) = f(ka+i) = f(i)+ka = (ni +k)a.
Besides the zero function, this is the general solution of the given func-tional equation. To verify this, we plug in m = ka + i, n = la + j and obtain
f(m+f(n)) = f(ka+i+f(la+j)) = f((k +l +nj)a +i) = (k +l +nj +ni)a = f(m) +f(n).
9. From the definition of a(n) we obtain
. 1 if n ≡ 0 or n ≡ 3 (mod 4); −1 if n ≡ 1 or n ≡ 2 (mod 4).
Let n = bkbk−1 ...b1b0 be the binary representation of n, where we as-sume bk = 1. If we define p(n) and q(n) to be the number of indices i = 0,1,...,k − 1 with bi = bi+1 and the number of i = 0,1,...,k − 1 with bi = bi+1 respectively, we get
a(n) = p(n) −q(n). (1)
(a) The maximum value of a(n) for n ≤ 1996 is 9 when p(n) = 9 and q(n) = 0, i.e., in the case n = 11111111112 = 1023.
The minimum value is −10 and is attained when p(n) = 0 and q(n) = 10, i.e., only for n = 101010101012 = 1365.
(b) From (1) we have that a(n) = 0 is equivalent to p(n) = q(n) = k/2. Hence k must be even, and the k/2 indices i for which bi = bi+1 can be chosen in exactly k/2 ways. Thus the number of positive integers n < 211 = 2048 with a(n) = 0 is equal to
0 + 1 + 2 + 3 + 4 + 5 = 351.
But five of thesenumbers exceed1996:these are2002 = 111110100102, 2004 = 111110101002, 2006 = 111110101102, 2010 = 111110110102, 2026 = 111111010102. Therefore there are 346 numbers n ≤ 1996 for which a(n) = 0.
10. We first show that H is the common orthocenter of the triangles ABC and AQR.
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