12th International Mathematics Competition for University StudentsBlagoevgrad, July 22 - July 28, 2005 First Day Problem 1.. |A| denotes the number of elements of the set A... 12th Inter
Trang 112th International Mathematics Competition for University Students
Blagoevgrad, July 22 - July 28, 2005
First Day
Problem 1 Let A be the n × n matrix, whose (i, j)th entry is i + j for all i, j = 1, 2, , n What is the rank of A?
Solution 1 For n = 1 the rank is 1 Now assume n ≥ 2 Since A = (i)n
i,j=1+ (j)n
i,j=1, matrix A is the sum
of two matrixes of rank 1 Therefore, the rank of A is at most 2 The determinant of the top-left 2 × 2 minor is −1, so the rank is exactly 2
Therefore, the rank of A is 1 for n = 1 and 2 for n ≥ 2
Solution 2 Consider the case n ≥ 2 For i = n, n − 1, , 2, subtract the (i − 1)th row from the nth row Then subtract the second row from all lower rows
rank
n + 1 n + 2 2n
= rank
2 3 n + 1
1 1 1
.
1 1 1
= rank
1 2 n
1 1 1
0 0 0
0 0 0
= 2
Problem 2 For an integer n ≥ 3 consider the sets
Sn = {(x1, x2, , xn) : ∀i xi ∈ {0, 1, 2}}
An = {(x1, x2, , xn) ∈ Sn: ∀i ≤ n − 2 |{xi, xi+1, xi+2}| 6= 1}
and
Bn = {(x1, x2, , xn) ∈ Sn: ∀i ≤ n − 1 (xi = xi+1⇒ xi 6= 0)} Prove that |An+1| = 3 · |Bn|
(|A| denotes the number of elements of the set A.)
Solution 1 Extend the definitions also for n = 1, 2 Consider the following sets
A0n = {(x1, x2, , xn) ∈ An: xn−1 = xn} , A00n = An\ A0
n,
Bn0 = {(x1, x2, , xn) ∈ Bn: xn = 0} , Bn00= Bn\ Bn0 and denote an = |An|, a0
n= |A0n|, a00
n = |A00n|, bn = |Bn|, b0
n = |Bn0|, b00
n = |Bn00|
It is easy to observe the following relations between the a–sequences
an = a0n+ a00n
a0n+1 = a00n
a00n+1 = 2a0n+ 2a00n
,
which lead to an+1 = 2an+ 2an−1
For the b–sequences we have the same relations
bn = b0n+ b00n
b0n+1 = b00n
b00n+1 = 2b0n+ 2b00n
,
therefore bn+1 = 2bn+ 2bn−1
By computing the first values of (an) and (bn) we obtain
a1 = 3, a2 = 9, a3 = 24
b1 = 3, b2 = 8
Trang 212th International Mathematics Competition for University
Students
Blagoevgrad, July 22 - July 28, 2005
Second Day
Problem 1 Let f (x) = x2 + bx + c, where b and c are real numbers, and let
M = {x ∈ R : |f (x)| < 1}
Clearly the set M is either empty or consists of disjoint open intervals Denote the sum of their lengths by |M | Prove that
|M | ≤ 2√2
Solution Write f (x) = x + 2b2+ d where d = c −b42 The absolute minimum of f is d
If d ≥ 1 then f (x) ≥ 1 for all x, M = ∅ and |M | = 0
If −1 < d < 1 then f (x) > −1 for all x,
−1 <
x + b 2
2
+ d < 1 ⇐⇒
x + b 2
<√
1 − d so
M =
−b
2 −√1 − d, −b
2 +
√
1 − d
and
|M | = 2√1 − d < 2√
2
If d ≤ −1 then
−1 <
x + b 2
2
+ d < 1 ⇐⇒ p|d| − 1 <
x + b 2
<p|d| + 1 so
M = −p|d| + 1, −p|d| − 1 ∪ p|d| − 1, p|d| + 1
and
|M | = 2p|d| + 1 −p|d| − 1= 2 (|d| + 1) − (|d| − 1)
p|d| + 1 + p|d| − 1 ≤ 2
2
√
1 + 1 +√
1 − 0 = 2
√ 2
Problem 2 Let f : R → R be a function such that (f (x))n is a polynomial for every
n = 2, 3, Does it follow that f is a polynomial?
Solution 1 Yes, it is even enough to assume that f2 and f3 are polynomials
Let p = f2 and q = f3 Write these polynomials in the form of
p = a · pa1
1 · · pak
k , q = b · qb1
1 · · qbl
l ,
Trang 3where a, b ∈ R, a1, , ak, b1, bl are positive integers and p1, , pk, q1, , ql are irre-ducible polynomials with leading coefficients 1 For p3 = q2 and the factorisation of p3 = q2
is unique we get that a3 = b2, k = l and for some (i1, , ik) permutation of (1, , k) we have p1 = qi1, , pk = qik and 3a1 = 2bi1, , 3ak = 2bik Hence b1, , bl are divisible by
3 let r = b1/3· qb1 /3
1 · · qbl /3
l be a polynomial Since r3 = q = f3 we have f = r
Solution 2 Let pq be the simplest form of the rational function ff32 Then the simplest form
of its square is pq22 On the other hand pq22 =ff32
2
= f2 is a polynomial therefore q must
be a constant and so f = ff32 = pq is a polynomial
Problem 3 In the linear space of all real n × n matrices, find the maximum possible dimension of a linear subspace V such that
∀X, Y ∈ V trace(XY ) = 0
(The trace of a matrix is the sum of the diagonal entries.)
Solution If A is a nonzero symmetric matrix, then trace(A2) = trace(AtA) is the sum of the squared entries of A which is positive So V cannot contain any symmetric matrix but 0
Denote by S the linear space of all real n × n symmetric matrices; dim V = n(n+1)2 Since V ∩ S = {0}, we have dim V + dim S ≤ n2 and thus dim V ≤ n2− n(n+1)2 = n(n−1)2 The space of strictly upper triangular matrices has dimension n(n−1)2 and satisfies the condition of the problem
Therefore the maximum dimension of V is n(n−1)2
Problem 4 Prove that if f : R → R is three times differentiable, then there exists a real number ξ ∈ (−1, 1) such that
f000(ξ)
f (1) − f (−1)
2 − f0(0)
Solution 1 Let
g(x) = −f (−1)
2(x − 1) − f (0)(x2− 1) + f (1)
2(x + 1) − f0(0)x(x − 1)(x + 1)
It is easy to check that g(±1) = f (±1), g(0) = f (0) and g0(0) = f0(0)
Apply Rolle’s theorem for the function h(x) = f (x) − g(x) and its derivatives Since h(−1) = h(0) = h(1) = 0, there exist η ∈ (−1, 0) and ϑ ∈ (0, 1) such that h0(η) =
h0(ϑ) = 0 We also have h0(0) = 0, so there exist % ∈ (η, 0) and σ ∈ (0, ϑ) such that
h00(%) = h00(σ) = 0 Finally, there exists a ξ ∈ (%, σ) ⊂ (−1, 1) where h000(ξ) = 0 Then
f000(ξ) = g000(ξ) = −f (−1)
2 · 6 − f (0) · 0 + f (1)
2 · 6 − f0(0) · 6 = f (1) − f (−1)
2 − f0(0)
Trang 4Solution 2 The expression f (1) − f (−1)
2 − f0(0) is the divided difference f [−1, 0, 0, 1] and there exists a number ξ ∈ (−1, 1) such that f [−1, 0, 0, 1] = f
000(ξ) 3! . Problem 5 Find all r > 0 such that whenever f : R2 → R is a differentiable function such that |grad f (0, 0)| = 1 and |grad f (u) − grad f (v)| ≤ |u − v| for all u, v ∈ R2, then the maximum of f on the disk {u ∈ R2 : |u| ≤ r} is attained at exactly one point (grad f (u) = (∂1f (u), ∂2f (u)) is the gradient vector of f at the point u For a vector
u = (a, b), |u| =√
a2+ b2.) Solution To get an upper bound for r, set f (x, y) = x − x
2
y2
2 This function satisfies the conditions, since grad f (x, y) = (1 − x, y), grad f (0, 0) = (1, 0) and |grad f (x1, y1) − grad f (x2, y2)| = |(x2− x1, y1− y2)| = |(x1, y1) − (x2, y2)|
In the disk Dr = {(x, y) : x2+ y2 ≤ r2}
f (x, y) = x
2+ y2
x − 1 2
2
+1
4 ≤ r
2
2 +
1
4.
If r > 12 then the absolute maximum is r22 + 14, attained at the points 12, ±qr2− 1
4
Therefore, it is necessary that r ≤ 12 because if r > 12 then the maximum is attained twice Suppose now that r ≤ 1/2 and that f attains its maximum on Dr at u, v, u 6= v Since
|grad f (z) − grad f (0)| ≤ r, |grad f (z)| ≥ 1 − r > 0 for all z ∈ Dr Hence f may attain its maximum only at the boundary of Dr, so we must have |u| = |v| = r and grad f (u) = au and grad f (v) = bv, where a, b ≥ 0 Since au = grad f (u) and bv = grad f (v) belong
to the disk D with centre grad f (0) and radius r, they do not belong to the interior of
Dr Hence |grad f (u) − grad f (v)| = |au − bv| ≥ |u − v| and this inequality is strict since D ∩ Dr contains no more than one point But this contradicts the assumption that
|grad f (u) − grad f (v)| ≤ |u − v| So all r ≤ 1
2 satisfies the condition
Problem 6 Prove that if p and q are rational numbers and r = p + q√
7, then there exists
a matrix a b
c d
6= ±1 0
0 1
with integer entries and with ad − bc = 1 such that
ar + b
cr + d = r.
Solution First consider the case when q = 0 and r is rational Choose a positive integer t such that r2t is an integer and set
a b
c d
=1 + rt −r2t
t 1 − rt
Then
deta b
c d
= 1 and ar + b
cr + d =
(1 + rt)r − r2t
tr + (1 − rt) = r.
Trang 5Now assume q 6= 0 Let the minimal polynomial of r in Z[x] be ux2+ vx + w The other root of this polynomial is r = p−q√
7, so v = −u(r+r) = −2up and w = urr = u(p2−7q2) The discriminant is v2− 4uw = 7 · (2uq)2 The left-hand side is an integer, implying that also ∆ = 2uq is an integer
The equation ar+b
cr+d = r is equivalent to cr2+ (d − a)r − b = 0 This must be a multiple
of the minimal polynomial, so we need
c = ut, d − a = vt, −b = wt for some integer t 6= 0 Putting together these equalities with ad − bc = 1 we obtain that
(a + d)2 = (a − d)2 + 4ad = 4 + (v2− 4uw)t2 = 4 + 7∆2t2 Therefore 4 + 7∆2t2 must be a perfect square Introducing s = a + d, we need an integer solution (s, t) for the Diophantine equation
such that t 6= 0
The numbers s and t will be even Then a + d = s and d − a = vt will be even as well and a and d will be really integers
Let (8±3√
7)n = kn±ln√7 for each integer n Then k2
n−7l2
n= (kn+ln√
7)(kn−ln√7) = ((8 + 3√
7)n(8 − 3√
7))n = 1 and the sequence (ln) also satisfies the linear recurrence
ln+1 = 16ln− ln−1 Consider the residue of ln modulo ∆ There are ∆2 possible residue pairs for (ln, ln+1) so some are the same Starting from such two positions, the recurrence shows that the sequence of residues is periodic in both directions Then there are infinitely many indices such that ln ≡ l0 = 0 (mod ∆)
Taking such an index n, we can set s = 2kn and t = 2ln/∆
Remarks 1 It is well-known that if D > 0 is not a perfect square then the Pell-like Diophantine equation
x2− Dy2 = 1 has infinitely many solutions Using this fact the solution can be generalized to all quadratic algebraic numbers
2 It is also known that the continued fraction of a real number r is periodic from a certain point if and only if r is a root of a quadratic equation This fact can lead to another solution
Trang 6which leads to
a2 = 3b1
a3 = 3b2 Now, reasoning by induction, it is easy to prove that an+1 = 3bn for every n ≥ 1
Solution 2 Regarding xi to be elements of Z3 and working “modulo 3”, we have that
(x1, x2, , xn) ∈ An ⇒ (x1+ 1, x2+ 1, , xn+ 1) ∈ An, (x1+ 2, x2+ 2, , xn+ 2) ∈ An
which means that 1/3 of the elements of An start with 0 We establish a bijection between the subset of all the vectors in An+1 which start with 0 and the set Bn by
(0, x1, x2, , xn) ∈ An+17−→ (y1, y2, , yn) ∈ Bn
y1 = x1, y2 = x2− x1, y3 = x3− x2, , yn= xn− xn−1 (if yk = yk+1 = 0 then xk− xk−1 = xk+1− xk = 0 (where x0 = 0), which gives xk−1 = xk = xk+1, which
is not possible because of the definition of the sets Ap; therefore, the definition of the above function is correct)
The inverse is defined by
(y1, y2, , yn) ∈ Bn7−→ (0, x1, x2, , xn) ∈ An+1
x1 = y1, x2 = y1+ y2, , xn= y1+ y2+ · · · + yn
Problem 3 Let f : R → [0, ∞) be a continuously differentiable function Prove that
Z 1
0
f3(x) dx − f2(0)
Z 1
0
f (x) dx
≤ max
0≤x≤1|f0(x)|
Z 1
0
f (x) dx
2
Solution 1 Let M = max
0≤x≤1|f0(x)| By the inequality −M ≤ f0(x) ≤ M, x ∈ [0, 1] it follows:
−M f (x) ≤ f (x) f0(x) ≤ M f (x) , x ∈ [0, 1]
By integration
−M
Z x
0
f (t) dt ≤ 1
2f
2(x) − 1
2f
2(0) ≤ M
Z x
0
f (t) dt, x ∈ [0, 1]
−M f (x)
Z x
0
f (t) dt ≤ 1
2f
3(x) − 1
2f
2(0) f (x) ≤ M f (x)
Z x
0
f (t) dt, x ∈ [0, 1] Integrating the last inequality on [0, 1] it follows that
−MR1
0 f (x)dx
2
≤R1
0 f3(x) dx − f2(0)R01f (x) dx ≤ MR01f (x)dx
2
⇔
R1
0 f3(x) dx − f2(0)R01f (x) dx
≤ MR1
0 f (x) dx
2
Solution 2 Let M = max
0≤x≤1|f0(x)| and F (x) = −Rx1f ; then F0 = f , F (0) = −R01f and F (1) = 0 Integrating by parts,
Z 1
0
f3 =
Z 1
0
f2· F0 = [f2F ]10−
Z 1
0
(f2)0F =
= f2(1)F (1) − f2(0)F (0) −
Z 1
0
2F f f0 = f2(0)
Z 1
0
f −
Z 1
0
2F f f0 Then
Z 1
0
f3(x) dx − f2(0)
Z 1
0
f (x) dx
=
Z 1
0
2F f f0
≤
Z 1
0
2F f |f0| ≤ M
Z 1
0
2F f = M · [F2]10 = M
Z 1
0
f
2
Trang 7
Problem 4 Find all polynomials P (x) = anxn+ an−1xn−1+ + a1x + a0 (an 6= 0) satisfying the following two conditions:
(i) (a0, a1, , an) is a permutation of the numbers (0, 1, , n)
and
(ii) all roots of P (x) are rational numbers
Solution 1 Note that P (x) does not have any positive root because P (x) > 0 for every x > 0 Thus, we can represent them in the form −αi, i = 1, 2, , n, where αi ≥ 0 If a0 6= 0 then there is a k ∈ N, 1 ≤ k ≤ n−1, with ak = 0, so using Viete’s formulae we get
α1α2 αn−k−1αn−k+ α1α2 αn−k−1αn−k+1+ + αk+1αk+2 αn−1αn= ak
an
= 0,
which is impossible because the left side of the equality is positive Therefore a0 = 0 and one of the roots of the polynomial, say αn, must be equal to zero Consider the polynomial Q(x) = anxn−1+an−1xn−2+ +a1
It has zeros −αi, i = 1, 2, , n − 1 Again, Viete’s formulae, for n ≥ 3, yield:
α1α2 αn−1= a1
α1α2 αn−2+ α1α2 αn−3αn−1+ + α2α3 αn−1 = a2
α1 + α2+ + αn−1 = an−1
Dividing (2) by (1) we get
1
α1 +
1
α2 + +
1
αn−1 =
a2
From (3) and (4), applying the AM-HM inequality we obtain
an−1 (n − 1)an =
α1 + α2+ + αn−1
α 1 +α1
2 + + α1
n−1
= (n − 1)a1
a2 , therefore a2 a n−1
a 1 a n ≥ (n − 1)2 Hence n2
2 ≥ a2 a n−1
a 1 a n ≥ (n − 1)2, implying n ≤ 3 So, the only polynomials possibly satisfying (i) and (ii) are those of degree at most three These polynomials can easily be found and they are P (x) = x, P (x) = x2+ 2x, P (x) = 2x2+ x, P (x) = x3+ 3x2+ 2x and P (x) = 2x3+ 3x2+ x 2
Solution 2 Consider the prime factorization of P in the ring Z[x] Since all roots of P are rational, P can
be written as a product of n linear polynomials with rational coefficients Therefore, all prime factor of P are linear and P can be written as
P (x) =
n
Y
k=1
(bkx + ck)
where the coefficients bk, ck are integers Since the leading coefficient of P is positive, we can assume bk > 0 for all k The coefficients of P are nonnegative, so P cannot have a positive root This implies ck ≥ 0 It
is not possible that ck = 0 for two different values of k, because it would imply a0 = a1 = 0 So ck > 0 in
at least n − 1 cases
Now substitute x = 1
P (1) = an+ · · · + a0 = 0 + 1 + · · · + n = n(n + 1)
n
Y
k=1
(bk+ ck) ≥ 2n−1;
therefore it is necessary that 2n−1 ≤ n(n+1)2 , therefore n ≤ 4 Moreover, the number n(n+1)2 can be written
as a product of n − 1 integers greater than 1
If n = 1, the only solution is P (x) = 1x + 0
If n = 2, we have P (1) = 3 = 1 · 3, so one factor must be x, the other one is x + 2 or 2x + 1 Both x(x + 2) = 1x2+ 2x + 0 and x(2x + 1) = 2x2+ 1x + 0 are solutions
Trang 8If n = 3, then P (1) = 6 = 1·2·3, so one factor must be x, another one is x+1, the third one is again x+2
or 2x+1 The two polynomials are x(x+1)(x+2) = 1x3+3x2+2x+0 and x(x+1)(2x+1) = 2x3+3x2+1x+0, both have the proper set of coefficients
In the case n = 4, there is no solution because n(n+1)2 = 10 cannot be written as a product of 3 integers greater than 1
Altogether we found 5 solutions: 1x+0, 1x2+2x+0, 2x2+1x+0, 1x3+3x2+2x+0 and 2x3+3x2+1x+0 Problem 5 Let f : (0, ∞) → R be a twice continuously differentiable function such that
|f00(x) + 2xf0(x) + (x2+ 1)f (x)| ≤ 1
for all x Prove that lim
x→∞f (x) = 0
Solution 1 Let g(x) = f0(x) + xf (x); then f00(x) + 2xf0(x) + (x2+ 1)f (x) = g0(x) + xg(x)
We prove that if h is a continuously differentiable function such that h0(x) + xh(x) is bounded then lim
∞ h = 0 Applying this lemma for h = g then for h = f , the statement follows
Let M be an upper bound for |h0(x)+xh(x)| and let p(x) = h(x)ex 2 /2 (The function e−x2/2is a solution
of the differential equation u0(x) + xu(x) = 0.) Then
|p0(x)| = |h0(x) + xh(x)|ex2/2 ≤ M ex 2 /2
and
|h(x)| =
p(x)
ex 2 /2
=
p(0) +R0xp0
ex 2 /2
≤ |p(0)| + M
Rx
0 ex2/2dx
ex 2 /2
Since lim
x→∞ex2/2 = ∞ and lim
Rx
0 ex2/2dx
ex 2 /2 = 0 (by L’Hospital’s rule), this implies limx→∞h(x) = 0
Solution 2 Apply L’Hospital rule twice on the fraction f (x)e
x 2 /2
ex 2 /2 (Note that L’Hospital rule is valid if the denominator converges to infinity, without any assumption on the numerator.)
lim
x→∞f (x) = lim
x→∞
f (x)ex 2 /2
ex 2 /2 = lim
x→∞
(f0(x) + xf (x))ex 2 /2
xex 2 /2 = lim
x→∞
(f00(x) + 2xf0(x) + (x2+ 1)f (x))ex 2 /2
= lim
x→∞
f00(x) + 2xf0(x) + (x2+ 1)f (x)
Problem 6 Given a group G, denote by G(m) the subgroup generated by the mth powers of elements of
G If G(m) and G(n) are commutative, prove that G(gcd(m, n)) is also commutative (gcd(m, n) denotes the greatest common divisor of m and n.)
Solution Write d = gcd(m, n) It is easy to see that hG(m), G(n)i = G(d); hence, it will suffice to check commutativity for any two elements in G(m) ∪ G(n), and so for any two generators am and bn Consider their commutator z = a−mb−nambn; then the relations
z = (a−mbam)−nbn = a−m(b−nabn)m show that z ∈ G(m) ∩ G(n) But then z is in the center of G(d) Now, from the relation ambn= bnamz, it easily follows by induction that
amlbnl= bnlamlzl2 Setting l = m/d and l = n/d we obtain z(m/d)2 = z(n/d)2 = e, but this implies that z = e as well
... |grad f (u) − grad f (v)| = |au − bv| ≥ |u − v| and this inequality is strict since D ∩ Dr contains no more than one point But this contradicts the assumption that|grad f (u)... set f (x, y) = x − x
2
y2
2 This function satisfies the conditions, since grad f (x, y) = (1 − x, y), grad f (0, 0) =... assume q 6= Let the minimal polynomial of r in Z[x] be ux2+ vx + w The other root of this polynomial is r = p−q√
7, so v = −u(r+r) = −2up and w = urr = u(p2−7q2)