Tuyển tập Olympic Toán sinh viên quốc tế 1994 - 2013

TOÁN SINH VIÊN QUỐC TẾInternational Mathematics Competition for University Students 1994-2013 Mục lục IMC 1994... International Competition in Mathematics forUniverstiy Students in Plovd

TOÁN SINH VIÊN QUỐC TẾ

International Mathematics Competition for

University Students

1994-2013 Mục lục

IMC 1994 1

IMC 1995 10

IMC 1996 21

IMC 1997 ngày 1 36

IMC 1997 ngày 2 44

IMC 1998 ngày 1 49

IMC 1998 ngày 2 52

IMC 1999 ngày 1 56

IMC 1999 ngày 2 59

IMC 2000 ngày 1 62

IMC 2000 ngày 2 66

IMC 2001 ngày 1 70

IMC 2001 ngày 2 74

IMC 2002 ngày 1 79

IMC 2002 ngày 2 84

IMC 2003 ngày 1 88

IMC 2003 ngày 2 92

IMC 2004 ngày 1 96

IMC 2004 ngày 2 100

IMC 2005 ngày 1 103

IMC 2005 ngày 2 107

IMC 2006 ngày 1 111

IMC 2006 ngày 2 114

IMC 2007 ngày 1 118

IMC 2007 ngày 2 121

IMC 2008 ngày 1 125

IMC 2008 ngày 2 127

IMC 2009 ngày 1 129

IMC 2009 ngày 2 133

IMC 2010 ngày 1 137

IMC 2010 ngày 2 141

IMC 2011 ngày 1 145

IMC 2011 ngày 2 147

IMC 2012 ngày 1 151

IMC 2012 ngày 2 155

IMC 2013 ngày 1 159

IMC 2013 ngày 2 162

International Competition in Mathematics for

Universtiy Students

in Plovdiv, Bulgaria

1994

First day — July 29, 1994

Problem 1 (13 points)

a) Let A be a n × n, n ≥ 2, symmetric, invertible matrix with realpositive elements Show that zn≤ n2− 2n, where zn is the number of zeroelements in A− 1

b) How many zero elements are there in the inverse of the n × n matrix

b1,1 = 2, bn,n = (−1)n, bi,i+1 = bi+1,i = (−1)i for i = 1, 2, , n − 1

1 + f2(x)+ 1 ≥ 0for x ∈ (a, b) Thus arctg f (x)+x is non-decreasing in the interval and usingthe limits we get π

2 + a ≤ −

π

2 + b Hence b − a ≥ π One has equality for

f (x) = cotg x, a = 0, b = π

Given a set S of 2n − 1, n ∈ N, different irrational numbers Provethat there are n different elements x1, x2, , xn ∈ S such that for all non-negative rational numbers a1, a2, , an with a1+ a2+ · · · + an> 0 we havethat a1x1+ a2x2+ · · · + anxn is an irrational number

Solution LetIbe the set of irrational numbers,Q– the set of rationalnumbers,Q+ =Q∩ [0, ∞) We work by induction For n = 1 the statement

is trivial Let it be true for n − 1 We start to prove it for n From theinduction argument there are n − 1 different elements x1, x2, , xn−1 ∈ Ssuch that

(1) a1x1+ a2x2+ · · · + an−1xn−1∈I

for all a1, a2, , an∈Q+ with a1+ a2+ · · · + an−1> 0

Denote the other elements of S by xn, xn+1, , x2n−1 Assume the ment is not true for n Then for k = 0, 1, , n − 1 there are rk ∈Q suchthat

i=1

bik+ ck> 0.Also

i=1

n−1 X

k=0

dkbik

!

xi∈Q,which contradicts (1) because of the conditions on b0s and d0s

Fk−i◦ (F ◦ G − G ◦ F ) ◦ Fi−1=

=

k X i=1

Z b 0

f (x)g(nx)dx = 1

b

Z b 0

f (x)dx ·

Z b 0g(x)dx

b) Find

limn→∞

In view of the uniform continuity of f we have ω(f, t) → 0 as t → 0 Usingthe periodicity of g we get

Z b

0 f (x)g(nx)dx =

n X

k=1

Z bk/n b(k−1)/nf (x)g(nx)dx

n X

k=1

Z bk/n b(k−1)/n{f (x) − f (bk/n)}g(nx)dx

This proves a) For b) we set b = π, f (x) = sin x, g(x) = (1 + 3cos2x)− 1.From a) and

0sin xdx = 2,

Let f ∈ C2[0, N ] and |f0(x)| < 1, f00(x) > 0 for every x ∈ [0, N ] Let

0 ≤ m0 < m1 < · · · < mk ≤ N be integers such that ni = f (mi) are alsointegers for i = 0, 1, , k Denote bi = ni− ni−1 and ai = mi− mi−1 for

ai < 1 From theconvexity of f we have that f0 is increasing and bi

ai = f

0(xi) < f0

(xi+1) =

bi+1

because of x < m < x

N ≥ mk− m0 =

k X

i=1

ai ≥ Xj∈S A

aj > A|SA|

and hence |SA| < N/A

c) All different fractions in (−1, 1) with denominators less or equal A are

no more 2A2 Using b) we get k < N/A + 2A2 Put A = N1/3 in the aboveestimate and get k < 3N2/3

Second day — July 30, 1994

Problem 1 (14 points)

Let f ∈ C1[a, b], f (a) = 0 and suppose that λ ∈R, λ > 0, is such that

|f0(x)| ≤ λ|f (x)|

for all x ∈ [a, b] Is it true that f (x) = 0 for all x ∈ [a, b]?

Solution Assume that there is y ∈ (a, b] such that f (y) 6= 0 Withoutloss of generality we have f (y) > 0 In view of the continuity of f there exists

c ∈ [a, y) such that f (c) = 0 and f (x) > 0 for x ∈ (c, y] For x ∈ (c, y] wehave |f0(x)| ≤ λf (x) This implies that the function g(x) = ln f (x) − λx isnot increasing in (c, y] because of g0(x) = f

b) Determine all points (x, y) such that ∂f

∂x(x, y) =

∂f

∂y(x, y) = 0 anddetermine for which of them f has global or local minimum or maximum.Solution We have f (1, 0) = e− 1, f (0, 1) = −e− 1 and te− t ≤ 2e− 2 for

maximum outside M Part a) follows from the compactness of M and thecontinuity of f Let (x, y) be a point from part b) From ∂f

∂x(x, y) =2x(1 − x2+ y2)e− x 2−y 2

at the points (1, 0) and (−1, 0) One has f (0, 1) = f (0, −1) = −e−1 and

f has global minimum at the points (0, 1) and (0, −1) The point (0, 0)

is not an extrema point because of f (x, 0) = x2e− x 2

> 0 if x 6= 0 and

f (y, 0) = −y2e−y2 < 0 if y 6= 0

Problem 3 (14 points)

Let f be a real-valued function with n + 1 derivatives at each point of

R Show that for each pair of real numbers a, b, a < b, such that

ln f (b) + f

0(b) + · · · + f(n)(b)

f (a) + f0(a) + · · · + f(n)(a)

!

= b − athere is a number c in the open interval (a, b) for which

f(n+1)(c) = f (c)

Note that ln denotes the natural logarithm

Solution Set g(x) = f (x) + f0(x) + · · · + f(n)(x)e− x From theassumption one get g(a) = g(b) Then there exists c ∈ (a, b) such that

g0

(c) = 0 Replacing in the last equality g0

(x) =f(n+1)(x) − f (x)e− x wefinish the proof

Problem 4 (18 points)

Let A be a n × n diagonal matrix with characteristic polynomial

(x − c1)d1(x − c2)d2 (x − ck)dk

,where c1, c2, , ckare distinct (which means that c1appears d1times on the

Let V be the space of all n × n matrices B such that AB = BA Prove thatthe dimension of V is

d21+ d22+ · · · + d2k.Solution Set A = (aij)ni,j=1, B = (bij)ni,j=1, AB = (xij)ni,j=1 and

Let x1, x2, , xkbe vectors of m-dimensional Euclidian space, such that

x1+x2+· · ·+xk= 0 Show that there exists a permutation π of the integers{1, 2, , k} such that

i=1

kxik2

! 1/2for each n = 1, 2, , k

Note that k · k denotes the Euclidian norm

Solution We define π inductively Set π(1) = 1 Assume π is definedfor i = 1, 2, , n and also

(1)

n X i=1

xπ(i)

2

≤

n X i=1

kxπ(i)k2.Note (1) is true for n = 1 We choose π(n + 1) in a way that (1) is fulfilledwith n + 1 instead of n Set y = Pn

i=1

xπ(i) and A = {1, 2, , k} \ {π(i) : i =

1, 2, , n} Assume that (y, xr) > 0 for all r ∈ A Then y, P

r∈A

xr

!

> 0and in view of y + P

≤

n X i=1

kxπ(i)k2+ kxrk2=

n+1 X i=1

kxπ(i)k2,which verifies (1) for n + 1 Thus we define π for every n = 1, 2, , k.Finally from (1) we get

n X i=1

xπ(i)

2

≤

n X i=1

kxπ(i)k2 ≤

k X i=1

N −2 X

N −2 X

AN = ln

2NN







M X

k=2+

N −M −1 X

k=M +1

+

N −2 X

ln2N



+ 1 to get(2) AN ≤

lim

N →∞

ln2NN

N −2

ln k · ln(N − k) = 1.

International Competition in Mathematics for

Universtiy Students

in Plovdiv, Bulgaria

1995

1PROBLEMS AND SOLUTIONS

First day

Problem 1 (10 points)

Let X be a nonsingular matrix with columns X1, X2, , Xn Let Y be amatrix with columns X2, X3, , Xn,0 Show that the matrices A = Y X−1and B = X−1Y have rank n − 1 and have only 0’s for eigenvalues

Solution Let J = (aij) be the n × n matrix where aij= 1 if i = j + 1and aij = 0 otherwise The rank of J is n − 1 and its only eigenvalues are

00s Moreover Y = XJ and A = Y X−1 = XJX−1, B = X−1Y = J Itfollows that both A and B have rank n − 1 with only 00s for eigenvalues

f2(t)dt ≥ 13.Solution From the inequality

0 ≤

Z 1

0 (f (x) − x)2dx=

Z 10

f2(x)dx − 2

Z 10

xf(x)dx +

Z 10

xf(x)dx −

Z 1 0

x2dx= 2

Z 1 0

xf(x)dx − 13

From the hypotheses we have

Z 10

Z 1x

f(t)dtdx ≥

Z 10

1 − x2

2 dxor

Z 10

tf(t)dt ≥1

3 This completes the proof.

Taking limits as x tends to 0+ we obtain

−x0 ≤ lim infx→0+ ff0(x)(x) ≤ lim sup

Solution From the definition we have

as x → ∞, it follows that the range of F is (F (1+), ∞) In order to determine

F(1+) we substitute t = ev in the definition of F and we get

Problem 5 (20 points)

Let A and B be real n × n matrices Assume that there exist n + 1different real numbers t1, t2, , tn+1 such that the matrices

Ci = A + tiB, i= 1, 2, , n + 1,are nilpotent (i.e Cin= 0)

Show that both A and B are nilpotent

Solution We have that

(A + tB)n= An+ tP1+ t2P2+ · · · + tn−1Pn−1+ tnBn

for some matrices P1, P2, , Pn−1 not depending on t

Assume that a, p1, p2, , pn−1, b are the (i, j)-th entries of the sponding matrices An, P1, P2, , Pn−1, Bn Then the polynomial

corre-btn+ pn−1tn−1+ · · · + p2t2+ p1t+ ahas at least n + 1 roots t1, t2, , tn+1 Hence all its coefficients vanish.Therefore An = 0, Bn= 0, Pi = 0; and A and B are nilpotent

Problem 6 (25 points)

Let p > 1 Show that there exists a constant Kp >0 such that for every

x, y∈R satisfying |x|p+ |y|p = 2, we have

2 

Since Dδ is compact it is enough to show that f is continuous on Dδ.For this we show that the denominator of f is different from zero Assumethe contrary Then |x + y| = 2, and

x+ y

2

p = 1 Since p > 1, the function

g(t) = |t|p is strictly convex, in other words

4−(x+y)2=4−(2−(p−1)t2+o(t2))2=4−4+4(p−1)t2+o(t2)=4(p−1)t2+o(t2)

So there exists δp>0 such that if |t| < δpwe have (x−y)2 <5t2, 4−(x+y)2>3(p − 1)t2 Then

if |x − 1| < δp From the symmetry we have that (∗) also holds when

|y − 1| < δp

To finish the proof it is enough to show that |x − y| ≥ 2δp whenever

|x − 1| ≥ δp, |y − 1| ≥ δp and xp+ yp = 2 Indeed, since xp+ yp= 2 we havethat max{x, y} ≥ 1 So let x − 1 ≥ δp Since

Second day

Problem 1 (10 points)

Let A be 3 × 3 real matrix such that the vectors Au and u are orthogonalfor each column vector u ∈R3 Prove that:

a) A>= −A, where A> denotes the transpose of the matrix A;

b) there exists a vector v ∈ R3 such that Au = v × u for every u ∈R3,where v × u denotes the vector product inR3

Solution a) Set A = (aij), u = (u1, u2, u3)> If we use the ity condition

with ui= δik we get akk= 0 If we use (1) with ui = δik+ δim we get

akk+ akm+ amk+ amm = 0and hence akm= −amk

b) Set v1 = −a23, v2 = a13, v3 = −a12 Then

bn2n

N X n=1[a2n2n− an2n+1+ 2n]

=

N X n=1[(an−1− 1)2n− (an− 1)2n+1]

= (a0− 1)21− (aN − 1)2N +1= 2 − 22

2 −

− 1

2−N Put x = 2−N Then x → 0 as N → ∞ and so



2 − 22

x− 1x

2zP0(z) − nP (z)lie on the same circle

Solution It is enough to consider only polynomials with leading ficient 1 Let P (z) = (z − α1)(z − α2) (z − αn) with |αj| = 1, where thecomplex numbers α1, α2, , αn may coincide

coef-We have

e

P(z) ≡ 2zP0(z) − nP (z) = (z + α1)(z − α2) (z − αn) +

+(z − α1)(z + α2) (z − αn) + · · · + (z − α1)(z − α2) (z + αn).Hence, Pe(z)

P(z) =

n X k=1

|z − α|2 for all complex z,

α, z 6= α, we deduce that in our case RePPe(z)(z) =

n X k=1

|z|2− 1

|z − αk|2 From |z| 6= 1

x−

n X k=1

λkx2k+1

... the last column we get that

Subtracting the n-th row of the above matrix from the (n+1)-st one,

(n−1)-st from n-th, , fir(n−1)-st from second we obtain that