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TUYỂN TẬP ĐỀ THI OLYMPIC TOÁN SINH VIÊN QUỐC TẾ International Mathematics Competition for University Students 1994-2013 Mục lục IMC 1994... International Competition in Mathematics forUn

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TUYỂN TẬP ĐỀ THI OLYMPIC 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

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International Competition in Mathematics for

Universtiy Students

in Plovdiv, Bulgaria

1994

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PROBLEMS AND SOLUTIONS

First day — July 29, 1994

Problem 1 (13 points)

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

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

A =





1 1 1 1 1

1 2 2 2 2

1 2 1 1 1

1 2 1 2 2

1 2 1 2





?

Solution Denote by aij and bij the elements of A and A−1, respectively Then for k 6= m we have Pn

i=0

akibim = 0 and from the positivity of aij we conclude that at least one of {bim : i = 1, 2, , n} is positive and at least one is negative Hence we have at least two non-zero elements in every column of A−1 This proves part a) For part b) all bij are zero except

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

Let f ∈ C1(a, b), lim

x→a+f (x) = +∞, lim

x→b−f (x) = −∞ and

f′(x) + f2(x) ≥ −1 for x ∈ (a, b) Prove that b − a ≥ π and give an example where b − a = π

Solution From the inequality we get

d

dx(arctg f (x) + x) =

f′(x)

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

2 + a ≤ −π2 + b Hence b − a ≥ π One has equality for

f (x) = cotg x, a = 0, b = π

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Given a set S of 2n − 1, n ∈ N, different irrational numbers Prove that 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 have that a1x1+ a2x2+ · · · + anxn is an irrational number

Solution LetIbe the set of irrational numbers,Q– the set of rational numbers,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 the induction argument there are n − 1 different elements x1, x2, , xn−1 ∈ S such 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 state-ment is not true for n Then for k = 0, 1, , n − 1 there are rk ∈Q such that

(2)

n−1 X

i=1

bikxi+ ckxn+k = rk for some bik, ck ∈Q+,

n−1 X

i=1

bik+ ck> 0 Also

(3)

n−1 X

k=0

dkxn+k= R for some dk∈Q+,

n−1 X

k=0

dk> 0, R ∈Q

If in (2) ck = 0 then (2) contradicts (1) Thus ck 6= 0 and without loss of generality one may take ck = 1 In (2) also n−1P

i=1

bik > 0 in view of xn+k ∈I Replacing (2) in (3) we get

n−1 X

k=0

dk −

n−1 X

i=1

bikxi+ rk

!

= R or

n−1 X

i=1

n−1 X

k=0

dkbik

!

xi∈Q, which contradicts (1) because of the conditions on b′s and d′s

Let α ∈R\ {0} and suppose that F and G are linear maps (operators) fromRn intoRn satisfying F ◦ G − G ◦ F = αF

a) Show that for all k ∈None has Fk◦ G − G ◦ Fk= αkFk

b) Show that there exists k ≥ 1 such that Fk= 0

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Solution For a) using the assumptions we have

Fk◦ G − G ◦ Fk =

k

X

i=1

Fk−i+1◦ G ◦ Fi−1− Fk−i◦ G ◦ Fi=

=

k

X

i=1

Fk−i◦ (F ◦ G − G ◦ F ) ◦ Fi−1=

=

k

X

i=1

Fk−i◦ αF ◦ Fi−1= αkFk

b) Consider the linear operator L(F ) = F ◦G−G◦F acting over all n×n matrices F It may have at most n2 different eigenvalues Assuming that

Fk 6= 0 for every k we get that L has infinitely many different eigenvalues

αk in view of a) – a contradiction

a) Let f ∈ C[0, b], g ∈ C(R) and let g be periodic with period b Prove that

Z b

0 f (x)g(nx)dx has a limit as n → ∞ and

lim

n→∞

Z b

0

f (x)g(nx)dx = 1

b

Z b

0 f (x)dx ·

Z b

0

g(x)dx

b) Find

lim

n→∞

Z π 0

sin x

1 + 3cos2nxdx.

Solution Set kgk1 =

Z b

0 |g(x)|dx and ω(f, t) = sup {|f(x) − f(y)| : x, y ∈ [0, b], |x − y| ≤ t}

In view of the uniform continuity of f we have ω(f, t) → 0 as t → 0 Using the 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

f (bk/n)

Z bk/n b(k−1)/ng(nx)dx +

n

X

k=1

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

= 1

n

X

k=1

f (bk/n)

Z b

0 g(x)dx + O(ω(f, b/n)kgk1)

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= 1

b

n

X

k=1

Z bk/n

b(k−1)/n

f (x)dx

Z b 0

g(x)dx

+1

b

n

X

k=1

b

nf (bk/n) −

Z bk/n

b(k−1)/nf (x)dx

! Z b

= 1

b

Z b

0

f (x)dx

Z b

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

Z π 0

sin xdx = 2,

Z π 0

(1 + 3cos2x)−1dx = π

2

we get

lim

n→∞

Z π

0

sin x

1 + 3cos2nxdx = 1.

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

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

i = 1, 2, , k

a) Prove that

−1 < ab1

1

< b2

a2 < · · · < abk

k

< 1

b) Prove that for every choice of A > 1 there are no more than N/A indices j such that aj > A

c) Prove that k ≤ 3N2/3 (i.e there are no more than 3N2/3 integer points on the curve y = f (x), x ∈ [0, N])

Solution a) For i = 1, 2, , k we have

bi = f (mi) − f(mi−1) = (mi− mi−1)f′(xi) for some xi ∈ (mi−1, mi) Hence bi

ai = f

′(xi) and so −1 < abi

i

< 1 From the convexity of f we have that f′ is increasing and bi

ai = f

′(xi) < f′(xi+1) =

bi+1

ai+1 because of xi < mi < xi+1.

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b) Set SA= {j ∈ {0, 1, , k} : aj > A} Then

N ≥ mk− m0 =

k

X

i=1

ai ≥ X

j∈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 above estimate and get k < 3N2/3

Second day — July 30, 1994

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

|f′(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 Without loss 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] we have |f′(x)| ≤ λf(x) This implies that the function g(x) = ln f(x) − λx is not increasing in (c, y] because of g′(x) = f

′(x)

f (x)−λ ≤ 0 Thus ln f(x)−λx ≥

ln f (y) − λy and f(x) ≥ eλx−λyf (y) for x ∈ (c, y] Thus

0 = f (c) = f (c + 0) ≥ eλc−λyf (y) > 0

— a contradiction Hence one has f (x) = 0 for all x ∈ [a, b]

Let f :R2 →Rbe given by f (x, y) = (x2− y2)e−x 2

−y 2

a) Prove that f attains its minimum and its maximum

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

∂x(x, y) =

∂f

∂y(x, y) = 0 and determine 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

t ≥ 2 Therefore |f(x, y)| ≤ (x2 + y2)e−x 2

−y 2

≤ 2e−2 < e−1 for (x, y) /∈

M = {(u, v) : u2 + v2 ≤ 2} and f cannot attain its minimum and its

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

∂x(x, y) = 2x(1 − x2+ y2)e−x2−y2 we get

(1) x(1 − x2+ y2) = 0

Similarly

(2) y(1 + x2− y2) = 0

All solutions (x, y) of the system (1), (2) are (0, 0), (0, 1), (0, −1), (1, 0) and (−1, 0) One has f(1, 0) = f(−1, 0) = e−1 and f has global maximum

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−x2 > 0 if x 6= 0 and

f (y, 0) = −y2e−y2 < 0 if y 6= 0

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

′(b) + · · · + f(n)(b)

f (a) + f′(a) + · · · + f(n)(a)

!

= b − a

there 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) + f′(x) + · · · + f(n)(x)

e−x From the assumption one get g(a) = g(b) Then there exists c ∈ (a, b) such that

g′(c) = 0 Replacing in the last equality g′(x) =

f(n+1)(x) − f(x)e−x we finish the proof

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 diagonal, c2appears d2times on the diagonal, etc and d1+d2+· · ·+dk = n)

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

BA = (yij)n

i,j=1 Then xij = aiibij and yij = ajjbij Thus AB = BA is equivalent to (aii− ajj)bij = 0 for i, j = 1, 2, , n Therefore bij = 0 if

aii6= ajj and bij may be arbitrary if aii= ajj The number of indices (i, j) for which aii = ajj = cm for some m = 1, 2, , k is d2

m This gives the desired result

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

n

X

i=1

xπ(i) ≤

k

X

i=1

kxik2

! 1/2

for 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 defined for 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 fulfilled with 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

!

> 0 and in view of y + P

r∈A

xr = 0 one gets −(y, y) > 0, which is impossible Therefore there is r ∈ A such that

Put π(n + 1) = r Then using (2) and (1) we have

n+1 X

i=1

xπ(i)

2

= ky + xrk2= kyk2+ 2(y, xr) + kxrk2 ≤ kyk2+ kxrk2 ≤

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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

kxik2

Find lim

N →∞

ln2N N

N −2 X

k=2

1

ln k · ln(N − k) Note that ln denotes the natural logarithm

Solution Obviously

(1) AN = ln

2N N

N −2 X

k=2

1

ln k · ln(N − k) ≥

ln2N

N ·N − 3

ln2N = 1 − 3

N.

Take M , 2 ≤ M < N/2 Then using that 1

ln k · ln(N − k) is decreasing in [2, N/2] and the symmetry with respect to N/2 one get

AN = ln

2N N







M

X

k=2

+

N −M −1 X

k=M +1

+

N −2 X

k=N −M







1

ln k · ln(N − k) ≤

≤ ln

2N N

2 M − 1

ln 2 · ln(N − 2)+

N − 2M − 1

ln M · ln(N − M)

≤

≤ ln 22 ·M ln NN +

1 −2MN

ln N

ln M + O

ln N

Choose M =

ln2N

+ 1 to get

(2) AN ≤

1 − 2

N ln2N

ln N − 2 ln ln N+O

ln N

≤ 1+O

ln ln N

ln N

Estimates (1) and (2) give

lim

N →∞

ln2N N

N −2 X

k=2

1

ln k · ln(N − k) = 1.

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International Competition in Mathematics for

Universtiy Students

in Plovdiv, Bulgaria

1995

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PROBLEMS AND SOLUTIONS

First day

Let X be a nonsingular matrix with columns X1, X2, , Xn Let Y be a matrix with columns X2, X3, , Xn, 0 Show that the matrices A = Y X−1

and 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 + 1 and aij = 0 otherwise The rank of J is n − 1 and its only eigenvalues are

0′s Moreover Y = XJ and A = Y X−1 = XJX−1, B = X−1Y = J It follows that both A and B have rank n − 1 with only 0′s for eigenvalues

Let f be a continuous function on [0, 1] such that for every x ∈ [0, 1] we have

Z 1

x f (t)dt ≥ 1 − x

2

2 Show that

Z 1

0

f2(t)dt ≥ 13 Solution From the inequality

0 ≤

Z 1

0 (f (x) − x)2dx =

Z 1

0 f2(x)dx − 2

Z 1

0 xf (x)dx +

Z 1

0 x2dx

we get

Z 1

0 f2(x)dx ≥ 2

Z 1

0 xf (x)dx −

Z 1

0 x2dx = 2

Z 1

0 xf (x)dx − 13 From the hypotheses we have

Z 1

0

Z 1

x f (t)dtdx ≥

Z 1

0

1 − x2

2 dx or

Z 1

0 tf (t)dt ≥ 1

3 This completes the proof.

Let f be twice continuously differentiable on (0, +∞) such that

lim

x→0+f′(x) = −∞ and limx→0+f′′(x) = +∞ Show that

lim

x→0+

f (x)

f′(x) = 0.

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Solution Since f′ tends to −∞ and f′′ tends to +∞ as x tends to 0+, there exists an interval (0, r) such that f′(x) < 0 and f′′(x) > 0 for all

x ∈ (0, r) Hence f is decreasing and f′ is increasing on (0, r) By the mean value theorem for every 0 < x < x0< r we obtain

f (x) − f(x0) = f′(ξ)(x − x0) > 0, for some ξ ∈ (x, x0) Taking into account that f′ is increasing, f′(x) <

f′(ξ) < 0, we get

x − x0 < f

′(ξ)

f′(x)(x − x0) = f (x) − f(x0)

f′(x) < 0.

Taking limits as x tends to 0+ we obtain

−x0 ≤ lim infx→0+ ff (x)′(x) ≤ lim sup

x→0+

f (x)

f′(x) ≤ 0

Since this happens for all x0 ∈ (0, r) we deduce that lim

x→0+

f (x)

f′(x) exists and lim

x→0+

f (x)

f′(x) = 0.

Let F : (1, ∞) →Rbe the function defined by

F (x) :=

Z x2

x

dt

ln t. Show that F is one-to-one (i.e injective) and find the range (i.e set of values) of F

Solution From the definition we have

F′(x) = x − 1

ln x , x > 1.

Therefore F′(x) > 0 for x ∈ (1, ∞) Thus F is strictly increasing and hence one-to-one Since

F (x) ≥ (x2− x) min

ln t : x ≤ t ≤ x2

= x

2− x

ln x2 → ∞

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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

F (x) =

Z 2 ln x

ln x

ev

vdv.

Hence

F (x) < e2 ln x

Z 2 ln x

ln x

1

vdv = x

2ln 2 and similarly F (x) > x ln 2 Thus F (1+) = ln 2

Let A and B be real n × n matrices Assume that there exist n + 1 different 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 corre-sponding matrices An, P1, P2, , Pn−1, Bn Then the polynomial

btn+ pn−1tn−1+ · · · + p2t2+ p1t + a has 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

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

(x − y)2 ≤ Kp

4 − (x + y)2

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