12th IMC

IMC 2005

Blagoevgrad, Bulgaria · 22 – 28 July 2005 · 12 problems across 2 papers

Official result files →

Day 1

6 problems
  1. Problem 1

    Let AA be the n×nn \times n matrix, whose (i,j)(i,j)th entry is i+ji + j for all i,j=1,2,…,ni, j = 1, 2, \dots, n. What is the rank of AA?

  2. Problem 2

    For an integer n≥3n \ge 3 consider the sets

    Sn={(x1,x2,…,xn):∀i  xi∈{0,1,2}}S_n = \{(x_1, x_2, \dots, x_n) : \forall i \; x_i \in \{0,1,2\}\} An={(x1,x2,…,xn)∈Sn:∀i≤n−2  ∣{xi,xi+1,xi+2}∣≠1}A_n = \{(x_1, x_2, \dots, x_n) \in S_n : \forall i \le n-2 \; |\{x_i, x_{i+1}, x_{i+2}\}| \ne 1\}

    and

    Bn={(x1,x2,…,xn)∈Sn:∀i≤n−1  (xi=xi+1⇒xi≠0)}.B_n = \{(x_1, x_2, \dots, x_n) \in S_n : \forall i \le n-1 \; (x_i = x_{i+1} \Rightarrow x_i \ne 0)\}.

    Prove that ∣An+1∣=3⋅∣Bn∣|A_{n+1}| = 3 \cdot |B_n|.

    (∣A∣|A| denotes the number of elements of the set AA.)

  3. Problem 3

    Let f:R→[0,∞)f : \mathbb{R} \to [0,\infty) be a continuously differentiable function. Prove that

    ∣∫01f3(x) dx−f2(0)∫01f(x) dx∣≤max⁡0≤x≤1∣f′(x)∣(∫01f(x) dx)2.\left| \int_0^1 f^3(x)\,dx - f^2(0) \int_0^1 f(x)\,dx \right| \le \max_{0 \le x \le 1} |f'(x)| \left( \int_0^1 f(x)\,dx \right)^2.
  4. Problem 4

    Find all polynomials P(x)=anxn+an−1xn−1+⋯+a1x+a0P(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0 (an≠0a_n \ne 0) satisfying the following two conditions:

    (i) (a0,a1,…,an)(a_0, a_1, \dots, a_n) is a permutation of the numbers (0,1,…,n)(0, 1, \dots, n)

    and

    (ii) all roots of P(x)P(x) are rational numbers.

  5. Problem 5

    Let f:(0,∞)→Rf : (0,\infty) \to \mathbb{R} be a twice continuously differentiable function such that

    ∣f′′(x)+2xf′(x)+(x2+1)f(x)∣≤1|f''(x) + 2xf'(x) + (x^2+1)f(x)| \le 1

    for all xx. Prove that lim⁡x→∞f(x)=0\lim\limits_{x \to \infty} f(x) = 0.

  6. Problem 6

    Given a group GG, denote by G(m)G(m) the subgroup generated by the mmth powers of elements of GG. If G(m)G(m) and G(n)G(n) are commutative, prove that G(gcd⁡(m,n))G(\gcd(m,n)) is also commutative. (gcd⁡(m,n)\gcd(m,n) denotes the greatest common divisor of mm and nn.)

Day 2

6 problems
  1. Problem 1

    Let f(x)=x2+bx+cf(x) = x^2 + bx + c, where bb and cc are real numbers, and let

    M={x∈R:∣f(x)∣<1}.M = \{x \in \mathbb{R} : |f(x)| < 1\}.

    Clearly the set MM is either empty or consists of disjoint open intervals. Denote the sum of their lengths by ∣M∣|M|. Prove that

    ∣M∣≤22.|M| \le 2\sqrt{2}.
  2. Problem 2

    Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a function such that (f(x))n(f(x))^n is a polynomial for every n=2,3,…n = 2, 3, \dots. Does it follow that ff is a polynomial?

  3. Problem 3

    In the linear space of all real n×nn \times n matrices, find the maximum possible dimension of a linear subspace VV such that

    ∀X,Y∈Vtrace⁡(XY)=0.\forall X, Y \in V \quad \operatorname{trace}(XY) = 0.

    (The trace of a matrix is the sum of the diagonal entries.)

  4. Problem 4

    Prove that if f:R→Rf : \mathbb{R} \to \mathbb{R} is three times differentiable, then there exists a real number ξ∈(−1,1)\xi \in (-1,1) such that

    f′′′(ξ)6=f(1)−f(−1)2−f′(0).\frac{f'''(\xi)}{6} = \frac{f(1) - f(-1)}{2} - f'(0).
  5. Problem 5

    Find all r>0r > 0 such that whenever f:R2→Rf : \mathbb{R}^2 \to \mathbb{R} is a differentiable function such that ∣grad⁡f(0,0)∣=1|\operatorname{grad} f(0,0)| = 1 and ∣grad⁡f(u)−grad⁡f(v)∣≤∣u−v∣|\operatorname{grad} f(u) - \operatorname{grad} f(v)| \le |u - v| for all u,v∈R2u, v \in \mathbb{R}^2, then the maximum of ff on the disk {u∈R2:∣u∣≤r}\{u \in \mathbb{R}^2 : |u| \le r\} is attained at exactly one point.

    (grad⁡f(u)=(∂1f(u),∂2f(u))\operatorname{grad} f(u) = (\partial_1 f(u), \partial_2 f(u)) is the gradient vector of ff at the point uu. For a vector u=(a,b)u = (a,b), ∣u∣=a2+b2|u| = \sqrt{a^2 + b^2}.)

  6. Problem 6

    Prove that if pp and qq are rational numbers and r=p+q7r = p + q\sqrt{7}, then there exists a matrix (abcd)≠±(1001)\begin{pmatrix} a & b \\ c & d \end{pmatrix} \ne \pm \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} with integer entries and with ad−bc=1ad - bc = 1 such that

    ar+bcr+d=r.\frac{ar+b}{cr+d} = r.