Unofficial archive — problems, solutions & results © IMC, reproduced with permission.

Studolymp / IMC / 2011

IMC 2011
contestants 305 · problems 10 (5+5) · scale 0–10 · per-problem no

Only total scores were published for this edition — no per-problem breakdown.

Problems

Day 1

P1

Let f:RRf : \mathbb{R} \to \mathbb{R} be a continuous function. A point xx is called a shadow point if there exists a point yRy \in \mathbb{R} with y>xy > x such that f(y)>f(x)f(y) > f(x). Let a<ba < b be real numbers and suppose that

  • all the points of the open interval I=(a,b)I = (a, b) are shadow points;
  • aa and bb are not shadow points.
Prove that

a) f(x)f(b)f(x) \le f(b) for all a<x<ba < x < b;

b) f(a)=f(b)f(a) = f(b).

(José Luis Díaz-Barrero, Barcelona)

P2

Does there exist a real 3×33 \times 3 matrix AA such that tr(A)=0\operatorname{tr}(A) = 0 and A2+At=IA^2 + A^t = I? (tr(A)\operatorname{tr}(A) denotes the trace of AA, AtA^t is the transpose of AA, and II is the identity matrix.)

(Moubinool Omarjee, Paris)

P3

Let pp be a prime number. Call a positive integer nn interesting if xn1=(xpx+1)f(x)+pg(x)x^n - 1 = (x^p - x + 1) f(x) + p g(x) for some polynomials ff and gg with integer coefficients.

a) Prove that the number pp1p^p - 1 is interesting.

b) For which pp is pp1p^p - 1 the minimal interesting number?

(Eugene Goryachko and Fedor Petrov, St. Petersburg)

P4

Let A1,A2,,AnA_1, A_2, \dots, A_n be finite, nonempty sets. Define the function f(t)=k=1n1i1<i2<<ikn(1)k1tAi1Ai2Aik.f(t) = \sum_{k=1}^{n} \sum_{1 \le i_1 < i_2 < \dots < i_k \le n} (-1)^{k-1} t^{|A_{i_1} \cup A_{i_2} \cup \dots \cup A_{i_k}|}. Prove that ff is nondecreasing on [0,1][0, 1].

(A|A| denotes the number of elements in AA.)

(Levon Nurbekyan and Vardan Voskanyan, Yerevan)

P5

Let nn be a positive integer and let VV be a (2n1)(2n-1)-dimensional vector space over the two-element field. Prove that for arbitrary vectors v1,,v4n1Vv_1, \dots, v_{4n-1} \in V, there exists a sequence 1i1<<i2n4n11 \le i_1 < \dots < i_{2n} \le 4n-1 of indices such that vi1++vi2n=0v_{i_1} + \dots + v_{i_{2n}} = 0.

(Ilya Bogdanov, Moscow and Géza Kós, Budapest)

Day 2

P6

Let (an)n=0(a_n)_{n=0}^{\infty} be a sequence with 12<an<1\frac{1}{2} < a_n < 1 for all n0n \ge 0. Define the sequence (xn)n=0(x_n)_{n=0}^{\infty} by x0=a0,xn+1=an+1+xn1+an+1xn(n0).x_0 = a_0, \qquad x_{n+1} = \frac{a_{n+1} + x_n}{1 + a_{n+1} x_n} \quad (n \ge 0). What are the possible values of limnxn\lim\limits_{n \to \infty} x_n? Can such a sequence diverge?

(Johnson Olaleru, Lagos)

P7

An alien race has three genders: male, female, and emale. A married triple consists of three persons, one from each gender, who all like each other. Any person is allowed to belong to at most one married triple. A special feature of this race is that feelings are always mutual — if xx likes yy, then yy likes xx.

The race is sending an expedition to colonize a planet. The expedition has nn males, nn females, and nn emales. It is known that every expedition member likes at least kk persons of each of the two other genders. The problem is to create as many married triples as possible to produce healthy offspring so the colony could grow and prosper.

a) Show that if nn is even and k=n2k = \frac{n}{2}, then it might be impossible to create even one married triple.

b) Show that if k3n4k \ge \frac{3n}{4}, then it is always possible to create nn disjoint married triples, thus marrying all of the expedition members.

(Fedor Duzhin and Nick Gravin, Singapore)

P8

Determine the value of n=1ln(1+1n)ln(1+12n)ln(1+12n+1).\sum_{n=1}^{\infty} \ln \left( 1 + \frac{1}{n} \right) \cdot \ln \left( 1 + \frac{1}{2n} \right) \cdot \ln \left( 1 + \frac{1}{2n+1} \right).

(Gerhard Woeginger, Utrecht)

P9

Let f(x)f(x) be a polynomial with real coefficients of degree nn. Suppose that f(k)f(m)km\dfrac{f(k) - f(m)}{k - m} is an integer for all integers 0k<mn0 \le k < m \le n. Prove that aba - b divides f(a)f(b)f(a) - f(b) for all pairs of distinct integers aa and bb.

(Fedor Petrov, St. Petersburg)

P10

Let F=A0A1AnF = A_0 A_1 \dots A_n be a convex polygon in the plane. Define for all 1kn11 \le k \le n-1 the operation fkf_k which replaces FF with a new polygon fk(F)=A0Ak1AkAk+1An,f_k(F) = A_0 \dots A_{k-1} A_k' A_{k+1} \dots A_n, where AkA_k' is the point symmetric to AkA_k with respect to the perpendicular bisector of Ak1Ak+1A_{k-1} A_{k+1}. Prove that (f1f2fn1)n(F)=F(f_1 \circ f_2 \circ \dots \circ f_{n-1})^n (F) = F. We suppose that all operations are well-defined on the polygons, to which they are applied, i.e. results are convex polygons again. (A0A_0, A1,,AnA_1, \dots, A_n are the vertices of FF in consecutive order.)

(Mikhail Khristoforov, St. Petersburg)