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IMO Shortlist 1973 problem 12

Consider the two square matrices
$A=\begin{bmatrix}+1 &+1 &+1&+1 &+1\\+1 &+1 &+1&-1 &-1\\ +1 &-1&-1 &+1&+1\\ +1 &-1 &-1 &-1 &+1\\ +1 &+1&-1 &+1&-1\end{bmatrix}\quad\text{ and }\quad B=\begin{bmatrix}+1 &+1 &+1&+1 &+1\\+1 &+1 &+1&-1 &-1\\ +1 &+1&-1&+1&-1\\ +1 &-1&-1&+1&+1\\ +1 &-1&+1&-1 &+1\end{bmatrix}$

with entries $+1$ and $-1$ . The following operations will be called elementary:

(1) Changing signs of all numbers in one row;
(2) Changing signs of all numbers in one column;
(3) Interchanging two rows (two rows exchange their positions);
(4) Interchanging two columns.

Prove that the matrix $B$ cannot be obtained from the matrix $A$ using these operations.

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IMO Shortlist 1997 problem 13

In town $A,$ there are $n$ girls and $n$ boys, and each girl knows each boy. In town $B,$ there are $n$ girls $g_1, g_2, \ldots, g_n$ and $2n - 1$ boys $b_1, b_2, \ldots, b_{2n-1}.$ The girl $g_i,$ $i = 1, 2, \ldots, n,$ knows the boys $b_1, b_2, \ldots, b_{2i-1},$ and no others. For all $r = 1, 2, \ldots, n,$ denote by $A(r),B(r)$ the number of different ways in which $r$ girls from town $A,$ respectively town $B,$ can dance with $r$ boys from their own town, forming $r$ pairs, each girl with a boy she knows. Prove that $A(r) = B(r)$ for each $r = 1, 2, \ldots, n.$

IMO Shortlist 1997 problem 19

Let $a_1\geq \cdots \geq a_n \geq a_{n + 1} = 0$ be real numbers. Show that
$\sqrt {\sum_{k = 1}^n a_k} \leq \sum_{k = 1}^n \sqrt k (\sqrt {a_k} - \sqrt {a_{k + 1}}).$
Proposed by Romania

IMO Shortlist 1997 problem 26

For every integer $n \geq 2$ determine the minimum value that the sum $\sum^n_{i=0} a_i$ can take for nonnegative numbers $a_0, a_1, \ldots, a_n$ satisfying the condition $a_0 = 1,$ $a_i \leq a_{i+1} + a_{i+2}$ for $i = 0, \ldots, n - 2.$

skakavac 2012 drugo kolo ss2 2

Neka je $M$ podskup skupa $\{1, 2, ..., 15\}$ koji ne sadrži $3$ elementa čiji je umnožak potpun kvadrat. Odredi maksimalan broj elemenata skupa $M$ .

MEMO 2011 pojedinačno problem 1

Initially, only the integer $44$ is written on a board. An integer a on the board can be re- placed with four pairwise different integers $a_1, a_2, a_3, a_4$ such that the arithmetic mean $\frac 14 (a_1 + a_2 + a_3 + a_4)$ of the four new integers is equal to the number $a$ . In a step we simultaneously replace all the integers on the board in the above way. After $30$ steps we end up with $n = 4^{30}$ integers $b_1, b2,\ldots, b_n$ on the board. Prove that $\frac{b_1^2 + b_2^2+b_3^2+\cdots+b_n^2}{n}\geq 2011.$

skakavac 2013 prvo kolo ss2 2

Skup $S$ sastoji se od $14$ prirodnih brojeva. Pokažite da postoji $k\in\{1, \ldots, 7\}$ za koji je moguće naci $k$ -člane disjunktne podskupove $\{a_1, \ldots, a_k\}$ i $\{b_1, \ldots, b_k\}$ skupa $S$ tako da se sume
$A = \frac{1}{a_1} + \cdots + \frac{1}{a_k}, \quad B = \frac{1}{b_1} + \cdots + \frac{1}{b_k}$ razlikuju za manje od $0.001$ .