Koło matematyczne I LO w Białymstoku

 
Zadania PDF.

Źródło zadań w texu.

 
%        File: starsi.tex
%     Created: Tue Nov 01 09:00 PM 2011 C
% Last Change: Tue Nov 01 09:00 PM 2011 C
\documentclass[10pt]{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\textwidth 16cm
\textheight 24cm
\oddsidemargin 0cm
\topmargin 0pt
\headheight 0pt
\headsep 0pt
\usepackage[polish]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{polski}
\usepackage{import}
\usepackage{multicol}
%\usepackage{MnSymbol}
% ----------------------------------------------------------------
\vfuzz4pt % Don't report over-full v-boxes if over-edge is small
\hfuzz4pt % Don't report over-full h-boxes if over-edge is small
% THEOREMS -------------------------------------------------------
\newtheorem{thm}{Twierdzenie}[section]
\newtheorem{cor}[thm]{Wniosek}
\newtheorem{lem}[thm]{Lemat}
\newtheorem{defn}[thm]{Definicja}
\newtheorem{tozs}[thm]{Tożsamość}
\newtheorem{hyp}[thm]{Hipoteza}
\newtheorem{useless}[thm]{}
 
\newenvironment{sol}[1][Rozwiązanie. ]{
\vskip 3mm
\noindent\emph{#1}
 
}
{\hfill\par}
 
\newcounter{problem}
\newenvironment{problem}[1][Zadanie]{
\stepcounter{problem}
\vskip 3mm
\noindent{\textsc{\bfseries #1 \theproblem}}\\}
{\hfill\par}
 
\def\abs #1{\left\vert #1\right\vert}
 
\renewcommand{\angle}{\sphericalangle}
\renewcommand{\vec}[1]{\overrightarrow{#1}}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}
\renewcommand{\dots}{\ldots}
 
\subimport{../}{style.sty}
\def\sectionwidth{8cm}
%\include{style}
 
\def\headpicture{../micek-2cm.jpg}
\def\author{Joachim Jelisiejew}
\def\date{2 listopada}
 
\def\bareroger{\includegraphics[height=1em]{jolly-roger-mat}}
\def\roger{\ \ \hbox{\bareroger{}}\ \ }
 
\begin{document}
 
\section{Nierówności $\geq 2$}
 
Wszystkie zmienne są dodatnie. Znak \bareroger{} należy zastąpić jednym z~$\geq,
\leq, <>$, tak, jak na poprzednim kółku.
\begin{multicols}{2}
    \begin{enumerate}
        \item $\frac{1}{x_1 + 2008x_2} + \frac{1}{x_2 + 2008x_3} + \dots +
            \frac{1}{x_{2009} + 2008x_1}$\\ $\roger 2009\cdot (x_1 + \dots +
            x_{2009})^{-1}$
        \item $x_1^3 + \dots + x_7^3 + x_1^{-1} + \dots + x_7^{-1} \roger
            3(x_1 + \dots + x_7)$
        \item $x_1^2 + \dots + x_{10}^2 = \sqrt{10}$, $x_1 + \dots + x_{10}
            \roger 10^{3/4}$
        \item $\frac{a}{2b + c} + \frac{b}{2c + a} + \frac{c}{2a + b} \roger 1$
        \item $x_1 \dots x_7 = 1$,\\$x_1(1 +x_1^2) + \dots + x_7(1+x_7^2)
            \roger 14$
        \item $a+b+c=1$,\\$(1+a)(1+b)(1+c) \roger 8(1-a)(1-b)(1-c)$
    \end{enumerate}
\end{multicols}
 
\interskip
 
\section{Nierówności $\geq 2$}
 
Wszystkie zmienne są dodatnie. Znak \bareroger{} należy zastąpić jednym z~$\geq,
\leq, <>$, tak, jak na poprzednim kółku.
\begin{multicols}{2}
    \begin{enumerate}
        \item $\frac{1}{x_1 + 2008x_2} + \frac{1}{x_2 + 2008x_3} + \dots +
            \frac{1}{x_{2009} + 2008x_1}$\\ $\roger 2009\cdot (x_1 + \dots +
            x_{2009})^{-1}$
        \item $x_1^3 + \dots + x_7^3 + x_1^{-1} + \dots + x_7^{-1} \roger
            3(x_1 + \dots + x_7)$
        \item $x_1^2 + \dots + x_{10}^2 = \sqrt{10}$, $x_1 + \dots + x_{10}
            \roger 10^{3/4}$
        \item $\frac{a}{2b + c} + \frac{b}{2c + a} + \frac{c}{2a + b} \roger 1$
        \item $x_1 \dots x_7 = 1$,\\$x_1(1 +x_1^2) + \dots + x_7(1+x_7^2)
            \roger 14$
        \item $a+b+c=1$,\\$(1+a)(1+b)(1+c) \roger 8(1-a)(1-b)(1-c)$
    \end{enumerate}
\end{multicols}
 
\end{document}