135 lines
4.4 KiB
TeX
Executable File
135 lines
4.4 KiB
TeX
Executable File
%%Tema para beamer "Imunam", versión 1.0
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\documentclass{beamer}
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\usepackage[utf8]{inputenc}
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\usepackage[T1]{fontenc}
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\usepackage[spanish]{babel}
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\graphicspath{ {images/} }
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\usepackage{mathtools}
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\usepackage{tikz}
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%% Logos de restaurantes:
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\newcommand{\logomanekin}{images/logomanekin}
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%%Se define el "environment" teorema
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%% The "environment" theorem is defined
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\newtheorem{Formula}{Formula}
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%%Tema de beamer "Imunam"
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%% Theme of beamer "Imunam"
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\usetheme[cuernavaca]{Imunam}
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%\usetheme{Imunam}
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%%Si se omite "[cuernavaca]" en éste comando, el logotipo se imprime sin la
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%%leyenda "Unidad Cuernavaca" en la parte inferior.
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%% If "[cuernavaca]" is omitted in this command, the logo is printed without the
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%% legend "Cuernavaca Unit" in the lower part.
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\title{Proof of the Pythagoras theorem.}
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\author{Enrique Andrade Gonzalez %Nombre del autor % Author's name
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\texttt{e.andrade@udc.es}} %email
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\date{Computer tools in mathematican's work} %subject
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%\institute{Instituto de Matemáticas, unidad Cuernavaca}
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%%Instituto del ponenete, dado que el texto "Intituto de matemáticas" aparece
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%%en el logo, parece redundante incluirlo además con éste comando.
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\begin{document}
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\begin{frame}
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\titlepage %Necesario para generar la portada
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\end{frame}
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%%La siguiente diapositiva es opcional, si se quiere la tabla de contenidos
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%%Se sebe compilar dos veces el documento para que funcione
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%--------------------------------------------------------------------------
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\begin{frame}
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\tableofcontents %Imprime la tabla de contenido
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\end{frame}
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%--------------------------------------------------------------------------
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\section{Introduction} %%Título de la sección (Opcional)
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\begin{frame}
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\frametitle{Introduction}
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\framesubtitle{objective} %%Subtítulo de la diapositiva (opcional)
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In this presentation we try to show a proof of the Pythagorean theorem.
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There are many demonstrations, but this one is one of the simplest.
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\end{frame}
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\section{Proof of the Pythagorean theorem} %%Otra sección
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\begin{frame}
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\frametitle{Concept}
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Suppose we have a square of side \textbf{r} and on each of its sides we place a right triangle of legs \textbf{x} and \textbf{y}. As in this situation the hypotenuse of each of the triangles is \textbf{r} we want to prove that:
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\\\
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\begin{Formula} %%Uso del "environment" definido al inicio del documento.
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\[x^{2}+y^{2}=r^{2}\]
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\end{Formula}
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\end{frame}
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\begin{frame}
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\frametitle{The figure}
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The figure that is obtained is the following:
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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%\begin{tikzpicture}
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%\draw (0,0) circle (1cm);
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\draw [fill=green] (-2,-2) rectangle (2,2);
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\draw [fill=red] (-2,0) -- (0,2) -- (2,0) -- (0,-2) -- cycle;
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% r
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\draw [dotted]
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(0.75,1) node[black] {r}
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(0.75,-1) node[black] {r}
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(-0.75,1) node[black] {r}
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(-0.75,-1) node[black] {r};
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% x y
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\draw [dotted]
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(1,-2.2) node[black] {y}
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(1,2.2) node[black] {x}
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(-1,2.2) node[black] {y}
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(-1,-2.2) node[black] {x}
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(2.2,1) node[black] {y}
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(2.2,-1) node[black] {x}
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(-2.2,1) node[black] {x}
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(-2.2,-1) node[black] {y};
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\end{tikzpicture}
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\[x^{2}+y^{2}=r^{2}\]
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\end{center}
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\end{frame}
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%\section{Restaurants} %%Otra sección
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\begin{frame}
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\frametitle{Conclusions}
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\begin{itemize}
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\item Each side of the \textbf{green square} is the sum of \textbf{x} and \textbf{y}. Therefore, the area of the square is: \[(x+y)^{2}\]
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\item For the same reason, the area of the \textbf{red square} is: \[r^{2}\]
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\item The area of each of the \textbf{green triangles} (y, x and r) is:\[\frac{x+y}{2}\]
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\end{itemize}
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\end{frame}
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\section{Demonstration} %%Otra sección
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\begin{frame}
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\frametitle{Demonstration}
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\begin{itemize}
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\item The green square is formed by the red square and the four green triangles, so the sum of all the areas is: \[(x+y)^{2}=r^{2} + 4 (\frac{x+y}{2})\]
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\item We develop the left part of equality: \[(x+y)^{2}=x^{2} + 2xy + y^{2}\]
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\item We substitute in the first formula: \[x^{2} + 2xy + y^{2} = r^{2} + 2xy\]
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\item \textbf{2xy} is eliminated on both sides of the equality, and we obtain the desired result: \[x^{2} + y^{2} = r^{2}\]
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\end{itemize}
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\end{frame}
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\end{document} |