\documentclass[12pt,titlepage]{article} \usepackage{amsmath} \usepackage{mathrsfs} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsthm} \usepackage{mathtools} \usepackage{graphicx} \usepackage{color} \usepackage{ucs} \usepackage[utf8x]{inputenc} \usepackage{xparse} \usepackage{tikz} \usepackage{hyperref} %----Macros---------- % % Unresolved issues: % % \righttoleftarrow % \lefttorightarrow % % \color{} with HTML colorspec % \bgcolor % \array with options (without options, it's equivalent to the matrix environment) % Of the standard HTML named colors, white, black, red, green, blue and yellow % are predefined in the color package. Here are the rest. \definecolor{aqua}{rgb}{0, 1.0, 1.0} \definecolor{fuschia}{rgb}{1.0, 0, 1.0} \definecolor{gray}{rgb}{0.502, 0.502, 0.502} \definecolor{lime}{rgb}{0, 1.0, 0} \definecolor{maroon}{rgb}{0.502, 0, 0} \definecolor{navy}{rgb}{0, 0, 0.502} \definecolor{olive}{rgb}{0.502, 0.502, 0} \definecolor{purple}{rgb}{0.502, 0, 0.502} \definecolor{silver}{rgb}{0.753, 0.753, 0.753} \definecolor{teal}{rgb}{0, 0.502, 0.502} % Because of conflicts, \space and \mathop are converted to % \itexspace and \operatorname during preprocessing. % itex: \space{ht}{dp}{wd} % % Height and baseline depth measurements are in units of tenths of an ex while % the width is measured in tenths of an em. \makeatletter \newdimen\itex@wd% \newdimen\itex@dp% \newdimen\itex@thd% \def\itexspace#1#2#3{\itex@wd=#3em% \itex@wd=0.1\itex@wd% \itex@dp=#2ex% \itex@dp=0.1\itex@dp% \itex@thd=#1ex% \itex@thd=0.1\itex@thd% \advance\itex@thd\the\itex@dp% \makebox[\the\itex@wd]{\rule[-\the\itex@dp]{0cm}{\the\itex@thd}}} \makeatother % \tensor and \multiscript \makeatletter \newif\if@sup \newtoks\@sups \def\append@sup#1{\edef\act{\noexpand\@sups={\the\@sups #1}}\act}% \def\reset@sup{\@supfalse\@sups={}}% \def\mk@scripts#1#2{\if #2/ \if@sup ^{\the\@sups}\fi \else% \ifx #1_ \if@sup ^{\the\@sups}\reset@sup \fi {}_{#2}% \else \append@sup#2 \@suptrue \fi% \expandafter\mk@scripts\fi} \def\tensor#1#2{\reset@sup#1\mk@scripts#2_/} \def\multiscripts#1#2#3{\reset@sup{}\mk@scripts#1_/#2% \reset@sup\mk@scripts#3_/} \makeatother % \slash \makeatletter \newbox\slashbox \setbox\slashbox=\hbox{$/$} \def\itex@pslash#1{\setbox\@tempboxa=\hbox{$#1$} \@tempdima=0.5\wd\slashbox \advance\@tempdima 0.5\wd\@tempboxa \copy\slashbox \kern-\@tempdima \box\@tempboxa} \def\slash{\protect\itex@pslash} \makeatother % math-mode versions of \rlap, etc % from Alexander Perlis, "A complement to \smash, \llap, and lap" % http://math.arizona.edu/~aprl/publications/mathclap/ \def\clap#1{\hbox to 0pt{\hss#1\hss}} \def\mathllap{\mathpalette\mathllapinternal} \def\mathrlap{\mathpalette\mathrlapinternal} \def\mathclap{\mathpalette\mathclapinternal} \def\mathllapinternal#1#2{\llap{$\mathsurround=0pt#1{#2}$}} \def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}} \def\mathclapinternal#1#2{\clap{$\mathsurround=0pt#1{#2}$}} % Renames \sqrt as \oldsqrt and redefine root to result in \sqrt[#1]{#2} \let\oldroot\root \def\root#1#2{\oldroot #1 \of{#2}} \renewcommand{\sqrt}[2][]{\oldroot #1 \of{#2}} % Manually declare the txfonts symbolsC font \DeclareSymbolFont{symbolsC}{U}{txsyc}{m}{n} \SetSymbolFont{symbolsC}{bold}{U}{txsyc}{bx}{n} \DeclareFontSubstitution{U}{txsyc}{m}{n} % Manually declare the stmaryrd font \DeclareSymbolFont{stmry}{U}{stmry}{m}{n} \SetSymbolFont{stmry}{bold}{U}{stmry}{b}{n} % Manually declare the MnSymbolE font \DeclareFontFamily{OMX}{MnSymbolE}{} \DeclareSymbolFont{mnomx}{OMX}{MnSymbolE}{m}{n} \SetSymbolFont{mnomx}{bold}{OMX}{MnSymbolE}{b}{n} \DeclareFontShape{OMX}{MnSymbolE}{m}{n}{ <-6> MnSymbolE5 <6-7> MnSymbolE6 <7-8> MnSymbolE7 <8-9> MnSymbolE8 <9-10> MnSymbolE9 <10-12> MnSymbolE10 <12-> MnSymbolE12}{} % Declare specific arrows from txfonts without loading the full package \makeatletter \def\re@DeclareMathSymbol#1#2#3#4{% \let#1=\undefined \DeclareMathSymbol{#1}{#2}{#3}{#4}} \re@DeclareMathSymbol{\neArrow}{\mathrel}{symbolsC}{116} \re@DeclareMathSymbol{\neArr}{\mathrel}{symbolsC}{116} \re@DeclareMathSymbol{\seArrow}{\mathrel}{symbolsC}{117} \re@DeclareMathSymbol{\seArr}{\mathrel}{symbolsC}{117} \re@DeclareMathSymbol{\nwArrow}{\mathrel}{symbolsC}{118} \re@DeclareMathSymbol{\nwArr}{\mathrel}{symbolsC}{118} \re@DeclareMathSymbol{\swArrow}{\mathrel}{symbolsC}{119} \re@DeclareMathSymbol{\swArr}{\mathrel}{symbolsC}{119} \re@DeclareMathSymbol{\nequiv}{\mathrel}{symbolsC}{46} \re@DeclareMathSymbol{\Perp}{\mathrel}{symbolsC}{121} \re@DeclareMathSymbol{\Vbar}{\mathrel}{symbolsC}{121} 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\setbox\z@\hbox{\m@th$#1#2$}% \setbox\tw@\hbox{\m@th$#1% \widehat{% \vrule\@width\z@\@height\ht\z@ \vrule\@height\z@\@width\wd\z@}$}% \dp\tw@-\ht\z@ \@tempdima\ht\z@ \advance\@tempdima2\ht\tw@ \divide\@tempdima\thr@@ \setbox\tw@\hbox{% \raise\@tempdima\hbox{\scalebox{1}[-1]{\lower\@tempdima\box \tw@}}}% {\ooalign{\box\tw@ \cr \box\z@}}} \makeatother % \mathraisebox{voffset}[height][depth]{something} \makeatletter \NewDocumentCommand\mathraisebox{moom}{% \IfNoValueTF{#2}{\def\@temp##1##2{\raisebox{#1}{$\m@th##1##2$}}}{% \IfNoValueTF{#3}{\def\@temp##1##2{\raisebox{#1}[#2]{$\m@th##1##2$}}% }{\def\@temp##1##2{\raisebox{#1}[#2][#3]{$\m@th##1##2$}}}}% \mathpalette\@temp{#4}} \makeatletter % udots (taken from yhmath) \makeatletter \def\udots{\mathinner{\mkern2mu\raise\p@\hbox{.} \mkern2mu\raise4\p@\hbox{.}\mkern1mu \raise7\p@\vbox{\kern7\p@\hbox{.}}\mkern1mu}} \makeatother %% Fix array \newcommand{\itexarray}[1]{\begin{matrix}#1\end{matrix}} %% \itexnum is a noop \newcommand{\itexnum}[1]{#1} %% Renaming existing commands \newcommand{\underoverset}[3]{\underset{#1}{\overset{#2}{#3}}} \newcommand{\widevec}{\overrightarrow} \newcommand{\darr}{\downarrow} \newcommand{\nearr}{\nearrow} \newcommand{\nwarr}{\nwarrow} \newcommand{\searr}{\searrow} \newcommand{\swarr}{\swarrow} \newcommand{\curvearrowbotright}{\curvearrowright} \newcommand{\uparr}{\uparrow} \newcommand{\downuparrow}{\updownarrow} \newcommand{\duparr}{\updownarrow} \newcommand{\updarr}{\updownarrow} \newcommand{\gt}{>} \newcommand{\lt}{<} \newcommand{\map}{\mapsto} \newcommand{\embedsin}{\hookrightarrow} \newcommand{\Alpha}{A} \newcommand{\Beta}{B} \newcommand{\Zeta}{Z} \newcommand{\Eta}{H} \newcommand{\Iota}{I} \newcommand{\Kappa}{K} \newcommand{\Mu}{M} \newcommand{\Nu}{N} \newcommand{\Rho}{P} \newcommand{\Tau}{T} \newcommand{\Upsi}{\Upsilon} \newcommand{\omicron}{o} \newcommand{\lang}{\langle} \newcommand{\rang}{\rangle} \newcommand{\Union}{\bigcup} \newcommand{\Intersection}{\bigcap} 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\newcommand{\toggle}[2]{#2} % Theorem Environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{dissected hexagon solution} \hypertarget{dissected_hexagon}{}\section*{{Dissected Hexagon}}\label{dissected_hexagon} [[DissectedHexagon.png:pic]] \begin{quote}% What fraction of this regular hexagon is not covered by the two isosceles triangles? \end{quote} \hypertarget{solution_by_properties_of_a_regular_hexagon_and_area_of_a_triangle}{}\subsection*{{Solution by [[Properties of a Regular Hexagon]] and [[Area of a Triangle]]}}\label{solution_by_properties_of_a_regular_hexagon_and_area_of_a_triangle} [[DissectedHexagonLabelled.png:pic]] Drawing the line segment $E D$ in the above diagram shows that the unshaded region comprises six [[congruent]] triangles. Each has base one half of the side length of the hexagon and height one half of its height. The [[regular hexagon]] itself comprises six [[congruent]] [[equilateral triangles]], each with base one side length of the hexagon and height one half of its height. Therefore, each of the thinner triangles has area half of that of one of the equilateral triangles, and so the unshaded area comprises half the hexagon. \end{document}