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doc/elegantpaper.cls
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372
doc/elegantpaper.cls
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% Author: Dongsheng Deng
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% Homepage: https://ddswhu.me/
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% Email: elegantlatex2e@gmail.com
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% Lastest Version: https://github.com/ElegantLaTeX/ElegantPaper
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% --- Class structure: identification part
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\ProvidesClass{elegantpaper}[2022/12/31 v0.11 ElegantLaTeX Paper class]
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\NeedsTeXFormat{LaTeX2e}
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%%%
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\RequirePackage{kvoptions}
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\RequirePackage{etoolbox}
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\RequirePackage{calc}
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\SetupKeyvalOptions{family=ELEGANT, prefix=ELEGANT@, setkeys=\kvsetkeys}
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\newcommand{\ekv}[1]{\kvsetkeys{ELEGANT}{#1}}
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\DeclareStringOption[en]{lang}
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\DeclareVoidOption{cn}{\ekv{lang=cn}}
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\DeclareVoidOption{en}{\ekv{lang=en}}
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\DeclareStringOption[cm]{math}
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\DeclareStringOption[numeric-comp]{citestyle}
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\DeclareStringOption[numeric]{bibstyle}
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\DeclareStringOption[biber]{bibend}
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\DeclareVoidOption{biber}{\ekv{bibend=biber}}
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\DeclareVoidOption{bibtex}{\ekv{bibend=bibtex}}
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\DeclareStringOption[ctexfont]{chinesefont}
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\DeclareVoidOption{ctexfont}{\ekv{chinesefont=ctexfont}}
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\DeclareVoidOption{founder}{\ekv{chinesefont=founder}}
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\DeclareVoidOption{nofont}{\ekv{chinesefont=nofont}}
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\DeclareVoidOption{newtx}{\ekv{math=newtx}}
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\DeclareVoidOption{mtpro2}{\ekv{math=mtpro2}}
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\DeclareVoidOption{cm}{\ekv{math=cm}}
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\DeclareDefaultOption{\PassOptionsToClass{\CurrentOption}{article}}
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\ProcessKeyvalOptions*\relax
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\LoadClass{article}
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\RequirePackage{hyperref}
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\hypersetup{
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pdfborder={0 0 0},
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colorlinks=true,
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linkcolor={winered},
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urlcolor={winered},
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filecolor={winered},
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citecolor={winered},
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linktoc=all,
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}
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% settings for the hyperref and geometry
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\RequirePackage[
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left=1in,
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right=1in,
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top=1in,
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bottom=1in,
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headheight=0pt,
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headsep=0pt]{geometry}
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\setlength{\headsep}{5pt}
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\RequirePackage{amsthm}
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\RequirePackage{amsmath}
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\RequirePackage{amssymb}
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\RequirePackage{indentfirst}
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\RequirePackage{booktabs}
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\RequirePackage{multicol}
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\RequirePackage{multirow}
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% \RequirePackage{linegoal}
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\RequirePackage{xcolor}
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\RequirePackage{graphicx}
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\RequirePackage{fancyvrb}
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\RequirePackage{abstract}
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\RequirePackage{hologo}
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\linespread{1.35}
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\graphicspath{{image/}{figure/}{fig/}{img/}}
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% caption settings
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\RequirePackage[labelfont={bf}]{caption}
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\captionsetup[table]{skip=3pt}
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\captionsetup[figure]{skip=3pt}
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% list/itemize/enumerate setting
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\RequirePackage[shortlabels,inline]{enumitem}
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\setlist{nolistsep}
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% define the hyperref color
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\RequirePackage{xcolor}
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\definecolor{winered}{rgb}{0.5,0,0}
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\definecolor{lightgrey}{rgb}{0.95,0.95,0.95}
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\definecolor{commentcolor}{RGB}{0,100,0}
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\definecolor{frenchplum}{RGB}{190,20,83}
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% add the email cmd
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\newcommand\email[1]{\href{mailto:#1}{\nolinkurl{#1}}}
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% font settings
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\PassOptionsToPackage{no-math}{fontspec}
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\PassOptionsToPackage{quiet}{fontspec}
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\RequirePackage{iftex}
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\ifXeTeX
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\RequirePackage[no-math]{fontspec}
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\setmainfont{texgyretermes}[
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UprightFont = *-regular ,
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BoldFont = *-bold ,
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ItalicFont = *-italic ,
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BoldItalicFont = *-bolditalic ,
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Extension = .otf ,
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Scale = 1.0]
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\setsansfont{texgyreheros}[
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UprightFont = *-regular ,
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BoldFont = *-bold ,
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ItalicFont = *-italic ,
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BoldItalicFont = *-bolditalic ,
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Extension = .otf ,
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Scale = 0.9]
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\else
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\RequirePackage{newtxtext}
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\RequirePackage[scaled=.90]{helvet}
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\fi
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\ifdefstring{\ELEGANT@lang}{cn}{
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\ifdefstring{\ELEGANT@chinesefont}{founder}{
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\RequirePackage[UTF8,scheme=plain,fontset=none]{ctex}
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\setCJKmainfont[BoldFont={FZHei-B01},ItalicFont={FZKai-Z03}]{FZShuSong-Z01}
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\setCJKsansfont[BoldFont={FZHei-B01}]{FZKai-Z03}
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\setCJKmonofont[BoldFont={FZHei-B01}]{FZFangSong-Z02}
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\setCJKfamilyfont{zhsong}{FZShuSong-Z01}
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\setCJKfamilyfont{zhhei}{FZHei-B01}
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\setCJKfamilyfont{zhkai}[BoldFont={FZHei-B01}]{FZKai-Z03}
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\setCJKfamilyfont{zhfs}[BoldFont={FZHei-B01}]{FZFangSong-Z02}
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\newcommand*{\songti}{\CJKfamily{zhsong}}
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\newcommand*{\heiti}{\CJKfamily{zhhei}}
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\newcommand*{\kaishu}{\CJKfamily{zhkai}}
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\newcommand*{\fangsong}{\CJKfamily{zhfs}}}{\relax}
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\ifdefstring{\ELEGANT@chinesefont}{nofont}{
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\RequirePackage[UTF8,scheme=plain,fontset=none]{ctex}}{\relax}
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\ifdefstring{\ELEGANT@chinesefont}{ctexfont}{
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\RequirePackage[UTF8,scheme=plain]{ctex}}{\relax}
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\AfterEndPreamble{
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\setlength\parindent{2\ccwd}}
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}{\relax}
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\ifcsname kaishu\endcsname
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\newcommand{\citshape}{\kaishu}
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\else
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\newcommand{\citshape}{\itshape}
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\fi
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\ifcsname kaishu\endcsname
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\newcommand{\cnormal}{\kaishu}
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\else
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\newcommand{\cnormal}{\normalfont}
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\fi
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\ifcsname fangsong\endcsname
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\newcommand{\cfs}{\fangsong}
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\else
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\newcommand{\cfs}{\normalfont}
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\fi
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\ifdefstring{\ELEGANT@math}{newtx}{
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\RequirePackage{newtxmath}
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\let\Bbbk\relax
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\RequirePackage{esint}
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%%% use yhmath pkg, uncomment following code
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% \let\oldwidering\widering
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% \let\widering\undefined
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% \RequirePackage{yhmath}
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% \let\widering\oldwidering
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%%% use esvect pkg, uncomment following code
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% \RequirePackage{esvect}
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\DeclareSymbolFont{CMlargesymbols}{OMX}{cmex}{m}{n}
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\let\sumop\relax\let\prodop\relax
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\DeclareMathSymbol{\sumop}{\mathop}{CMlargesymbols}{"50}
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\DeclareMathSymbol{\prodop}{\mathop}{CMlargesymbols}{"51}
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}{\relax}
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\RequirePackage[
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backend=\ELEGANT@bibend,
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citestyle=\ELEGANT@citestyle,
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bibstyle=\ELEGANT@bibstyle,
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sorting=none]{biblatex}
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\RequirePackage{appendix}
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\ifdefstring{\ELEGANT@lang}{cn}{
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\theoremstyle{plain}% default
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\newtheorem{theorem}{定理}[section] %
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\newtheorem{lemma}[theorem]{引理} %
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\newtheorem{proposition}[theorem]{命题} %
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\newtheorem*{corollary}{推论} %
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\theoremstyle{definition} %
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\newtheorem{definition}{定义}[section] %
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\newtheorem{conjecture}{猜想}[section] %
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\newtheorem{example}{例}[section] %
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\theoremstyle{remark} %
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\newtheorem*{remark}{\normalfont\bfseries 评论} %
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\newtheorem*{note}{\normalfont\bfseries 注} %
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\newtheorem{case}{\normalfont\bfseries 案例} %
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\renewcommand*{\proofname}{\normalfont\bfseries 证明} %
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\renewcommand\contentsname{目录}
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\renewcommand\refname{参考文献} %
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\renewcommand\figurename{图} %
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\renewcommand\tablename{表}%
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\newcommand\versiontext{版本：}%
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\newcommand\updatetext{日期：}%
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\newcommand{\ebibname}{参考文献}
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\renewcommand\abstractname{摘\hspace{2em}要}
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\setlength\parindent{2\ccwd}
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\renewcommand{\abstracttextfont}{\small\citshape\noindent\ignorespaces}
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% 新定义命令
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\newcommand{\keywords}[1]{\vskip2ex\par\noindent\normalfont{\bfseries 关键词: } #1}
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\newcommand\figref[1]{{\bfseries 图~\ref{#1}}}
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\newcommand\tabref[1]{{\bfseries 表~\ref{#1}}}
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\renewcommand{\appendixtocname}{附录}
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\renewcommand{\appendixpagename}{附录}}{\relax}
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\ifdefstring{\ELEGANT@lang}{en}{
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\theoremstyle{plain}% default
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\newtheorem{theorem}{Theorem}[section] %
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\newtheorem{lemma}[theorem]{Lemma} %
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\newtheorem{proposition}[theorem]{Proposition} %
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\newtheorem*{corollary}{Corollary} %
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\theoremstyle{definition} %
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\newtheorem{definition}{Definition}[section] %
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\newtheorem{conjecture}{Conjecture}[section] %
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\newtheorem{example}{Example}[section] %
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\theoremstyle{remark} %
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\newtheorem*{remark}{Remark} %
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\newtheorem*{note}{Note} %
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\newtheorem{case}{Case} %
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\renewcommand*{\proofname}{\normalfont\bfseries Proof}%
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\newcommand\versiontext{\itshape Version: }%
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\newcommand\updatetext{\itshape Date: }%
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\setlength\parindent{2em}
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\renewcommand{\abstracttextfont}{\sffamily\small\noindent\ignorespaces}
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% newcommands defined in this template.
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\newcommand{\keywords}[1]{\vskip 2ex\par\noindent\normalfont{\bfseries Keywords: } #1}
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\newcommand\figref[1]{{\bfseries Figure~\ref{#1}}}
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\newcommand\tabref[1]{{\bfseries Table~\ref{#1}}}
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\newcommand{\ebibname}{Bibliography}}{\relax}
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\def\bibfont{\footnotesize}
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\AtBeginEnvironment{verbatim}{\microtypesetup{activate=false}}
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\BeforeBeginEnvironment{tabular}{\small}
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\AfterEndEnvironment{tabular}{}
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\AtBeginDocument{
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\setlength{\abovedisplayskip}{3pt}
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\setlength{\belowdisplayskip}{3pt}
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\RequirePackage[flushmargin]{footmisc}
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\setlength{\footnotesep}{12pt}}
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\def\IfEmpty#1{%
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\edef\1{\the#1}
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\ifx\1\empty
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}
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\newtoks\version
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\newtoks\institute
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\renewcommand\maketitle{\par
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\begingroup
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\renewcommand\thefootnote{\@fnsymbol\c@footnote}%
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\def\@makefnmark{\rlap{\@textsuperscript{\normalfont\@thefnmark}}}%
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\long\def\@makefntext##1{\parindent 1em\noindent
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\hb@xt@0.1em{%
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\hss\@textsuperscript{\normalfont\@thefnmark}}##1}%
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\if@twocolumn
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\ifnum \col@number=\@ne
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\@maketitle
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\else
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\twocolumn[\@maketitle]%
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\fi
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\else
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\newpage
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\global\@topnum\z@ % Prevents figures from going at top of page.
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\@maketitle
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\fi
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\thispagestyle{plain}\@thanks
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\endgroup
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\setcounter{footnote}{0}%
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\global\let\thanks\relax
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\global\let\maketitle\relax
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\global\let\@maketitle\relax
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\global\let\@thanks\@empty
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\global\let\@author\@empty
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\global\let\@ELEGANT@version\@empty
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\global\let\@date\@empty
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\global\let\@title\@empty
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\global\let\title\relax
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\global\let\author\relax
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\global\let\date\relax
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\global\let\and\relax
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\global\let\ELEGANT@version\relax
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}
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\def\@maketitle{%
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\newpage
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\null
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\vskip 2em%
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\begin{center}%
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\let \footnote \thanks
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{\LARGE\bfseries \@title \par}%
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\vskip 1.5em%
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{\large
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\lineskip .5em%
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\begin{tabular}[t]{c}%
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\@author\\[1ex]
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\end{tabular}\par}
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\the\institute%
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\vskip 0.5ex%
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\IfEmpty\version
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\else
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{\small\normalfont\citshape\versiontext\the\version}
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\fi
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\ifx\@date\empty
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\else
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\vskip 0.1em%
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{\small\normalfont\citshape\updatetext\@date}%
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\fi
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\end{center}%
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\par
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}
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\RequirePackage{listings}
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\renewcommand{\ttdefault}{cmtt}
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\lstdefinestyle{estyle}{
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basicstyle=%
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\ttfamily
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\lst@ifdisplaystyle\footnotesize\fi
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}
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\lstset{basicstyle=\scriptsize\ttfamily,style=estyle}
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\lstset{language=[LaTeX]TeX,
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texcsstyle=*\color{winered},
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numbers=none,
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breaklines=true,
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keywordstyle=\color{winered},
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frame=tlbr,framesep=4pt,framerule=0pt,
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commentstyle=\color{commentcolor},
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emph={elegantpaper,fontenc,fontspec,xeCJK,FiraMono,xunicode,newtxmath,figure,fig,image,img,table,itemize,enumerate,newtxtext,newtxtt,ctex,microtype,description,times,newtx,booktabs,tabular,PDFLaTeX,XeLaTeX,type1cm,BibTeX,cite,gbt7714,lang},
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emphstyle={\color{frenchplum}},
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morekeywords={DeclareSymbolFont,SetSymbolFont,toprule,midrule,bottomrule,institute,version,includegraphics,setmainfont,setsansfont,setmonofont ,setCJKmainfont,setCJKsansfont,setCJKmonofont,RequirePackage,figref,tabref,email,maketitle,keywords,zhdate,zhtoday},
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||||
tabsize=2,
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||||
backgroundcolor=\color{lightgrey}
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||||
}
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||||
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||||
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||||
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||||
% restore the tt default family to lmodern tt family
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||||
\renewcommand\ttdefault{lmtt}
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497
doc/main.tex
Normal file
497
doc/main.tex
Normal file
@@ -0,0 +1,497 @@
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%!TEX program = xelatex
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% 完整编译: xelatex -> biber/bibtex -> xelatex -> xelatex
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\documentclass[lang=cn,a4paper]{elegantpaper}
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\title{Pond}
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\author{作者 \\ Patrick}
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\date{\zhdate{2025/03/09}}
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||||
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||||
% 本文档命令
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||||
\usepackage{array}
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||||
\usepackage{pgfplots}
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||||
\usepackage{soul}
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||||
\usepackage{physics}
|
||||
\usepackage{amssymb}
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||||
\newcommand{\ccr}[1]{\makecell{{\color{#1}\rule{1cm}{1cm}}}}
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||||
\addbibresource[location=local]{reference.bib} % 参考文献，不要删除
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||||
\pgfplotsset{compat=1.15}
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||||
\usetikzlibrary{arrows}
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||||
\graphicspath{ {../images/} }
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||||
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||||
\begin{document}
|
||||
\vspace*{\fill}
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||||
\begin{figure}[h]
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||||
\begin{center}
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||||
\includegraphics[width=0.5\textwidth]{THE_LAST_PROOF.png}
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||||
\end{center}
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||||
\end{figure}
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||||
\begin{center}
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||||
\sffamily {
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||||
\textbf{\LARGE{Silence TLP}}
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||||
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||||
\Large{2025 / 03 / 09}
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||||
}
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||||
\end{center}
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||||
\vspace*{\fill}
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||||
\thispagestyle{empty}
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||||
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||||
DOI: 11.4514/sil.tlp.2025.0000001
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||||
\thispagestyle{empty}
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||||
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||||
\newpage
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||||
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||||
\setcounter{page}{1}
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||||
\maketitle
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||||
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||||
\begin{abstract}
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||||
Pond 是一个试验性项目，使用了 ElegantPaper 模板。
|
||||
\keywords{Nonsense，Bullshit}
|
||||
\end{abstract}
|
||||
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||||
\section{题目}
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||||
如图1，在$\triangle ABC$中，$BD$平分$\angle ABC$，$AD \bot BD$，$E$为$BC$中点，$EF$交射线$CA$于$F$，交$AB$于$G$，$GB = GE$，$DE = m$，$DF = n$.
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||||
\begin{enumerate}
|
||||
\item 在图1中找到与$\angle BAC$相等的角，并证明；
|
||||
|
||||
\item 求$BD$的长（用$m$，$n$表示）；
|
||||
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||||
\item 如图2，$AD$交$BC$的中垂线于$M$，连接$CM$，若$CB$平分$\angle ACM$，求$\dfrac{m}{n}$的值.
|
||||
\end{enumerate}
|
||||
|
||||
% 图1
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=4]
|
||||
\clip(-1.2,-0.1) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.15cm,yshift=0.1cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{original_pic1}
|
||||
\end{figure}
|
||||
|
||||
% 图2
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=4]
|
||||
\clip(-1.2,-0.5) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865476,0.7071067811865476)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865476,0.7071067811865476);
|
||||
\draw [line width=0.8pt] (-0.414213562373095,0.585786437626905)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.414213562373095,0.585786437626905)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (0.,-0.4142135623730955)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (0.,-0.4142135623730955);
|
||||
\draw [line width=0.8pt] (0.,0.)-- (0.,-0.4142135623730955);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.15cm,yshift=0.1cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (0, -0.414213562373095) node[xshift=0cm,yshift=-0.15cm] (M) {$M$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{original_pic2}
|
||||
\end{figure}
|
||||
|
||||
\section{解析}
|
||||
\subsection{第一小问}
|
||||
|
||||
$\angle BAC = \angle BDE$.
|
||||
|
||||
证明：如图3，以$AB$中点$H$为圆心，$HA$长度为半径作$\odot \mathrm{H}$，交$BC$于$K$，连接$HK$，$HE$，$HD$，
|
||||
|
||||
所以$HA=HB=HK=\dfrac{1}{2}AB$，
|
||||
|
||||
% 图3
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.2) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
|
||||
\draw (-0.414213562373095, 0) node[xshift=0.15cm,yshift=-0.15cm] (K) {$K$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{step_pic1}
|
||||
\end{figure}
|
||||
|
||||
$\because AD \bot BD$，$\therefore \angle ADB=\dfrac{\pi}{2}$，
|
||||
|
||||
$\because H$是$AB$中点，$\therefore HD=\dfrac{1}{2}AB$，
|
||||
|
||||
$\therefore HD=HA=HB=HK=\dfrac{1}{2}AB$，
|
||||
|
||||
$\therefore$点$D$在$\odot \mathrm{H}$上；
|
||||
|
||||
设$\angle EBD = \alpha$，
|
||||
|
||||
$\because BD$平分$\angle ABC$，
|
||||
|
||||
$\therefore \angle ABD = \angle EBD = \alpha$，
|
||||
|
||||
$\therefore \angle ABC = 2\alpha$，
|
||||
|
||||
$\because HB=HD$，
|
||||
|
||||
$\therefore \angle ABD = \angle HDB = \alpha$，
|
||||
|
||||
$\therefore \angle EBD = \angle HDB$，
|
||||
|
||||
$\therefore HD \mathop{//} BC$，
|
||||
|
||||
$\because GB=GE$，
|
||||
|
||||
$\therefore \angle ABC = \angle GEB = 2\alpha$，
|
||||
|
||||
$\because HB=HK$，
|
||||
|
||||
$\therefore \angle ABC = \angle HKB = 2\alpha$，
|
||||
|
||||
$\therefore \angle GEB = \angle HKB$，
|
||||
|
||||
$\therefore HK \mathop{//} EF$，
|
||||
|
||||
$\because HD \mathop{//} BC$，
|
||||
|
||||
$\therefore$四边形$HDEK$是平行四边形，
|
||||
|
||||
$\therefore HD=HK=EK$，
|
||||
|
||||
$\therefore \angle KHE = \angle KEH$，
|
||||
|
||||
$\because H$是$AB$的中点，$E$是$BC$的中点，
|
||||
|
||||
$\therefore HE \mathop{//} AC$，
|
||||
|
||||
$\therefore \angle KEH = \angle C$，
|
||||
|
||||
$\therefore \angle KHE = \angle KEH = \angle C$，
|
||||
|
||||
$\therefore \angle HKB = \angle KHE + \angle KEH = 2\angle C = 2\alpha$，
|
||||
|
||||
$\therefore \angle C = \alpha$，
|
||||
|
||||
$\because \angle GEB = \angle C + \angle F = \alpha + \angle F = 2\alpha$，
|
||||
|
||||
$\therefore \angle F = \alpha$，
|
||||
|
||||
$\therefore \angle C = \angle F = \alpha$；
|
||||
|
||||
$
|
||||
\begin{aligned}
|
||||
\because & \angle BDE = \pi - \angle EBD - \angle GEB = \pi - \alpha - 2\alpha = \pi - 3\alpha \\
|
||||
& \angle BAC = \pi - \angle ABC - \angle C = \pi - 2\alpha - \alpha = \pi - 3\alpha
|
||||
\end{aligned}
|
||||
$，
|
||||
|
||||
$\therefore \angle BAC = \angle BDE$. $\square$
|
||||
|
||||
\subsection{第二小问}
|
||||
如图4，连接$AK$，
|
||||
% 图4
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.2) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.41421356237309503,0.5857864376269049)-- (-0.41421356237309503,0.);
|
||||
% \draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-1.,0.);
|
||||
% \draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
|
||||
\draw (-0.414213562373095, 0) node[xshift=0.15cm,yshift=-0.15cm] (K) {$K$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{step_pic2}
|
||||
\end{figure}
|
||||
|
||||
$\because AB$是$\odot \mathrm{H}$直径，
|
||||
|
||||
$\therefore \angle AKB = \dfrac{\pi}{2}$，
|
||||
|
||||
$\because HD = HK$，
|
||||
|
||||
$\therefore$平行四边形$HDEK$是菱形，
|
||||
|
||||
$\because DE=m$，
|
||||
|
||||
$\therefore HD = EK = DE = m$，
|
||||
|
||||
$\therefore HF = HA = HB = HD = HK = m$，
|
||||
|
||||
$\therefore AB = 2m$，
|
||||
|
||||
$\because DF = n$，
|
||||
|
||||
$\therefore EF = DE + DF = m + n$，
|
||||
|
||||
$\because EB = EK + BK$，$EK = m$，
|
||||
|
||||
$\therefore EB = m + BK$，
|
||||
|
||||
$\therefore BK = n$，
|
||||
|
||||
$\therefore \cos \angle ABC = \cos 2\alpha = \dfrac{BK}{AB} = \dfrac{n}{2m}$，
|
||||
|
||||
$\therefore \cos \alpha = \cos \dfrac{2\alpha}{2} = \sqrt{\dfrac{1+\dfrac{n}{2m}}{2}} = \sqrt{\dfrac{2m+n}{4m}}$，
|
||||
|
||||
$\therefore BD = AB \times \cos \angle ABD = AB \times \cos\alpha = 2m \times \sqrt{\dfrac{2m+n}{4m}} = \sqrt{2m^2 + mn}$. $\square$
|
||||
|
||||
\subsection{第三乐章}
|
||||
% 图5
|
||||
如图5，记$AM$交$BC$于R，
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.5) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.41421356237309503,0.5857864376269049)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt] (-0.2928932188134525,0.2928932188134525)-- (0.,-0.414213562373095);
|
||||
\draw [line width=0.8 pt] (0.,-0.414213562373095)-- (1.,0.);
|
||||
\draw [line width=0.8 pt] (0.,0.)-- (0.,-0.414213562373095);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
|
||||
\draw (-0.414213562373095, 0) node[xshift=0cm,yshift=-0.2cm] (K) {$K$};
|
||||
\draw (0, -0.414213562373095) node[xshift=0cm,yshift=-0.15cm] (M) {$M$};
|
||||
\draw (-0.17157287525381, 0) node[xshift=-0.2cm,yshift=-0.2cm] (R) {$R$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{step_pic3}
|
||||
\end{figure}
|
||||
|
||||
$\because BD \bot AR$，
|
||||
|
||||
$\therefore \angle ADB = \angle RDB = \dfrac{\pi}{2}$，
|
||||
|
||||
在$\mathrm{Rt}\triangle ADB$与$\mathrm{Rt}\triangle RDB$中
|
||||
|
||||
$$
|
||||
\begin{cases}
|
||||
\angle ADB = \angle RDB \\
|
||||
BD = BD \\
|
||||
\angle ABD = \angle EBD
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
$\therefore \triangle ADB \cong \triangle RDB (ASA)$
|
||||
|
||||
$\therefore AB = RB = 2m$，
|
||||
|
||||
$\because EB = BK + EK = m + n$，
|
||||
|
||||
$\because E$是$BC$的中点，
|
||||
|
||||
$\therefore BC = 2EB = 2m + 2n$，
|
||||
|
||||
$\therefore RC = BC - RB = 2m + 2n - 2m = 2n$，$EC = BC - EB = 2m + 2n - (m + n) = m + n$，
|
||||
|
||||
$\because EM$是$BC$的中垂线，
|
||||
|
||||
$\therefore EM \bot BC$，
|
||||
|
||||
$\therefore \angle MEC = \dfrac{\pi}{2}$，
|
||||
|
||||
$\because CB$平分$\angle ACM$，
|
||||
|
||||
$\therefore \angle ACB = \angle ECM = \alpha$，
|
||||
|
||||
$\therefore CM = EC \times \sec\alpha = RC \times \cos\alpha$，
|
||||
|
||||
$\because \cos\alpha = \sqrt{\dfrac{2m+n}{4m}}$
|
||||
|
||||
$\therefore (m + n)\sec\alpha = 2n\sqrt{\dfrac{2m+n}{4m}}$，
|
||||
|
||||
整理得
|
||||
\[
|
||||
\qty(\dfrac{4m^2}{2n^2})=1
|
||||
\]
|
||||
\[
|
||||
\dfrac{m^2}{n^2}=\dfrac{1}{2}
|
||||
\]
|
||||
\[
|
||||
\dfrac{m}{n}=\pm\dfrac{\sqrt{2}}{2}
|
||||
\]
|
||||
|
||||
因为$m>0$，$n>0$，所以
|
||||
$$
|
||||
\dfrac{m}{n}=\dfrac{\sqrt{2}}{2}
|
||||
$$
|
||||
|
||||
$\square$
|
||||
|
||||
\subsection{点\texorpdfstring{$F$}{}在\texorpdfstring{$\odot \mathrm{H}$}{}上的证明}
|
||||
|
||||
如图6，连接$HF$，$FB$，
|
||||
% 图6
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.2) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.41421356237309503,0.5857864376269049)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
|
||||
\draw (-0.414213562373095, 0) node[xshift=0.15cm,yshift=-0.15cm] (K) {$K$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{step_pic4}
|
||||
\end{figure}
|
||||
|
||||
设$\angle ABF = \beta$，
|
||||
|
||||
$\because E$是$BC$的中点，
|
||||
|
||||
$\therefore EB = EC$，
|
||||
|
||||
$\because \angle C = \angle EFC = \alpha$，
|
||||
|
||||
$\therefore EF = EC$，
|
||||
|
||||
$\therefore EF = EB$，
|
||||
|
||||
$\therefore \angle BFE = \angle EBF$，
|
||||
|
||||
$\because \angle EBF = \angle ABC + \angle ABF = 2\alpha + \beta$，
|
||||
|
||||
$\therefore \angle BFE = 2\alpha + \beta$，
|
||||
|
||||
$\therefore \angle BFC = \angle BFE + \angle EFC = 2\alpha + \beta + \alpha = 3\alpha + \beta$，
|
||||
|
||||
$\because \angle EBF + \angle C = \angle ABF + \angle ABC + \angle C = \beta + 2\alpha + \alpha = 3\alpha + \beta$，
|
||||
|
||||
$\therefore \angle BFC = \angle EBF + \angle C$，
|
||||
|
||||
$\because \angle BFC + \angle EBF + \angle C = \pi$，
|
||||
|
||||
$\therefore 2\angle BFC = \pi$，
|
||||
|
||||
$\therefore \angle BFC = \dfrac{\pi}{2}$，
|
||||
|
||||
$\because H$是$AB$中点，$\therefore HF=\dfrac{1}{2}AB$，
|
||||
|
||||
$\therefore HF = HA = HB = HD = HK = \dfrac{1}{2}AB$，
|
||||
|
||||
$\therefore$点$F$在$\odot \mathrm{H}$上. $\square$
|
||||
|
||||
\nocite{*}
|
||||
\printbibliography[heading=bibintoc, title=\ebibname]
|
||||
|
||||
\appendix
|
||||
%\appendixpage
|
||||
\addappheadtotoc
|
||||
|
||||
\end{document}
|
||||
0
doc/reference.bib
Normal file
0
doc/reference.bib
Normal file
Reference in New Issue
Block a user