commit 93a2b5164e458fce01af2d48aec899eca0687972 Author: Patrick Date: Mon Sep 29 00:22:13 2025 +0800 First commit diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..5a99e2a --- /dev/null +++ b/.gitignore @@ -0,0 +1,251 @@ +## Core latex/pdflatex auxiliary files: +*.aux +*.lof +*.log +*.lot +*.fls +*.out +*.toc +*.fmt +*.fot +*.cb +*.cb2 +.*.lb +.DS_Store +clean.bat +elegant*-cn.pdf +elegant*-en.pdf +*.dates + +## Intermediate documents: +*.dvi +*.xdv +*-converted-to.* +# these rules might exclude image files for figures etc. +# *.ps +# *.eps +# *.pdf + +## Generated if empty string is given at "Please type another file name for output:" +.pdf + +## Bibliography auxiliary files (bibtex/biblatex/biber): +*.bbl +*.bcf +*.blg +*-blx.aux +*-blx.bib +*.run.xml + +## Build tool auxiliary files: +*.fdb_latexmk +*.synctex +*.synctex(busy) +*.synctex.gz +*.synctex.gz(busy) +*.pdfsync + +## Auxiliary and intermediate files from other packages: +# algorithms +*.alg +*.loa + +# achemso +acs-*.bib + +# amsthm +*.thm + +# beamer +*.nav +*.pre +*.snm +*.vrb + +# changes +*.soc + +# cprotect +*.cpt + +# elsarticle (documentclass of Elsevier journals) +*.spl + +# endnotes +*.ent + +# fixme +*.lox + +# feynmf/feynmp +*.mf +*.mp +*.t[1-9] +*.t[1-9][0-9] +*.tfm + +#(r)(e)ledmac/(r)(e)ledpar +*.end +*.?end +*.[1-9] +*.[1-9][0-9] +*.[1-9][0-9][0-9] +*.[1-9]R +*.[1-9][0-9]R +*.[1-9][0-9][0-9]R +*.eledsec[1-9] +*.eledsec[1-9]R +*.eledsec[1-9][0-9] +*.eledsec[1-9][0-9]R +*.eledsec[1-9][0-9][0-9] +*.eledsec[1-9][0-9][0-9]R + +# glossaries +*.acn +*.acr +*.glg +*.glo +*.gls +*.glsdefs + +# gnuplottex +*-gnuplottex-* + +# gregoriotex +*.gaux +*.gtex + +# htlatex +*.4ct +*.4tc +*.idv +*.lg +*.trc +*.xref + +# hyperref +*.brf + +# knitr +*-concordance.tex +# TODO Comment the next line if you want to keep your tikz graphics files +*.tikz +*-tikzDictionary + +# listings +*.lol + +# makeidx +*.idx +*.ilg +*.ind +*.ist + +# minitoc +*.maf +*.mlf +*.mlt +*.mtc[0-9]* +*.slf[0-9]* +*.slt[0-9]* +*.stc[0-9]* + +# minted +_minted* +*.pyg + +# morewrites +*.mw + +# nomencl +*.nlg +*.nlo +*.nls + +# pax +*.pax + +# pdfpcnotes +*.pdfpc + +# sagetex +*.sagetex.sage +*.sagetex.py +*.sagetex.scmd + +# scrwfile +*.wrt + +# sympy +*.sout +*.sympy +sympy-plots-for-*.tex/ + +# pdfcomment +*.upa +*.upb + +# pythontex +*.pytxcode +pythontex-files-*/ + +# thmtools +*.loe + +# TikZ & PGF +*.dpth +*.md5 +*.auxlock + +# todonotes +*.tdo + +# easy-todo +*.lod + +# xmpincl +*.xmpi + +# xindy +*.xdy + +# xypic precompiled matrices +*.xyc + +# endfloat +*.ttt +*.fff + +# Latexian +TSWLatexianTemp* + +## Editors: +# WinEdt +*.bak +*.sav + +# Texpad +.texpadtmp + +# Kile +*.backup + +# KBibTeX +*~[0-9]* + +# auto folder when using emacs and auctex +./auto/* +*.el + +# expex forward references with \gathertags +*-tags.tex + +# standalone packages +*.sta + +# generated if using elsarticle.cls +*.spl + +# build +./build +./output diff --git a/License b/License new file mode 100644 index 0000000..2244313 --- /dev/null +++ b/License @@ -0,0 +1,416 @@ +The LaTeX Project Public License +=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- + +LPPL Version 1.3c 2008-05-04 + +Copyright 1999 2002-2008 LaTeX3 Project + Everyone is allowed to distribute verbatim copies of this + license document, but modification of it is not allowed. + + +PREAMBLE +======== + +The LaTeX Project Public License (LPPL) is the primary license under +which the LaTeX kernel and the base LaTeX packages are distributed. + +You may use this license for any work of which you hold the copyright +and which you wish to distribute. 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You may use the text +of this license as a model for your own license, but your license +should not refer to the LPPL or otherwise give the impression that +your work is distributed under the LPPL. + +The document `modguide.tex' in the base LaTeX distribution explains +the motivation behind the conditions of this license. It explains, +for example, why distributing LaTeX under the GNU General Public +License (GPL) was considered inappropriate. Even if your work is +unrelated to LaTeX, the discussion in `modguide.tex' may still be +relevant, and authors intending to distribute their works under any +license are encouraged to read it. + +A Recommendation on Modification Without Distribution +----------------------------------------------------- + +It is wise never to modify a component of the Work, even for your own +personal use, without also meeting the above conditions for +distributing the modified component. While you might intend that such +modifications will never be distributed, often this will happen by +accident -- you may forget that you have modified that component; or +it may not occur to you when allowing others to access the modified +version that you are thus distributing it and violating the conditions +of this license in ways that could have legal implications and, worse, +cause problems for the community. It is therefore usually in your +best interest to keep your copy of the Work identical with the public +one. Many works provide ways to control the behavior of that work +without altering any of its licensed components. + +How to Use This License +----------------------- + +To use this license, place in each of the components of your work both +an explicit copyright notice including your name and the year the work +was authored and/or last substantially modified. Include also a +statement that the distribution and/or modification of that +component is constrained by the conditions in this license. + +Here is an example of such a notice and statement: + + %% pig.dtx + %% Copyright 2005 M. Y. Name + % + % This work may be distributed and/or modified under the + % conditions of the LaTeX Project Public License, either version 1.3 + % of this license or (at your option) any later version. + % The latest version of this license is in + % http://www.latex-project.org/lppl.txt + % and version 1.3 or later is part of all distributions of LaTeX + % version 2005/12/01 or later. + % + % This work has the LPPL maintenance status `maintained'. + % + % The Current Maintainer of this work is M. Y. Name. + % + % This work consists of the files pig.dtx and pig.ins + % and the derived file pig.sty. + +Given such a notice and statement in a file, the conditions +given in this license document would apply, with the `Work' referring +to the three files `pig.dtx', `pig.ins', and `pig.sty' (the last being +generated from `pig.dtx' using `pig.ins'), the `Base Interpreter' +referring to any `LaTeX-Format', and both `Copyright Holder' and +`Current Maintainer' referring to the person `M. Y. Name'. + +If you do not want the Maintenance section of LPPL to apply to your +Work, change `maintained' above into `author-maintained'. +However, we recommend that you use `maintained', as the Maintenance +section was added in order to ensure that your Work remains useful to +the community even when you can no longer maintain and support it +yourself. + +Derived Works That Are Not Replacements +--------------------------------------- + +Several clauses of the LPPL specify means to provide reliability and +stability for the user community. They therefore concern themselves +with the case that a Derived Work is intended to be used as a +(compatible or incompatible) replacement of the original Work. If +this is not the case (e.g., if a few lines of code are reused for a +completely different task), then clauses 6b and 6d shall not apply. + + +Important Recommendations +------------------------- + + Defining What Constitutes the Work + + The LPPL requires that distributions of the Work contain all the + files of the Work. It is therefore important that you provide a + way for the licensee to determine which files constitute the Work. + This could, for example, be achieved by explicitly listing all the + files of the Work near the copyright notice of each file or by + using a line such as: + + % This work consists of all files listed in manifest.txt. + + in that place. In the absence of an unequivocal list it might be + impossible for the licensee to determine what is considered by you + to comprise the Work and, in such a case, the licensee would be + entitled to make reasonable conjectures as to which files comprise + the Work. + diff --git a/README.md b/README.md new file mode 100644 index 0000000..3705c20 --- /dev/null +++ b/README.md @@ -0,0 +1,10 @@ +# Pond +Pond is an experimental project that uses the ElegantPaper template. + +Pond 是一个试验性项目，使用了 ElegantPaper 模板。 + +## License + +This project is released under the LaTeX Project Public License, v1.3c or later. + +本项目发布遵循 LaTeX 项目公共许可证 1.3c 或更高版本。 diff --git a/doc/elegantpaper.cls b/doc/elegantpaper.cls new file mode 100644 index 0000000..c5c64d6 --- /dev/null +++ b/doc/elegantpaper.cls @@ -0,0 +1,372 @@ +% Author: Dongsheng Deng +% Homepage: https://ddswhu.me/ +% Email: elegantlatex2e@gmail.com +% Lastest Version: https://github.com/ElegantLaTeX/ElegantPaper + +% --- Class structure: identification part +\ProvidesClass{elegantpaper}[2022/12/31 v0.11 ElegantLaTeX Paper class] +\NeedsTeXFormat{LaTeX2e} + +%%% +\RequirePackage{kvoptions} +\RequirePackage{etoolbox} +\RequirePackage{calc} + +\SetupKeyvalOptions{family=ELEGANT, prefix=ELEGANT@, setkeys=\kvsetkeys} +\newcommand{\ekv}[1]{\kvsetkeys{ELEGANT}{#1}} + +\DeclareStringOption[en]{lang} +\DeclareVoidOption{cn}{\ekv{lang=cn}} +\DeclareVoidOption{en}{\ekv{lang=en}} + +\DeclareStringOption[cm]{math} + +\DeclareStringOption[numeric-comp]{citestyle} +\DeclareStringOption[numeric]{bibstyle} + +\DeclareStringOption[biber]{bibend} +\DeclareVoidOption{biber}{\ekv{bibend=biber}} +\DeclareVoidOption{bibtex}{\ekv{bibend=bibtex}} + + +\DeclareStringOption[ctexfont]{chinesefont} +\DeclareVoidOption{ctexfont}{\ekv{chinesefont=ctexfont}} +\DeclareVoidOption{founder}{\ekv{chinesefont=founder}} +\DeclareVoidOption{nofont}{\ekv{chinesefont=nofont}} + +\DeclareVoidOption{newtx}{\ekv{math=newtx}} +\DeclareVoidOption{mtpro2}{\ekv{math=mtpro2}} +\DeclareVoidOption{cm}{\ekv{math=cm}} + +\DeclareDefaultOption{\PassOptionsToClass{\CurrentOption}{article}} +\ProcessKeyvalOptions*\relax +\LoadClass{article} + +\RequirePackage{hyperref} +\hypersetup{ + pdfborder={0 0 0}, + colorlinks=true, + linkcolor={winered}, + urlcolor={winered}, + filecolor={winered}, + citecolor={winered}, + linktoc=all, +} +% settings for the hyperref and geometry +\RequirePackage[ + left=1in, + right=1in, + top=1in, + bottom=1in, + headheight=0pt, + headsep=0pt]{geometry} +\setlength{\headsep}{5pt} + +\RequirePackage{amsthm} +\RequirePackage{amsmath} +\RequirePackage{amssymb} +\RequirePackage{indentfirst} +\RequirePackage{booktabs} +\RequirePackage{multicol} +\RequirePackage{multirow} +% \RequirePackage{linegoal} +\RequirePackage{xcolor} +\RequirePackage{graphicx} +\RequirePackage{fancyvrb} +\RequirePackage{abstract} +\RequirePackage{hologo} + +\linespread{1.35} +\graphicspath{{image/}{figure/}{fig/}{img/}} + +% caption settings +\RequirePackage[labelfont={bf}]{caption} +\captionsetup[table]{skip=3pt} +\captionsetup[figure]{skip=3pt} + +% list/itemize/enumerate setting +\RequirePackage[shortlabels,inline]{enumitem} +\setlist{nolistsep} + + +% define the hyperref color +\RequirePackage{xcolor} +\definecolor{winered}{rgb}{0.5,0,0} + +\definecolor{lightgrey}{rgb}{0.95,0.95,0.95} +\definecolor{commentcolor}{RGB}{0,100,0} +\definecolor{frenchplum}{RGB}{190,20,83} + + +% add the email cmd +\newcommand\email[1]{\href{mailto:#1}{\nolinkurl{#1}}} + +% font settings + + +\PassOptionsToPackage{no-math}{fontspec} +\PassOptionsToPackage{quiet}{fontspec} +\RequirePackage{iftex} + +\ifXeTeX + \RequirePackage[no-math]{fontspec} + \setmainfont{texgyretermes}[ + UprightFont = *-regular , + BoldFont = *-bold , + ItalicFont = *-italic , + BoldItalicFont = *-bolditalic , + Extension = .otf , + Scale = 1.0] + + \setsansfont{texgyreheros}[ + UprightFont = *-regular , + BoldFont = *-bold , + ItalicFont = *-italic , + BoldItalicFont = *-bolditalic , + Extension = .otf , + Scale = 0.9] +\else + \RequirePackage{newtxtext} + \RequirePackage[scaled=.90]{helvet} +\fi + +\ifdefstring{\ELEGANT@lang}{cn}{ + \ifdefstring{\ELEGANT@chinesefont}{founder}{ + \RequirePackage[UTF8,scheme=plain,fontset=none]{ctex} + \setCJKmainfont[BoldFont={FZHei-B01},ItalicFont={FZKai-Z03}]{FZShuSong-Z01} + \setCJKsansfont[BoldFont={FZHei-B01}]{FZKai-Z03} + \setCJKmonofont[BoldFont={FZHei-B01}]{FZFangSong-Z02} + \setCJKfamilyfont{zhsong}{FZShuSong-Z01} + \setCJKfamilyfont{zhhei}{FZHei-B01} + \setCJKfamilyfont{zhkai}[BoldFont={FZHei-B01}]{FZKai-Z03} + \setCJKfamilyfont{zhfs}[BoldFont={FZHei-B01}]{FZFangSong-Z02} + \newcommand*{\songti}{\CJKfamily{zhsong}} + \newcommand*{\heiti}{\CJKfamily{zhhei}} + \newcommand*{\kaishu}{\CJKfamily{zhkai}} + \newcommand*{\fangsong}{\CJKfamily{zhfs}}}{\relax} + + \ifdefstring{\ELEGANT@chinesefont}{nofont}{ + \RequirePackage[UTF8,scheme=plain,fontset=none]{ctex}}{\relax} + + \ifdefstring{\ELEGANT@chinesefont}{ctexfont}{ + \RequirePackage[UTF8,scheme=plain]{ctex}}{\relax} + + \AfterEndPreamble{ + \setlength\parindent{2\ccwd}} +}{\relax} + + +\ifcsname kaishu\endcsname + \newcommand{\citshape}{\kaishu} +\else + \newcommand{\citshape}{\itshape} +\fi +\ifcsname kaishu\endcsname + \newcommand{\cnormal}{\kaishu} +\else + \newcommand{\cnormal}{\normalfont} +\fi + +\ifcsname fangsong\endcsname + \newcommand{\cfs}{\fangsong} +\else + \newcommand{\cfs}{\normalfont} +\fi + +\ifdefstring{\ELEGANT@math}{newtx}{ + \RequirePackage{newtxmath} + \let\Bbbk\relax + \RequirePackage{esint} + %%% use yhmath pkg, uncomment following code + % \let\oldwidering\widering + % \let\widering\undefined + % \RequirePackage{yhmath} + % \let\widering\oldwidering + + %%% use esvect pkg, uncomment following code + % \RequirePackage{esvect} + + \DeclareSymbolFont{CMlargesymbols}{OMX}{cmex}{m}{n} + \let\sumop\relax\let\prodop\relax + \DeclareMathSymbol{\sumop}{\mathop}{CMlargesymbols}{"50} + \DeclareMathSymbol{\prodop}{\mathop}{CMlargesymbols}{"51} +}{\relax} + + +\RequirePackage[ + backend=\ELEGANT@bibend, + citestyle=\ELEGANT@citestyle, + bibstyle=\ELEGANT@bibstyle, + sorting=none]{biblatex} + +\RequirePackage{appendix} + +\ifdefstring{\ELEGANT@lang}{cn}{ + \theoremstyle{plain}% default + \newtheorem{theorem}{定理}[section] % + \newtheorem{lemma}[theorem]{引理} % + \newtheorem{proposition}[theorem]{命题} % + \newtheorem*{corollary}{推论} % + \theoremstyle{definition} % + \newtheorem{definition}{定义}[section] % + \newtheorem{conjecture}{猜想}[section] % + \newtheorem{example}{例}[section] % + \theoremstyle{remark} % + \newtheorem*{remark}{\normalfont\bfseries 评论} % + \newtheorem*{note}{\normalfont\bfseries 注} % + \newtheorem{case}{\normalfont\bfseries 案例} % + \renewcommand*{\proofname}{\normalfont\bfseries 证明} % + \renewcommand\contentsname{目录} + \renewcommand\refname{参考文献} % + \renewcommand\figurename{图} % + \renewcommand\tablename{表}% + \newcommand\versiontext{版本：}% + 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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}, + emphstyle={\color{frenchplum}}, + morekeywords={DeclareSymbolFont,SetSymbolFont,toprule,midrule,bottomrule,institute,version,includegraphics,setmainfont,setsansfont,setmonofont ,setCJKmainfont,setCJKsansfont,setCJKmonofont,RequirePackage,figref,tabref,email,maketitle,keywords,zhdate,zhtoday}, + tabsize=2, + backgroundcolor=\color{lightgrey} +} + + + +% restore the tt default family to lmodern tt family +\renewcommand\ttdefault{lmtt} diff --git a/doc/main.tex b/doc/main.tex new file mode 100644 index 0000000..23493c7 --- /dev/null +++ b/doc/main.tex @@ -0,0 +1,497 @@ +%!TEX program = xelatex +% 完整编译: xelatex -> biber/bibtex -> xelatex -> xelatex +\documentclass[lang=cn,a4paper]{elegantpaper} + +\title{Pond} +\author{作者 \\ Patrick} +\date{\zhdate{2025/03/09}} + +% 本文档命令 +\usepackage{array} +\usepackage{pgfplots} +\usepackage{soul} +\usepackage{physics} +\usepackage{amssymb} +\newcommand{\ccr}[1]{\makecell{{\color{#1}\rule{1cm}{1cm}}}} +\addbibresource[location=local]{reference.bib} % 参考文献，不要删除 +\pgfplotsset{compat=1.15} +\usetikzlibrary{arrows} +\graphicspath{ {../images/} } + +\begin{document} +\vspace*{\fill} +\begin{figure}[h] + \begin{center} + \includegraphics[width=0.5\textwidth]{THE_LAST_PROOF.png} + \end{center} +\end{figure} +\begin{center} + \sffamily { + \textbf{\LARGE{Silence TLP}} + + \Large{2025 / 03 / 09} + } +\end{center} +\vspace*{\fill} +\thispagestyle{empty} + +DOI: 11.4514/sil.tlp.2025.0000001 +\thispagestyle{empty} + +\newpage + +\setcounter{page}{1} +\maketitle + +\begin{abstract} + Pond 是一个试验性项目，使用了 ElegantPaper 模板。 + \keywords{Nonsense，Bullshit} +\end{abstract} + +\section{题目} +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$G$，$GB = GE$，$DE = m$，$DF = n$. +\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）； + + \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} diff --git a/doc/reference.bib b/doc/reference.bib new file mode 100644 index 0000000..e69de29 diff --git a/geogebra/original_pic1.ggb b/geogebra/original_pic1.ggb new file mode 100644 index 0000000..55a3dd6 Binary files /dev/null and b/geogebra/original_pic1.ggb differ diff --git a/geogebra/original_pic2.ggb b/geogebra/original_pic2.ggb new file 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