├── TODO.tex ├── cover ├── P22-Underground-Reg.ttf ├── P22UndergroundCYBookSC.ttf └── cover.tex ├── LICENSE.md ├── .gitignore ├── title.tex ├── README.md ├── all-lectures.tex ├── lectures.sty ├── library.bib ├── list-of-files.txt ├── intro.tex ├── similar.tex ├── cong.tex ├── axioms.tex ├── triangle.tex ├── klein.tex ├── half-planes.tex ├── affine.tex ├── perp.tex ├── sphere.tex ├── absolute.tex └── inscribed-angle.tex /TODO.tex: -------------------------------------------------------------------------------- 1 | -------------------------------------------------------------------------------- /cover/P22-Underground-Reg.ttf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/anton-petrunin/birkhoff/HEAD/cover/P22-Underground-Reg.ttf -------------------------------------------------------------------------------- /cover/P22UndergroundCYBookSC.ttf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/anton-petrunin/birkhoff/HEAD/cover/P22UndergroundCYBookSC.ttf -------------------------------------------------------------------------------- /LICENSE.md: -------------------------------------------------------------------------------- 1 | This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International License. 2 | To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/4.0/. 3 | -------------------------------------------------------------------------------- /.gitignore: -------------------------------------------------------------------------------- 1 | *.pdf 2 | .kile/* 3 | lib/* 4 | www/* 5 | *.backup 6 | *.log 7 | *.aux 8 | *.dvi 9 | *.gz 10 | *.idx 11 | *.ind 12 | *.out 13 | *.ilg 14 | *.toc 15 | *.patch 16 | *.kilepr 17 | */*.zip 18 | *.mps 19 | *.mpx 20 | 21 | -------------------------------------------------------------------------------- /title.tex: -------------------------------------------------------------------------------- 1 | \title{\tt Euclidean plane and its relatives} 2 | \subtitle{\tt A minimalist introduction} 3 | \author{\tt Anton Petrunin} 4 | \date{} 5 | \maketitle 6 | 7 | 8 | 9 | %The book is available at \texttt{arXiv:1302.1630}. The version \texttt{v12} is the current second printing and \texttt{v11} is the first printing. The numbering of some statements has been changed. 10 | 11 | 12 | \null\vfill\noindent{\includegraphics[scale=0.5]{pics/by-sa} 13 | \vspace*{1mm} 14 | \\ 15 | \hbox{\parbox{.7\textwidth} 16 | {This work is licensed under the Creative Commons Attribution-ShareAlike 17 | 4.0 International License. 18 | To view a copy of this license, visit\\ 19 | \texttt{http://creativecommons.org/licenses/by-sa/4.0/}}} 20 | 21 | %\ 22 | 23 | %\noindent{\tt 24 | %http://www.createspace.com/4122116 25 | %\\ 26 | %\\ 27 | %ISBN-13: 978-1481918473 28 | %\\ 29 | %ISBN-10: 1481918478 30 | %} 31 | 32 | 33 | \thispagestyle{empty} 34 | \newpage 35 | { 36 | \cleardoublepage 37 | \phantomsection 38 | \pdfbookmark[0]{\contentsname}{bm:toc} 39 | \sloppy 40 | \tableofcontents 41 | } 42 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Euclidean plane and its relatives 2 | ## A minimalistic introduction 3 | 4 | * http://anton-petrunin.github.io/birkhoff/ 5 | 6 | The book is designed for a semester-long course in [Foundations of Geometry](http://en.wikipedia.org/wiki/Foundations_of_geometry) and meant to be rigorous, conservative, elementary and minimalist. 7 | 8 | This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International License. 9 | To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/4.0/. 10 | 11 | ## How to build 12 | 13 | The following command will create a local copy of the source code for you. 14 | 15 | `git clone https://github.com/anton-petrunin/birkhoff.git` 16 | 17 | Go to the crated folder, and run `mpost`, `pdflatex` and `makeindex` few times; it worked with full installation of TeX Live on Kubuntu 20.04. 18 | 19 | `cd birkhoff/mpics/`
20 | `mpost pic.mp`
21 | `mpost pic-hints.mp`
22 | `cd ..`
23 | `pdflatex all-lectures.tex`
24 | `makeindex all-lectures`
25 | `pdflatex all-lectures.tex`
26 | `biber all-lectures`
27 | `makeindex all-lectures`
28 | `pdflatex all-lectures.tex`
29 | 30 | the following command will extract all needed files in arXiv.tar: 31 | 32 | `tar -cvf arXiv.tar --files-from list-of-files.txt` 33 | 34 | The following command produces PDF version 1.3 from eps-file 35 | 36 | `epstopdf --gsopt=-dCompatibilityLevel=1.3 foo.eps` 37 | 38 | -------------------------------------------------------------------------------- /all-lectures.tex: -------------------------------------------------------------------------------- 1 | \documentclass[twoside]{book} 2 | \usepackage{geometry} 3 | \usepackage{lectures} 4 | \hypersetup{ 5 | breaklinks=true, 6 | pdftitle={Euclidean plane and its relatives; a minimalist introduction.} 7 | } 8 | 9 | \newcommand{\arxiv}[2]{#1} %for arXiv 10 | %\newcommand{\arxiv}[2]{#2} %for kdp 11 | \arxiv{}{\pdfminorversion=3} 12 | 13 | \newcommand{\spell}[2]{#1} %for book 14 | %\newcommand{\spell}[2]{#2} %for spelling 15 | 16 | \geometry{top=0.9in, bottom=0.9in,inner=0.9in, outer=0.7in, paperwidth=6in, paperheight=9in} 17 | %\geometry{top=1.025in, bottom=1.025in,inner=0.9in, outer=0.825in, paperwidth=6.125in, paperheight=9.25in}%with bleed 18 | 19 | \makeindex 20 | %\overfullrule=100mm 21 | 22 | \spell{}{\makeatletter\usepackage{environ}\RenewEnviron{figure}[1][]{}\RenewEnviron{figure*}[1][]{}\RenewEnviron{wrapfigure}[1][]{}\RenewEnviron{Figure}[1][]{}\pagestyle{empty}\let\ps@plain\ps@empty\makeatother\usepackage[none]{hyphenat}\def\emph{\textit}\renewcommand{\footnote}[1]{\textup{\ (#1)}}\renewcommand{\frac}[2]{(#1)/(#2)}\renewcommand{\tfrac}[2]{(#1/#2)}\everydisplay{\textstyle}} 23 | 24 | 25 | \begin{document} 26 | 27 | \input{title.tex} 28 | \input{intro.tex} 29 | 30 | \input{metric.tex} 31 | \input{axioms.tex} 32 | \input{half-planes.tex} 33 | \input{cong.tex} 34 | \input{perp.tex} 35 | \input{similar.tex} 36 | \input{parallel.tex} 37 | \input{triangle.tex} 38 | 39 | \input{inscribed-angle.tex} 40 | \input{inversion.tex} 41 | 42 | \input{absolute.tex} 43 | \input{poincare.tex} 44 | \input{h-plane.tex} 45 | 46 | \input{affine.tex} 47 | \input{proj.tex} 48 | 49 | \input{sphere.tex} 50 | \input{klein.tex} 51 | \input{complex.tex} 52 | \input{car.tex} 53 | \input{area.tex} 54 | 55 | \newgeometry{top=0.9in, bottom=0.9in,inner=0.55in, outer=0.45in} 56 | %\newgeometry{top=1.025in, bottom=1.025in,inner=0.5in, outer=0.625in, paperwidth=6.125in, paperheight=9.25in}%with bleed 57 | {\footnotesize 58 | 59 | \input{hints.tex} 60 | 61 | } 62 | \newpage 63 | \phantomsection 64 | {\scriptsize 65 | \input{all-lectures.ind} 66 | } 67 | \renewcommand{\bibname}{Used resources} 68 | { 69 | 70 | \def\emph{\textit} 71 | 72 | \printbibliography[heading=bibintoc] 73 | \fussy 74 | } 75 | \end{document} 76 | -------------------------------------------------------------------------------- /cover/cover.tex: -------------------------------------------------------------------------------- 1 | %Lualatex+needs P22UndergroundCYBookSC.tt and P22-Underground-Reg.ttf 2 | \documentclass[ 3 | coverheight=9in, 4 | coverwidth=6in, 5 | spinewidth=0.53in, 6 | bleedwidth=.125in, 7 | marklength=0in, 8 | markcolor=black]{bookcover} 9 | 10 | \usepackage{xcolor} 11 | \usepackage{background} 12 | \usepackage{blindtext} 13 | \usepackage{fontspec} 14 | 15 | \newbookcoverpart{pfront}{ 16 | \setpartposx{\marklength+\bleedwidth+\spinewidth+\coverwidth+10mm} 17 | \setpartposy{\marklength+\bleedwidth+10mm} 18 | \setpartheight{\coverheight-20mm} 19 | \setpartwidth{\coverwidth-20mm} 20 | \settrimmedpart{0mm}{0mm}{0pt}{0pt} 21 | } 22 | 23 | \newbookcoverpart{pback}{ 24 | \setpartposx{\marklength+\bleedwidth+10mm} 25 | \setpartposy{\marklength+\bleedwidth+10mm}\setpartheight{\coverheight-20mm} 26 | \setpartwidth{\coverwidth-20mm} 27 | \settrimmedpart{0mm}{0mm}{0pt}{0pt} 28 | } 29 | 30 | \begin{document} 31 | 32 | \begin{bookcover} 33 | 34 | \definecolor{mycolor}{HTML}{41758A}%{92A42C}%{5E4BA4}%{3A73D5}%{250,205,25} 35 | \bookcovercomponent{color}{bg whole}{mycolor} 36 | %\bookcovercomponent{center}{bg whole}{\includegraphics{fon}} 37 | 38 | \setmainfont{P22-Underground-Reg} 39 | 40 | \bookcovercomponent{center}{spine} 41 | {\rotatebox[origin=c]{-90}{ 42 | \color{white} 43 | \fontsize{18}{25}\selectfont\fontdimen2\font=1ex 44 | EUCLIDEAN PLANE AND ITS RELATIVES 45 | \hskip30mm 46 | ANTON PETRUNIN 47 | }} 48 | 49 | 50 | \bookcovercomponent{normal}{pfront}{ 51 | \begin{flushright} 52 | {\color{white}\fontsize{42}{48}\selectfont 53 | Euclidean plane 54 | \\[3mm] 55 | } 56 | {\color{white}\fontsize{41.3}{48}\selectfont 57 | and its relatives} 58 | \\[5mm] 59 | \setmainfont{P22UndergroundCYBookSC} 60 | {\color{white}\fontsize{21.5}{48}\selectfont 61 | a minimalist introduction\hskip.2mm 62 | } 63 | \\[4mm] 64 | {\color{white}\fontsize{15}{48}\selectfont 65 | Anton Petrunin\hskip.4mm 66 | } 67 | \\[120mm] 68 | \includegraphics{../mppics/pic-351} 69 | \end{flushright} 70 | } 71 | 72 | \setmainfont{P22UndergroundCYBookSC} 73 | 74 | \bookcovercomponent{normal}{pfront}{ 75 | \begin{flushleft} 76 | \rotatebox[origin=c]{-90} 77 | { 78 | \includegraphics{../mppics/pic-187} 79 | } 80 | \vfill 81 | {\color{white}\fontsize{10}{48}\selectfont 82 | third edition\\ 83 | tenth printing} 84 | \end{flushleft} 85 | 86 | } 87 | 88 | \setmainfont{P22-Underground-Reg} 89 | 90 | \bookcovercomponent{normal}{pback}{ 91 | \begin{flushleft} 92 | \parbox{.35\textwidth}{ 93 | \color{white}\fontsize{10}{14}\selectfont 94 | The book is designed for a semester-long course in Foundations of geometry and meant to be rigorous, conservative, elementary, and minimalist. 95 | \medskip 96 | \\ 97 | Introduction\\ 98 | \hbox{\ \phantom{1}}1. Preliminaries 99 | \medskip 100 | \\ 101 | Euclidean geometry\\ 102 | \hbox{\ \phantom{1}}2. The Axioms\\ 103 | \hbox{\ \phantom{1}}3. Half-planes\\ 104 | \hbox{\ \phantom{1}}4. Congruent triangles\\ 105 | \hbox{\ \phantom{1}}5. Perpendicular lines\\ 106 | \hbox{\ \phantom{1}}6. Similar triangles\\ 107 | \hbox{\ \phantom{1}}7. Parallel lines\\ 108 | \hbox{\ \phantom{1}}8. Triangle geometry 109 | \medskip 110 | \\ 111 | Inversive geometry\\ 112 | \hbox{\ \phantom{1}}9. Inscribed angles\\ 113 | \hbox{\ }10. Inversion 114 | \medskip 115 | \\ 116 | Non-Euclidean geometry\\ 117 | \hbox{\ }11. Neutral plane\\ 118 | \hbox{\ }12. Hyperbolic plane\\ 119 | \hbox{\ }13. Geometry of h-plane 120 | \medskip 121 | \\ 122 | Additional topics 123 | \\ 124 | \hbox{\ }14. Affine geometry\\ 125 | \hbox{\ }15. Projective geometry\\ 126 | \hbox{\ }16. Spherical insights\\ 127 | \hbox{\ }17. Projective model\\ 128 | \hbox{\ }18. Complex coordinates\\ 129 | \hbox{\ }19. Geometric constructions\\ 130 | \hbox{\ }20. Area} 131 | \\[10mm] 132 | \includegraphics{../mppics/pic-277} 133 | \\[3mm] 134 | \color{white}\fontsize{10}{14}\selectfont 135 | arXiv:1302.1630v25 136 | \\[1mm] 137 | anton-petrunin.github.io/birkhoff/ 138 | \end{flushleft} 139 | } 140 | 141 | \end{bookcover} 142 | 143 | \end{document} 144 | -------------------------------------------------------------------------------- /lectures.sty: -------------------------------------------------------------------------------- 1 | \usepackage{amssymb, amsfonts, amsmath, amsthm} 2 | \usepackage{mathtools} 3 | \usepackage{enumerate} 4 | \usepackage{bm} 5 | \usepackage{color} 6 | \usepackage{pifont} 7 | 8 | \usepackage[pdfauthor={Anton Petrunin}, 9 | colorlinks=true, 10 | citecolor=black, 11 | linkcolor=black, 12 | anchorcolor=black, 13 | filecolor=black, 14 | menucolor=black, 15 | urlcolor=black, 16 | unicode, 17 | linktocpage 18 | ]{hyperref} 19 | \usepackage{bookmark} 20 | 21 | \usepackage{caption} 22 | \usepackage{paracol} 23 | 24 | \usepackage[utf8]{inputenc} 25 | \usepackage[T1,T2A]{fontenc} 26 | \usepackage[russian,german,english]{babel} 27 | \DeclareUnicodeCharacter{2010}{-}% support older LaTeX versions 28 | 29 | 30 | 31 | \usepackage[ 32 | sorting=debug, 33 | bibencoding=auto, 34 | backend=biber, 35 | maxnames=10, 36 | %babel=other, 37 | autolang=other, 38 | doi=false, 39 | %url=false, 40 | style=numeric-comp, 41 | isbn=false]{biblatex} 42 | \usepackage{csquotes} 43 | \AtEveryBibitem{\clearlist{language}} 44 | \AtEveryBibitem{\clearlist{note}} 45 | \renewbibmacro{in:}{} 46 | \renewcommand*{\bibfont}{\scriptsize} 47 | \addbibresource{library.bib} 48 | 49 | \DeclareFieldFormat{postnote}{#1} 50 | \DeclareFieldFormat{multipostnote}{#1} 51 | \DeclareFieldFormat{pages}{#1} 52 | %removes p. in the cites 53 | 54 | 55 | 56 | %\usepackage{multibib} 57 | \usepackage{epigraph} 58 | \usepackage{framed} 59 | 60 | \usepackage{tikz} 61 | \usetikzlibrary{arrows} 62 | 63 | \usepackage{wrapfig} 64 | 65 | \usepackage{multicol} 66 | \newenvironment{Figure} 67 | {\par\medskip\noindent\minipage{\linewidth}} 68 | {\endminipage\par\medskip} 69 | 70 | \usepackage{titling} 71 | \newcommand{\subtitle}[1]{% 72 | \posttitle{% 73 | \par\end{center} 74 | \begin{center}\large#1\end{center} 75 | \vskip0.5em}% 76 | } 77 | 78 | 79 | \usepackage[nottoc]{tocbibind} 80 | 81 | \csname @openrightfalse\endcsname 82 | 83 | \usepackage{titletoc}% http://ctan.org/pkg/titletoc 84 | \titlecontents*{section}[.5em] 85 | {\small} 86 | {\thecontentslabel.~{}} 87 | {} 88 | { \textbf{\thecontentspage}} 89 | [; ][.] 90 | 91 | \usepackage{accsupp} 92 | \usepackage{soul} 93 | \sodef\an{}{.1em}{.5em plus.1em}{.5em plus.1em minus.1em} 94 | 95 | \newcommand\ann[1]{% 96 | \BeginAccSupp{method=escape,ActualText={\detokenize{#1}}}% 97 | \an{#1}% 98 | \EndAccSupp{}% 99 | } 100 | 101 | \def\emph{\ann} 102 | 103 | % Fix latex 104 | \def\smallskip{\vskip\smallskipamount} 105 | \def\medskip{\vskip\medskipamount} 106 | \def\bigskip{\vskip\bigskipamount} 107 | 108 | %RUSSIAN STYLE 109 | \def\emptyset{\varnothing} 110 | \def\cosh{\operatorname{ch}} 111 | \def\sinh{\operatorname{sh}} 112 | \def\tanh{\operatorname{th}} 113 | \def\tan{\operatorname{tg}} 114 | \def\limsup{\varlimsup} 115 | \def\liminf{\varliminf} 116 | 117 | %MY STYLE 118 | \def\parit#1{\medskip\noindent{\it #1}} 119 | \def\parbf#1{\medskip\noindent{\bf #1}} 120 | \def\qeds{\qed\par\medskip}%END OF PROOF 121 | \def\qedsf{\vskip-6mm\qeds}%END OF PROOF BY FORMULA 122 | 123 | %THEOREM 124 | \newcounter{thm}[chapter] 125 | \renewcommand{\thethm}{\thechapter.\arabic{thm}} 126 | \newcommand\claim[2][.]{\par\medskip\noindent\refstepcounter{thm}\hbox{\bf \arabic{chapter}.\arabic{thm}. #2#1} 127 | \it\ %\ignorespaces 128 | } 129 | \newcommand\nonumclaim[2][{}]{\par\medskip\noindent\refstepcounter{thm}\hbox{\bf #2.#1} 130 | \it\ %\ignorespaces 131 | } 132 | \def\endclaim{ 133 | \par\medskip} 134 | \newenvironment{thm}{\claim}{\endclaim} 135 | \newenvironment{thm*}{\nonumclaim}{\endclaim} 136 | 137 | \def\a{\textit{a}} 138 | %\def\abs{$\!\!{}^{{}^{\a}}$} 139 | \def\a{\checkmark} 140 | \def\abs{.$\!{}^{\a}$} 141 | 142 | \def\epsilon{\varepsilon} 143 | \def\phi{\varphi} 144 | %\def\theta{\vartheta} 145 | 146 | \def\ge{\geqslant} 147 | \def\le{\leqslant} 148 | 149 | \def\df{:=} 150 | %{\buildrel \mathrm{def} \over{=\!\!=}} 151 | 152 | \def\:{\colon} 153 | 154 | \def\defect{\mathop{\rm defect}\nolimits} 155 | \def\area{\mathop{\rm area}\nolimits} 156 | \def\circum{\mathop{\rm circum}\nolimits} 157 | \def\Re{\mathop{\rm Re}\nolimits} 158 | \def\Im{\mathop{\rm Im}\nolimits} 159 | 160 | \def\solidtriangle{\text{\ding{115}}} 161 | \def\solidsquare{\blacksquare} 162 | 163 | \newcommand*{\z}[1]{#1\nobreak\discretionary{}% 164 | {\hbox{$\mathsurround=0pt #1$}}{}} 165 | 166 | 167 | \let\oldcdot\cdot 168 | \def\cdot{{\hskip0.5pt\z\oldcdot\hskip0.5pt}} 169 | 170 | \hyphenation{% 171 | col-lin-e-ar 172 | geo-de-sic 173 | half-space 174 | Loba-chev-skian 175 | Py-thag-o-re-an 176 | re-pa-ra-met-ri-za-tion 177 | semi-pe-ri-me-ter} 178 | 179 | 180 | %LIST STILES 181 | \renewcommand{\@listI}{% 182 | \leftmargin=25pt 183 | \rightmargin=0pt 184 | \labelsep=5pt 185 | \labelwidth=20pt 186 | \itemindent=0pt 187 | \listparindent=0pt 188 | \topsep=0pt plus 1pt 189 | \partopsep=\topsep 190 | \parsep=\topsep 191 | \itemsep=\topsep} 192 | 193 | %ITEMIZE-first order sign 194 | \renewcommand{\labelitemi}{$\mathsurround=0pt \diamond$} 195 | \renewcommand{\labelitemii}{$\mathsurround=0pt \circ$} 196 | %chaptername 197 | %\renewcommand{\chaptername}{} 198 | 199 | %NUMBERONCIRC 200 | \newcommand*{\numform}[1]{\@numform{\@nameuse{c@#1}}} 201 | \newcommand*{\@numform}[1]% 202 | {\ifcase #1\or \ding{202}\or \ding{203}\or \ding{204}\or \ding{205}\or \ding{206}\or \ding{207}\or \ding{208}\or \ding{209}\or \ding{210}\or \ding{211}\else\@ctrerr \fi} 203 | 204 | %EQNUMBER 205 | \newcounter{eqtn}[chapter] 206 | \renewcommand{\@eqnnum}{\theeqtn} 207 | \renewcommand{\theeqtn}{\numform{eqtn}} 208 | \newcommand{\eqlbl}[1]{\refstepcounter{eqtn}\leqno\text{\numform{eqtn}\label{#1}}} 209 | 210 | 211 | \makeatletter\renewcommand{\p@section}{\thechapter}\makeatother 212 | \renewcommand{\thesection}{\Alph{section}} 213 | -------------------------------------------------------------------------------- /library.bib: -------------------------------------------------------------------------------- 1 | @book{akopyan, 2 | title={Geometry in figures}, 3 | author={Akopyan, A.}, 4 | year={2017}, 5 | addendum={[Translated to Bulgarian, Chinese, French, Hebrew, Polish, Russian, and Spanish.]} 6 | } 7 | 8 | @article {akopyan-fedorov, 9 | AUTHOR = {Akopyan, A. and Fedorov, R.}, 10 | TITLE = {Two circles and only a straightedge}, 11 | JOURNAL = {Proc. Amer. Math. Soc.}, 12 | FJOURNAL = {Proceedings of the American Mathematical Society}, 13 | VOLUME = {147}, 14 | YEAR = {2019}, 15 | NUMBER = {1}, 16 | PAGES = {91--102}, 17 | ISSN = {0002-9939}, 18 | MRCLASS = {51M15 (14A25 14H52 51M09)}, 19 | MRNUMBER = {3876733}, 20 | MRREVIEWER = {Victor V. Pambuccian}, 21 | DOI = {10.1090/proc/14240}, 22 | } 23 | 24 | @incollection{engeler, 25 | title={Remarks on the theory of geometrical constructions}, 26 | author={Engeler, E.}, 27 | booktitle={The syntax and semantics of infinitary languages}, 28 | pages={64--76}, 29 | year={1968} 30 | } 31 | 32 | @article {alexandrov, 33 | AUTHOR = {Aleksandrov, A. D.}, 34 | TITLE = {Minimal foundations of geometry}, 35 | JOURNAL = {Siberian Math. J.}, 36 | VOLUME = {35}, 37 | YEAR = {1994}, 38 | NUMBER = {6}, 39 | PAGES = {1057--1069}, 40 | } 41 | 42 | @book{bachmann, 43 | title={Aufbau der Geometrie aus dem Spiegelungsbegriff}, 44 | author={Bachmann, F.}, 45 | year={1959}, 46 | addendum={[Translated to Russian.]} 47 | } 48 | 49 | @online{arnold-rogness, 50 | title={M{\"o}bius transformations revealed}, 51 | author={Arnold, D. and Rogness, J.}, 52 | year={2008}, 53 | url={http://www-users.math.umn.edu/~arnold/moebius/} 54 | } 55 | 56 | @online{euclidea, 57 | title={Euclidea}, 58 | url={https://www.euclidea.xyz} 59 | } 60 | 61 | @article{beltrami, 62 | title={Teoria fondamentale degli spazii di curvatura costante}, 63 | author={Beltrami, E.}, 64 | journal={Annali. di Mat., ser II}, 65 | volume={2}, 66 | pages={232--255}, 67 | year={1868}, 68 | addendum={[Translated by J. Stillwell in \textit{Sources of Hyperbolic Geometry}, pp. 41--62 (1996).]} 69 | } 70 | 71 | @article {birkhoff, 72 | AUTHOR = {Birkhoff, G. D.}, 73 | TITLE = {A set of postulates for plane geometry, based on scale and 74 | protractor}, 75 | JOURNAL = {Ann. of Math. (2)}, 76 | FJOURNAL = {Annals of Mathematics. Second Series}, 77 | VOLUME = {33}, 78 | YEAR = {1932}, 79 | NUMBER = {2}, 80 | PAGES = {329--345}, 81 | } 82 | 83 | @book {bolyai, 84 | AUTHOR = {Bolyai, J.}, 85 | TITLE = {Appendix}, 86 | YEAR ={1832}, 87 | addendum={[Translated by Ferenc Kárteszi in \textit{Appendix. 88 | The theory of space}, (1987).]} 89 | } 90 | 91 | @book {euclid, 92 | TITLE = {Euclid's Elements} 93 | } 94 | 95 | @book{byrne, 96 | title={The first six books of the Elements of Euclid: in which coloured diagrams and symbols are used instead of letters for the greater ease of learners}, 97 | author={Byrne, O.}, 98 | year={1847}, 99 | url={https://github.com/jemmybutton/byrne-euclid/releases} 100 | } 101 | 102 | @book{hadamard, 103 | title={Le{\c{c}}ons de g{\'e}om{\'e}trie {\'e}l{\'e}mentaire: G{\'e}om{\'e}trie plane}, 104 | author={Hadamard, J.}, 105 | year={1906}, 106 | addendum={[Translated by M. Saul in \textit{Lessons in geometry. I. 107 | Plane geometry}, (2008).]} 108 | } 109 | 110 | @book{kiselev, 111 | title={Элементарная геометрия}, 112 | author={Киселёв, А. П.}, 113 | addendum={[Translated by A. Givental in \textit{Kiselev's geometry}, (2006).]} 114 | } 115 | 116 | @article{lambert, 117 | title={Theorie der parallellinien}, 118 | author={Lambert, J. H.}, 119 | journal={Leipziger Magazin f{\"u}r reine und angewandte Mathematik}, 120 | volume={1}, 121 | number={2}, 122 | pages={137--164}, 123 | year={1786} 124 | } 125 | 126 | @article{legendre, 127 | title={El{\'e}ments de g{\'e}om{\'e}trie}, 128 | author={Legendre, A.-M.}, 129 | year={1794} 130 | } 131 | 132 | @article{lobachevsky, 133 | author={Лобачевский, Н. И.}, 134 | title={О началах геометрии}, 135 | journal={Казанский вестник}, 136 | number={25---28}, 137 | year={1829---1830} 138 | } 139 | 140 | @incollection{lobachevsky-1840, 141 | title={Geometrische Untersuchungen zur Theorie der Parallellinien}, 142 | author={Lobatschewsky, N. I.}, 143 | year={1840}, 144 | addendum={[Translated by G.~B. Halsted in \textit{The Theory of Parallels}, (2015).]} 145 | } 146 | 147 | @book{prasolov, 148 | title={Задачи по планиметрии}, 149 | author={Прасолов, В. В.}, 150 | year={1986}, 151 | addendum={[Translated by D. Leites in \textit{Problems in plane and solid geometry}, (2006).]} 152 | } 153 | 154 | @book{saccheri, 155 | title={Euclides ab omni n\ae vo vindicatus}, 156 | author={Saccheri, G. G.}, 157 | year={1733}, 158 | addendum={[Translated by G. B. Halsted in \textit{Euclides vindicatus}, (1986).]}, 159 | } 160 | 161 | @book{sharygin, 162 | title={Геометрия 7--9}, 163 | author={Шарыгин, И. Ф.}, 164 | year={1997} 165 | } 166 | 167 | @book{staudt, 168 | title={Geometrie der lage}, 169 | author={von Staudt, K. G. C.}, 170 | year={1847}, 171 | } 172 | 173 | @inproceedings {tarski, 174 | AUTHOR = {Tarski, A.}, 175 | TITLE = {What is elementary geometry?}, 176 | BOOKTITLE = {The axiomatic method (edited by {L}. {H}enkin, {P}. {S}uppes and {A}. 177 | {T}arski)}, 178 | PAGES = {16--29}, 179 | YEAR = {1959}, 180 | } 181 | 182 | @book{taurinus, 183 | title={Geometriae: Prima elementa. Recensuit et novas observationes adjecit}, 184 | author={Taurinus, F. A.}, 185 | year={1826}, 186 | publisher={Bachem} 187 | } 188 | 189 | @online{zadachi, 190 | title={Задачи}, 191 | url={https://www.problems.ru/} 192 | } 193 | -------------------------------------------------------------------------------- /list-of-files.txt: -------------------------------------------------------------------------------- 1 | list-of-files.txt 2 | all-lectures.tex 3 | all-lectures.bbl 4 | lectures.sty 5 | title.tex 6 | intro.tex 7 | metric.tex 8 | axioms.tex 9 | half-planes.tex 10 | cong.tex 11 | perp.tex 12 | similar.tex 13 | parallel.tex 14 | triangle.tex 15 | inscribed-angle.tex 16 | inversion.tex 17 | absolute.tex 18 | poincare.tex 19 | h-plane.tex 20 | affine.tex 21 | proj.tex 22 | sphere.tex 23 | klein.tex 24 | complex.tex 25 | car.tex 26 | area.tex 27 | hints.tex 28 | all-lectures.ind 29 | library.bib 30 | pics/H2checkers_334.eps 31 | pics/by-sa.eps 32 | pics/by-sa-eps-converted-to.pdf 33 | pics/H2checkers_334-eps-converted-to.pdf 34 | mppics/pic-128.mps 35 | 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mppics/pic-35.mps 239 | mppics/pic-189.mps 240 | mppics/pic-225.mps 241 | mppics/pic-245.mps 242 | mppics/pic-319.mps 243 | mppics/pic-333.mps 244 | mppics/pic-337.mps 245 | mppics/pic-349.mps 246 | mppics/pic-353.mps 247 | mppics/pic-69.mps 248 | mppics/pic-915.mps 249 | mppics/pic-838.mps 250 | mppics/pic-445.mps 251 | mppics/pic-446.mps 252 | mppics/pic-447.mps 253 | mppics/pic-448.mps 254 | mppics/pic-449.mps 255 | mppics/pic-251.mps 256 | mppics/pic-1114.mps 257 | -------------------------------------------------------------------------------- /intro.tex: -------------------------------------------------------------------------------- 1 | \chapter*{Preface} 2 | \addcontentsline{toc}{chapter}{Preface} 3 | 4 | This book is meant to be 5 | rigorous, 6 | conservative, 7 | elementary, 8 | and minimalist. 9 | At the same time, it includes about the maximum that students can absorb in one semester. 10 | 11 | The present book is based 12 | on courses given by the author 13 | at Pennsylvania State University. 14 | These lectures were oriented toward sophomore and senior university students who are familiar with real numbers and continuity. 15 | This makes it possible to cover the material faster and more rigorously than in high school. 16 | Approximately one-third of the material used to be covered in high school, but not anymore. 17 | 18 | \section{Prerequisites} 19 | 20 | The students should be familiar 21 | with the following topics: 22 | \begin{itemize} 23 | \item Elementary set theory: 24 | $\in$, 25 | $\cup$, 26 | $\cap$, 27 | $\backslash$, 28 | $\subset$,~$\times$. 29 | \item Real numbers: intervals, inequalities, algebraic identities. 30 | \item Limits, continuous functions, and the intermediate value theorem. 31 | \item Standard functions: 32 | absolute value, 33 | natural logarithm, 34 | exponential function. 35 | Occasionally, trigonometric functions are used. 36 | \item Chapter~\ref{chap:trans} uses basic vector algebra. 37 | \item To read Chapter~\ref{chap:sphere}, it is better to have some previous experience with the \textit{scalar product}, also known as the \textit{dot product}. 38 | \item To read Chapter~\ref{chap:complex}, it is better to have some previous experience with complex numbers. 39 | \end{itemize} 40 | 41 | \section{Overview} 42 | 43 | We use the so-called \textit{metric approach} introduced by Birkhoff. 44 | It means that we define the Euclidean plane as a \textit{metric space} that satisfies a list of properties (\textit{axioms}). 45 | This way we minimize the tedious parts 46 | which are unavoidable in the more classical Hilbert's approach. 47 | At the same time, the students have a chance to learn the geometry of metric spaces. 48 | 49 | Here is a dependency graph of the chapters. 50 | 51 | \begin{figure}[!ht] 52 | \centering 53 | 54 | \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,scale=1.4, 55 | thick,main node/.style={circle,draw,font=\sffamily\bfseries,minimum size=8mm}] 56 | 57 | \node[main node] (1) at (-1.5,10/6) {\ref{chap:metr}}; 58 | \node[main node] (2) at (-.5,10/6){\ref{chap:axioms}}; 59 | \node[main node] (3) at (.5,10/6) {\ref{chap:half-planes}}; 60 | \node[main node] (4) at (1.5,10/6) {\ref{chap:cong}}; 61 | \node[main node] (5) at (2.5,10/6) {\ref{chap:perp}}; 62 | \node[main node] (6) at (3.5,10/6) {\ref{chap:parallel}}; 63 | \node[main node] (7) at (4.5,10/6) {\ref{chap:angle-sum}}; 64 | \node[main node] (8) at (5.5,10/6) {\ref{chap:triangle}}; 65 | \node[main node] (9) at (5,5/6){\ref{chap:inscribed-angle}}; 66 | \node[main node] (10) at (4.5,0) {\ref{chap:inversion}}; 67 | \node[main node] (11) at (3,5/6) {\ref{chap:non-euclid}}; 68 | \node[main node] (12) at (3.6,0){\ref{chap:poincare}}; 69 | \node[main node] (13) at (3,-5/6) {\ref{chap:h-plane}}; 70 | \node[main node] (14) at (5,-5/6) {\ref{chap:trans}}; 71 | \node[main node] (15) at (4.5,-10/6) {\ref{chap:proj}}; 72 | \node[main node] (16) at (4,-5/6) {\ref{chap:sphere}}; 73 | \node[main node] (17) at (3.5,-10/6) {\ref{chap:klein}}; 74 | \node[main node] (18) at (5.5,0) {\ref{chap:complex}}; 75 | \node[main node] (19) at (4,15/6) {\ref{chap:car}}; 76 | \node[main node] (20) at (5,15/6) {\ref{chap:area}}; 77 | 78 | \path[every node/.style={font=\sffamily\small}] 79 | (1) edge node{}(2) 80 | (2) edge node{}(3) 81 | (3) edge node{}(4) 82 | (4) edge node{}(5) 83 | (5) edge node{}(6) 84 | (6) edge node{}(7) 85 | (7) edge node{}(9) 86 | (7) edge node{}(8) 87 | (7) edge node{}(20) 88 | (10) edge node{}(14) 89 | (14) edge node{}(15) 90 | (7) edge node{}(19) 91 | (9) edge node{}(10) 92 | (10) edge node{}(12) 93 | (10) edge node{}(16) 94 | (10) edge node{}(18) 95 | (5) edge node{}(11) 96 | (11) edge node{}(12) 97 | (12) edge node{}(13) 98 | (15) edge node{}(17) 99 | (16) edge[dashed] node{}(17) 100 | (13) edge node{}(17); 101 | \end{tikzpicture} 102 | \end{figure} 103 | 104 | In (\ref{chap:metr}) we introduce the concepts necessary to formulate the axioms; 105 | it includes metric spaces, 106 | lines, 107 | angle measure, 108 | continuous maps, 109 | and congruent triangles. 110 | 111 | Next, we dive into Euclidean geometry: 112 | (\ref{chap:axioms}) Axioms and immediate corollaries; 113 | (\ref{chap:half-planes}) Half-planes and continuity; 114 | (\ref{chap:cong}) Congruent triangles; 115 | (\ref{chap:perp}) Circles, motions, and perpendicular lines; 116 | (\ref{chap:parallel}) Similar triangles and (\ref{chap:angle-sum}) Parallel lines 117 | --- these are the first two chapters where we use Axiom~\ref{def:birkhoff-axioms:4}, an equivalent of Euclid's parallel postulate. 118 | In (\ref{chap:triangle}) we give the most classical theorems of triangle geometry; 119 | this chapter is included mainly as an illustration. 120 | 121 | In the following two chapters, we discuss the geometry of circles on the Euclidean plane: 122 | (\ref{chap:inscribed-angle}) Inscribed angles; (\ref{chap:inversion}) Inversion. 123 | Inversion will be used to construct the model of the hyperbolic plane. 124 | 125 | Next, we discuss non-Euclidean geometry: 126 | (\ref{chap:non-euclid}) 127 | Neutral geometry --- geometry without the parallel postulate; 128 | (\ref{chap:poincare}) 129 | Conformal disc model --- 130 | this is a construction of the hyperbolic plane, 131 | an example of a neutral plane that is not Euclidean. 132 | In (\ref{chap:h-plane}) we discuss the geometry of the constructed hyperbolic plane --- this is the high point in the book. 133 | 134 | In the remaining chapters, we discuss additional topics: 135 | (\ref{chap:trans}) Affine geometry; 136 | (\ref{chap:proj}) Projective geometry; 137 | (\ref{chap:sphere}) Spherical geometry; 138 | (\ref{chap:klein}) Projective model of the hyperbolic plane; 139 | (\ref{chap:complex}) Complex coordinates; 140 | (\ref{chap:car}) Geometric constructions; 141 | (\ref{chap:area}) Area. 142 | The proofs in these chapters are not completely rigorous. 143 | 144 | We encourage you to use the visual assignments on the author's website. 145 | 146 | \textbf{Classroom exercises} usually illustrate newly introduced concepts and are intended to be discussed in class. 147 | Exercises that are used later in the sequel are marked with an exclamation point: \textbf{Exercise!} 148 | A statement is marked with ``$\a$''\label{a-mark} (for example, ``\textbf{Theorem\abs}'') if its proof does not use Axiom~\ref{def:birkhoff-axioms:4}. 149 | This will become important starting from Chapter~\ref{chap:non-euclid}. 150 | 151 | 152 | \section{Disclaimer} 153 | 154 | Most of the proofs in the book have already appeared in Euclid's Elements. 155 | I did not try to find original references; it is an impossible task. 156 | 157 | \section{Recommended resources} 158 | 159 | \begin{itemize} 160 | \item {}\emph{Byrne's Euclid} \cite{byrne} --- a colored version of the first six books of Euclid's Elements edited by Oliver Byrne. 161 | 162 | \item {}\emph{Kiselyov's geometry} \cite{kiselev} --- 163 | a classical textbook for school students written by Andrey Kiselyov; it should help if you have trouble following this book. 164 | 165 | \item {}\emph{Lessons in Geometry} by Jacques Hadamard \cite{hadamard} --- an encyclopedia of elementary geometry originally written for school teachers. 166 | 167 | %\item Moise's book, \cite{moise} --- should be good for further study. 168 | 169 | %\item Greenberg's book \cite{greenberg} --- a historical tour in the axiomatic systems of various geometries. 170 | 171 | \item {}\emph{Problems in geometry} by Victor Prasolov \cite{prasolov} is perfect for mastering your problem-solving skills. 172 | 173 | \item {}\emph{Geometry in figures} by Arseniy Akopyan \cite{akopyan} --- an encyclopedia of Euclidean geometry with barely any words. 174 | 175 | \item {}\emph{Euclidea} \cite{euclidea} --- a fun and challenging way to learn geometric constructions. 176 | 177 | \item {}\emph{Geometry} by Igor Sharygin \cite{sharygin} --- the greatest textbook in geometry for school students. 178 | I recommend it to anyone who can read Russian. 179 | 180 | \item {}\emph{problems.ru} \cite{zadachi} --- an on-line collection of problems in Russian. 181 | 182 | \end{itemize} 183 | 184 | \section{Acknowledgments} 185 | 186 | {\sloppy 187 | 188 | Let me thank 189 | Stephanie Alexander, 190 | Thomas Barthelme, 191 | Berk Ceylan, 192 | Matthew Chao, 193 | Quinn Culver, 194 | Svetlana Katok, 195 | Nina Lebedeva, 196 | Alexander Lytchak, 197 | Alexei Novikov, 198 | and Lukeria Petrunina 199 | for useful suggestions and for correcting misprints. 200 | 201 | This work was partially supported by 202 | NSF grants 203 | DMS-0103957, 204 | DMS-0406482, 205 | DMS-0905138, 206 | DMS-1309340, 207 | DMS-2005279, 208 | and Simons Foundation grants 209 | 245094, 584781. 210 | 211 | } 212 | -------------------------------------------------------------------------------- /similar.tex: -------------------------------------------------------------------------------- 1 | \chapter[Similar triangles]{Similar triangles}\label{chap:parallel} 2 | 3 | Two triangles $A' B' C'$ and $A B C$ are called 4 | \index{triangle!similar triangles}\index{similar triangles}\emph{similar} (briefly \index{30@$\sim$}$\triangle A' B' C'\z\sim\triangle A B C$) if (1) their sides are proportional; 5 | that is, 6 | $$A' B' 7 | = 8 | k\cdot A B, 9 | \quad 10 | B' C'=k\cdot B C 11 | \quad 12 | \text{and} 13 | \quad 14 | C' A' 15 | = 16 | k\cdot C A 17 | \eqlbl{dist} 18 | $$ 19 | for some coefficient $k>0$ (which is called the \index{similarity coefficient}\emph{similarity coefficient}), and (2) the corresponding angles are equal up to sign: 20 | $$ 21 | \begin{aligned} 22 | \measuredangle A' B' C'&=\pm\measuredangle A B C, 23 | \\ 24 | \measuredangle B' C' A'&=\pm\measuredangle B C A, 25 | \\ 26 | \measuredangle C' A' B'&=\pm\measuredangle CAB. 27 | \end{aligned} 28 | \eqlbl{angles} 29 | $$ 30 | 31 | \parbf{Remarks.} 32 | \begin{itemize} 33 | \item According to \ref{thm:signs-of-triug}, in the above three equalities, the signs can be assumed to be the same. 34 | 35 | \item The equations in \ref{dist} can be rewritten as 36 | \[\frac{A'B'}{AB}=\frac{B'C'}{BC}=\frac{C'A'}{CA}.\] 37 | 38 | \item If $\triangle A' B' C'\sim\triangle A B C$ with the coefficient $k=1$, 39 | then $\triangle A' B' C'\z\cong\triangle A B C$. 40 | 41 | \item Notice that ``$\sim$'' is an 42 | \index{equivalence relation}\emph{equivalence relation}. 43 | That is, 44 | \begin{enumerate}[(i)] 45 | \item $\triangle A B C\sim\triangle A B C$ 46 | for every $\triangle A B C$. 47 | \item If $\triangle A' B' C'\sim\triangle A B C$, then 48 | $$\triangle A B C\sim\triangle A' B' C'.$$ 49 | \item If $\triangle A'' B'' C''\sim\triangle A' B' C'$ and $\triangle A' B' C'\z\sim\triangle A B C$, then 50 | $$\triangle A'' B'' C''\sim\triangle A B C.$$ 51 | \end{enumerate} 52 | \end{itemize} 53 | 54 | \section{Similarity conditions} 55 | 56 | Using the new notation ``$\sim$'', we can reformulate Axiom~\ref{def:birkhoff-axioms:4}: 57 | 58 | \begin{thm}{Reformulation of Axiom~\ref{def:birkhoff-axioms:4}} 59 | If for 60 | $\triangle ABC$, 61 | $\triangle AB'C'$, 62 | and $k>0$ we have 63 | $B'\in [AB)$, 64 | $C'\in [AC)$, 65 | $AB'=k\cdot AB$ and 66 | $AC'=k\cdot AC$, 67 | then $\triangle ABC\sim\triangle AB'C'$. 68 | \end{thm} 69 | 70 | In other words, Axiom~\ref{def:birkhoff-axioms:4} provides 71 | a condition which guarantees that two triangles are similar. 72 | Let us formulate three more such {}\emph{similarity conditions}. 73 | 74 | \begin{thm}{Similarity conditions}\label{prop:sim} 75 | Two triangles 76 | $\triangle ABC$ and $\triangle A'B'C'$ 77 | are similar if one of the following conditions holds: 78 | 79 | (SAS)\index{SAS similarity condition} 80 | For some constant $k>0$ we have 81 | \[A B=k\cdot A' B', 82 | \quad 83 | A C=k\cdot A' C', 84 | \quad 85 | \text{and} 86 | \quad 87 | \measuredangle B A C=\pm\measuredangle B' A' C'.\] 88 | 89 | (AA)\index{AA similarity condition} The triangle $A' B' C'$ is nondegenerate 90 | and 91 | $$\measuredangle A B C 92 | = 93 | \pm\measuredangle A' B' C', 94 | \quad 95 | \measuredangle B A C 96 | = 97 | \pm\measuredangle B' A' C'.$$ 98 | 99 | (SSS)\index{SSS similarity condition} For some constant $k>0$ we have 100 | $$A B=k\cdot A' B', 101 | \quad 102 | A C=k\cdot A' C', 103 | \quad 104 | CB=k\cdot C'B'.$$ 105 | 106 | \end{thm} 107 | 108 | Each of these conditions is proved by applying Axiom~\ref{def:birkhoff-axioms:4} with the SAS, ASA, and SSS congruence conditions respectively 109 | (see Axiom~\ref{def:birkhoff-axioms:3} and the conditions \ref{thm:ASA} and \ref{thm:SSS}). 110 | 111 | 112 | \parit{Proof.} 113 | Let $k=\tfrac{AB}{A'B'}$. 114 | Choose points $B''\in [A'B')$ and $C''\in [A'C')$, 115 | so that $A'B''=k\cdot A'B'$ and $A'C''=k\cdot A'C'$. 116 | By Axiom~\ref{def:birkhoff-axioms:4}, 117 | $\triangle A'B'C'\z\sim \triangle A'B''C''$. 118 | 119 | Applying the SAS, ASA, or SSS congruence condition, depending on the case, 120 | we get that $\triangle A'B''C''\cong \triangle ABC$. 121 | Hence the result. 122 | \qeds 123 | 124 | \begin{thm}{Exercise}\label{ex:mid-triangle} 125 | Let $A'$, $B'$, and $C'$ be the midpoints of sides $[BC]$, $[CA]$, and $[AB]$ of $\triangle ABC$. 126 | Show that $\triangle A'B'C'\sim\triangle ABC$ and find the similarity coefficient. 127 | \end{thm} 128 | 129 | \begin{thm}{Exercise}\label{ex:k*triangle} 130 | Let $O$, $A$, $B$, $C$, $A'$, $B'$, and $C'$ be distinct points. 131 | Suppose that $A'\in [OA)$, $B'\in[OB)$, $C'\in [OC)$, $OA'=k\cdot OA$, $OB'=k\cdot OB$, and $OC'=k\cdot OC$ for some $k>0$. 132 | Show that $\triangle A'B'C'\sim\triangle ABC$. 133 | \end{thm} 134 | 135 | A bijection $X\leftrightarrow X'$ from a plane to itself is called an \index{angle-preserving transformation}\emph{angle-preserving transformation} if 136 | \[\measuredangle ABC= \measuredangle A'B'C'\] 137 | for every $\triangle ABC$ and its image $\triangle A'B'C'$. 138 | 139 | (The term \index{transformation}\emph{transformation} is used for a bijection of the plane to itself that preserves a specified geometric structure. 140 | For example, {}\emph{motions} are \textit{distance-preserving transformations}.) 141 | 142 | 143 | \begin{thm}{Exercise}\label{ex:angle-preserving-euclid} 144 | Show that every angle-preserving transformation of the plane multiplies all distances by a fixed constant. 145 | \end{thm} 146 | 147 | \section{The Pythagorean theorem}\index{Pythagorean theorem} 148 | 149 | Recall that a triangle is called \index{right!triangle}\index{triangle!right triangle}\emph{right} if one of its angles is right. 150 | The side opposite the right angle is called the \index{hypotenuse}\emph{hypotenuse}. 151 | The sides adjacent to the right angle are called \index{leg}\emph{legs}. 152 | 153 | 154 | \begin{thm}{Theorem}\label{thm:pyth} 155 | Assume $\triangle ABC$ has a right angle at~$C$. 156 | Then 157 | $$AC^2+BC^2=AB^2.$$ 158 | 159 | \end{thm} 160 | 161 | \parit{Proof.} 162 | Let $D$ be the footpoint of $C$ on~$(AB)$. 163 | 164 | \begin{wrapfigure}[4]{r}{40mm} 165 | \vskip-2mm 166 | \centering 167 | \includegraphics{mppics/pic-66} 168 | \end{wrapfigure} 169 | 170 | According to Lemma~\ref{lem:perp0$ and $BC>0$; 184 | otherwise, $A=B$ or $B=C$, and the statement follows. 185 | In this case, $B'$ lies between $A$ and $C$. 186 | By Theorem~\ref{thm:straight-angle}, $\measuredangle AB'C=\pi$. 187 | 188 | By SSS, 189 | \[\triangle ABC\cong \triangle AB'C.\] 190 | Therefore, $\measuredangle ABC=\pi$. 191 | By Theorem~\ref{thm:straight-angle}, $B$ lies between $A$ and $C$. 192 | \qeds 193 | 194 | 195 | 196 | \begin{thm}{Advanced exercise}\label{ex:SMS} 197 | Consider two triangles $A B C$ and $A' B' C'$. 198 | Let $M$ be the midpoint of $[A B]$, and 199 | $M'$ be the midpoint of $[A' B']$. 200 | Assume $C' A'=C A$, $C' B'= C B$, and $C' M'\z= C M$. 201 | Prove that 202 | \[\triangle A' B' C'\cong\triangle A B C.\] 203 | 204 | \end{thm} 205 | 206 | { 207 | 208 | \begin{wrapfigure}[6]{r}{26mm} 209 | \vskip-0mm 210 | \centering 211 | \includegraphics{mppics/pic-42} 212 | \end{wrapfigure} 213 | 214 | \begin{thm}{Exercise}\label{ex:isos-sides} 215 | Let $\triangle A B C$ be an isosceles triangle with the base $[A B]$. 216 | Suppose that point $A'$ lies between $B$ and $C$, 217 | point $B'$ lies between $A$ and $C$, 218 | and $C A'\z=C B'$. 219 | Show that 220 | \begin{enumerate}[(a)] 221 | \item $\triangle A A' C\cong \triangle B B' C$; 222 | \item $\triangle A B B'\cong \triangle B A A'$. 223 | \end{enumerate} 224 | 225 | \end{thm} 226 | 227 | \begin{thm}[!]{Exercise}\label{ex:ABC-motion} 228 | Let $\triangle ABC$ be a nondegenerate triangle, and 229 | let $f$ be a distance-preserving map from the plane to itself 230 | such that 231 | $$f(A)=A, 232 | \quad 233 | f(B)=B, 234 | \quad 235 | \text{and} 236 | \quad 237 | f(C)=C.$$ 238 | 239 | Show that $f$ is the identity map; 240 | that is, $f(X)=X$ for every point $X$ in the plane. 241 | \end{thm} 242 | 243 | The statement in the exercise is called the \emph{three-nail lemma}, since three nails fix the plane in place. %(Unlike in Jerusalem, nobody gets hurt here.) 244 | 245 | } 246 | 247 | %{ 248 | 249 | %\begin{wrapfigure}{r}{28mm}\vskip-11mm\centering\includegraphics{mppics/pic-43}\end{wrapfigure} 250 | 251 | %\begin{thm}{Exercise}\label{ex:3-isos} 252 | %Suppose that $\triangle ABC'$, $\triangle BCA'$, and $\triangle CAB'$ are isosceles triangles that lie outside of $\triangle ABC$; 253 | %that is, 254 | %$A$ and $A'$ lie on the opposite sides of $(BC)$, 255 | %$B$ and $B'$ lie on the opposite sides of $(CA)$, 256 | %$C$ and $C'$ lie on the opposite sides of $(AB)$. 257 | 258 | %Show that $AA'=BB'=CC'$. 259 | %\end{thm} 260 | 261 | %} 262 | 263 | \section{On side-side-angle and side-angle-angle} 264 | 265 | In each of the conditions SAS, ASA, and SSS we specify three corresponding parts of the triangles. 266 | Now, let us discuss other triples of corresponding parts. 267 | 268 | The first triple is called {}\emph{side-side-angle}, or briefly SSA; 269 | it specifies two sides and a non-included angle. 270 | This condition is not sufficient for congruence. 271 | In other words, there exist two nondegenerate triangles $ABC$ and $A'B'C'$ such that 272 | \[AB=A'B',\quad BC=B'C',\quad \measuredangle BAC\equiv\pm \measuredangle B'A'C',\] 273 | but $\triangle ABC\not\cong\triangle A'B'C'$ and $AC\ne A'C'$. 274 | 275 | \begin{wrapfigure}{r}{35mm} 276 | \vskip-6mm 277 | \centering 278 | \includegraphics{mppics/pic-44} 279 | \end{wrapfigure} 280 | 281 | We will not use this negative statement, so there is no need to prove it formally. 282 | You can guess an example from the picture. 283 | 284 | The second triple is {}\emph{side-angle-angle}, or briefly SAA; 285 | it specifies one side and two angles, one of which is opposite to the side. 286 | This provides a congruence condition; 287 | that is, if one of the triangles $ABC$ or $A'B'C'$ is nondegenerate and 288 | $AB=A'B'$, $\measuredangle ABC\equiv\pm \measuredangle A'B'C'$, $\measuredangle BCA\z\equiv\pm \measuredangle B'C'A'$, 289 | then $\triangle ABC\cong\triangle A'B'C'$. 290 | 291 | The SAA condition will not be used directly in the sequel. 292 | One proof of this condition can be obtained from ASA and the theorem on the sum of angles of a triangle, which is proved below (see~\ref{thm:3sum}). 293 | For a more direct proof, see Exercise~\ref{ex:SAA}. 294 | 295 | Another triple is called {}\emph{angle-angle-angle}, or briefly AAA. 296 | By Axiom~\ref{def:birkhoff-axioms:4}, it is not a congruence condition in the Euclidean plane. 297 | However, in the hyperbolic plane it is; see \ref{thm:AAA}. 298 | 299 | { 300 | 301 | \begin{wrapfigure}{r}{33mm} 302 | \vskip-8mm 303 | \centering 304 | \includegraphics{mppics/pic-445} 305 | \bigskip 306 | \includegraphics{mppics/pic-446} 307 | \bigskip 308 | \includegraphics{mppics/pic-447} 309 | \bigskip 310 | \includegraphics{mppics/pic-448} 311 | \bigskip 312 | \includegraphics{mppics/pic-449} 313 | \end{wrapfigure} 314 | 315 | \section{Constructions} 316 | 317 | The construction problems provide a valuable source of exercises in geometry, 318 | which we will use further in the book. 319 | Chapter~\ref{chap:car} will be devoted to the subject. 320 | 321 | A \index{ruler-and-compass construction}\emph{ruler-and-compass construction} is a construction with an idealized ruler and compass. 322 | The idealized ruler can be used only to draw a line thru two given points. 323 | The idealized compass can be used only to draw a circle with a given center and radius. 324 | That is, given three points $A$, $B$, and $O$ 325 | we can draw the circle of radius $AB$ centered at~$O$. 326 | We may also mark new points in the plane, 327 | as well as on the constructed lines, circles, 328 | and their intersections (assuming that such points exist). 329 | 330 | We may also look at the different sets of construction tools. 331 | For example, 332 | we may only use the ruler, 333 | or we may invent a new tool, 334 | say a tool that produces a midpoint for two given points, 335 | or the reflection of one point across the other. 336 | 337 | \begin{thm}{Problem} 338 | Given two lines $\ell$ and $m$ and a point $M$ not on these lines, 339 | construct a line segment with ends on $\ell$ and $m$ that is bisected by~$M$. 340 | \end{thm} 341 | 342 | The stages of the following construction are shown in the picture. 343 | 344 | \parit{Construction.} 345 | \begin{enumerate}[1.] 346 | \item Draw a circle centered at $M$ that crosses line $\ell$ at two points, say $P$ and $Q$. 347 | \item Draw the line $(PM)$ and mark by $P'$ its other point of intersection with the circle. 348 | Similarly, draw $(QM)$ and mark the other intersection point by $Q'$. 349 | \item Draw the line $\ell'=(P'Q')$ and mark by $A$ its point of intersection with $m$ (if it exists). 350 | \end{enumerate} 351 | 352 | } 353 | 354 | \begin{enumerate}[1.]\setcounter{enumi}{3} 355 | \item Draw the line $(MA)$; 356 | mark by $B$ its point of intersection with $\ell$. 357 | Then $[AB]$ is the required segment. 358 | \end{enumerate} 359 | 360 | 361 | It is often required to do \emph{problem analysis}: 362 | to prove that the result of the construction satisfies the required conditions and also to determine when the construction is possible and how many solutions may exist. 363 | Let us do it for the example above. 364 | 365 | 366 | \parit{Problem analysis.} 367 | Notice that $P'$ and $Q'$ are reflections of $P$ and $Q$ across~$M$. 368 | Therefore, the line $\ell'=(P'Q')$ is the reflection of $\ell=(PQ)$ across~$M$. 369 | 370 | Since $A$ lies on $\ell'$, 371 | the point $B$ is the reflection of $A$ across $M$. 372 | Indeed, $B$ is the (necessarily unique) point of intersection of $(MA)$ and $\ell$, and $\ell'$ is the reflection of $\ell$. 373 | By the definition of reflection, $M$ is the midpoint of $[AB]$. 374 | 375 | The line $\ell'$ may have no points of intersection with $m$. 376 | In this case, the problem has no solution. 377 | Also, it might happen that $\ell'=m$; 378 | in this case, we have an infinite set of solutions --- any point on $m$ can be marked as~$A$. 379 | In all the remaining cases, we have exactly one solution. 380 | \qeds 381 | 382 | \begin{thm}{Exercise}\label{ex:motion} 383 | Suppose $\triangle ABC$ and $\triangle A'B'C'$ are congruent nondegenerate triangles; 384 | in particular, there is a motion $f$ such that 385 | \[f\colon A\mapsto A',\quad f\colon B\mapsto B',\quad\text{and}\quad f\colon C\mapsto C'.\] 386 | 387 | Given a point $X$, construct $X'=f(X)$. 388 | \end{thm} 389 | 390 | 391 | -------------------------------------------------------------------------------- /axioms.tex: -------------------------------------------------------------------------------- 1 | %\part*{Euclidean geometry} 2 | \addtocontents{toc}{\protect\contentsline{part}{\protect\numberline{}Euclidean geometry}{}{}} 3 | 4 | \chapter{Axioms} 5 | \label{chap:axioms} 6 | 7 | \vfill 8 | 9 | A system of axioms already appears in Euclid's ``Elements'' --- the most successful and influential textbook ever written. 10 | 11 | The systematic study of geometries as axiomatic systems 12 | was 13 | triggered by the discovery of non-Euclidean geometry. 14 | The branch of mathematics that emerged this way is called the foundations of geometry. 15 | 16 | The most popular system of axioms 17 | was proposed by David Hilbert in 1899. 18 | This is also the first rigorous system by modern standards. 19 | It consists of twenty axioms in five groups, six ``primitive notions'', and three ``primitive terms''; 20 | these are not defined in terms of previously defined concepts. 21 | 22 | Later, many different systems were proposed. 23 | It is worth mentioning 24 | the system developed by Alexandr Alexandrov \cite{alexandrov}, which is intuitive and elementary, 25 | the system by Friedrich Bachmann \cite{bachmann} based on the concept of symmetry, 26 | and the system by Alfred Tarski \cite{tarski} --- a minimalist system designed for analysis using mathematical logic. 27 | 28 | We will use another system close to the one proposed by George Birkhoff \cite{birkhoff}. 29 | It is based on the key observations (\ref{preaxiomI})--(\ref{preaxiomV}) listed in Section~\ref{preaxioms}. 30 | The axioms use the notions of 31 | metric space, 32 | lines, 33 | angles, 34 | triangles, 35 | equalities modulo $2\cdot\pi$ ($\equiv$), 36 | the continuity of maps between metric spaces, 37 | and the congruence of triangles ($\cong$). 38 | All of these concepts are discussed in the preliminaries. 39 | 40 | Our system is built on metric spaces. 41 | In particular, we use real numbers as a building block. 42 | For this reason our approach is not purely axiomatic --- we build the theory upon something else; 43 | it resembles a model-based introduction to Euclidean geometry discussed in Section~\ref{page:model}. 44 | We use this approach to minimize the tedious parts, which are unavoidable in purely axiomatic foundations. 45 | 46 | \newpage 47 | 48 | \section{The axioms} 49 | \label{sec:axioms} 50 | 51 | 52 | \begin{framed} 53 | \begin{enumerate}[I.] 54 | \item\label{def:birkhoff-axioms:0} The \index{plane!Euclidean plane}\index{Euclidean!plane}\emph{Euclidean plane} is a metric space with at least two points. 55 | 56 | 57 | \item\label{def:birkhoff-axioms:1} 58 | There is one and only one line that contains any two given distinct points $P$ and $Q$ in the Euclidean plane. 59 | 60 | \item\label{def:birkhoff-axioms:2} 61 | Any angle $\angle AOB$ in the Euclidean plane 62 | defines a real number in the interval $(-\pi,\pi]$. 63 | This number is called the \index{angle!measure}\emph{angle measure of $\angle AOB$} 64 | and is denoted by $\measuredangle A O B$. 65 | It satisfies the following conditions: 66 | \begin{enumerate}[(a)] 67 | \item\label{def:birkhoff-axioms:2a} 68 | Given a half-line $[O A)$ and $\alpha\in(-\pi,\pi]$, 69 | there is a unique half-line $[O B)$, 70 | such that $\measuredangle A O B= \alpha$. 71 | \item\label{def:birkhoff-axioms:2b} 72 | For any points $A$, $B$, and $C$, distinct from $O$, we have 73 | $$\measuredangle A O B+\measuredangle B O C 74 | \equiv\measuredangle A O C.$$ 75 | \item\label{def:birkhoff-axioms:2c} 76 | The function 77 | $$\measuredangle\:(A,O,B)\mapsto\measuredangle A O B$$ 78 | is continuous at any triple of points $(A,O,B)$ 79 | such that $O\ne A$ and $O\ne B$ and $\measuredangle A O B\ne\pi$. 80 | 81 | \end{enumerate} 82 | 83 | \item\label{def:birkhoff-axioms:3} 84 | In the Euclidean plane, we have 85 | $\triangle A B C\cong\triangle A' B' C'$ 86 | if and only if 87 | \begin{align*} 88 | A' B'&=A B, & A' C'&= A C, &&\text{and} 89 | &\measuredangle C' A' B'&=\pm\measuredangle C A B. 90 | \end{align*} 91 | \item\label{def:birkhoff-axioms:4} 92 | If for two triangles $\triangle ABC$, $\triangle AB'C'$ in the Euclidean plane 93 | and for $k>0$ we have 94 | \begin{align*} 95 | B'&\in [AB), 96 | & C'&\in [AC), 97 | \\ 98 | AB'&=k\cdot AB,& 99 | AC'&=k\cdot AC, 100 | \end{align*} 101 | then 102 | \begin{align*} 103 | B'C'&=k\cdot BC,& 104 | \measuredangle AB'C'&=\measuredangle ABC, 105 | & 106 | \measuredangle AC'B'&=\measuredangle ACB. 107 | \end{align*} 108 | \end{enumerate} 109 | \end{framed} 110 | 111 | From now on, 112 | we can use no information about the Euclidean plane that does not follow from the five axioms above. 113 | 114 | \begin{thm}{Classroom exercise}\label{ex:infinite} 115 | Show that there is (a) an infinite set of points and (b) an infinite set of lines on the plane. 116 | \end{thm} 117 | 118 | \section{Lines and half-lines} 119 | 120 | \begin{thm}[\abs]{Proposition}\label{lem:line-line} 121 | \let\thefootnote\relax\footnotetext{${}^\a$ The mark ``$\a$'' indicates that Axiom~\ref{def:birkhoff-axioms:4} is not used in the proof. 122 | Please ignore this mark for now; it will become important in Chapter~\ref{chap:non-euclid}.} 123 | Any two distinct lines intersect in at most one point. 124 | \end{thm} 125 | 126 | \parit{Proof.} 127 | Assume that two lines $\ell$ and $m$ intersect at two distinct points $P$ and~$Q$. 128 | Applying Axiom~\ref{def:birkhoff-axioms:1}, we get that $\ell=m$. 129 | \qeds 130 | 131 | \begin{thm}{Exercise}\label{ex:[OA)=[OA')} 132 | Suppose $A'\in[OA)$ and $A'\not=O$. 133 | Show that 134 | \[[O A)\z=[O A').\] 135 | 136 | \end{thm} 137 | 138 | \section{A zero angle} 139 | 140 | \begin{thm}[\abs]{Proposition}\label{lem:AOA=0} 141 | $\measuredangle A O A= 0$ for every $A\not=O$. 142 | \end{thm} 143 | 144 | \parit{Proof.} 145 | According to Axiom~\ref{def:birkhoff-axioms:2b}, 146 | $$\measuredangle A O A 147 | + 148 | \measuredangle A O A 149 | \equiv 150 | \measuredangle A O A.$$ 151 | Subtracting $\measuredangle A O A$ from both sides, we get that 152 | $\measuredangle A O A \equiv 0$. 153 | 154 | By Axiom~\ref{def:birkhoff-axioms:2}, $-\pi<\measuredangle A O A\le \pi$; 155 | therefore $\measuredangle A O A \z= 0$. 156 | \qeds 157 | 158 | \begin{thm}[!]{Exercise}\label{ex:[OA)=[OB)} 159 | Assume $\measuredangle A O B= 0$. 160 | Show that $[OA)=[OB)$. 161 | \end{thm} 162 | 163 | \begin{thm}[\abs]{Proposition}\label{lem:AOB+BOA=0} 164 | For any points $A$ and $B$ distinct from $O$, 165 | we have 166 | $$\measuredangle A O B\equiv-\measuredangle B O A.$$ 167 | 168 | \end{thm} 169 | 170 | \parit{Proof.} 171 | According to Axiom~\ref{def:birkhoff-axioms:2b}, 172 | $$\measuredangle A O B+\measuredangle B O A \equiv\measuredangle A O A.$$ 173 | By Proposition~\ref{lem:AOA=0}, $\measuredangle A O A=0$. 174 | Hence the result. 175 | \qeds 176 | 177 | \section{A straight angle} 178 | 179 | If $\measuredangle A O B=\pi$, 180 | we say that $\angle A O B$ is a 181 | \index{angle!straight angle}\emph{straight angle}. 182 | By Proposition~\ref{lem:AOB+BOA=0}, 183 | if $\angle A O B$ is straight, 184 | then so is $\angle B O A$. 185 | 186 | We say that point $O$ \index{between}\emph{lies between} points $A$ and $B$ 187 | if $O\not= A$, $O\not= B$, and $O\in[A B]$. 188 | 189 | \begin{thm}[\abs]{Theorem}\label{thm:straight-angle} 190 | An angle $A O B$ is straight 191 | if and only if $O$ 192 | lies between $A$ and~$B$. 193 | \end{thm} 194 | 195 | \begin{wrapfigure}{r}{40mm} 196 | \centering 197 | \vskip-0mm 198 | \includegraphics{mppics/pic-8} 199 | \end{wrapfigure} 200 | 201 | \parit{Proof.} 202 | By Exercise~\ref{ex:trig==}, we may assume that 203 | $O A = O B = 1$. 204 | 205 | \parit{``If'' part.} 206 | Assume $O$ 207 | lies between $A$ and~$B$. 208 | Set $\alpha=\measuredangle A O B$. 209 | 210 | Applying Axiom~\ref{def:birkhoff-axioms:2a}, 211 | we get a half-line $[OA')$ such that $\alpha\z=\measuredangle B O A'$. 212 | By Exercise~\ref{ex:trig==}, we can assume that $OA'=1$. 213 | According to Axiom~\ref{def:birkhoff-axioms:3}, 214 | \[\triangle AOB\z\cong\triangle BOA'.\] 215 | Suppose that $f$ denotes the corresponding motion of the plane; 216 | that is, $f$ is a motion such that $f(A)=B$, $f(O)=O$, and $f(B)=A'$. 217 | 218 | \begin{wrapfigure}{o}{40mm} 219 | \centering 220 | \vskip-2mm 221 | \includegraphics{mppics/pic-10} 222 | \end{wrapfigure} 223 | 224 | Then 225 | \[(A'B)=f((AB))\ni f(O)=O.\] 226 | Therefore, both lines $(AB)$ and $(A'B)$ contain $B$ and~$O$. 227 | By Axiom~\ref{def:birkhoff-axioms:1}, $(AB)=(A'B)$. 228 | 229 | By the definition of a line, $(AB)$ is isometric to the real line $\mathbb{R}$. 230 | Therefore, $(AB)$ contains exactly two points, $A$ and $B$, that are at distance $1$ from~$O$. 231 | Since $OA' = 1$, it follows that $A' = A$ or $A' = B$. 232 | Since an isometry is injective, $A' = f(B) \ne f(A) = B$. 233 | Thus, $A' \ne B$, and hence $A = A'$. 234 | 235 | By Axiom~\ref{def:birkhoff-axioms:2b} and Proposition~\ref{lem:AOA=0}, we get that 236 | \begin{align*} 237 | 2\cdot\alpha&= 238 | \measuredangle AOB+\measuredangle BOA'= 239 | \\ 240 | &=\measuredangle AOB+\measuredangle BOA\equiv 241 | \\ 242 | &\equiv\measuredangle AOA= 243 | \\ 244 | &= 0. 245 | \end{align*} 246 | Therefore, by Exercise~\ref{ex:2a=0}, $\alpha$ is either $0$ or~$\pi$. 247 | 248 | Since $[OA)\ne [OB)$, 249 | we have that $\alpha\ne 0$; see Exercise~\ref{ex:[OA)=[OB)}. 250 | Therefore, $\alpha=\pi$. 251 | 252 | 253 | \parit{``Only if'' part.} 254 | Suppose that $\measuredangle A O B= \pi$. 255 | Consider the line $(OA)$ and choose a point $B'$ on $(OA)$ so that $O$ lies between $A$ and~$B'$. 256 | 257 | From above, we have that $\measuredangle AOB'=\pi$. 258 | Applying Axiom~\ref{def:birkhoff-axioms:2a}, 259 | we get that $[O B)\z=[O B')$. 260 | In particular, $O$ lies between $A$ and~$B$. 261 | \qeds 262 | 263 | A triangle $ABC$ is called 264 | \index{triangle!degenerate triangle}\index{degenerate! triangle}\emph{degenerate} 265 | if $A$, $B$, and $C$ lie on one line. 266 | The following corollary is just a reformulation of Theorem~\ref{thm:straight-angle}. 267 | 268 | \begin{thm}[\abs]{Corollary}\label{cor:degenerate=pi} 269 | A triangle is degenerate if and only if one of its angles is equal to $\pi$ or~$0$. 270 | Moreover, in a degenerate triangle, the angle measures are $0$, $0$, and $\pi$. 271 | \end{thm} 272 | 273 | \begin{thm}{Exercise}\label{ex:lineAOB} 274 | Show that three distinct points $A$, $O$, and $B$ lie on one line if and only if 275 | $$2\cdot \measuredangle AOB\equiv 0.$$ 276 | 277 | \end{thm} 278 | 279 | \begin{thm}[!]{Exercise}\label{ex:ABCO-line} 280 | Let $A$, $B$, and $C$ be three points distinct from~$O$. 281 | Show that $B$, $O$, and $C$ lie on one line if and only if 282 | $$2\cdot \measuredangle AOB\equiv 2\cdot \measuredangle AOC.$$ 283 | 284 | \end{thm} 285 | 286 | \begin{thm}{Exercise}\label{ex:infinite-number-of-lines} 287 | Show that there is a nondegenerate triangle. 288 | \end{thm} 289 | 290 | \section{Vertical angles} 291 | 292 | A pair of angles $AOB$ and $A'OB'$ 293 | is called \index{angle!vertical angles}\index{vertical angles}\emph{vertical} 294 | if point $O$ 295 | lies between $A$ and $A'$ 296 | and between $B$ and $B'$ at the same time. 297 | 298 | 299 | \begin{thm}[\abs]{Proposition}\label{prop:vert} 300 | Vertical angles have equal measures. 301 | \end{thm} 302 | 303 | 304 | \parit{Proof.} 305 | Assume that the angles $AOB$ and $A'OB'$ are vertical. 306 | Note that $\angle AOA'$ and $\angle BOB'$ are straight. 307 | Therefore, $\measuredangle AOA'\z=\measuredangle BOB'=\pi$. 308 | 309 | { 310 | 311 | \begin{wrapfigure}[4]{o}{24mm} 312 | \vskip-2mm 313 | \centering 314 | \includegraphics{mppics/pic-12} 315 | \end{wrapfigure} 316 | 317 | It follows that 318 | \begin{align*} 319 | 0&=\measuredangle AOA'-\measuredangle BOB'\equiv 320 | \\ 321 | &\equiv 322 | \measuredangle AOB+\measuredangle BOA'-\measuredangle BOA'-\measuredangle A'OB' 323 | \equiv 324 | \\ 325 | &\equiv\measuredangle AOB-\measuredangle A'OB'. 326 | \end{align*} 327 | Since $-\pi<\measuredangle AOB\le \pi$ and $-\pi<\measuredangle A'OB'\le \pi$, we get that $\measuredangle AOB\z=\measuredangle A'OB'$. 328 | \qeds 329 | 330 | } 331 | 332 | \section{Reflection across a point} 333 | 334 | \begin{thm}[\abs]{Lemma}\label{lem:point-reflection} 335 | Given distinct points $X$ and $O$, there is a unique point $X'$ such that $O$ is the midpoint of a line segment $[XX']$. 336 | \end{thm} 337 | 338 | \begin{wrapfigure}{o}{33mm} 339 | \vskip-0mm 340 | \centering 341 | \includegraphics{mppics/pic-915} 342 | \end{wrapfigure} 343 | 344 | \parit{Proof.} 345 | The point $O$ is the midpoint of a line segment $[XX']$ if and only if $X'\in [XO)$ and $XX'=2\cdot XO$. 346 | 347 | By Axiom~\ref{def:birkhoff-axioms:1}, the line $(XO)$ and therefore the half-line $[XO)$ are uniquely defined. 348 | Further, by Exercise~\ref{ex:trig==} there is a unique point $X'\in [XO)$ such that $XX'=2\cdot XO$. 349 | \qeds 350 | 351 | The point $X'$ provided by the lemma is called the \index{reflection across a point}\emph{reflection} of point $X$ across $O$. 352 | We also assume that $O'=O$; that is, $O$ is a reflection of itself across itself. 353 | 354 | The point $O$ is called the \index{center!of reflection}\emph{center of reflection}. 355 | 356 | \begin{thm}[\abs]{Proposition}\label{prop:point-reflection} 357 | A reflection across a point is a motion of the plane. 358 | \end{thm} 359 | 360 | \parit{Proof.} 361 | Let $O$ be the center of reflection. 362 | Observe that if $X'$ is a reflection of $X$ across $O$, 363 | then $X$ is a reflection of $X'$. 364 | In other words, the composition of the reflection with itself is the identity map. 365 | In particular, the reflection is a bijection. 366 | 367 | \begin{wrapfigure}{o}{33mm} 368 | \vskip-6mm 369 | \centering 370 | \includegraphics{mppics/pic-76} 371 | \end{wrapfigure} 372 | 373 | Now choose two points $X$ and $Y$; 374 | let $X'$ and $Y'$ be their reflections across $O$. 375 | To check that the reflection is distance preserving, we need to show that $X'Y'=XY$. 376 | 377 | We may assume that $X$, $Y$, and $O$ are distinct; otherwise, the statement is trivial. 378 | By the definition of the reflection, we have that $OX=OX'$, $OY=OY'$. 379 | Note also that the angles $XOY$ and $X'OY'$ are vertical. 380 | By \ref{prop:vert}, $\measuredangle XOY\z=\measuredangle X'OY'$. 381 | 382 | Now Axiom~\ref{def:birkhoff-axioms:3} implies that $\triangle XOY\cong\triangle X'OY'$ and $X'Y'=XY$. 383 | \qeds 384 | 385 | \begin{thm}{Exercise}\label{ex:refelection-of-line} 386 | Show that the reflection across a point maps any line to a line. 387 | \end{thm} 388 | -------------------------------------------------------------------------------- /triangle.tex: -------------------------------------------------------------------------------- 1 | \chapter{Triangle geometry}\label{chap:triangle} 2 | 3 | Triangle geometry is the study of the properties of triangles, including associated centers and circles. 4 | 5 | We discuss the most basic results in triangle geometry, 6 | mostly to show that we have developed sufficient machinery to prove things. 7 | 8 | \section{Circumcircle and circumcenter} 9 | 10 | \begin{thm}{Theorem}\label{thm:circumcenter} 11 | Perpendicular bisectors to the sides of every nondegenerate triangle intersect at one point. 12 | \end{thm} 13 | 14 | The point of intersection of the perpendicular bisectors is called the \index{circumcenter}\emph{circumcenter}. 15 | It is the center of the \index{circumcircle}\emph{circumcircle} of the triangle; 16 | that is, a circle that passes thru all three vertices of the triangle. 17 | The circumcenter of the triangle is usually denoted by~$O$. 18 | 19 | \begin{wrapfigure}{o}{29mm} 20 | \centering 21 | \includegraphics{mppics/pic-102} 22 | \end{wrapfigure} 23 | 24 | 25 | \parit{Proof.} 26 | Let $\triangle ABC$ be nondegenerate. 27 | Let $\ell$ and $m$ be perpendicular bisectors to sides $[AB]$ and $[AC]$ respectively. 28 | 29 | Assume $\ell$ intersects $m$, say, at $O$. 30 | 31 | Let us apply Theorem~\ref{thm:perp-bisect}. 32 | Since $O\in\ell$, we have $OA\z=OB$ and since $O\in m$, we have $OA\z=OC$. 33 | It follows that $OB\z=OC$; 34 | that is, $O$ lies on the perpendicular bisector of~$[B C]$. 35 | 36 | It remains to show that $\ell\nparallel m$. 37 | Suppose, to the contrary, that $\ell\parallel m$. 38 | Since $\ell\perp(AB)$ and $m\z\perp(AC)$, Exercise~\ref{ex:perp-perp}\textit{\ref{ex:perp-perp:b}} implies that 39 | $(AC)\parallel(AB)$, and by Theorem~\ref{thm:parallel} we have $(AC)=(AB)$. 40 | Hence, $\triangle ABC$ is degenerate --- a contradiction. 41 | \qeds 42 | 43 | \begin{thm}[!]{Exercise}\label{ex:unique-cline} 44 | Show that there is a unique circle that passes thru the vertices of a given nondegenerate triangle in the Euclidean plane. 45 | \end{thm} 46 | 47 | 48 | 49 | \section{Altitudes and orthocenter} 50 | 51 | An \index{altitude}\emph{altitude} of a triangle is a line thru a vertex and perpendicular to the line containing the opposite side. 52 | The term \index{altitude}\emph{altitude} may also be used for the distance from the vertex to its footpoint on the line containing the opposite side. 53 | 54 | \begin{thm}{Theorem}\label{thm:orthocenter} 55 | Three altitudes of every nondegenerate triangle intersect at a single point. 56 | \end{thm} 57 | 58 | The point of intersection of altitudes is called the \index{orthocenter}\emph{orthocenter}; 59 | it is usually denoted by~$H$. 60 | 61 | { 62 | 63 | \begin{wrapfigure}{o}{34mm} 64 | \vskip-4mm 65 | \centering 66 | \includegraphics{mppics/pic-104} 67 | \end{wrapfigure} 68 | 69 | \parit{Proof.} 70 | Fix a nondegenerate triangle $A B C$. 71 | Consider three lines $\ell$, $m$, and $n$ 72 | such that 73 | \begin{align*} 74 | \ell&\parallel(BC), 75 | & 76 | m&\parallel(CA), 77 | & 78 | n&\parallel(AB), 79 | \\ 80 | \ell&\ni A, 81 | & 82 | m&\ni B, 83 | & 84 | n&\ni C. 85 | \end{align*} 86 | Since $\triangle A B C$ is nondegenerate, 87 | no pair of the lines $\ell$, $m$, and $n$ is parallel. 88 | Let $A'$, $B'$, and $C'$ be the points of intersection of 89 | $m$ and $n$, $n$ and $\ell$, and $\ell$ and $m$, respectively. 90 | 91 | } 92 | 93 | Note that $\square A B C B'$, $\square B C A C'$, and $\square C A B A'$ are parallelograms. 94 | Applying Lemma~\ref{lem:parallelogram} we get that $\triangle ABC$ is the \index{medial triangle}\emph{medial triangle} of $\triangle A' B' C'$; 95 | that is, $A$, $B$, and $C$ are the midpoints of $[B' C']$, $[C' A']$, and $[A' B']$ respectively. 96 | 97 | Since $(B' C')\parallel (BC)$, Exercise~\ref{ex:perp-perp}\textit{\ref{ex:perp-perp:a}} implies that the altitude from $A$ is perpendicular to $[B' C']$, and from above, it bisects~$[B' C']$. 98 | 99 | Hence the altitudes of $\triangle A B C$ 100 | are also perpendicular bisectors of $\triangle A' B' C'$. 101 | Applying Theorem~\ref{thm:circumcenter}, we get that altitudes of $\triangle ABC$ intersect at one point. 102 | \qeds 103 | 104 | \begin{thm}{Exercise}\label{ex:orthic-4} 105 | Assume that the orthocenter $H$ of $\triangle ABC$ is distinct from its vertices. 106 | Show that $A$ is the orthocenter of $\triangle H B C$. 107 | \end{thm} 108 | 109 | \begin{thm}{Exercise}\label{ex:orthic-sim} 110 | Let $A'$, $B'$, and $C'$ be the footpoints of the corresponding altitudes of an acute triangle $ABC$. 111 | Show that 112 | \[ 113 | \triangle ABC \sim \triangle A'B'C \sim \triangle AB'C' \sim \triangle A'BC'. 114 | \] 115 | \end{thm} 116 | 117 | 118 | 119 | \section{Medians and centroid} 120 | 121 | A median of a triangle is the segment joining a vertex to the midpoint of the opposite side. 122 | 123 | \begin{thm}{Theorem}\label{thm:centroid} 124 | Three medians of every nondegenerate triangle intersect at a single point. 125 | Moreover, the point of intersection divides each median in the ratio 2:1. 126 | \end{thm} 127 | 128 | The point of intersection of medians is called the \index{centroid}\emph{centroid} of the triangle; 129 | it is usually denoted by~$M$. 130 | In the proof, we will apply exercises \ref{ex:chevinas} and \ref{ex:smililar+parallel}; their complete solutions are given in the hints. 131 | 132 | \parit{Proof.} 133 | Consider a nondegenerate triangle $A B C$. 134 | Let $[A A']$ and $[B B']$ be its medians. 135 | According to Exercise~\ref{ex:chevinas}, 136 | $[A A']$ and $[B B']$ have a point of intersection; 137 | denote it by $M$. 138 | 139 | \begin{wrapfigure}{o}{36mm} 140 | \vskip-4mm 141 | \centering 142 | \includegraphics{mppics/pic-106} 143 | \end{wrapfigure} 144 | 145 | Draw a line $\ell$ thru $A'$ parallel to $(BB')$. 146 | Applying Exercise~\ref{ex:smililar+parallel} for $\triangle BB'C$ and $\ell$, we get that $\ell$ crosses $[B'C]$ at a point, say $X$, and 147 | \[\frac{CX}{CB'}=\frac{CA'}{CB}=\frac12;\] 148 | that is, $X$ is the midpoint of $[CB']$. 149 | 150 | Since $B'$ is the midpoint of $[AC]$ and $X$ is the midpoint of $[B'C]$, we get that 151 | \[\frac{AB'}{AX}=\frac23.\] 152 | 153 | Applying Exercise~\ref{ex:smililar+parallel} for $\triangle XA'A$ and the line $(BB')$, we get that 154 | \[\frac{AM}{AA'}=\frac{AB'}{AX}=\frac23;\eqlbl{eq:2/3}\] 155 | that is, $M$ divides $[AA']$ in the ratio 2:1. 156 | 157 | Condition \ref{eq:2/3} uniquely defines $M$ on $[AA']$. 158 | Repeating the same argument for medians $[AA']$ and $[CC']$, we get that they intersect at~$M$ as well, 159 | hence the result. 160 | \qeds 161 | 162 | 163 | \begin{thm}{Exercise}\label{ex:midle} 164 | Let $\square ABCD$ be a nondegenerate quadrangle 165 | and $X$, $Y$, $V$, and~$W$ be midpoints of 166 | $[AB]$, $[BC]$, $[CD]$, and~$[DA]$ respectively. 167 | Show that $\square XYVW$ is a parallelogram. 168 | \end{thm} 169 | 170 | \begin{thm}{Advanced exercise}\label{ex:euler-line} 171 | Show that the orthocenter $H$, centroid $M$, and circumcenter $O$ of every nondegenerate triangle $ABC$ lie on a single line. 172 | Moreover, the centroid divides the segment $[HO]$ in the ratio $2:1$. 173 | \end{thm} 174 | 175 | The line in this exercise is called \emph{Euler's line}. 176 | 177 | 178 | \section{Angle bisectors} 179 | 180 | If $\measuredangle A B X\equiv-\measuredangle C B X$, 181 | then we say that the line $(BX)$ {}\emph{bisects} $\angle ABC$, 182 | or the line $(BX)$ is a \index{bisector!angle bisector}\emph{bisector} of $\angle ABC$. 183 | If $\measuredangle A B X\equiv\pi-\measuredangle C B X$, then the line $(BX)$ is called the \index{bisector!external bisector}\emph{external bisector} of $\angle ABC$. 184 | 185 | 186 | \begin{wrapfigure}{o}{42mm} 187 | \centering 188 | \includegraphics{mppics/pic-108} 189 | \end{wrapfigure} 190 | 191 | If $\measuredangle ABA'=\pi$; 192 | that is, if $B$ lies between $A$ and $A'$, 193 | then the bisector of $\angle ABC$ is the external bisector of $\angle A' B C$ and the other way around. 194 | 195 | Note that the bisector and the external bisector are uniquely defined by the angle. 196 | 197 | \begin{thm}{Classroom exercise}\label{ex:perp-bisectors} 198 | Show that for every angle, its bisector and external bisector are perpendicular. 199 | \end{thm} 200 | 201 | The bisectors of $\angle ABC$, $\angle BCA$, and $\angle CAB$ of a nondegenerate triangle $A B C$ 202 | are called \index{bisector!of a triangle}\emph{bisectors of the triangle} $A B C$ at vertices $A$, $B$, and $C$ respectively. 203 | 204 | \begin{thm}{Exercise}\label{ex:bisect=altitude} 205 | Assume that, at one vertex of a nondegenerate triangle, its bisector coincides with its altitude. 206 | Show that the triangle is isosceles. 207 | \end{thm} 208 | 209 | \begin{thm}{Lemma}\label{lem:bisect-ratio} 210 | Let $\triangle A B C$ be a nondegenerate triangle. 211 | Assume that the bisector at $A$ 212 | intersects $[BC]$ at~$D$. 213 | Then 214 | $$\frac{AB}{AC}=\frac{DB}{DC}. 215 | \eqlbl{bisect-ratio}$$ 216 | 217 | \end{thm} 218 | 219 | \begin{wrapfigure}{r}{28mm} 220 | \vskip-6mm 221 | \centering 222 | \includegraphics{mppics/pic-110} 223 | \end{wrapfigure} 224 | 225 | \parit{Proof.} 226 | Let $\ell$ be a line passing thru $C$ that is parallel to~$(AB)$. 227 | Note that the lines $\ell$ and $(AD)$ are not parallel; 228 | let $E$ be their point of intersection. 229 | 230 | Note also that $B$ and $C$ lie on opposite sides of~$(AD)$. 231 | By the transversal property (\ref{thm:parallel-2}), 232 | $$\measuredangle BAD=\measuredangle CED.\eqlbl{eq:0$. 37 | 38 | For a point $P'\in \Sigma$, consider its footpoint $\hat P$ 39 | on $\Pi$; 40 | this is the closest point to~$P'$. 41 | 42 | The composition of these two maps $P\z\leftrightarrow P'\z\leftrightarrow\hat P$ 43 | gives a bijection from the h-plane to itself. 44 | Furthermore, $P=\hat P$ 45 | if and only if $P\in \Omega$ or $P=O$. 46 | 47 | %\begin{thm}{Exercise}\label{ex:P-->hat-P} 48 | %Suppose that $P\leftrightarrow \hat P$ is the bijection described above. 49 | %Assume that $P$ is a point of the h-plane distinct from the center of the absolute and $Q$ is its inverse across the absolute. 50 | %Show that the inversion of $\hat P$ across the absolute is the midpoint of $[PQ]$. 51 | %\end{thm} 52 | 53 | \begin{thm}{Lemma}\label{lem:P-hat-chord} 54 | Let $P\leftrightarrow\hat P$ be the bijection of the h-plane described above. 55 | Suppose $(PQ)_h$ is an h-line with the ideal points $A$ and~$B$. 56 | Then $\hat P,\hat Q\in[AB]$. 57 | 58 | Moreover, 59 | $$\frac{A\hat Q\cdot B\hat P}{\hat QB\cdot \hat PA} 60 | = 61 | \left(\frac{AQ\cdot BP}{QB\cdot PA}\right)^2. 62 | \eqlbl{eq:lem:P-hat-chord}$$ 63 | In particular, if $A,P,Q,B$ appear in the same order, then 64 | $$PQ_h=\tfrac12\cdot\ln\frac{A\hat Q\cdot B\hat P}{\hat QB\cdot \hat PA}.$$ 65 | \end{thm} 66 | 67 | \parit{Proof.} 68 | Consider the stereographic projection $\Pi\to \Sigma$ from the south pole~$S$. 69 | It fixes $A$ and $B$; 70 | denote by $P'$ and $Q'$ the images of $P$ and~$Q$; 71 | 72 | According to Theorem~\ref{thm:inversion-3d}\textit{\ref{thm:inversion-3d:cross-ratio}}, 73 | $$\frac{AQ\cdot BP}{QB\cdot PA}=\frac{AQ'\cdot BP'}{Q'B\cdot P'A}.\eqlbl{eq:(AB;PQ)=(AB;P'Q')}$$ 74 | 75 | By Theorem~\ref{thm:inversion-3d}\textit{\ref{thm:inversion-3d:angle}}, 76 | each circline in $\Pi$ that is perpendicular to $\Omega$ 77 | is mapped to a circle in $\Sigma$ that is still perpendicular to~$\Omega$. 78 | It follows that the stereographic projection sends $(PQ)_h$ to the intersection of the northern half-sphere of $\Sigma$ with a plane perpendicular to~$\Pi$. 79 | 80 | Denote this plane by $\Lambda$; 81 | it contains the points $A$, $B$, $P'$, $\hat P$ and the circle $\Gamma=\Sigma\cap\Lambda$. 82 | (It also contains $Q'$ and $\hat Q$ but we will not use these points for a while.) 83 | 84 | { 85 | 86 | \begin{wrapfigure}{r}{30mm} 87 | \vskip-0mm 88 | \centering 89 | \includegraphics{mppics/pic-258} 90 | \caption*{Plane $\Lambda$.} 91 | \end{wrapfigure} 92 | 93 | Note that 94 | \begin{itemize} 95 | \item 96 | $A,B,P'\in\Gamma$, 97 | \item $[AB]$ is a diameter of $\Gamma$, 98 | \item $(AB)=\Pi\cap\Lambda$, 99 | \item $\hat P\in [AB]$, 100 | \item $(P'\hat P)\perp (AB)$. 101 | \end{itemize} 102 | 103 | 104 | 105 | Since $[AB]$ is the diameter of $\Gamma$, 106 | by Corollary~\ref{cor:right-angle-diameter}, 107 | the angle $AP'B$ is right. 108 | Hence $\triangle A\hat PP'\z\sim \triangle AP'B\z\sim \triangle P'\hat PB$. 109 | In particular 110 | $$\frac{AP'}{BP'}=\frac{A\hat P}{P'\hat P}=\frac{P'\hat P}{B\hat P}.$$ 111 | Therefore 112 | $$\frac{A\hat P}{B\hat P}=\left(\frac{AP'}{BP'}\right)^2 113 | \quad\text{and similarly}\quad 114 | \frac{A\hat Q}{B\hat Q}=\left(\frac{AQ'}{BQ'}\right)^2. 115 | \eqlbl{eq:AP/BP}$$ 116 | Finally, 117 | \ref{eq:(AB;PQ)=(AB;P'Q')}+\ref{eq:AP/BP} imply \ref{eq:lem:P-hat-chord}. 118 | 119 | } 120 | 121 | The last statement follows from \ref{eq:lem:P-hat-chord} and the definition of the h-distance. 122 | Indeed, 123 | \begin{align*} 124 | PQ_h&\df\ln\frac{A Q\cdot B P}{QB\cdot PA}= 125 | \\ 126 | &=\ln\left(\frac{A \hat Q\cdot B \hat P}{\hat QB\cdot \hat PA}\right)^{\frac12}= 127 | \\ 128 | &=\tfrac12\cdot\ln\frac{A \hat Q\cdot B \hat P}{\hat QB\cdot \hat PA}. 129 | \end{align*} 130 | \qedsf 131 | 132 | { 133 | 134 | \begin{wrapfigure}[8]{r}{36mm} 135 | \centering 136 | \vskip-8mm 137 | \includegraphics{mppics/pic-260} 138 | \end{wrapfigure} 139 | 140 | \begin{thm}{Exercise}\label{ex:hex} 141 | Let $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ 142 | be three circles perpendicular to a circle~$\Omega$. 143 | Let $[A_1B_1]$, $[A_2B_2]$, and $[A_3B_3]$ denote 144 | the common chords of $\Omega$ and $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ respectively. 145 | Show that the chords $[A_1B_1]$, $[A_2B_2]$, and $[A_3B_3]$ intersect at one point inside $\Omega$ if and only if $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ intersect at two points. 146 | \end{thm} 147 | 148 | \begin{thm}{Exercise}\label{ex:P<->hatP} 149 | Show that $2\cdot OP_h=O\hat P_h$, where $O$ is the center of the absolute, and $P\leftrightarrow \hat P$ is the bijection of the h-plane described above. 150 | \end{thm} 151 | 152 | 153 | } 154 | 155 | \section{The projective model} 156 | \label{sec:proj-model} 157 | 158 | The following picture illustrates the bijection $P\mapsto \hat P$ of the h-plane described in the previous section --- 159 | \begin{figure}[!ht] 160 | \centering 161 | \includegraphics{mppics/pic-262} 162 | \end{figure} 163 | if you take the picture on the left and apply the map $P\z\mapsto \hat P$, 164 | you get the picture on the right. 165 | The pictures are {}\emph{conformal} and \index{projective!model}\emph{projective models} of the hyperbolic plane respectively. 166 | The map $P\mapsto \hat P$ serves as a \textit{translation} between the models. 167 | 168 | In the projective model, things look different; 169 | some become simpler, and others become more complex. 170 | 171 | \parbf{Lines.} 172 | In the projective model, the h-lines are represented as chords of the absolute, more precisely, chords without their endpoints. 173 | 174 | This observation can be used to transfer statements about lines and points from the Euclidean plane to the h-plane. 175 | As an example let us state a version of Pappus' theorem for h-plane. 176 | 177 | \begin{thm}{Hyperbolic Pappus' theorem}\label{thm:pappus-h}\index{Pappus' theorem} 178 | Let $A$, $B$, $C$ and $A'$, $B'$, $C'$ be two triples of h-collinear h-points, and let $X$, $Y$, and $Z$ be the intersection points of $(BC')_h$ and $(B'C)_h$, $(CA')_h$ and $(C'A)_h$, 179 | and $(AB')_h$ and $(A'B)_h$, respectively. 180 | Then the points $X$, $Y$, $Z$ are h-collinear. 181 | \end{thm} 182 | 183 | In the projective model, this statement follows immediately from the original Pappus' theorem \ref{thm:pappus}. 184 | The same can be done for Desargues' theorem \ref{thm:desargues}. 185 | The same argument shows that the construction of a tangent line with a ruler only described in Exercise~\ref{ex:tangent ruler} works in the h-plane as well. 186 | These statements are not at all evident in the conformal model. 187 | 188 | \parbf{Circles and equidistants.} 189 | In the projective model, the h-circles and equidistants are ellipses and their open arcs. 190 | This follows since the stereographic projection sends circles on the plane to circles on the unit sphere and the footpoint projection of the circle back to the plane is an ellipse. 191 | (One may define an \index{ellipse}\emph{ellipse} as a footpoint projection of a circle.) 192 | 193 | 194 | 195 | \parbf{Distance.} 196 | Consider a pair of h-points $P$ and $Q$. 197 | Let $A$ and $B$ be the ideal points of the h-line in the projective model; 198 | that is, $A$ and $B$ are the intersections of the Euclidean line $(PQ)$ with the absolute. 199 | 200 | \begin{wrapfigure}{o}{36mm} 201 | \vskip-2mm 202 | \centering 203 | \includegraphics{mppics/pic-264} 204 | \vskip-2mm 205 | \end{wrapfigure} 206 | 207 | Then by Lemma~\ref{lem:P-hat-chord}, 208 | $$PQ_h=\tfrac12\cdot\ln\frac{AQ\cdot BP}{QB\cdot PA},\eqlbl{eq:proj-h-dist}$$ 209 | assuming the points $A, P, Q, B$ appear on the line in the same order. 210 | 211 | \parbf{Angles.} 212 | The hyperbolic angle measure in the projective model is very different from Euclidean, making it hard to determine from the picture. 213 | \label{klein-angles} 214 | For example, all the intersecting h-lines on the picture 215 | are perpendicular. 216 | 217 | To find the angle measure, 218 | one may apply a motion of the h-plane that moves 219 | the vertex of the angle to the center of the absolute; 220 | once done, the hyperbolic and Euclidean angles have the same measure. 221 | In particular, if $O$ is the center of the absolute, then 222 | $$\measuredangle_hAOB=\measuredangle AOB.$$ 223 | 224 | \begin{thm}{Observation}\label{obs:h-p-perp} 225 | If $O$ is the center of the absolute, then 226 | \[(OA)\z\perp (AB)\qquad\Longleftrightarrow\qquad(OA_h)\perp (AB)_h.\] 227 | 228 | \end{thm} 229 | 230 | \parit{Proof.} 231 | The Euclidean reflection across $(OA)$ induces the h-reflection across $(OA)_h$. 232 | Let $B'$ be the reflection of $B$. 233 | Observe that $(OA)\perp (AB)$ $\Longleftrightarrow$ $A\in(BB')$ $\Longleftrightarrow$ $A\in(BB')_h$ $\Longleftrightarrow$ $(OA_h)\perp (AB)_h$. 234 | \qeds 235 | 236 | \parbf{Motions.} 237 | The motions of the h-plane in the conformal and projective models are relevant to inversive transformations and projective transformations in the same way. 238 | Namely: 239 | \begin{itemize} 240 | \item Inversive transformations that preserve the h-plane describe motions of the h-plane in the conformal model. 241 | \item Projective transformations that preserve the h-plane describe motions of the h-plane in the projective model.\footnote{Exercise~\ref{ex:cone} 242 | gives a rich source of such projective transformations, which will be used in the proof of \ref{thm:circle-center-proj}.} 243 | \end{itemize} 244 | 245 | The following exercise is a hyperbolic analog of Exercise~\ref{ex:s-medians}. 246 | 247 | \begin{thm}{Exercise}\label{ex:h-median} 248 | Let $P$ and $Q$ be points in the h-plane that lie at the same distance from the center of the absolute. 249 | Observe that in the projective model, the h-midpoint of $[PQ]_h$ coincides with the Euclidean midpoint of $[PQ]$. 250 | 251 | Conclude that if an h-triangle is inscribed in an h-circle, then its medians meet at one point. 252 | 253 | Recall that an h-triangle might also be inscribed in a horocycle or an equidistant. 254 | Think about how to prove the statement in this case. 255 | \end{thm} 256 | 257 | \begin{thm}{Exercise}\label{ex:h-altitudes} 258 | Show that the altitudes of a hyperbolic triangle either intersect at one point or are pairwise disjoint. 259 | \end{thm} 260 | 261 | 262 | { 263 | 264 | \begin{wrapfigure}[4]{r}{34mm} 265 | \vskip-11mm 266 | \centering 267 | \includegraphics{mppics/pic-266} 268 | \end{wrapfigure} 269 | 270 | \begin{thm}{Exercise}\label{ex:klein-perp} 271 | Let $\ell$ and $m$ be h-lines in the projective model. 272 | Let $s$ and $t$ denote the Euclidean lines tangent to the absolute 273 | at the ideal points of $\ell$. 274 | Show that 275 | if the lines $s$, $t$, and the extension of $m$ meet at one point, then $\ell\perp m$. 276 | \end{thm} 277 | 278 | } 279 | 280 | \begin{thm}{Exercise}\label{ex:klein-for-angle-parallelism} 281 | Use the projective model to derive the formula for the angle of parallelism (Proposition~\ref{prop:angle-parallelism}). 282 | \end{thm} 283 | 284 | \begin{thm}{Exercise}\label{ex:klein-inradius} 285 | Use the projective model to find an inradius of an ideal triangle. 286 | \end{thm} 287 | 288 | \begin{thm}{Advanced exercise}\label{ex:pyth-h-proj} 289 | Prove the hyperbolic Pythagorean theorem (\ref{thm:pyth-h-poincare}) using the following idea. 290 | \end{thm} 291 | 292 | Recall that the hyperbolic Pythagorean theorem\index{Pythagorean theorem} (\ref{thm:pyth-h-poincare}) 293 | states that if $\triangle_hACB$ is an h-triangle with a right angle at~$C$, then 294 | \[\cosh c=\cosh a\cdot\cosh b, 295 | \eqlbl{eq:hyp-pyth-proj}\] 296 | where $a\z=BC_h$, $b=CA_h$, and $c=AB_h$. 297 | 298 | \begin{wrapfigure}{r}{34mm} 299 | \vskip-12mm 300 | \centering 301 | \includegraphics{mppics/pic-268} 302 | \end{wrapfigure} 303 | 304 | \parit{Idea.} 305 | We can assume that $A$ is the center of the absolute. 306 | By \ref{obs:h-p-perp} the Euclidean triangle $ABC$ is right. 307 | Set 308 | $s=BC$, $t =CA$, $u\z= AB$. 309 | According to the Euclidean Pythagorean theorem (\ref{thm:pyth}), we have 310 | $$u^2=s^2+t^2.\eqlbl{eq:hyp-proj}$$ 311 | It remains to express $a$, $b$, and $c$ using $s$, $u$, and $t$ and show that \ref{eq:hyp-proj} implies~\ref{eq:hyp-pyth-proj}. 312 | 313 | \section{Bolyai's construction} 314 | 315 | Assume we need to construct a line thru $P$ that is asymptotically parallel to the given line $\ell$ in the h-plane. 316 | 317 | If $A$ and $B$ are ideal points of $\ell$ in the projective model, 318 | then we could simply draw the Euclidean line $(PA)$. 319 | However, the ideal points do not lie in the h-plane; therefore there is no way to use them in the construction. 320 | 321 | In the following construction we assume that you know a ruler-and-compass construction of the perpendicular line; see Exercise~\ref{ex:construction-perpendicular}. 322 | 323 | \begin{thm}{Bolyai's construction} 324 | \begin{enumerate} 325 | \item Drop a perpendicular from $P$ to~$\ell$; denote it by~$m$. 326 | Let $Q$ be the footpoint of $P$ on~$\ell$. 327 | \item Erect a perpendicular from $P$ to~$m$; denote it by~$n$. 328 | \item Mark a point $R$ on $\ell$ distinct from $Q$. 329 | \item Drop a perpendicular from $R$ to~$n$; denote it by~$k$. 330 | \item Draw the circle $\Gamma$ with center $P$ and the radius $QR$. 331 | Mark a point of intersection of $\Gamma$ with~$k$; denote it by $T$. 332 | \item The line $(PT)_h$ is asymptotically parallel to~$\ell$. 333 | \end{enumerate} 334 | \end{thm} 335 | 336 | \begin{thm}{Exercise}\label{ex:Boyai-in-Euclid} 337 | Explain what happens if one performs the Bolyai construction in the Euclidean plane. 338 | \end{thm} 339 | 340 | The following proposition implies that Bolyai's construction works. 341 | 342 | \begin{thm}{Proposition}\label{prop:boyai} 343 | Assume $P$, $Q$, $R$, $S$, $T$ are points in the h-plane 344 | such that 345 | $S\in (RT)_h$, 346 | $(PQ)_h\perp (QR)_h$, 347 | $(PS)_h\perp(PQ)_h$, 348 | $(RT)_h\perp (PS)_h$ and 349 | $(PT)_h$ and $(QR)_h$ are asymptotically parallel. 350 | Then $QR_h=PT_h$. 351 | \end{thm} 352 | 353 | 354 | \parit{Proof.} 355 | We will use the projective model. 356 | Without loss of generality, we may assume that $P$ is the center of the absolute. 357 | By~\ref{obs:h-p-perp}, 358 | the corresponding Euclidean lines are also perpendicular; 359 | that is, $(PQ)\perp (QR)$, $(PS)\perp(PQ)$, and $(RT)\z\perp (PS)$. 360 | 361 | Let $A$ be the common ideal point of $(QR)_h$ and $(PT)_h$. 362 | Let $B$ and $C$ denote the remaining ideal points of $(QR)_h$ and $(PT)_h$ respectively. 363 | 364 | The Euclidean lines $(PQ)$, $(TR)$, and $(CB)$ are parallel. 365 | Therefore, 366 | \[\triangle AQP\sim \triangle ART \sim\triangle ABC.\] 367 | 368 | \begin{wrapfigure}{o}{55mm} 369 | \vskip-0mm 370 | \centering 371 | \includegraphics{mppics/pic-270} 372 | \vskip2mm 373 | \end{wrapfigure} 374 | 375 | In particular, 376 | \[\frac{AC}{AB}=\frac{AT}{AR}=\frac{AP}{AQ}.\] 377 | It follows that 378 | \[\frac{AT}{AR}=\frac{AP}{AQ}=\frac{BR}{CT}=\frac{BQ}{CP}.\] 379 | In particular, 380 | \[\frac{AT\cdot CP}{TC\cdot PA}=\frac{AR\cdot BQ}{RB\cdot QA}.\] 381 | Applying the formula for h-distance \ref{eq:proj-h-dist}, we get that $QR_h=PT_h$. 382 | \qeds 383 | 384 | \begin{wrapfigure}{r}{35mm} 385 | \vskip-6mm 386 | \centering 387 | \includegraphics{mppics/pic-271} 388 | \vskip2mm 389 | \end{wrapfigure} 390 | 391 | \begin{thm}{Advanced exercise}\label{ex:common-perp} 392 | Look at the picture and recover the construction of a common perpendicular $n$ to the h-lines $\ell$ and $m$. 393 | 394 | Prove that it works. 395 | \end{thm} 396 | -------------------------------------------------------------------------------- /half-planes.tex: -------------------------------------------------------------------------------- 1 | \chapter{Half-planes}\label{chap:half-planes} 2 | 3 | This chapter contains lengthy proofs of intuitively evident statements. 4 | It is okay to skip it, but make sure you understand the definitions of positive/negative angles and that your intuition agrees with statements \ref{thm:signs-of-triug}, \ref{prop:half-plane}, \ref{cor:half-plane}, \ref{thm:pasch}, and \ref{thm:abc}. 5 | 6 | \section{Sign of an angle} 7 | 8 | Positive and negative angles can be visualized as {}\emph{counterclockwise} and {}\emph{clockwise} directions; they are defined the following way: 9 | \begin{itemize} 10 | \item The angle $A O B$ is called \index{angle!positive and negative angles}\emph{positive} 11 | if $0<\measuredangle A O B<\pi$; 12 | \item The angle $A O B$ is called {}\emph{negative} 13 | if $\measuredangle A O B<0$. 14 | \end{itemize} 15 | 16 | According to the above definitions the straight angle, as well as the zero angle, 17 | are neither positive nor negative. 18 | 19 | \begin{thm}{Classroom exercise}\label{ex:AOB+<=>BOA-} 20 | Show that $\angle A O B$ is positive if and only if $\angle B O A$ is negative. 21 | \end{thm} 22 | 23 | \begin{thm}{Lemma}\label{lem:straight-sign} 24 | Let $\angle AOB$ be straight. 25 | Then $\angle AOX$ is positive 26 | if and only if $\angle BOX$ is negative. 27 | \end{thm} 28 | 29 | \parit{Proof.} 30 | Set $\alpha=\measuredangle AOX$ 31 | and 32 | $\beta=\measuredangle BOX$. 33 | Since $\angle AOB$ is straight, 34 | $$\alpha-\beta\equiv \pi.\eqlbl{eq:alpha-beta}$$ 35 | 36 | It follows that $\alpha=\pi$ $\Leftrightarrow$ $\beta=0$ 37 | and $\alpha=0$ $\Leftrightarrow$ $\beta=\pi$. 38 | In these two cases, the signs of $\angle AOX$ and $\angle BOX$ are undefined. 39 | 40 | In the remaining cases we have that $|\alpha|<\pi$ and $|\beta|<\pi$. 41 | If $\alpha$ and $\beta$ have the same sign, then $|\alpha-\beta|<\pi$; 42 | the latter contradicts \ref{eq:alpha-beta}. 43 | Hence the statement follows. 44 | \qeds 45 | 46 | \begin{thm}{Exercise}\label{ex:PP(PN)} 47 | Assume that angles $ABC$ and $A'B'C'$ have the same sign 48 | and 49 | \[2\cdot \measuredangle ABC\equiv 2\cdot \measuredangle A'B'C'.\] 50 | Show that $\measuredangle ABC= \measuredangle A'B'C'$. 51 | \end{thm} 52 | 53 | \section{Intermediate value theorem} 54 | 55 | \begin{thm}{Intermediate value theorem}\label{thm:intermidiate} 56 | Let $f\:[a,b]\to \mathbb{R}$ be a continuous function. 57 | Assume 58 | $f(a)$ and $f(b)$ have opposite signs. 59 | Then $f(t_0)=0$ for some $t_0\in[a,b]$. 60 | \end{thm} 61 | 62 | \begin{wrapfigure}{r}{38mm} 63 | \vskip-6mm 64 | \centering 65 | \includegraphics{mppics/pic-14} 66 | \end{wrapfigure} 67 | 68 | The intermediate value theorem is assumed to be known; 69 | it should be covered in any calculus course. 70 | We will use only the following corollary: 71 | 72 | \begin{thm}[\abs]{Corollary}\label{cor:intermidiate} 73 | Assume that for any $t\z\in [0,1]$ we have three points $O_t$, $A_t$, and $B_t$ in the plane such that 74 | \begin{enumerate}[(a)] 75 | \item Each function $t\mapsto O_t$, $t\mapsto A_t$, and $t\z\mapsto B_t$ is continuous. 76 | \end{enumerate} 77 | 78 | \begin{enumerate}[(a)]\addtocounter{enumi}{1} 79 | \item For any $t\in [0,1]$, the points $O_t$, $A_t$, and $B_t$ do not lie on one line. 80 | \end{enumerate} 81 | Then $\angle A_0O_0B_0$ and $\angle A_1O_1B_1$ have the same sign. 82 | \end{thm} 83 | 84 | \parit{Proof.} 85 | Consider the function 86 | $f(t)=\measuredangle A_tO_tB_t$. 87 | 88 | Since the points $O_t$, $A_t$, and $B_t$ do not lie on one line, Theorem~\ref{thm:straight-angle} implies that $f(t)=\measuredangle A_tO_tB_t$ is neither $0$ nor $\pi$ for any $t\in[0,1]$. 89 | 90 | Therefore, by Axiom~\ref{def:birkhoff-axioms:2c} and Exercise~\ref{ex:comp+cont}, 91 | $f$ is a continuous function. 92 | By the intermediate value theorem, $f(0)$ and $f(1)$ have the same sign; 93 | hence the result follows. 94 | \qeds 95 | 96 | 97 | \section{Same sign lemmas} 98 | 99 | 100 | 101 | \begin{thm}[\abs]{Lemma}\label{lem:signs} 102 | Assume $Q'\in [PQ)$ and $Q'\z\ne P$. 103 | Then for every $X\z\notin (PQ)$ the angles $PQX$ and $PQ'X$ have the same sign. 104 | \end{thm} 105 | 106 | { 107 | 108 | \begin{wrapfigure}{o}{33mm} 109 | \centering 110 | \vskip-5mm 111 | \includegraphics{mppics/pic-16} 112 | \end{wrapfigure} 113 | 114 | \parit{Proof.} 115 | By Exercise~\ref{ex:trig==}, 116 | for any $t\in [0,1]$ there is a unique point $Q_t\in[PQ)$ 117 | such that 118 | \[PQ_t= (1-t)\cdot PQ+t\cdot PQ'.\] 119 | The map $t\mapsto Q_t$ is continuous, 120 | \begin{align*} 121 | Q_0&=Q, 122 | & 123 | Q_1&=Q' 124 | \end{align*} 125 | and for any $t\in [0,1]$, 126 | we have that $P\z\ne Q_t$. 127 | 128 | } 129 | 130 | Applying Corollary~\ref{cor:intermidiate}, 131 | for $P_t=P$, $Q_t$, and $X_t=X$, we get that $\angle PQX$ has the same sign as $\angle PQ'X$. 132 | \qeds 133 | 134 | 135 | {\sloppy 136 | 137 | \begin{thm}[\abs]{Signs of angles of a triangle}\label{thm:signs-of-triug} 138 | In any nondegenerate triangle $ABC$, the angles $ABC$, $BCA$, and $CAB$ have the same sign. 139 | \end{thm} 140 | 141 | } 142 | 143 | { 144 | 145 | \begin{wrapfigure}{o}{33mm} 146 | \vskip-4mm 147 | \centering 148 | \includegraphics{mppics/pic-18} 149 | \end{wrapfigure} 150 | 151 | \parit{Proof.} 152 | Choose a point $Z\in (AB)$ so that $A$ lies between $B$ and~$Z$. 153 | 154 | 155 | According to Lemma~\ref{lem:signs}, 156 | the angles $ZBC$ and $ZAC$ have the same sign. 157 | 158 | 159 | Note that $\measuredangle ABC=\measuredangle ZBC$ 160 | and 161 | $$\measuredangle ZAC+\measuredangle CAB\equiv \pi.$$ 162 | Therefore, $\angle CAB$ has the same sign as $\angle ZAC$ 163 | which in turn has the same sign as $\measuredangle ABC\z=\measuredangle ZBC$. 164 | 165 | } 166 | 167 | Repeating the same argument for $\angle BCA$ and $\angle CAB$, 168 | we get the result. 169 | \qeds 170 | 171 | \begin{wrapfigure}{r}{26mm} 172 | \vskip-2mm 173 | \centering 174 | \includegraphics{mppics/pic-838} 175 | \end{wrapfigure} 176 | 177 | \begin{thm}[!]{Classroom exercise}\label{ex:between} 178 | Let $A$, $B$, $C$, and $P$ be points such that $B$ lies between $A$ and $C$, and let $P \not\in (AC)$. 179 | Show that the angles $\angle APB$, $\angle BPC$, and $\angle APC$ have the same sign. 180 | Conclude that 181 | \[|\measuredangle APB|+|\measuredangle BPC|=|\measuredangle APC|.\] 182 | \end{thm} 183 | 184 | \begin{thm}[\abs]{Lemma}\label{lem:signsXY} 185 | Assume $[XY]$ does not intersect $(PQ)$. 186 | Then the angles $PQX$ and $PQY$ have the same sign. 187 | \end{thm} 188 | 189 | \begin{wrapfigure}{o}{26mm} 190 | \vskip-4mm 191 | \centering 192 | \includegraphics{mppics/pic-20} 193 | \end{wrapfigure} 194 | 195 | The proof is nearly identical to the one above. 196 | 197 | \parit{Proof.} 198 | According to Exercise~\ref{ex:trig==}, 199 | for any $t\z\in [0,1]$ there is a point $X_t\in[XY]$, 200 | such that 201 | \[XX_t= t\cdot XY.\] 202 | The map $t\mapsto X_t$ is continuous. 203 | Moreover, $X_0=X$, $X_1=Y$, and $X_t\notin(QP)$ for any $t\in [0,1]$. 204 | 205 | Applying Corollary~\ref{cor:intermidiate}, 206 | for $P_t\z=P$, $Q_t\z=Q$, and $X_t$, we get that 207 | $\angle PQX$ has the same sign as $\angle PQY$. 208 | \qeds 209 | 210 | \section{Half-planes} 211 | 212 | \begin{thm}{Proposition}\label{prop:half-plane} 213 | Assume $X,Y\notin(PQ)$. 214 | Then the angles $PQX$ and $PQY$ have the same sign if and only if $[XY]$ does not intersect $(PQ)$. 215 | \end{thm} 216 | 217 | \begin{wrapfigure}{o}{30mm} 218 | \includegraphics{mppics/pic-22} 219 | \centering 220 | \end{wrapfigure} 221 | 222 | \parit{Proof.} The if-part follows from Lemma~\ref{lem:signsXY}. 223 | 224 | Assume $[XY]$ intersects $(PQ)$; 225 | suppose that $Z$ denotes the point of intersection. 226 | Without loss of generality, we can assume $Z\ne P$. 227 | 228 | Note that $Z$ lies between $X$ and $Y$. 229 | According to Lemma~\ref{lem:straight-sign}, $\angle PZX$ and $\angle PZY$ have opposite signs. 230 | This proves the statement if $Z=Q$. 231 | 232 | If $Z\ne Q$, then $\angle ZQX$ and $\angle QZX$ have opposite signs by \ref{thm:signs-of-triug}. 233 | In the same way, we get that $\angle ZQY$ and $\angle QZY$ have opposite signs. 234 | 235 | If $Q$ lies between $Z$ and $P$, then by Lemma~\ref{lem:straight-sign} two pairs of angles $\angle PQX$, $\angle ZQX$ and $\angle PQY$, $\angle ZQY$ have opposite signs. 236 | It follows that $\angle PQX$ and $\angle PQY$ have opposite signs as required. 237 | 238 | In the remaining case $[QZ)=[QP)$ and therefore $\angle PQX=\angle ZQX$ and $\angle PQY=\angle ZQY$. 239 | Therefore again $\angle PQX$ and $\angle PQY$ have opposite signs as required. 240 | \qeds 241 | 242 | \begin{thm}[\abs]{Corollary}\label{cor:half-plane} 243 | The complement of a line $(PQ)$ in the plane 244 | can be presented in a unique way as a union of two disjoint subsets 245 | called \index{half-plane}\emph{half-planes} 246 | such that 247 | \begin{enumerate}[(a)] 248 | \item\label{cor:half-plane:angle} Two points $X,Y\notin(PQ)$ lie in the same half-plane if and only if the angles $PQX$ and $PQY$ have the same sign. 249 | \item\label{cor:half-plane:intersect} Two points $X,Y\notin(PQ)$ lie in the same half-plane if and only if $[XY]$ does not intersect~$(PQ)$. 250 | \end{enumerate} 251 | 252 | \end{thm} 253 | 254 | { 255 | 256 | \begin{wrapfigure}{r}{26mm} 257 | \vskip-4mm 258 | \centering 259 | \includegraphics{mppics/pic-24} 260 | \end{wrapfigure} 261 | 262 | We say that $X$ and $Y$ lie on {}\emph{one side of} $(PQ)$ if they lie in one of the half-planes of $(PQ)$ and we say that $P$ and $Q$ lie on {}\emph{opposite sides of} $\ell$ if they lie in different half-planes of~$\ell$. 263 | 264 | 265 | \begin{thm}{Exercise}\label{ex:vert-intersect} 266 | Suppose that angles $AOB$ and $A'OB'$ are vertical and $B\notin (OA)$. 267 | Show that the line $(AB)$ does not intersect the segment~$[A'B']$. 268 | \end{thm} 269 | 270 | } 271 | 272 | Consider triangle $ABC$. 273 | Recall that its vertices are points $A$, $B$ and $C$; 274 | segments $[AB]$, $[BC]$, and $[CA]$ will be called the \index{side!of triangle}\emph{sides of the triangle}. 275 | 276 | { 277 | 278 | \begin{wrapfigure}{r}{27mm} 279 | \vskip-4mm 280 | \centering 281 | \includegraphics{mppics/pic-26} 282 | \end{wrapfigure} 283 | 284 | \begin{thm}[\abs]{Pasch's theorem}\label{thm:pasch}\index{Pasch's theorem} 285 | Assume that a line $\ell$ does not pass thru any vertex of a triangle. 286 | Then it intersects either two or zero sides of the triangle. 287 | \end{thm} 288 | 289 | \parit{Proof.} 290 | Assume that line $\ell$ intersects side $[AB]$ of the triangle $ABC$ and does not pass thru $A$, $B$, and $C$. 291 | 292 | By Corollary~\ref{cor:half-plane}, the vertices $A$ and $B$ lie on opposite sides of~$\ell$. 293 | 294 | } 295 | 296 | The vertex $C$ may lie on the same side as $A$ and on the opposite side from $B$ or the other way around. 297 | By Corollary~\ref{cor:half-plane}, in the first case, $\ell$ intersects side $[BC]$ and does not intersect $[AC]$; in the second case, $\ell$ intersects side $[AC]$ and does not intersect $[BC]$. 298 | Hence the statement follows. 299 | \qeds 300 | 301 | { 302 | 303 | \begin{wrapfigure}[5]{r}{29mm} 304 | \vskip-4mm 305 | \centering 306 | \includegraphics{mppics/pic-28} 307 | \bigskip 308 | \includegraphics{mppics/pic-30} 309 | \end{wrapfigure} 310 | 311 | \begin{thm}[!]{Exercise}\label{ex:signs-PXQ-PYQ} 312 | Show that two points $X,Y\z\notin(PQ)$ lie on the same side of $(PQ)$ 313 | if and only if angles $PXQ$ and $PYQ$ have the same sign. 314 | \end{thm} 315 | 316 | \begin{thm}[!]{Exercise}\label{ex:chevinas} 317 | Let $\triangle ABC$ be a nondegenerate triangle, 318 | $A'\in[BC]$ and 319 | $B'\in [AC]$. 320 | Show that $[AA']$ intersects $[BB']$. 321 | \end{thm} 322 | 323 | \begin{thm}{Exercise}\label{ex:Z} 324 | Assume that points $X$ and $Y$ lie on opposite sides of the line~$(PQ)$. 325 | Show that $[PX)$ does not intersect~$[QY)$. 326 | \end{thm} 327 | 328 | } 329 | 330 | \begin{thm}{Advanced exercise}\label{ex:angle-measures} 331 | The following quantity 332 | $$\tilde\measuredangle ABC= 333 | \begin{cases} 334 | \pi&\text{if}\ \measuredangle ABC=\pi 335 | \\ 336 | -\measuredangle ABC&\text{if}\ \measuredangle ABC<\pi 337 | \end{cases} 338 | $$ 339 | can serve as the angle measure; 340 | that is, the axioms hold if one replaces $\measuredangle$ by $\tilde\measuredangle$ everywhere. 341 | 342 | Show that $\measuredangle$ and $\tilde\measuredangle$ are the only possible angle measures on the plane. 343 | 344 | Show that without Axiom~\ref{def:birkhoff-axioms:2c}, this is no longer true. 345 | \end{thm} 346 | 347 | \section{Triangle with given sides} 348 | 349 | Given $\triangle ABC$, set 350 | \begin{align*} 351 | a&=BC, 352 | & 353 | b&=CA, 354 | & 355 | c&=AB. 356 | \end{align*} 357 | Without loss of generality, we may assume that 358 | \[a\le b \le c.\] 359 | Then all three triangle inequalities for $\triangle ABC$ 360 | hold if and only if 361 | \[c\le a+b.\] 362 | The following theorem states that this is the only restriction on $a$, $b$, and~$c$. 363 | 364 | \begin{thm}[\abs]{Theorem}\label{thm:abc} 365 | Assume that $00$ 375 | and two distinct points $A$ and~$B$. 376 | Then for 377 | any real number $\beta\in [0,\pi]$, 378 | there is a unique point $C_\beta$ such that $BC_\beta=r$ 379 | and $\measuredangle ABC_\beta=\beta$. 380 | Moreover, $\beta\mapsto C_\beta$ 381 | is a continuous map from $[0,\pi]$ to the plane. 382 | \end{thm} 383 | 384 | \parit{Proof of Theorem~\ref{thm:abc} modulo Proposition~\ref{prop:C-cont}.}\label{page:proof:thm:abc} 385 | Fix the points $A$ and $B$ such that $AB=c$. 386 | Given $\beta\in [0,\pi]$, 387 | suppose that $C_\beta$ denotes the point such that $BC_\beta\z=a$ and $\measuredangle ABC_\beta=\beta$. 388 | 389 | According to \ref{prop:C-cont}, 390 | the map 391 | $\beta\mapsto C_\beta$ is continuous. 392 | Therefore, the function $b\:\beta\mapsto AC_\beta$ is continuous 393 | (formally, it follows from Exercise~\ref{ex:dist-cont} and Exercise~\ref{ex:comp+cont}). 394 | 395 | Note that $b(0)=c-a$ and $b(\pi)=c+a$. 396 | Since $c-a\le b\le c+a$, 397 | by the intermediate value theorem (\ref{thm:intermidiate}) 398 | there is $\beta_0\in[0,\pi]$ such that 399 | $b(\beta_0)=b$, 400 | hence the result. 401 | \qeds 402 | 403 | The proof of Proposition~\ref{prop:C-cont} relies on the following lemma. 404 | 405 | 406 | { 407 | 408 | \begin{wrapfigure}{o}{22mm} 409 | \vskip-2mm 410 | \centering 411 | \includegraphics{mppics/pic-32} 412 | \end{wrapfigure} 413 | 414 | Assume $r>0$ and $0<\beta<\pi$. 415 | Consider the triangle $ABC$ such that 416 | $AB=BC=r$ and $\measuredangle ABC=\beta$. 417 | The existence of such a triangle follows from Axiom~\ref{def:birkhoff-axioms:2a} and Exercise~\ref{ex:trig==}. 418 | By Axiom~\ref{def:birkhoff-axioms:3}, 419 | the values 420 | $\beta$ and~$r$ determine $\triangle ABC$ up to congruence. 421 | In particular, the distance $AC$ depends only on $\beta$ and~$r$; 422 | thus, we can consider the function $s$ defined by 423 | $$s(r,\beta)\df AC.$$ 424 | 425 | } 426 | 427 | \begin{thm}[\abs]{Lemma}\label{lem:f(r,a)} 428 | Given $r>0$ and $\epsilon>0$, there is $\delta>0$ such that 429 | \[0<\beta<\delta\quad\Longrightarrow\quad s(r,\beta)<\epsilon.\] 430 | 431 | \end{thm} 432 | 433 | 434 | 435 | { 436 | 437 | \begin{wrapfigure}{l}{36mm} 438 | \vskip-8mm 439 | \centering 440 | \includegraphics{mppics/pic-34} 441 | \end{wrapfigure} 442 | 443 | \parit{Proof.} 444 | Fix two points $A$ and $B$ such that $AB\z=r$. 445 | 446 | Choose a point $X$ such that $\measuredangle ABX$ is positive. 447 | Let $Y\in [AX)$ be the point such that $AY=\tfrac\epsilon2$; 448 | it exists by Exercise~\ref{ex:trig==}. 449 | 450 | Points $X$ and $Y$ lie on the same side of $(AB)$; 451 | therefore, $\angle ABY$ is positive. 452 | Set $\delta\z=\measuredangle ABY$. 453 | 454 | Suppose $0<\beta<\delta$; 455 | by Axiom~\ref{def:birkhoff-axioms:2a}, we can choose $C$ so that $\measuredangle ABC\z=\beta$ and $BC\z=r$. 456 | Furthermore, we can choose a half-line $[BZ)$ such that $\measuredangle ABZ\z=\tfrac12\cdot \beta$. 457 | 458 | } 459 | 460 | Points $A$ and $Y$ lie on opposite sides of~$(BZ)$ and $\measuredangle ABZ\z\equiv -\measuredangle CBZ$. 461 | In particular, $(BZ)$ intersects $[AY]$; 462 | denote by $D$ the point of intersection. 463 | 464 | Since $D$ lies between $A$ and $Y$, we have $AD0=>b>0}\] 288 | 289 | \columnbreak 290 | 291 | By \ref{eq:f-x}, 292 | we also get 293 | \[a\le 0\quad \Longrightarrow\quad f(a)\le 0.\eqlbl{a<0=>b<0}\] 294 | \end{multicols} 295 | \setlength{\columnseprule}{0pt} 296 | 297 | Now assume $f(a)\ne a$ for some $a\in\mathbb{R}$. 298 | Then there is a rational number $\tfrac{m}{n}$ that lies between $a$ and $f(a)$; 299 | that is, 300 | the numbers 301 | \[x\z=a-\tfrac{m}{n}\quad\text{and}\quad y=f(a)-\tfrac{m}{n}\] 302 | have opposite signs. 303 | 304 | By \ref{eq:m/n}, 305 | \begin{align*} 306 | y+\tfrac{m}{n}&=f(a)= 307 | \\ 308 | &=f(x+\tfrac{m}{n})= 309 | \\ 310 | &=f(x)+f(\tfrac{m}{n})= 311 | \\ 312 | &=f(x)+\tfrac{m}{n}; 313 | \end{align*} 314 | that is, $f(x)=y$. 315 | By \ref{a>0=>b>0} and \ref{a<0=>b<0} the values $x$ and $y$ cannot have opposite signs --- a contradiction. 316 | \qeds 317 | 318 | \section{Three theorems} 319 | 320 | We discussed several statements in Euclidean geometry that have an affine nature; 321 | that is, their assumptions and conclusions survive under affine transformations. 322 | The examples include 323 | \ref{thm:parallel-point-reflection}, 324 | \ref{lem:parallelogram}\ref{lem:parallelogram:midpoint}, 325 | \ref{ex:4parallels}, 326 | \ref{thm:centroid}, 327 | and \ref{ex:midle}. 328 | In this section, we present three more examples of that type. 329 | 330 | Suppose that $A$, $B$, and $X$ are distinct points on one line. 331 | Note that 332 | \[\overrightarrow{AX}=t\cdot \overrightarrow{BX},\] 333 | where $t=\pm\frac{AX}{BX}$. 334 | 335 | Choose an affine transformation $P\mapsto P'$. 336 | By \ref{prop:affine-linear}, we have 337 | \[\overrightarrow{A'X'}=t\cdot \overrightarrow{B'X'}.\] 338 | It follows that $\frac{A'X'}{B'X'}=\frac{AX}{BX}$; 339 | that is, \textit{if $A$, $B$, and $X$ are distinct points on one line, then the ratio $\frac{AX}{BX}$ is preserved under affine transformations}. 340 | 341 | \begin{thm}{Menelaus's theorem}\index{Menelaus's theorem} 342 | Let $ABC$ be a nondegenerate triangle. 343 | Suppose a line $\ell$ crosses the lines $(BC)$, $(CA)$, and $(AB)$ at three distinct points $A'$, $B'$, and $C'$. 344 | Then 345 | \[\frac{AC'}{BC'}\cdot\frac{BA'}{CA'}\cdot \frac{CB'}{AB'}=1.\] 346 | \end{thm} 347 | 348 | As we saw, the ratios $\frac{AC'}{BC'}$, $\frac{BA'}{CA'}$, $\frac{CB'}{AB'}$ are preserved under affine transformations. 349 | Therefore Menelaus's theorem belongs to affine geometry. 350 | However, in the proof we can (and will) use Euclidean-type arguments. 351 | 352 | \parit{Proof.} 353 | Let $X$, $Y$ and $Z$ be the footpoints of $A$, $B$ and $C$ on $\ell$. 354 | By the AA similarity condition, $\triangle AXC'\sim \triangle BYC'$. 355 | \begin{figure}[!ht] 356 | \centering 357 | \includegraphics{mppics/pic-227} 358 | \vskip-10mm 359 | \end{figure} 360 | Therefore, 361 | \[\frac{AC'}{BC'}=\frac{AX}{BY}.\] 362 | The same way we get 363 | \[\frac{BA'}{CA'}=\frac{BY}{CZ} 364 | \quad\text{and}\quad 365 | \frac{CB'}{AB'}=\frac{CZ}{AX}.\] 366 | Therefore, 367 | \[\frac{AC'}{BC'}\cdot\frac{BA'}{CA'}\cdot \frac{CB'}{AB'}=\frac{AX}{BY}\cdot \frac{BY}{CZ}\cdot \frac{CZ}{AX}=1.\] 368 | \qedsf 369 | 370 | 371 | { 372 | 373 | \begin{thm}{Exercise}\label{thm:ceva-affine}\label{ex:ceva-affine} 374 | Let $ABC$ be a nondegenerate triangle. 375 | Suppose three distinct points $A'$, $B'$, and $C'$ lie on the lines $(BC)$, $(CA)$, and $(AB)$ respectively. 376 | Assume that the lines $(AA')$, $(BB')$ and $(CC')$ meet at a point $X$. 377 | 378 | \begin{wrapfigure}{r}{31mm} 379 | \centering 380 | \vskip-2mm 381 | \includegraphics{mppics/pic-229} 382 | \end{wrapfigure} 383 | 384 | Use Menelaus's theorem to prove the following. 385 | 386 | \begin{enumerate}[(a)] 387 | \item \index{Ceva's theorem}\textbf{Ceva's theorem:} 388 | \[\frac{AC'}{BC'}\cdot\frac{BA'}{CA'}\cdot \frac{CB'}{AB'}=1.\] 389 | \item \index{Van Aubel's theorem}\textbf{Van Aubel's theorem:} If $A'$ lies between $B$ and $C$, then 390 | \[\frac{AC'}{BC'}+\frac{AB'}{CB'}=\frac{AX}{A'X}.\] 391 | \end{enumerate} 392 | \end{thm} 393 | 394 | } 395 | 396 | \parbf{Signed versions.} 397 | These three theorems can be formulated in a more precise way by taking the ratios $\frac{AC'}{BC'}$, $\frac{BA'}{CA'}$, and $\frac{CB'}{AB'}$ with sign. 398 | Namely, given distinct points $A$, $B$, and $X$ on a line, let us define the \index{signed ratio}\emph{signed ratio} $(\frac{AX}{BX})\df\pm\frac{AX}{BX}$, where the sign is chosen so that 399 | $\overrightarrow{AX}=\left(\frac{AX}{BX}\right)\cdot \overrightarrow{BX}$. 400 | Equivalently, $(\frac{AX}{BX})<0$ if and only if $X$ lies between $A$ and $B$. 401 | 402 | The more exact identities of Menelaus's and Ceva's theorems will be 403 | \[\left(\tfrac{AC'}{BC'}\right)\cdot\left(\tfrac{BA'}{CA'}\right)\cdot \left(\tfrac{CB'}{AB'}\right)=1 404 | \quad\text{and}\quad 405 | \left(\tfrac{AC'}{BC'}\right)\cdot\left(\tfrac{BA'}{CA'}\right)\cdot \left(\tfrac{CB'}{AB'}\right)=-1 406 | \] 407 | respectively. 408 | In Van Aubel's theorem we have 409 | \[\left(\tfrac{AC'}{BC'}\right)+\left(\tfrac{AB'}{CB'}\right)=\left(\tfrac{AX}{A'X}\right)\] 410 | and this identity holds without the assumption that $A'$ lies between $B$ and~$C$. 411 | 412 | Moreover, the converse of each of the three signed versions also holds. 413 | 414 | -------------------------------------------------------------------------------- /perp.tex: -------------------------------------------------------------------------------- 1 | \chapter{Perpendicular lines}\label{chap:perp} 2 | 3 | \section{Right, acute and obtuse angles} 4 | 5 | \begin{itemize} 6 | \item If $|\measuredangle A O B|=\tfrac\pi2$, we say that $\angle A O B$ is \index{right!angle}\emph{right}; 7 | %\item If $\measuredangle A O B\ne\pm\tfrac\pi2$, we say that the angle $\angle A O B$ is \index{angle!oblique angle}\emph{oblique}; 8 | \item If $|\measuredangle A O B|<\tfrac\pi2$, we say that $\angle A O B$ is 9 | \index{acute}\emph{acute}; 10 | \item If $|\measuredangle A O B|>\tfrac\pi2$, we say that $\angle A O B$ is \index{angle!right, acute, and obtuse angles}\index{obtuse}\emph{obtuse}. 11 | \end{itemize} 12 | 13 | \begin{wrapfigure}[2]{r}{25mm} 14 | \vskip-25mm 15 | \centering 16 | \includegraphics{mppics/pic-46} 17 | \end{wrapfigure} 18 | 19 | Right angles will be marked with a small square, as shown. 20 | 21 | If $\angle A O B$ is right, we may also say that $[O A)$ is \index{perpendicular}\emph{perpendicular} to $[O B)$. 22 | This can be written as \index{38@$\perp$}$[O A)\z\perp [O B)$. 23 | 24 | Theorem~\ref{thm:straight-angle} allows us to call two lines $(O A)$ and $(O B)$ {}\emph{perpendicular} if $[O A)\z\perp [O B)$; so, we can write $(O A)\z\perp (O B)$. 25 | 26 | \begin{thm}{Exercise}\label{ex:acute-obtuce} 27 | Assume that point $O$ lies between $A$ and $B$ and $X\ne O$. 28 | Show that 29 | $\angle XOA$ is acute if and only if 30 | $\angle XOB$ is obtuse. 31 | \end{thm} 32 | 33 | \section{Perpendicular bisector} 34 | 35 | Assume $M$ is the midpoint of the segment $[AB]$; 36 | equivalently, $M\in(A B)$ and $A M \z= M B$. 37 | 38 | The line $\ell$ thru $M$ and perpendicular to $(AB)$ is called the \index{bisector!perpendicular bisector}\index{perpendicular!bisector}\emph{perpendicular bisector} of the segment~$[AB]$. 39 | 40 | \begin{thm}[\abs]{Theorem}\label{thm:perp-bisect} 41 | Given distinct points $A$ and $B$, 42 | all points that are equidistant from $A$ and $B$ and no 43 | others lie on the perpendicular bisector of~$[A B]$. 44 | \end{thm} 45 | 46 | \parit{Proof.} Let $M$ be the midpoint of~$[A B]$. 47 | Assume $P A= P B$ and $P\ne M$. 48 | According to SSS (\ref{thm:SSS}), 49 | $\triangle A M P \z\cong\triangle B M P$. 50 | Hence 51 | $\measuredangle A M P=\pm \measuredangle B M P$. 52 | Since $A\not=B$, we have ``$-$'' in this formula. 53 | 54 | \begin{wrapfigure}[8]{o}{34mm} 55 | \centering 56 | \includegraphics{mppics/pic-48} 57 | \end{wrapfigure} 58 | 59 | Furthermore, 60 | \begin{align*} 61 | \pi 62 | &= 63 | \measuredangle A M B 64 | \equiv 65 | \\ 66 | &\equiv\measuredangle A M P+\measuredangle P M B 67 | \equiv 68 | \\ 69 | &\equiv 70 | 2\cdot \measuredangle A M P. 71 | \end{align*} 72 | That is, $\measuredangle A M P 73 | = 74 | \pm 75 | \tfrac\pi2$. 76 | Therefore, $P$ lies on the perpendicular bisector. 77 | 78 | 79 | To prove the converse, 80 | suppose $P$ 81 | is any point on the perpendicular bisector of $[A B]$ and $P\z\ne M$. 82 | Then $\measuredangle A M P=\pm \tfrac\pi2$, 83 | $\measuredangle B M P=\pm \tfrac\pi2$ and 84 | $A M\z=B M$. 85 | By SAS, $\triangle A M P\cong \triangle B M P$; 86 | in particular, $A P\z= B P$. 87 | \qeds 88 | 89 | 90 | \begin{thm}[!]{Exercise}\label{ex:pbisec-side} 91 | Let $\ell$ be a perpendicular bisector of $[A B]$, and $X$ be an arbitrary point in the plane. 92 | Show that 93 | $AXBC\iff|\measuredangle ABC|>|\measuredangle CAB|.\] 100 | \end{thm} 101 | 102 | \begin{thm}{Exercise}\label{ex:pbisec-motion} 103 | A motion has a fixed point $F$ and maps a point $X$ to another point $Y$. 104 | Show that $F$ lies on the perpendicular bisector of $[XY]$. 105 | \end{thm} 106 | 107 | 108 | 109 | 110 | \section{Uniqueness of a perpendicular} 111 | 112 | \begin{thm}[\abs]{Theorem}\label{perp:ex+un} 113 | There is one and only one line that passes thru a given point $P$ and is perpendicular to a given line~$\ell$. 114 | \end{thm} 115 | 116 | According to the theorem above, 117 | there is a unique point $Q\in\ell$ such that $(QP)\perp\ell$. 118 | This point $Q$ is called the \index{footpoint}\emph{footpoint} of $P$ on~$\ell$. 119 | 120 | \parit{Proof.} 121 | If $P\in\ell$, then both existence and uniqueness follow from Axiom~\ref{def:birkhoff-axioms:2}. 122 | 123 | { 124 | 125 | \begin{wrapfigure}{r}{30mm} 126 | \vskip-4mm 127 | \centering 128 | \includegraphics{mppics/pic-50} 129 | \end{wrapfigure} 130 | 131 | \parit{Existence for $P\not\in\ell$.} 132 | Let $A$ and $B$ be two distinct points on~$\ell$. 133 | Choose $P'$ so that $AP'\z=AP$ and $\measuredangle BAP' \equiv -\measuredangle BAP$. 134 | According to Axiom~\ref{def:birkhoff-axioms:3}, $\triangle A P' B\z\cong\triangle A P B$. 135 | In particular, $A P= A P'$ and $B P= B P'$. 136 | 137 | According to Theorem~\ref{thm:perp-bisect}, $A$ and $B$ lie on the perpendicular bisector of~$[P P']$. 138 | In particular, $(P P')\perp (A B)=\ell$. 139 | 140 | } 141 | 142 | 143 | \parit{Uniqueness for $P\not\in\ell$.} 144 | From above we can choose a point $P'$ in such a way that $\ell$ is the perpendicular bisector of~$[PP']$. 145 | 146 | Assume $m\perp \ell$ and $m\ni P$. 147 | Then $m$ is a perpendicular bisector of some segment $[Q Q']$ of $\ell$; 148 | in particular, $P Q= P Q'$. 149 | 150 | { 151 | 152 | \begin{wrapfigure}{r}{37mm} 153 | \centering 154 | \includegraphics{mppics/pic-52} 155 | \end{wrapfigure} 156 | 157 | Since $\ell$ is the perpendicular bisector of $[P P']$, 158 | we get that $PQ= P'Q$ and $PQ' \z= P'Q'$. 159 | Therefore, 160 | $$P' Q=P Q=P Q'= P' Q'.$$ 161 | By Theorem~\ref{thm:perp-bisect}, 162 | $P'$ lies on the perpendicular bisector of $[QQ']$, which is~$m$. 163 | By Axiom~\ref{def:birkhoff-axioms:1}, $m=(P P')$. 164 | \qeds 165 | 166 | } 167 | 168 | \begin{thm}[!]{Exercise}\label{ex:construction-perpendicular} 169 | Describe a ruler-and-compass construction of a line thru a given point that is perpendicular to a given line. 170 | \end{thm} 171 | 172 | \section{Reflection across a line} 173 | 174 | Assume the point $P$ and the line $(AB)$ are given. 175 | To find the \index{reflection!across a line}\emph{reflection} $P'$ of $P$ across $(AB)$, 176 | one drops a perpendicular from $P$ onto $(AB)$, 177 | and continues it to the same distance on the other side. 178 | 179 | { 180 | 181 | \begin{wrapfigure}{o}{37mm} 182 | \vskip-2mm 183 | \centering 184 | \includegraphics{mppics/pic-54} 185 | \end{wrapfigure} 186 | 187 | According to Theorem~\ref{perp:ex+un}, $P'$ is uniquely determined by~$P$. 188 | Note that $P=P'$ if and only if $P\in(AB)$. 189 | 190 | \begin{thm}[\abs]{Proposition}\label{prop:reflection} 191 | Assume $P'$ is the reflection of $P$ across $(AB)$. 192 | Then $AP'=AP$, 193 | and if $A\ne P$, 194 | then 195 | $\measuredangle BAP'\equiv -\measuredangle BAP$. 196 | \end{thm} 197 | 198 | 199 | \parit{Proof.} 200 | If $P\in (AB)$, 201 | then $P\z=P'$. 202 | By Corollary~\ref{cor:degenerate=pi}, $\measuredangle BAP\z=0$ or~$\pi$. 203 | Hence the statement follows. 204 | 205 | } 206 | 207 | If $P\notin (AB)$, then~$P'\ne P$. 208 | By the construction of $P'$, 209 | the line $(AB)$ is a perpendicular bisector of~$[PP']$. 210 | Therefore, according to Theorem~\ref{thm:perp-bisect}, $AP'\z=AP$ and $BP'\z=BP$. 211 | In particular, 212 | $\triangle ABP'\z\cong \triangle ABP$. 213 | Therefore, $\measuredangle BAP'=\pm \measuredangle BAP$. 214 | 215 | Since $P'\ne P$ and $AP'=AP$, 216 | we get that $\measuredangle BAP'\ne \measuredangle BAP$. 217 | That is, we are left with the case $\measuredangle BAP'\equiv-\measuredangle BAP$. 218 | \qeds 219 | 220 | 221 | 222 | \begin{thm}{Exercise}\label{ex:2-reflections} 223 | Let $X$ and $Y$ be the reflections of $P$ across the lines $(AB)$ and $(BC)$ respectively. 224 | Show that $\measuredangle XBY\equiv 2\cdot \measuredangle ABC$. 225 | 226 | \end{thm} 227 | 228 | 229 | \begin{thm}[\abs]{Corollary}\label{cor:reflection+angle} 230 | A reflection across a line is a motion of the plane. 231 | Moreover, 232 | $\measuredangle Q'P'R'\equiv -\measuredangle QPR$ 233 | if $\triangle P'Q'R'$ is the reflection of $\triangle PQR$. 234 | \end{thm} 235 | 236 | 237 | \parit{Proof.} 238 | The composition of two reflections across the same line 239 | is the identity map. 240 | In particular, any reflection is a bijection. 241 | 242 | Fix a line $(AB)$ and two points $P$ and $Q$; 243 | denote their reflections across $(AB)$ by $P'$ and $Q'$. 244 | Let us show that 245 | $$P'Q'=PQ;\eqlbl{eq:P'Q'=PQ}$$ 246 | that is, the reflection is distance-preserving, 247 | 248 | \begin{wrapfigure}{r}{33mm} 249 | \centering 250 | \vskip-15mm 251 | \includegraphics{mppics/pic-56} 252 | \end{wrapfigure} 253 | 254 | Without loss of generality, we may assume that the points $P$ and $Q$ are distinct from $A$ and~$B$. 255 | By Proposition~\ref{prop:reflection}, we get that 256 | \begin{align*} 257 | \measuredangle BAP'&\equiv -\measuredangle BAP, 258 | & 259 | \measuredangle BAQ'&\equiv -\measuredangle BAQ, 260 | \\ 261 | AP'&=AP, 262 | & 263 | AQ'&=AQ. 264 | \end{align*} 265 | It follows that 266 | \[\measuredangle P'AQ'\equiv -\measuredangle PAQ.\eqlbl{eq:P'AQ'=PAQ}\] 267 | By SAS, 268 | $\triangle P'AQ'\cong\triangle PAQ$ 269 | and \ref{eq:P'Q'=PQ} follows. 270 | Moreover, we also get that 271 | \[\measuredangle AP'Q'\equiv\pm\measuredangle APQ.\] 272 | From \ref{eq:P'AQ'=PAQ} and the theorem on the signs of angles of triangles (\ref{thm:signs-of-triug}) we get 273 | \[\measuredangle AP'Q'\equiv-\measuredangle APQ.\eqlbl{eq:AP'Q'=APQ}\] 274 | 275 | Repeating the same argument for a pair of points $P$ and $R$, 276 | we get that 277 | $$\measuredangle AP'R'\equiv-\measuredangle APR.\eqlbl{eq:AP'R'=APR}$$ 278 | Subtracting \ref{eq:AP'R'=APR} from \ref{eq:AP'Q'=APQ}, 279 | we get that 280 | $\measuredangle Q'P'R'\equiv-\measuredangle QPR$. 281 | \qedsf 282 | 283 | 284 | \section{Direct and indirect motions} 285 | \label{direct motion} 286 | 287 | \begin{thm}[!]{Classroom exercise}\label{ex:3-reflections} 288 | Show that every motion of the plane can be presented as a 289 | composition of at most three reflections across lines. 290 | \end{thm} 291 | 292 | A motion $X\mapsto X'$ is called \index{direct motion}\emph{direct} if 293 | $$\measuredangle Q'P'R'= \measuredangle QPR$$ 294 | for any triangle $PQR$; 295 | if instead we always have 296 | $$\measuredangle Q'P'R'\equiv -\measuredangle QPR,$$ 297 | then the motion $f$ is called \index{indirect motion}\emph{indirect}. 298 | 299 | By Corollary~\ref{cor:reflection+angle}, any reflection across a line is an indirect motion. 300 | Note that the composition of two reflections is a direct motion. 301 | More generally, the composition of two indirect motions is direct, 302 | the composition of two direct motions is direct, 303 | and the composition of direct and indirect motions is indirect. 304 | In particular, the exercise above implies the following. 305 | 306 | \begin{thm}{Proposition}\label{prop:direct-indirect} 307 | Any motion of the plane is either direct or indirect. 308 | \end{thm} 309 | 310 | 311 | \section{The perpendicular is shortest} 312 | \label{sec:perpPQ$$ 318 | for every point $X$ on $\ell$ distinct from~$Q$. 319 | \end{thm} 320 | 321 | If $P$, $Q$, and $\ell$ are as above, 322 | then $PQ$ is called the \label{distance!from a point to a line}\index{distance!from a point to a line}\emph{distance from $P$ to~$\ell$}. 323 | 324 | \parit{Proof.} 325 | If $P\in \ell$, 326 | then the result follows since $PQ=0$. 327 | So, we can assume that $P\notin \ell$. 328 | 329 | \begin{wrapfigure}{o}{24mm} 330 | \centering 331 | \vskip-4mm 332 | \includegraphics{mppics/pic-58} 333 | \vskip0mm 334 | \end{wrapfigure} 335 | 336 | Let $P'$ be the reflection of $P$ across the line~$\ell$. 337 | Note that $Q$ is the midpoint of $[PP']$ 338 | and $\ell$ is the perpendicular bisector of $[PP']$. 339 | Therefore 340 | $$PX=P'X 341 | \quad 342 | \text{and} 343 | \quad 344 | PQ=P'Q=\tfrac12\cdot PP'.$$ 345 | 346 | The line $\ell$ meets $[PP']$ only at the point $Q$. 347 | Therefore, $X\notin [PP']$. 348 | By the triangle inequality and Corollary~\ref{cor:degenerate-trig}, we have 349 | $$PX+P'X>PP',$$ 350 | and hence the result: $PX>PQ$. 351 | \qeds 352 | 353 | \begin{thm}{Classroom exercise}\label{ex:right-acute} 354 | Show that 355 | \begin{enumerate}[(a)] 356 | \item If $\triangle ABC$ has a right angle at $C$, then its other angles are acute. 357 | \item If $\triangle ABC$ has an obtuse angle at $C$, then its other angles are acute. 358 | \end{enumerate} 359 | \end{thm} 360 | 361 | This exercise is closely related to Proposition~\ref{prop:2sum}. 362 | 363 | If a triangle has an obtuse or right angle, then it is called \index{triangle!right, acute, and obtuse triangle}\index{obtuse}\emph{obtuse} or \index{triangle!right triangle}\index{right!triangle}\emph{right}, respectively. 364 | Otherwise, if all its angles are acute, it is called an \index{acute}\emph{acute triangle}. 365 | 366 | \begin{thm}[!]{Exercise}\label{ex:obtuce} 367 | Suppose that point $X$ lies between $V$ and $W$. 368 | Show that for every point $P$, we have $PX < PV$ or $PX < PW$. 369 | \end{thm} 370 | 371 | \begin{thm}{Exercise}\label{ex:PMQ} 372 | Assume that points $P$ and $Q$ lie on one side of line~$\ell$. 373 | Let $M\in \ell$ be the point that minimizes the sum $PM+MQ$. 374 | Show that $M\in (P'Q)$, where $P'$ is the reflection of $P$ across $\ell$. 375 | \end{thm} 376 | 377 | 378 | 379 | \section{Circles} 380 | 381 | Recall that a circle with radius $r$ and center $O$ is the set of all points at distance $r$ from $O$. 382 | We say that a point $P$ lies \index{inside!a circle}\emph{inside} the circle if $OPr$, we say that $P$ lies \index{outside a circle}\emph{outside} of the circle. 384 | \label{def:circle} 385 | 386 | \begin{thm}[!]{Exercise}\label{ex:inside-outside} 387 | Let $\Gamma$ be a circle and $P\notin \Gamma$. 388 | Assume a line $\ell$ passes thru the point $P$ 389 | and intersects $\Gamma$ at two distinct points, $X$ and~$Y$. 390 | Show that $P$ is inside $\Gamma$ if and only if $P$ lies between $X$ and~$Y$. 391 | \end{thm} 392 | 393 | A segment between two points on a circle is called a \index{chord}\emph{chord} of the circle. 394 | A chord passing thru the center of the circle is called its \index{diameter}\emph{diameter}. 395 | 396 | \begin{thm}{Exercise}\label{ex:chord-perp} 397 | Assume two distinct circles $\Gamma$ and $\Gamma'$ have a common chord~$[A B]$. 398 | Show that the line between the centers of $\Gamma$ and $\Gamma'$ is a perpendicular bisector of~$[A B]$. 399 | \end{thm} 400 | 401 | \begin{thm}{Exercise}\label{ex:center} 402 | Describe a ruler-and-compass construction of the center 403 | of a given circle. 404 | \end{thm} 405 | 406 | { 407 | 408 | \begin{wrapfigure}[4]{r}{54mm} 409 | \vskip-8mm 410 | \centering 411 | \includegraphics{mppics/pic-60} 412 | \end{wrapfigure} 413 | 414 | \begin{thm}[\abs]{Lemma}\label{lem:line-circle} 415 | A line and a circle can have at most two points of intersection. 416 | \end{thm} 417 | 418 | 419 | \parit{Proof.} Assume $A$, $B$, and $C$ are distinct points that lie on a line $\ell$ and a circle $\Gamma$ with the center~$O$. 420 | Then $OA=OB=OC$; in particular, $O$ lies on the perpendicular bisectors 421 | $m$ and $n$ to $[A B]$ and $[B C]$ respectively. 422 | Note that the midpoints of $[AB]$ and $[BC]$ are distinct. 423 | Therefore, $m$ and $n$ are distinct. 424 | The latter contradicts the uniqueness of the perpendicular (Theorem~\ref{perp:ex+un}). 425 | \qeds 426 | 427 | } 428 | 429 | \begin{thm}[!]{Exercise}\label{ex:two-circ} 430 | Show that two distinct circles can have at most two points of intersection. 431 | \end{thm} 432 | 433 | As a consequence of the above lemma, 434 | a line $\ell$ and a circle $\Gamma$ might have 2, 1, or 0 points of intersection. 435 | In the first two cases, the line is called \index{secant line}\emph{secant} or \index{tangent!line}\emph{tangent} respectively; 436 | if $P$ is the only point of intersection of $\ell$ and $\Gamma$, 437 | we say that \textit{$\ell$ is tangent to $\Gamma$ at $P$}. 438 | 439 | Similarly, according to Exercise~\ref{ex:two-circ}, 440 | two distinct circles might have 2, 1, or 0 points of intersection. 441 | If $P$ is the only point of intersection of circles $\Gamma$ and $\Gamma'$, 442 | we say that \index{tangent!circles}\emph{$\Gamma$ is tangent to $\Gamma'$ at $P$}; we also assume that a circle is tangent to itself at each of its points. 443 | 444 | \begin{thm}[\abs]{Lemma}\label{lem:tangent} 445 | Let $\ell$ be a line and $\Gamma$ be a circle centered at~$O$. 446 | Assume $P$ is a common point of $\ell$ and~$\Gamma$. 447 | Then $\ell$ is tangent to $\Gamma$ at $P$ if and only if $(PO)\perp \ell$. 448 | \end{thm} 449 | 450 | \parit{Proof.} 451 | Let $Q$ be the footpoint of $O$ on~$\ell$. 452 | 453 | Assume~$P\ne Q$. 454 | Let $P'$ be the reflection of $P$ across~$(OQ)$. 455 | Note that $P'\in\ell$ and $(OQ)$ is the perpendicular bisector of~$[PP']$. 456 | Therefore, $OP=OP'$. 457 | Hence $P,P'\in \Gamma\cap \ell$; 458 | that is, $\ell$ is secant to~$\Gamma$. 459 | 460 | If $P=Q$, 461 | then according to Lemma~\ref{lem:perp0$; it will be denoted by~$\Sigma^+$. 135 | 136 | { 137 | 138 | \begin{wrapfigure}{o}{40mm} 139 | \centering 140 | \vskip-7mm 141 | \includegraphics{mppics/pic-254} 142 | \end{wrapfigure} 143 | 144 | Given a point $P\in \Sigma^+$, consider the half-line $[OP)$. 145 | Suppose that $P'$ denotes the intersection of $[OP)$ and~$\Pi^+$. 146 | If $P=(x,y,z)$, then $P'=(\tfrac xz,\tfrac yz,1)$. 147 | It follows that $P\leftrightarrow P'$ is a bijection between $\Sigma^+$ and~$\Pi^+$. 148 | 149 | } 150 | 151 | The described bijection $\Sigma^+\leftrightarrow \Pi^+$ is called the \index{central projection}\emph{central projection} of 152 | the half-sphere~$\Sigma^+$. 153 | 154 | The central projection sends the intersections of the great circles with $\Sigma^+$ to the lines in~$\Pi^+$. 155 | This follows since the great circles are intersections of $\Sigma$ with planes passing thru the origin, 156 | and the lines in $\Pi^+$ are the intersection of $\Pi^+$ with these planes. 157 | 158 | The following exercise 159 | is analogous to Exercise~\ref{ex:h-median}. 160 | 161 | \begin{thm}{Exercise}\label{ex:s-medians} 162 | Let $\triangle_sABC$ be a nondegenerate spherical triangle. 163 | Assume that the plane $\Pi^+$ is parallel to the plane passing thru $A$, $B$, and~$C$. 164 | Let $A'$, $B'$, and $C'$ denote the central projections of $A$, $B$, and~$C$. 165 | \begin{enumerate}[(a)] 166 | \item\label{ex:s-medians:a} Show that the midpoints of $[A'B']$, $[B'C']$, and $[C'A']$ 167 | are central projections of the midpoints of $[AB]_s$, $[BC]_s$, and $[CA]_s$ respectively. 168 | \item\label{ex:s-medians:b} Use part (\ref{ex:s-medians:a}) to show that the medians of a spherical triangle intersect at one point. 169 | \end{enumerate} 170 | 171 | \end{thm} 172 | 173 | 174 | \begin{thm}{Exercise}\label{ex:s-altitudes} 175 | Let $P\leftrightarrow P'$ be the central projection described above 176 | and $N$ be the north pole; so, $N'=N$. 177 | Show that $|\measuredangle_s NPQ|=\tfrac\pi2$ if and only if $|\measuredangle N'P'Q'|=\tfrac\pi2$. 178 | \end{thm} 179 | 180 | \section{The stereographic projection} 181 | 182 | In this section, we define stereographic projection; 183 | it is closely related to the conformal model of the hyperbolic plane, which is discussed in Chapter~\ref{chap:poincare}. 184 | 185 | { 186 | 187 | \begin{wrapfigure}{r}{48mm} 188 | \vskip-6mm 189 | \centering 190 | \includegraphics{mppics/pic-252} 191 | \caption*{The plane thru $P$, $O$, and~$S$.} 192 | \end{wrapfigure} 193 | 194 | Consider the unit sphere $\Sigma$ 195 | centered at the origin $(0,0,0)$. 196 | This sphere can be described by the equation $x^2+y^2\z+z^2\z=1$. 197 | 198 | Suppose that $\Pi$ denotes the $(x,y)$-plane; 199 | it is defined by the equation $z \z= 0$. 200 | Clearly, $\Pi$ 201 | runs thru the center of~$\Sigma$. 202 | 203 | Let $N = (0, 0, 1)$ and $S\z=(0, 0, -1)$ denote the ``north'' and ``south'' poles of $\Sigma$; 204 | these are the points on the sphere that have extremal distances to~$\Pi$. 205 | Suppose that $\Omega$ denotes the ``equator'' of $\Sigma$; 206 | it is the intersection $\Sigma\cap\Pi$. 207 | 208 | } 209 | 210 | For a given point $P\ne S$ on $\Sigma$, 211 | consider the line $(SP)$ in the space. 212 | This line intersects $\Pi$ at exactly one point, denoted by~$P'$. 213 | Set $S'=\infty$. 214 | 215 | 216 | The map $\xi_s\: P\mapsto P'$ is called the \index{stereographic projection}\emph{stereographic projection from $\Sigma$ to $\Pi$ with respect to the south pole}. 217 | The inverse of this map $\xi^{-1}_s\: P'\z\mapsto P$ is called the {}\emph{stereographic projection from $\Pi$ to $\Sigma$ with respect to the south pole}. 218 | 219 | In the same way, one can define the 220 | {}\emph{stereographic projections $\xi_n$ and $\xi^{-1}_n$ with respect to the north pole}~$N$. 221 | 222 | Note that $P=P'$ if and only if $P\in\Omega$. 223 | 224 | Exercise~\ref{ex:stereographic-inversion} below states that the stereographic projection preserves 225 | the angles between arcs; 226 | more precisely, \textit{the absolute value of the angle measure} between arcs on the sphere. 227 | This is a useful property in cartography; 228 | the curves on the sphere meet at the same angles as their stereographic projections. 229 | 230 | 231 | \section[The inversion]{Inversion across a sphere} 232 | 233 | The inversion across a sphere is defined the same way as the inversion across a circle. 234 | 235 | Formally, let $\Sigma$ be the sphere with center $O$ and radius~$r$. 236 | The \index{inversion!inversion across a sphere}\emph{inversion} across $\Sigma$ of a point $P$ is the point $P'\in[OP)$ such that 237 | $$OP\cdot OP'=r^2.$$ 238 | In this case, the sphere $\Sigma$ will be called the 239 | \index{inversion!sphere of inversion}\emph{sphere of inversion}, 240 | and its center is called the \index{inversion!center of inversion}\emph{center of inversion}. 241 | 242 | We also add $\infty$ to the space and assume that the center of inversion is mapped to $\infty$ and the other way around. 243 | The space $\mathbb{R}^3$ with the point $\infty$ will be called \index{inversive!space}\emph{inversive space}. 244 | 245 | The inversion of space shares many properties with the inversion of the plane. 246 | Most importantly, analogs of theorems \ref{lem:inverse-4-angle}, \ref{thm:inverse-cline}, \ref{thm:angle-inversion} can be summarized as follows: 247 | 248 | \begin{thm}{Theorem}\label{thm:inversion-3d} 249 | The inversion across the sphere has the following properties: 250 | \begin{enumerate}[(a)] 251 | \item\label{thm:inversion-3d:a} Inversion maps a sphere or a plane into a sphere or a plane. 252 | \item\label{thm:inversion-3d:b} Inversion maps a circline into a circline. 253 | \item\label{thm:inversion-3d:cross-ratio} Inversion preserves the cross-ratio; 254 | that is, if $A'$, $B'$, $C'$, and $D'$ are the inverses of the points $A$, $B$, $C$, and $D$ respectively, 255 | then 256 | $$\frac{AB\cdot CD}{BC\cdot DA}= \frac{A'B'\cdot C'D'}{B'C'\cdot D'A'}.$$ 257 | \item Inversion maps arcs into arcs. 258 | \item\label{thm:inversion-3d:angle} 259 | Inversion preserves the absolute value of the angle 260 | measure between tangent half-lines to the arcs. 261 | \end{enumerate} 262 | \end{thm} 263 | 264 | In the following exercises, we apply the theorem above to tie the stereographic projection with the inversion. 265 | We assume that $\Sigma$, $\Pi$, $\Omega$, $O$, $S$, $N$ and $\xi_s$ are as in the previous section. 266 | 267 | { 268 | 269 | \begin{wrapfigure}{r}{32mm} 270 | \vskip-0mm 271 | \centering 272 | \includegraphics{mppics/pic-253} 273 | \end{wrapfigure} 274 | 275 | \begin{thm}{Exercise}\label{ex:stereographic-inversion} 276 | Show that the stereographic projections 277 | $\xi_s\: \Sigma\to\Pi$ and $\xi^{-1}_s\: \Pi\to\Sigma$ 278 | are the restrictions to $\Sigma$ and $\Pi$, respectively, of the inversion across the sphere $\Upsilon$ with the center $S$ and radius $\sqrt{2}$. 279 | 280 | Conclude that the stereographic projection preserves 281 | the angles between arcs; 282 | more precisely, the absolute value of the angle measure between arcs on the sphere. 283 | \end{thm} 284 | 285 | \begin{thm}{Exercise}\label{ex:great-circ} 286 | Show that the stereographic projection $\xi_s\:\Sigma\to\Pi$ 287 | sends the great circles to plane circlines that intersect $\Omega$ at opposite points. 288 | \end{thm} 289 | 290 | } 291 | 292 | \begin{thm}{Exercise}\label{ex:conform-sphere} 293 | Fix a point $P\in \Pi$ and let $Q$ be another point in~$\Pi$. 294 | Let $P'$ and $Q'$ denote their stereographic projections to~$\Sigma$. 295 | Set $x=PQ$ and $y=P'Q'_s$. 296 | Show that 297 | $$\lim_{x\to 0}\, \frac{y}{x}=\frac{2}{1+OP^2}.$$ 298 | \end{thm} 299 | 300 | The last exercise is analogous to Lemma~\ref{lem:conformal}. 301 | 302 | The proof of \ref{thm:inversion-3d} resembles the corresponding proofs in plane geometry. 303 | Let us give a couple of hints to the reader who wants to reconstruct its proof. 304 | To prove \ref{thm:inversion-3d}\textit{\ref{thm:inversion-3d:a}}, one needs the following lemma; 305 | its proof is left to the reader. 306 | 307 | \begin{thm}{Lemma} 308 | Let $\Sigma$ be a subset of the Euclidean space 309 | that contains at least two points. 310 | Fix a point $O$ in the space. 311 | 312 | Then $\Sigma$ is 313 | a sphere 314 | if and only if 315 | for every plane $\Pi$ passing thru $O$, 316 | the intersection $\Pi\cap \Sigma$ is either an empty set, 317 | a one-point set, or a circle. 318 | \end{thm} 319 | 320 | The following observation reduces part~\textit{(\ref{thm:inversion-3d:b})} to part~\textit{(\ref{thm:inversion-3d:a})}. 321 | 322 | \begin{thm}{Observation} 323 | Any circle in the space is an intersection of two spheres. 324 | \end{thm} 325 | 326 | { 327 | 328 | \begin{wrapfigure}{o}{25mm} 329 | \centering 330 | \includegraphics{mppics/pic-250} 331 | \end{wrapfigure} 332 | 333 | Let us define a \index{circular cone}\emph{circular cone} as a set formed by line segments from a fixed point, called the \index{tip of cone}\emph{tip} of the cone, to all the points on a fixed circle, called the \index{base!of cone}\emph{base} of the cone; 334 | we always assume that the base does not lie in the same plane as the tip. 335 | We say that the cone is \index{right!circular cone}\emph{right} 336 | if the center of the base circle is the footpoint of the tip on the base plane; 337 | otherwise, we call it \index{oblique circular cone}\emph{oblique}. 338 | 339 | } 340 | 341 | \begin{thm}[!]{Exercise}\label{ex:cone} 342 | Let $K$ be an oblique circular cone. 343 | Show that there is a plane $\Pi$ that is not parallel to the base plane of $K$ such that the intersection $\Pi\cap K$ is a circle. 344 | \end{thm} 345 | 346 | \section{The Pythagorean theorem} 347 | 348 | Here is an analog of the Pythagorean theorems (\ref{thm:pyth} and \ref{thm:pyth-h-poincare}) in spherical geometry. 349 | 350 | \begin{thm}{Spherical Pythagorean theorem}\label{thm:s-pyth}\index{Pythagorean theorem} 351 | Let $\triangle_sABC$ be a spherical triangle with a right angle at~$C$. 352 | Set $a=BC_s$, $b=CA_s$, and $c=AB_s$. 353 | Then 354 | $$\cos c=\cos a\cdot\cos b.$$ 355 | 356 | \end{thm} 357 | 358 | In the proof, we will use the notion of the scalar product which we are about to discuss. 359 | 360 | Let $v_A\z=(x_A,y_A,z_A)$ and $v_B=(x_B,y_B,z_B)$ denote the position vectors of points $A$ and $B$. 361 | The \index{scalar product}\emph{scalar product} of the two vectors $v_A$ and $v_B$ in $\mathbb{R}^3$ 362 | is defined as 363 | $$\langle v_A,v_B\rangle 364 | \df 365 | x_A\cdot x_B+y_A\cdot y_B+z_A\cdot z_B.\eqlbl{eq:scal-def}$$ 366 | 367 | Assume both vectors $v_A$ and $v_B$ are nonzero; 368 | suppose that $\phi$ denotes the angle measure between them. 369 | Then the scalar product can be expressed the following way: 370 | $$\langle v_A,v_B\rangle=|v_A|\cdot|v_B|\cdot\cos\phi, 371 | \eqlbl{eq:scal-angle}$$ 372 | where 373 | \begin{align*} 374 | |v_A|&=\sqrt{x_A^2+y_A^2+z_A^2}, 375 | & 376 | |v_B|&=\sqrt{x_B^2+y_B^2+z_B^2}. 377 | \end{align*} 378 | 379 | Now, assume that the points $A$ and $B$ 380 | lie on the unit sphere $\Sigma$ in $\mathbb{R}^3$ centered at the origin. 381 | In this case, $|v_A|=|v_B|=1$. 382 | By \ref{eq:scal-angle} we get that 383 | $$\cos AB_s=\langle v_A,v_B\rangle. 384 | \eqlbl{eq:scalar-s-dist}$$ 385 | 386 | \parit{Proof of the spherical Pythagorean theorem.} 387 | Since the angle at $C$ is right, 388 | we can choose the coordinates in $\mathbb{R}^3$ so that 389 | $v_C\z=(0,0,1)$, $v_A$ lies in the $(x,y)$-plane, so $v_A\z=(x_A,0,z_A)$, 390 | and $v_B$ lies in the $(y,z)$-plane, so $v_B=(0,y_B,z_B)$. 391 | 392 | { 393 | 394 | \begin{wrapfigure}{r}{40mm} 395 | \vskip-6mm 396 | \centering 397 | \includegraphics{mppics/pic-248} 398 | \end{wrapfigure} 399 | 400 | 401 | Applying \ref{eq:scalar-s-dist}, 402 | we get that 403 | \begin{align*} 404 | z_A&=\langle v_C,v_A\rangle 405 | =\cos b, 406 | \\ 407 | z_B&=\langle v_C,v_B\rangle 408 | =\cos a. 409 | \end{align*} 410 | 411 | Applying \ref{eq:scal-def} and \ref{eq:scalar-s-dist}, we get that 412 | \begin{align*} 413 | \cos c &=\langle v_A,v_B\rangle= 414 | \\ 415 | &=x_A\cdot 0+0\cdot y_B+z_A\cdot z_B= 416 | \\ 417 | &=\cos b\cdot\cos a. 418 | \end{align*} 419 | \qedsf 420 | 421 | } 422 | 423 | \begin{thm}{Exercise}\label{ex:2(pi/4)=pi/3} 424 | Show that 425 | if $\triangle_sABC$ is a spherical triangle with a right angle at $C$, 426 | and $AC_s=BC_s=\tfrac\pi4$, then $AB_s=\tfrac\pi3$. 427 | \end{thm} 428 | 429 | 430 | 431 | \section{Imaginary distance} 432 | 433 | Recall that 434 | \[ 435 | \cosh x=\frac {e^{x}+e^{-x}}2 436 | \qquad\text{and}\qquad 437 | \cos x=\frac {e^{i\cdot x}+e^{-i\cdot x}}2. 438 | \] 439 | The first formula is the definition of the hyperbolic cosine (see \ref{sec:hyp-trig}); 440 | the second one is called Euler's formula (see \ref{sec:Euler's formula}). 441 | It follows that 442 | \[\cosh x=\cos (i\cdot x) 443 | \qquad\text{and}\qquad 444 | \cos x=\cosh (i\cdot x).\] 445 | 446 | Let us compare the formulas in hyperbolic and spherical Pythagorean theorems (see \ref{thm:s-pyth} and \ref{thm:pyth-h-poincare}): 447 | \begin{align*} 448 | \cosh c&=\cosh a\cdot \cosh b, 449 | & 450 | \cos c&=\cos a\cdot \cos b. 451 | \end{align*} 452 | Observe that if we change $a$, $b$, and $c$ to $i\cdot a$, $i\cdot b$, and $i\cdot c$, 453 | then the first formula transforms into the second one, and the other way around. 454 | 455 | This is not a coincidence; the same holds for all analytic formulas: changing every distance $d$ to $i\cdot d$ transforms a valid spherical formula into a valid hyperbolic formula (the angle measures must remain the same). 456 | This magic substitution was found by Franz Taurinus \cite{taurinus}; we are not going to prove it. 457 | 458 | \begin{thm}{Advanced exercise}\label{ex:taurinus} 459 | Consider a spherical triangle $ABC$; set 460 | \begin{align*} 461 | a&=BC_s,& b&=CA_s,& c&=AB_s, 462 | \\ 463 | \alpha&=|\measuredangle_sCAB|,& \beta&=|\measuredangle_sABC|,& \gamma&=|\measuredangle_sBCA|. 464 | \end{align*} 465 | Use the Taurinus substitution to write the hyperbolic analogs to the following formulas in spherical geometry: 466 | 467 | \begin{enumerate}[(a)] 468 | \item The spherical cosine rule: 469 | \[\cos c=\cos a \cdot \cos b+\sin a\cdot \sin b\cdot \cos\gamma.\] 470 | \item The dual spherical cosine rule: 471 | \[\cos \gamma=-\cos \alpha \cdot \cos \beta+\sin \alpha\cdot \sin \beta \cdot \cos c.\] 472 | \item 473 | The spherical sine rule: 474 | \[\frac{\sin \alpha}{\sin a}=\frac{\sin \beta}{\sin b}=\frac{\sin \gamma}{\sin c}.\] 475 | \end{enumerate} 476 | 477 | \end{thm} 478 | -------------------------------------------------------------------------------- /absolute.tex: -------------------------------------------------------------------------------- 1 | %\part*{Non-Euclidean geometry} 2 | \addtocontents{toc}{\protect\contentsline{part}{\protect\numberline{}Non-Euclidean geometry}{}{}} 3 | 4 | \chapter{Neutral plane}\label{chap:non-euclid} 5 | 6 | Let us remove Axiom~\ref{def:birkhoff-axioms:4} from our axiomatic system (Section~\ref{sec:axioms}). 7 | This way we define a new object called the 8 | \index{plane!neutral plane}\index{neutral plane}\emph{neutral plane} or \index{plane!absolute plane}\index{absolute plane}\emph{absolute plane}. 9 | (In a neutral plane, Axiom~\ref{def:birkhoff-axioms:4} may or may not hold.) 10 | 11 | Every theorem in neutral geometry holds in Euclidean geometry. 12 | In other words, the Euclidean plane is an example of a neutral plane. 13 | In the next chapter, we will construct an example of a neutral plane that is not Euclidean. 14 | 15 | We use Axiom~\ref{def:birkhoff-axioms:4} starting from Chapter~\ref{chap:parallel}, 16 | and all the statements before that hold in neutral geometry. 17 | This makes all the discussed results 18 | about 19 | half-planes, 20 | signs of angles, 21 | congruence conditions, 22 | perpendicular lines, 23 | and reflections 24 | true in neutral geometry. 25 | Recall that a statement is marked with ``$\a$'' (for example, ``\textbf{Theorem\abs}'') if it holds in every neutral plane, and the same proof works. 26 | 27 | 28 | Let us give an example of a theorem in neutral geometry that admits a simpler proof in Euclidean geometry. 29 | 30 | \begin{thm}{Hypotenuse-leg congruence condition}\label{thm:hypotenuse-leg} 31 | Assume that triangles $ABC$ and $A'B'C'$ 32 | have right angles at $C$ and $C'$ respectively, 33 | $AB\z=A'B'$ and $AC\z=A'C'$. 34 | Then $\triangle ABC\cong\triangle A'B'C'$. 35 | \end{thm} 36 | 37 | 38 | \parit{Euclidean proof.} 39 | By the Pythagorean theorem $BC=B'C'$. 40 | Then the statement follows from the SSS congruence condition. 41 | \qeds 42 | 43 | The proof of the Pythagorean theorem used properties of similar triangles, which in turn used Axiom~\ref{def:birkhoff-axioms:4}. 44 | Therefore this proof does not work in a neutral plane. 45 | 46 | \parit{Neutral proof.} 47 | Suppose that $D$ denotes the reflection of $A$ across $(BC)$ 48 | and $D'$ denotes the reflection of $A'$ across $(B'C')$. 49 | Note that 50 | $$ 51 | AD=2\cdot AC=2\cdot A'C'=A'D',\qquad 52 | BD=BA=B'A'=B'D'. 53 | $$ 54 | 55 | { 56 | 57 | \begin{wrapfigure}{r}{28mm} 58 | \vskip-0mm 59 | \centering 60 | \includegraphics{mppics/pic-180} 61 | \end{wrapfigure} 62 | 63 | By the SSS congruence condition (\ref{thm:SSS}), 64 | we get that $\triangle ABD\cong \triangle A'B'D'$. 65 | 66 | The statement follows since $C$ is the midpoint of $[AD]$ 67 | and $C'$ is the midpoint of $[A'D']$. 68 | \qeds 69 | 70 | \begin{thm}{Exercise}\label{ex:tangent-angle-neutral} 71 | Read the proof of Theorem~\ref{thm:tangent-angle} and identify the first statement that does not work in the neutral plane. 72 | \end{thm} 73 | 74 | } 75 | 76 | 77 | \begin{thm}{Exercise}\label{ex:abs-bisect=median} 78 | Give a proof of Exercise~\ref{ex:bisect=median} 79 | that works in the neutral plane. 80 | \end{thm} 81 | 82 | 83 | 84 | \begin{thm}{Exercise}\label{ex:abs-inscibed} 85 | Let $ABCD$ be an inscribed quadrangle in the neutral plane. 86 | Show that 87 | $$\measuredangle ABC+\measuredangle CDA\equiv \measuredangle BCD+\measuredangle DAB.$$ 88 | 89 | \end{thm} 90 | 91 | One cannot use Corollary~\ref{cor:inscribed-quadrangle} to solve the exercise above since it uses Theorems~\ref{thm:tangent-angle} and \ref{thm:inscribed-angle}, 92 | which in turn use Theorem~\ref{thm:3sum}. 93 | 94 | 95 | \section{Two angles of a triangle} 96 | 97 | In this section, we will prove a weaker form of Theorem~\ref{thm:3sum} 98 | which holds in every neutral plane. 99 | 100 | \begin{thm}{Proposition}\label{prop:2sum} 101 | Let $\triangle ABC$ be a nondegenerate triangle in the neutral plane. 102 | Then 103 | $$|\measuredangle CAB|+|\measuredangle ABC|< \pi.$$ 104 | 105 | \end{thm} 106 | 107 | \begin{wrapfigure}{r}{23mm} 108 | \vskip-1mm 109 | \centering 110 | \includegraphics{mppics/pic-182} 111 | \end{wrapfigure} 112 | 113 | According to \ref{thm:signs-of-triug}, the angles $ABC$, $BCA$, and $CAB$ 114 | have the same sign. 115 | Therefore, in the Euclidean plane, the theorem follows immediately from Theorem~\ref{thm:3sum}. 116 | 117 | \parit{Proof.} 118 | Let $X$ be the reflection of $C$ across the midpoint $M$ of $[AB]$. 119 | Applying \ref{ex:between}, we get 120 | \[|\measuredangle CAX|=|\measuredangle CAB|+|\measuredangle ABC|.\eqlbl{eq:CAX=CAB+ABC}\] 121 | 122 | By Proposition~\ref{prop:point-reflection+} 123 | $\measuredangle BAX\z=\measuredangle ABC$. 124 | 125 | Note that $\measuredangle CAX\z{\not\equiv} \pi$; otherwise, $X$ would lie on $(AC)$. 126 | Therefore the identity \ref{eq:CAX=CAB+ABC} implies that 127 | \[|\measuredangle CAB|+|\measuredangle ABC|=|\measuredangle CAX|<\pi.\] 128 | \qedsf 129 | 130 | \begin{thm}[!]{Exercise}\label{ex:parallel-abs} 131 | Assume $A$, $B$, $C$, and $D$ are points in a neutral plane 132 | such that 133 | $$2\cdot \measuredangle ABC+2\cdot\measuredangle BCD\equiv 0.$$ 134 | Show that $(AB)\parallel (CD)$. 135 | \end{thm} 136 | 137 | Note that one cannot apply the transversal property (\ref{thm:parallel-2}) here. 138 | 139 | 140 | \begin{thm}{Exercise}\label{ex:SAA} 141 | Prove the \index{SAA congruence condition}\emph{side-angle-angle} congruence condition in neutral geometry. 142 | 143 | 144 | In other words, let $ABC$ and $A'B'C'$ be two triangles in a neutral plane; 145 | suppose that $\triangle A'B'C'$ is nondegenerate. 146 | Show that $\triangle ABC\z\cong \triangle A'B'C'$ 147 | if 148 | $$AB=A'B', 149 | \quad 150 | \measuredangle ABC=\pm\measuredangle A'B'C', 151 | \quad 152 | \text{and} 153 | \quad 154 | \measuredangle BCA=\pm\measuredangle B'C'A'.$$ 155 | 156 | \end{thm} 157 | 158 | In the Euclidean plane, the above exercise follows from ASA and the theorem on the sum of angles of a triangle (\ref{thm:3sum}). 159 | However, Theorem~\ref{thm:3sum} cannot be used here, since its proof uses Axiom~\ref{def:birkhoff-axioms:4}. 160 | Later (Theorem~\ref{thm:3sum-h}) 161 | we will show that Theorem~\ref{thm:3sum} does not hold in a neutral plane. 162 | 163 | \begin{thm}{Exercise}\label{ex:chev