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\pdfrefximage\pdflastximage}}$}}#8$}} \else \special{header=mdrlib.ps} \def\RGBColor#1#2{\special{color push rgb #1}#2\special{color pop}} \long\def\InsertPSFigure#1 #2 #3 #4\EndFig{\par \hbox{\hskip #1mm$\vbox to#2mm{\vfil\special{" #3}}#4$}} \long\def\InsertPSFile#1 #2 #3 #4 #5 #6\EndFig{\par \hbox{\hskip #1mm$\vbox to#2mm{% \vfil\special{psfile=#5 hscale=#3 vscale=#4}}#6$}} \long\def\InsertImage#1 #2 #3 #4 #5 #6 #7 #8\EndFig{\par \hbox{\hskip #1mm$\vbox to#2mm{% \vfil\special{psfile="`img2eps file #7 height #4 mm width #3 mm gamma #5 angle #6}}#8$}} \fi \long\def\LabelTeX#1 #2 #3\ELTX{\rlap{\kern#1mm\raise#2mm\hbox{#3}}} % bibibliography \def\bibitem#1&% {\hangindent=1cm\hangafter=1 \noindent\rlap{\hbox{\eightpointbf #1}}\kern1cm{\rm #2}, {\it #3}, {\rm #4}} \openauxfile \notthispage=\pageno \title{A rigorous deductive approach to} \title{elementary Euclidean geometry} \titlerunning{A rigorous deductive approach to elementary Euclidean geometry} \bigskip \centerline{\twelvebf Jean-Pierre Demailly}\medskip \centerline{\twelverm Universit\'e Joseph Fourier Grenoble I} \smallskip \centerline{\tt http://www-fourier.ujf-grenoble.fr/$\,\tilde{}\,$demailly/documents.html} \vskip0.6cm \noindent {\bf R\'esum\'e~:} L'objectif de ce texte est de proposer une piste pour un enseignement logiquement rigoureux et cependant assez simple de la g\'eom\'etrie euclidienne au coll\`ege et au lyc\'ee. La g\'eom\'etrie euclidienne se trouve \^etre un domaine tr\`es privil\'egi\'e des math\'ematiques, \`a l'int\'erieur duquel il est possible de mettre en œuvre d\`es le d\'epart des raisonnements riches, tout en faisant appel de mani\`ere remarquable \`a la vision et \`a l'intuition. Notre pr\'eoccupation est d'autant plus grande que l'\'evolution des programmes scolaires depuis 3 ou 4 d\'ecen\-nies r\'ev\`ele une diminution tr\`es marqu\'ee des contenus g\'eom\'etriques enseign\'es, en m\^eme temps qu'un affaiblissement du raisonnement math\'ematique auquel l'enseignement de la g\'eom\'etrie permettait pr\'ecis\'e\-ment de contribuer de fa\c{c}on essentielle. Nous esp\'erons que ce texte sera utile aux professeurs et aux auteurs de manuels de math\'ematiques qui ont la possibilit\'e de s'affranchir des contraintes et des prescriptions trop indigentes des programmes officiels. Les premi\`eres sections devraient id\'ealement \^etre ma\^{\i}tris\'ees aussi par tous les professeurs d'\'ecole, car il est \`a l'\'evidence tr\`es utile d'avoir du recul sur toutes les notions que l'on doit enseigner~! \noindent {\bf Mots-cl\'es~:} g\'eom\'etrie euclidienne \bigskip \noindent {\bf Resumen~:} El objetivo de este art\'{\i}culo es presentar un enfoque riguroso y a\'un razonablemente simples para la ense\~nanza de la geometr\'{\i}a euclidiana elemental a nivel de educaci\'on secundaria. La geometr\'{\i}a euclidiana es una \'area privilegiada de las matem\'aticas, ya que permite desde un primer nivel practicar razonamientos rigurosos y ejercitar la visi\'on y la intuici\'on. Nuestra preocupaci\'on es que las numerosas reformas de planes de estudio en las \'ultimas 3 d\'ecadas en Francia, y posiblemente en otros pa\'{\i}ses occidentales, han llevado a una disminuci\'on preocupante de la geometr\'{\i}a, junto con un generalizado debilitamiento del razonamiento matem\'atico al que la geometr\'{\i}a contribuye espec\'{\i}ficamente de manera esencial. Esperamos que este punto de vista sea de inter\'es para los autores de libros de texto y tambi\'en para los profesores que tienen la posibilidad de no seguir exactamente las prescripciones sobre los contenidos menos relevantes, cuando est\'an por desgracia impuestos por las autoridades educativas y por los planes de estudios. El contenido de las primeras secciones, en principio, deber\'{\i}a tambi\'en ser dominado por los profesores de la escuela primaria, ya que siempre es recomendable conocer m\'as de lo que uno tiene que ense\~nar, a cualquier nivel~! \noindent {\bf Palabras clave~:} Geometr\'{\i}a euclidiana \section{0. Introduction} The goal of this article is to explain a rigorous and still reasonably simple approach to teaching elementary Euclidean geometry at the secondary education levels. Euclidean geometry is a privileged area of mathematics, since it allows from an early stage to practice rigorous reasonings and to exercise vision and intuition. Our concern is that the successive reforms of curricula in the last 3 decades in France, and possibly in other western countries as well, have brought a worrying decline of geometry, along with a weakening of mathematical reasoning which geometry specifically contributed to in an essential way. We hope that these views will be of some interest to textbook authors and to teachers who have a possibility of not following too closely the prescriptions for weak contents, when they are unfortunately enforced by education authorities and curricula. The first sections should ideally also be mastered by primary school teachers, as it is always advisable to know more than what one has to teach at any given level~! \noindent {\bf Keywords~:} Euclidean geometry \section{1. On axiomatic approaches to geometry} As a formal discipline, geometry originates in Euclid's list of axioms and the work of his successors, even though substantial geometric knowledge existed before. \InsertImage 27 64 0 60 1.0 0 {Euclide_Elements.png} \EndFig \centerline{\it An excerpt of Euclid's book} \medskip The traditional teaching of geometry that took place in France during the period 1880-1970 was directly inspired by Euclid's axioms, stating first the basic properties of geometric objects and using the ``triangle isometry criteria'' as the starting point of geometric reasoning. This approach had the advantage of being very effective and of quickly leading to rich contents. It also adequately reflected the intrinsic nature of geometric properties, without requiring extensive algebraic calculations. These choices echoed a mathematical tradition that was firmly rooted in the nineteenth century, aiming to develop ``pure geometry'', the highlight of which was the development of projective geometry by Poncelet. Euclid's axioms, however, were neither complete nor entirely satisfactory from a logical perspective, leading mathematicians as Hilbert and Pasch to develop the system of axioms now attributed to Hilbert, that was settled in his famous memoir Grundlagen der Geometrie in 1899. \InsertImage 47.5 54 0 50 1.0 0 {Hilbert.jpg} \EndFig \centerline{\it David Hilbert $($1862--1943$\,)$, in 1912} \medskip It should be observed, though, that the complexity of Hilbert's system of axioms makes it actually unpractical to teach geometry at an elementary level\note{Even the improved and simplified version of Hilbert's axioms presented by Emil Artin in his famous book ``Geometric Algebra'' can hardly be taught before the $3^{\rm rd}$ or $4^{\rm th}$ year at university.}. The result, therefore, was that only a very partial axiomatic approach was taught, leading to a situation where a large number of properties that could have been proved formally had to be stated without proof, with the mere justification that they looked intutively true. This was not necessarily a major handicap, since pupils and their teachers may not even have noticed the logical gaps. However, such an approach, even though it was in some sense quite successful, meant that a substantial shift had to be accepted with more contemporary developments in mathematics, starting already with Descartes' introduction of analytic geometry. The drastic reforms implemented in France around 1970 (with the introduction of ``modern mathematics'', under the direction of Andr\'e Lichnerowicz) swept away all these concerns by implementing an entirely new paradigm~: according to Jean Dieudonn\'e, one of the Bourbaki founders, geometry should be taught as a corollary of linear algebra, in a completely general and formal setting. The first step of the reform implemented this approach from ``classe de seconde'' (grade 10) on. A major problem, of course, is that the linear algebra viewpoint completely departs from the physical intuition of Euclidean space, where the group of invariance is the group of Euclidean motions and not the group of affine transformations. \InsertImage 2 55 0 50 1.0 0 {Descartes.jpg} \EndFig \vskip -55.3mm \InsertImage 48 55 0 50 1.0 0 {Dieudonne0.jpg} \EndFig \vskip -55.3mm \InsertImage 91 55 0 50 1.0 0 {Lichnerowicz0.jpg} \EndFig \centerline{\it from Descartes $($1596-1650$\,)$ to Dieudonn\'e $($1906-1992$\,)$ and Lichnerowicz $($1915-1998$\,)$} \medskip The reform could still be followed in a quite acceptable way for about one decade, as long as pupils had a solid background in elementary geometry from their earlier grades, but became more and more unpractical when primary school and junior high school curricula were themselves (quite unfortunately) downgraded. All mathematical contents of high school were then severely axed around 1986, resulting in curricula prescriptions that in fact did not allow any more the introduction of substantial deductive activity, at least in a systematic way. \iflongertext As far as geometry is concerned, it is therefore essential to return to teaching practices that clearly introduce the geometric nature of objects under consideration. This means that precise definitions should be explicitly given and that curricula should be organized in a way that allows to state and prove a rich set of properties from those taken as the starting point. In a word, one should return, in a possibly not entirely explicit form, to an ``axiomatic'' presentation of geometry. We do not mean here that the order of presentation of concepts must necessarily follow the order implied by the internal logical constraints of ``axioms'' (whatever they are), but one should at least provide a clear framework that can be adopted by teachers and that serves as a firm guide for designing curricula. \InsertImage 15 55 0 50 1.0 0 {Viete0.jpg} \EndFig \vskip -55mm \InsertImage 81 55 0 50 1.0 0 {Stevin0.png} \EndFig \medskip \centerline{\it Fran\c{c}ois Vi\`ete $($1540--1603$\,)$\kern25mm Simon Stevin $($1548--1620$\,)$} \medskip To justify our desire to go beyond the traditional Euclidean approach that was in use 50 or 60 years ago, our observation is that one now has a considerable advantage over the Greek, which is to have at our disposal an efficient and universally accepted algebraic notation thanks to Simon Stevin, and the possibility to investigate geometry though coordinates and analytic calculations, thanks to Descartes. In Greek geometry, real numbers were only thought as ratios of quantities of the same kind, rather than ``abstract numbers'' in the modern sense. The concept of a polynomial function (or of a general function) was then much harder to tackle, although the Greeks certainly had methods for solving problems involving second and third degree equations through geometric constructions. The approach we propose here is somehow a synthesis of the viewpoint of Pythagoras and Euclid with that of Descartes. Euclidean geometry is characterized by the expression of the distance through Pythagoras' theorem. All objects involved in Euclidean geometry can then be derived merely from the concept of length\note{It is well known today that a metric structure determines many other geometrical invariants, such as Hausdorff measures, curvature, tensors, etc. This viewpoint has been extensively developed by Mikhail Gromov in the last 2 or Last 3 decades. See for example M.~Gromov, {\it Metric structures for Riemannian manifolds} by J.~Lafontaine and P.~Pansu, Cedic/Fernand Nathan, Paris, 1981.}. In this context, our reconstruction of Euclidean geometry derives Thales' theorem from Pythagoras' property. Another advantage is that all concepts can be defined using a minimal and intuitive formalism. In fact, the theory can be set to use a single axiom, directly related to Pythagoras' theorem, that can moreover be easily ``justified'' by simple visual considerations. However, the crucial point is certainly not the use of a single axiom -- however what we call the ``Pythagoras/Descartes'' axiom, rather than a genuine axiom, is rather a concise description of a model of Euclidean geometry. Unlike the approach based on linear algebra, we start from points and affine concepts rather than from the much less intuitive concepts of vectors and vector spaces, and here the notion of vector will be constructed a posteriori. Another goal is to invalidate the occasional argument that elementary traditional geometry does not constitute a serious or useful part of mathematics, because is cannot be exposed in a rigorous or formalized way, in the modern sense. However, there are certainly some disadvantages that are inherent to our exposition. One of them is that it is only a ``modern reconstruction'', and in spite of being somehow obvious to contemporary mathematicians (and most probably so to Klein or Hilbert), it has probably never been taught in this form on a large scale. Another feature is that giving from scratch a ``rigid model'' of Euclidean geometry may not be the most appropriate framework to introduce other subjects such as affine or projective geometry, which rely rather on incidence properties. Finally, real numbers are deeply embedded in the model, so, at a later stage, introducing geometries over other fields will require a different approach. Nevertheless, for a potential use at secondary school level, we have deliberately preferred simplicity to generality, and have therefore chosen to focus on the Euclidean and Archimedean model which is also the one of Newtonian mechanics$\,\ldots$ \else We believe however that it is necessary to introduce the basic language of mathematics, e.g. the basic concepts of sets, inclusion, intersection, etc, as soon as needed, most certainly already at the beginning of junior high school. Geometry is a very appropriate groundfield for using this language in a concrete way. \fi \section{2. Geometry, numbers and arithmetic operations} \iflongertext According to the above discussion, teaching geometry is inseparable from teaching numbers. This is especially true for the study of contemporary versions of the Euclidean model, which are based on real numbers. We first describe fundamental facts about numbers that should be taught accordingly. \subsection{2.1. Primary school and the 4 arithmetic operations} It is essential that handwritten calculations be taught again thoroughly at the primary school level, so as to gain a complete control of operation algorithms -- calculators should be used only when students have already acquired a very good acquaintance to arithmetic operations. Safe and effective practice of handwritten calculations requires a fluid knowledge of addition and multiplication tables (as well as ``subtraction and division tables'' -- by this, we mean being able to read addition and multiplication tables in the opposite direction). Some educators inclined to a ``minimalist approach'' have insisted on the fact that it is enough to mentally perform approximate calculations and estimated orders of magnitude, but the following facts are unavoidable$\,$: {\parindent=4mm \item{$\bullet$} although mental arithmetic implies a fluid knowledge of the multiplication table, just as handwritten calculations do, its procedures are different, since intermediate results have to be memorized. This is done by handling units, tens, hundreds, thousands, rather than digits taken separately, and one usually starts as well by taking into account higher weight digits rather than lower weight digits as in the handwritten algorithms. In addition to this, the smaller size of numbers involved does not usually allow to reach the level of generality that is necessary to acquire a complete understanding of the algorithms of handwritten calculations. \item{$\bullet$} even if young children may develop some sort of intuitive perception of the size of numbers before being able to perform exact calculations (this ability should of course not be neglected or negated), reaching a sufficient reliability in performing just approximate estimates is attained only by means of certain exact procedures, such as calculating powers of ten combined with a knowledge of the multiplication table. \item{$\bullet$} Finally, developing an ability to performing approximate calculations (just in case this would be the major target) is greatly enhanced when one masters e.g.\ the division algorithm$\,$: when the divider has two or more digits, guessing the relevant digit in the quotient requires a very effective ability to intuitively grasp the approximate value of the product of a several digit umber by a single digit number. In this circumstance, it is clear that children need precisely defined objectives to build their mental models, it is certainly not sufficient that the curricula declare the ability to perform approximate calculation a worthy target to hope that this goal will be achieved to some sort of spontaneous perception.\vskip0pt } Of course, mastering the arithmetic algorithms is very far from being sufficient to understand numbers$\,$; children can access the real meaning of operations only by solving numerous problems involving everyday life objects (counting apples, currency, lengths, weights$\,\ldots $). Contrary to what has been stated in certain ``modern'' pedagogical theories, this intuitive capability can be built much more efficiently when all four arithmetic operations are introduced simultaneously, so that children can compare (and oppose) the use of different operations. Therefore introducing the four arithmetic operations should be done rather early, e.g.\ in the first grade, and certain related activities can be considered even at pre-school levels (kindergarten). \subsection{2.2. Synergy of numeracy and geometry} Geometric activities can already be organized at kindergarten, for example through drawing or coloring. It is certainly very useful to ask children to draw simple geometric patterns, friezes, etc. During the first years of primary school, geometric activities can be organized in close connection with arithmetic calculations. For instance, calculating the perimeter of a rectangle involves addition, while calculating the area involves multiplication, and possibly changes of units (when dealing with the case where sides are decimal numbers). Also, the geometric representation of the rectangle immediately gives a proof of the commutativity of multiplication\?: \InsertPSFigure 5.000 32.000 {%%conditional gsave /a 7 def /b 5 def /w 15 def 0.3 setlinewidth 0 1 a {/k exch def k w mul 0 moveto 0 w b mul rlineto stroke } for 0 1 b {/l exch def 0 l w mul moveto w a mul 0 rlineto stroke } for 0 1 a 1 sub {/k exch def 0 1 b 1 sub {/l exch def w 2 div dup w k mul add exch w l mul add moveto currentpoint 3 0 360 arc fill } for } for 1 0 0 setrgbcolor 0 1 b 1 sub {/l exch def w 2 div dup l w mul add moveto w a 1 sub mul 0 rlineto stroke } for 0 0 0 setrgbcolor w a 2.5 add mul 0 translate 0 1 a {/k exch def k w mul 0 moveto 0 w b mul rlineto stroke } for 0 1 b {/l exch def 0 l w mul moveto w a mul 0 rlineto stroke } for 0 1 a 1 sub {/k exch def 0 1 b 1 sub {/l exch def w 2 div dup w k mul add exch w l mul add moveto currentpoint 3 0 360 arc fill } for } for 1 0 0 setrgbcolor 0 1 a 1 sub {/k exch def w 2 div dup k w mul add exch moveto 0 w b 1 sub mul rlineto stroke } for grestore } \LabelTeX 100 12 $7\times 5=5\times 7$\ELTX \EndFig \medskip Certain numerical identities such as $6\times 8=7\times 7-1$ or $5\times 9=7\times 7-4$ can be viewed and explained through geometric considerations (in this direction, it would probably be appropriate to bring pupils to manipulate wooden pieces, so that vision can be consolidated by hand work$\,\ldots$) \InsertPSFigure 5.000 43.000 {%%conditional gsave /a 7 def /b 7 def /w 15 def 0.3 setlinewidth 0 0 moveto a 1 sub w mul 0 rlineto 0 w rlineto a 1 sub w mul neg 0 rlineto closepath 0.8 0.8 1 setrgbcolor fill 0 w moveto a w mul 0 rlineto 0 b 1 sub w mul rlineto a w mul neg 0 rlineto closepath 1 0.8 0.8 setrgbcolor fill w a 1 sub mul 0 moveto w 0 rlineto 0 w rlineto w neg 0 rlineto closepath 0.8 1 0.8 setrgbcolor fill 0 setgray 0 1 a 1 add {/k exch def k w mul 0 moveto 0 w b mul rlineto stroke } for 0 1 b {/l exch def 0 l w mul moveto w a 1 add mul 0 rlineto stroke } for w a 1 add 2.5 add mul 0 translate a w mul w moveto w 0 rlineto 0 w b 1 sub mul rlineto w neg 0 rlineto closepath 0.8 0.8 1 setrgbcolor fill 0 w moveto a w mul 0 rlineto 0 b 1 sub w mul rlineto a w mul neg 0 rlineto closepath 1 0.8 0.8 setrgbcolor fill w a 1 sub mul 0 moveto w 0 rlineto 0 w rlineto w neg 0 rlineto closepath 0.8 1 0.8 setrgbcolor fill 0 setgray 0 1 a 1 add {/k exch def k w mul 0 moveto 0 w b mul rlineto stroke } for 0 1 b {/l exch def 0 l w mul moveto w a 1 add mul 0 rlineto stroke } for grestore } \LabelTeX 15 -5 $7\times 7$\ELTX \LabelTeX 65 -5 $6\times 8=7\times 7-1$\ELTX \EndFig \vskip3mm \InsertPSFigure 5.000 41.000 {%%conditional gsave /a 7 def /b 7 def /w 15 def 0.3 setlinewidth 0 0 moveto a 2 sub w mul 0 rlineto 0 w 2 mul rlineto a 2 sub w mul neg 0 rlineto closepath 0.8 0.8 1 setrgbcolor fill 0 w 2 mul moveto a w mul 0 rlineto 0 b 2 sub w mul rlineto a w mul neg 0 rlineto closepath 1 0.8 0.8 setrgbcolor fill w a 2 sub mul 0 moveto w 2 mul 0 rlineto 0 w 2 mul rlineto w -2 mul 0 rlineto closepath 0.8 1 0.8 setrgbcolor fill 0 setgray 0 1 a 2 add {/k exch def k w mul 0 moveto 0 w b mul rlineto stroke } for 0 1 b {/l exch def 0 l w mul moveto w a 2 add mul 0 rlineto stroke } for w a 2 add 2.5 add mul 0 translate a w mul w 2 mul moveto w 2 mul 0 rlineto 0 w b 2 sub mul rlineto w -2 mul 0 rlineto closepath 0.8 0.8 1 setrgbcolor fill 0 w 2 mul moveto a w mul 0 rlineto 0 b 2 sub w mul rlineto a w mul neg 0 rlineto closepath 1 0.8 0.8 setrgbcolor fill w a 2 sub mul 0 moveto w 2 mul 0 rlineto 0 w 2 mul rlineto w -2 mul 0 rlineto closepath 0.8 1 0.8 setrgbcolor fill 0 setgray 0 1 a 2 add {/k exch def k w mul 0 moveto 0 w b mul rlineto stroke } for 0 1 b {/l exch def 0 l w mul moveto w a 2 add mul 0 rlineto stroke } for grestore } \LabelTeX 15 -5 $7\times 7$\ELTX \LabelTeX 65 -5 $5\times 9=7\times 7-2\times 2$\ELTX \EndFig \vskip3mm These simple reasonings are actually genuine mathematical proofs, probably among the first ones that can be presented to pupils. Let us come to the problem of calculating areas. It is natural to start by the area of a rectangle whose sides are integer multiples of the unit of length. Later on, the problem of computing the area of a rectangle whose sides are not whole numbers, such as $1.2\,$m by $0.7\,$m, is solved by converting lengths into decimeters. This yields $$12\,\hbox{dm}\times 7\,\hbox{dm}=84\,\hbox{dm}^2=0.84\,\hbox{m}^2,$$ once it is realized that $1\,\hbox{m}^2=100\,\hbox{dm}^2$, and thus $1\,\hbox{dm}^2=0.01\,\hbox{m}^2$. Therefore, one sees that the area of a rectangle is always the product of the lengths of its sides, even when the side lengths are decimal numbers. The distributivity of multiplication with respect to addition can also be seen geometrically\?: \InsertPSFigure 5.000 36.000 {%%conditional gsave 0 0 moveto 120 0 lineto 120 90 lineto 0 90 lineto closepath 1 0.8 0.8 setrgbcolor fill 120 0 moveto 180 0 lineto 180 90 lineto 120 90 lineto closepath 0.8 0.8 1 setrgbcolor fill 0 setgray 0 0 moveto 120 0 lineto 120 90 lineto 0 90 lineto closepath stroke 120 0 moveto 180 0 lineto 180 90 lineto 120 90 lineto closepath stroke grestore} \LabelTeX -3 15 $a$\ELTX \LabelTeX 20 -4 $b$\ELTX \LabelTeX 52 -4 $c$\ELTX \LabelTeX 75 15 $a\times(b+c)=a\times b+a\times c$\ELTX \EndFig \vskip3mm Calculating areas and volumes thus allows to firmly consolidate the understanding of the arithmetic operations, in relation e.g.\ with physical units and their handling in practical problems. It is possible to give, already at primary school level, genuine (i.e.\ non entirely trivial) mathematical proofs -- for instance in grade~5, one can give a complete justification of the formula that expresses the area of a disk\?: \InsertPSFigure 11.000 55.000 {%%conditional gsave 0.5 0.5 scale -60 0 translate /R 120 def /steps 16 def /delta 360 steps div def R 1.1 mul R 1.1 mul translate 0 1 steps 1 sub { /k exch def 0 0 moveto k delta mul dup cos R mul exch sin R mul lineto 0 0 R k delta mul k 1 add delta mul arc 0 0 lineto closepath gsave k 2 mul steps lt { 0.8 0.8 1 setrgbcolor } { 1 0.8 0.8 setrgbcolor } ifelse fill grestore stroke} for gsave 0.4 setlinewidth [ 5 2 ] 0 setdash 0 0 moveto 180 delta 0.5 mul sub dup cos R mul exch sin R mul lineto stroke grestore /delta delta 2 div def /dx delta sin R mul 2 mul def R 1.6 mul R -0.5 mul translate gsave 0 1 steps 2 div 1 sub { /k exch def 0 0 moveto delta sin R mul delta cos R mul lineto 0 0 R 90 delta sub 90 delta add arc currentpoint /y exch def /x exch def 0 0 lineto closepath gsave 0.8 0.8 1 setrgbcolor fill grestore stroke gsave x dx add y translate 1 -1 scale 0 0 moveto delta sin R mul delta cos R mul lineto 0 0 R 90 delta sub 90 delta add arc 0 0 lineto closepath gsave 1 0.8 0.8 setrgbcolor fill grestore stroke grestore dx 0 translate} for grestore gsave 0.4 setlinewidth [ 5 2 ] 0 setdash 0 0 moveto 0 R lineto stroke dx steps mul 0.5 mul 0 translate 0 0 moveto 0 R lineto stroke 0 0 R 90 90 delta add arc stroke grestore 0 -30 moveto 0 180 10 vector stroke 0 -30 moveto dx steps mul 0.5 mul 0 10 vector stroke dx steps mul 0.5 mul 35 add 0 translate 0 0 moveto 0 -90 10 vector stroke 0 0 moveto R 90 10 vector stroke grestore } \LabelTeX 7 48 Disk\kern1.6cm $\longrightarrow$\kern1.6cm parallelogram (or rectangle)\ELTX \LabelTeX -9 -5 $\displaystyle \pi={P\over D}~~\Rightarrow~~ P=\pi\times D=2\times \pi\times R$\ELTX \LabelTeX 72 0 base$\displaystyle{}\simeq{P\over 2}=\pi\times R$\ELTX \LabelTeX 115 22 $R$\ELTX \EndFig In the limit, by increasing the number of triangular sectors, one sees that the area of a disk is given by $\pi\times R\times R=\pi R^2$. Of course, this activity presupposes that the pupils have been explained before how to compute the area of rectangles, triangles and parallelograms, again through a visualization of the classical geometric decompositions used to justify the formulas. The conceptual position of formula $P=\pi D=2\pi R$ is different, in this case it should rather be viewed as a {\it definition} of number~$\pi$\?: this is the ratio between the perimeter and the diameter, which is independent of the size of the circle under consideration (one can intuitively mention that if the diameter doubles or triple, then the perimeter changes in the same way$\,$; of course, this follows formally from Thales theorem, by approximating circle arcs with polygons$\,\ldots$). If time allows, it is of course advisable to turn this observation into a practical experiment, using a pipe with a known diameter, and putting around a few loops of string, so as to get an approximate value of~$\pi$. As a general rule, geometry should be taught through a combination of reasoning with concrete mani\-pulations\?: paper and scissor decompositions, usage of instruments (ruler, compasses, protractor...), elementary constructions (midpoints, medians, bissectors,~$\ldots$). Working with graph paper helps in developing the intuitive understanding of cartesian coordinates$\,$; it would thus be extremely useful to start such activities already from grade 2 at latest. \subsection{2.3. Negative numbers, square roots, real numbers} While developing a better understanding elementary arithmetic operations, children can be introduced simple ``arithmetic and geometric'' progressions $$ \eqalign{ &0,\quad a,\quad 2a,\quad 3a,\quad 4a,\quad 5a,\quad 6a,\quad 7a, \quad \ldots\cr &1,\quad a,\kern10.7pt a^2,\kern10.7pt a^3,\kern10.7pt a^4,\kern10.7pt a^5,\kern10.7pt a^6,\kern10.7pt a^7,\kern10pt \ldots\cr} $$ where $n\,a=a+a+\ldots+a$ (repeated $n$ times) and $a^n=a\times a\times\ldots\times a$ (repeated $n$ times). The special case of squares, cubes and powers of 10 should already appear at the primary school level. Here again, suitable practical considerations (ruler, thermometer, altitude) are the appropriate context for the introduction of negative numbers, especially when dealing with negative multiples of the unit of length\?: $$ \ldots~~-5u,\quad -4u,\quad -3u,\quad -2u,\quad -u,\quad 0, \quad u,\quad 2u,\quad 3u,\quad 4u,\quad 5u,\quad 6u,\quad 7u~~\ldots $$ At the beginning of junior high school, one can then pursue with negative decimal numbers and real numbers, e.g.\ in relation with length measurements. It~is natural to extend the distributivity property of multiplication with respect to addition to numbers of arbitrary positive or negative sign, and the ``sign rule'' follows\?: $$a\times(b+(-b))=a\times b + a\times(-b),$$ as the left hand side is equal to $a\times 0=0$, one must have $a\times(-b)=-(a\times b)$. At the end of primary chool, pupils should normally reach a rather safe and effective pratice of handwritten divisions. Such a practice should allow them to observe the periodicity of remainders, and therefore the periodicity of the decimal expansion of fractions of integers. This is especially noticeable on many fractions that have a small denominator leading to a very short periodicity cycle (e.g.\ fractions of denominator 3,~7, 9, 11, 21, 27, 33, 37, 41, 63, 77, 99, 101,~271 ($\ldots$) and their multiples by 2 and 5, which lead to a period of length~6 at most). Real numbers appear in a natural way as non periodic decimal expansions of certain numerical values such as square roots. However, a premature use of calculators and an insufficient practice of approximate decimal (handwritten) calculations, e.g.\ of divisions, may lead to a very limited and formal perception of the concept of square root. It is absolutely necessary that pupils be faced to the task of extracting square roots numerically. For instance, to start with, one can explain the numerical approximation of $\sqrt{2}\,$: $$ \eqalign{ &(1.4)^2= 1.96,\kern40pt (1.5)^2= 2.25\kern40pt\hbox{thus}\kern19pt 1.4<\sqrt{2}<1.5,\cr &(1.41)^2= 1.9881,\kern25pt (1.42)^2= 2.0164\kern25pt\hbox{thus}\kern14pt 1.41<\sqrt{2}<1.42,\cr &(1.414)^2= 1.999396,\quad (1.415)^2= 2.002225\quad\hbox{thus} \kern9pt 1.414<\sqrt{2}<1.415\quad\ldots\cr} $$ We recommend that such concepts be introduced at latest in grade~7, at the same time Pythagoras' theorem is explained, so that the geometric necessity of square roots clearly appeals to pupils (of course, such contents can only be introduced in grade 6 or 7 if one assumes that the severe deficiencies of the current French primary curricula have been alleviated$\,\ldots$). When these concepts are understood, it becomes possible to give a precise general definition of the concept of real number -- at the same time, this is a good opportunity to approach implicitly the important concept of limit$\,$: \medskip {\bf (2.3.1) Definition.} {\it Real numbers are numbers that are used to measure physical quantities with an unlimited accuracy. A real number is therefore expressed by an arbitrary $($non necessarily periodic$)$ decimal expansion, in other words a sequence $\pm\;\sqsymb\sqsymb\ldots\sqsymb\sqsymb\,\hbox{\twelvebf .}\, \sqsymb\sqsymb\sqsymb\sqsymb\sqsymb\ldots$ of~digits $0,1,2,3,4,5,6,7,8,9$ that is finite on the left side of the dot, and infinite on the right side, with a $+$ or~$-$ initial sign $($the absence of any sign implicitly means that we have a $+{}$sign, except for the null value which has no definite sign, actually one simply writes $+\,0.000\ldots={}-\,0.000\ldots=0)$.\\ From a geometric viewpoint, a real number corresponds to a point on an oriented axis, that would be set by means of a ``metric ruler with infinite accuracy''.} \medskip We firmly recommend that the manual algorithm to extract square roots be taught at school at the same period in time$\,$; in fact, mastering such an algorithm ``enpowers'' children by giving them a lot of control on their numerical environment -- and at the same time it provides a strong hint that a square root is going to lead to a non periodic expansion, since one sees no reason that the algorithm would exhibit periodicity, as was the case with the division of integers. This would be an excellent way of consolidating the practice of handwritten claculations. Experiments have shown that well trained pupils can learn the square root algorithm in only one or two hours, assuming that they already have a good practice of handwritten division. Unfortunately, it appears to be almost impossible to test these issues with the immense majority of French pupils, at a time when the mathematical menu confines to extreme deficiency and the algorithms are not sufficiently mastered (of course we do not mean that being able to perform the formal algorithms suffices in any way to grasp primary mathematics$\,$; understanding the meaning of operations and their use in solving problems is even more important.) To turn Definition (2.3.1) into a precise and mathematically adequate statement, one must also explain what are proper and improper decimal expansions$\,$\note{Once this is done, Definition (2.3.1) can be considered as a perfectly acceptable formal definition of real numbers -- even though it has the draw-back, in reality more a lack of elegance than a genuine inconvenience, to seemingly depend on the numeration system (base 10 here).}. First one should make pupils realize that $0,999999\ldots = 1$, in fact, if one sets $x=0.999999\ldots\;$, then \hbox{$10\,x = 9.999999\ldots\;$}, hence \hbox{$10\,x-x=9$}, and therefore one is led to admit that necessarily $x=1$, at least under the assumption that the usual rules applied to decimal numbers still apply to infinite decimal expansions. More generally, one has to take e.g. $$ 0.34999999\ldots{} = 0.35 = 0.35000000\ldots $$ These observations appear as a complementary specification to add in Definition~(2.3.1). \medskip {\bf (2.3.2) Complementary specification in the definition of real numbers.} {\it Decimal numbers have two distinct expansions, one that is finite $($or, eqivalently, one that contains an infinite sequence of consecutive $0$ digits$)$, called ``proper expansion'', the other one containing an infinite sequence of consecutive $9$ digits $($and the previous digit decreased by~$1)$, called the ``improper expansion''. Real numbers that are not decimal numbers have only one infinite decimal expansion.} \medskip Arithmetic operations on decimal numbers can be extended to find the expansion of the sum and product of two real numbers with any accuracy given in advance -- and therefore to calculate the sum and product of two real numbers, at least in principle\note{For a complete theoretical justification of this claim, we need the theorem that asserts the convergence of bounded increasing sequences of real numbers, a theorem that is at best seen at high school level, cf.\ our manuscript {\it Puissances, exponentielles, logarithmes,$\,\ldots$} for details.}. The order relation is obtained by comparing digit one by one in ``lexicographic'' order. At this point, say in the 7th grade, one should be able to reach the following important characterizations (provided that curricula of all previous levels have been upgraded to provide a sufficient background$\,$!) \medskip \noindent {\bf (2.3.3) Characterization of rational and decimal numbers.} {\it {\parindent=6.5mm \item{\rm (a)} An infinite decimal expansion represents a rational number $($i.e.\ a fraction of integers$)$ if and only if the development is periodic from a certain index. \item{\rm (b)} Among the rational numbers, decimal numbers are those for which the expansion has an infinite sequence of consecutive $0$ digits $($``proper decimal expansion''$)$ or an infinite sequence of consecutive $9$ digits $($``improper decimal expansion''$)$. \item{\rm (c)} Real numbers that are not decimal are characterized by the fact that their expansion does not possess an infinite sequence of consecutive digits which are all equal to $0$ or all equal to $9\;$; in other words, either they have an infinite number of digits that are neither $0$ or $9$, or an infinite alternating sequence $($possibly irregular$)$ of digits $0$ and $9$, starting from a certain index.\vskip0pt}} {\it Proof.} (a) Indeed, given a simplified fraction $p/q$ which is not a decimal number (i.e., $q$ has at least one prime factor distinct from $2$ and $5$), the division algorithm of $p$ by~$q$ never terminates and leads to remainders that fall in the finite sequence $1$, $2,\,\ldots\,$, $q-1$. After at most $q-1$ steps beyond the dot, the remainder necessarily reproduces a value that has been already found. This implies that the expansion is periodic and that the period length is at most $q-1$. Conversely, if the expansion is periodic, say of length~$5$, and if we have e.g. $$ x=0.107\RGBColor{1 0 0}{23114}\RGBColor{0 0 1}{23114}\RGBColor{1 0 0}{23114}\RGBColor{0 0 1}{23114}\ldots, $$ we observe that the quotient $1:99999$ yields $$ {1\over 99999}=0.\RGBColor{1 0 0}{00001}\RGBColor{0 0 1}{00001}\RGBColor{1 0 0}{00001}\RGBColor{0 0 1}{00001}\ldots, $$ therefore $$ \eqalign{ &{23114\over 99999}=23114\times{1\over 99999} =0.\RGBColor{1 0 0}{23114}\RGBColor{0 0 1}{23114}\RGBColor{1 0 0}{23114}\RGBColor{0 0 1}{23114}\ldots\cr \noalign{\vskip6pt} &{23114\over 99999000} =0.000\RGBColor{1 0 0}{23114}\RGBColor{0 0 1}{23114}\RGBColor{1 0 0}{23114}\RGBColor{0 0 1}{23114}\ldots\cr } $$ Finally, since $0.107=\displaystyle{107\over 1000}$, we get $$ x={107\over 1000}+{23114\over 99999000} ={107\times 99999+23114\over 99999000} ={10723007\over 99999000} $$ which is actually a rational number. This process is easily generalized to convert any periodic decimal expansion into a fraction. Statement (b) is just a reformulation of Definition (2.3.2), and (c) is also an equivalent statement. The final situation described in (c) occurs for instance when one considers the rational number $$1/11=0.09090909\,\ldots$$ or an irrational number like $0.909009000900009\,\ldots$ (a clearly aperiodic sequence).\qed \medskip All these considerations can be strengthened by introducing calculations with polynomials, by manipulating inequalities, approximations and algebraic identities. It is important to visualize geometrically $(a+b)^2$, $(a+b)(a-b)$. In fact, all primary schools and junior high schools should have wooden pieces designed to visualize quantities such as $(a+b)^2$, $(a+b)^3$ (such matters can even be approached at the end of primary school, in a non necessarily formalized manner, on the occasion of the introduction of areas and volumes). The identity $(10a+b)^2-100a^2=(20a+b)b$ occurs in the justification of the square root algorithm. At a more basic level, say in grade~5, and possibly with a justification relying only on a use of squares drawn on graph paper (take a side which is e.g.\ $75\,$mm), the formula $$ (10a+5)^2=100\,a(a+1)+25 $$ can be used to perform efficient mental calculations of squares of numbers ending by the digit 5 in decimal base$\,$: for instance $(75)^2=5625$, where the number $56$ is obtained by computing $a(a+1)=7\times 8$. \else An important issue is the relation between geometry and numbers. Greek mathematicians already had the fundamental idea that ratios of lengths with a given unit length were in one to one correspondence with numbers~: in modern terms, there is a natural distance preserving bijection between points of a line and the set of real numbers. This viewpoint is of course not at all in contradiction with elementary education since measuring lengths in integer (and then decimal) values with a ruler is one of of the first important facts taught at primary school. However, at least in France, several reforms have put forward the extremely toxic idea that the emergence of electronic calculators would somehow free pupils from learning elementary arithmetic algorithms for addition, subtraction, multiplication and division, and that mastering magnitude orders and the ``meaning'' of arithmetic operations would be more than enough to understand society and even to pursue in science. The fact is that one cannot conceptually separate numbers from the operations that can be performed on them, and that mastering algorithms mentally and in written form is instrumental to realizing magnitude orders and the relation of numbers with physical quantities. The first contact that pupils will have with ``elementary physics'', again at primary school level, is probably through measuring lengths, areas, volumes, weights, densities, etc. Understanding the link with arithmetic operations is the basic knowledge that will be involved later to connect physics with mathematics. The idea of a real number as a possibly infinite decimal expansion then comes in a natural way when measuring a given physical quantity with greater and greater accuracy. Square roots are forced upon us by Pythagoras' theorem, and computing their numerical values is also a very good introduction to the concept of real number. I would certainly recommend to (re)introduce from the very start of junior high school (not later than grade 6 and 7), the observation that fractions of integers produce periodic decimal expansions, e.g.\ $1/7 = 0.\underline{142857}142857\,...~$, while no visible period appears when computing the square root of~$2$. In order to understand this (and before any formal proof can be given, as they are conceptually harder to grasp), it is again useful to learn here the hand and paper algorithm for computing square roots, which is only slightly more involved than the division algorithm and makes it immediately clear that there is no reason the result has to be periodic -- unfortunately, this algorithm is no longer taught in France since a long time. When all this work is correctly done, it becomes really possible to give a precise meaning to the concept of real number at junior high school -- of course many more details have to be explained, such as the identification of proper and improper decimal expansions, e.g.\ $1=0.99999\,...~$, the natural order relation on such expansions, decimal approximations with at given accuracy, etc. In what follows, we propose an approach to geometry based on the assumption that pupils have a reasonable understanding of numbers, arithmetic operations and physical quantities from their primary school years -- with consolidation about things such infinite decimal expansions and square roots in the first two years of junior high school~; this was certainly the situation that prevailed in France before 1970, but things have unfortunately changed for the worse since then. Our small experimental network of schools SLECC (``Savoir Lire Ecrire Compter Calculer''), which accepts random pupils and operates in random parts of the country, shows that such knowledge can still be reached today for a very large majority of pupils, provided appropriate curricula are enforced. For the others, our views will probably remain a bit utopistic, or will have to be delayed and postponed at a later stage. \fi \bigskip \section{3. First steps of the introduction of Euclidean geometry} \sectionrunning{3. First steps of the introduction of Euclidean geometry} \subsection{3.1. Fundamental concepts} The primitive concepts we are going to use freely are~: {\parindent=4mm \item{$\bullet$} real numbers, with their properties already discussed above~; \item{$\bullet$} points and geometric objects as sets of points~: a point should be thought of as a geometric object with no extension, as can be represented with a sharp pencil~; a line or a curve are infinite sets of points (at this point, this is given only for intuition, but will not be needed formally)~; \item{$\bullet$} distances between points. } Let us mention that the language of set theory has been for more than one century the universal language of mathematicians. Although excessive abstraction should be avoided at early stages, we feel that it is appropriate to introduce at the beginning of junior high school the useful concepts of sets, of inclusion, the notation $x\in E$, operations on sets such as union, intersection and difference~; geometry and numbers already provide rich and concrete illustrations. A geometric figure is simply an ordered finite collection of points $A_j$ and sets $S_k$ (vertices, segments, circles, arcs, ...) Given two points $A$, $B$ of the plane or of space, we denote by $d(A, B)$ (or simply by~$AB$) their distance, which is in general a positive number, equal to zero when the points $A$ and $B$ coincide -- concretely, this distance can be mesured with a ruler. A fundamental property of distances is~: \medskip {\bf 3.1.1. Triangular inequality.} {\it For any triple of points $A,B,C$, their mutual distances always satisfy the inequality $AC\le AB+BC$, in other words the length of any side of a triangle is always at most equal to the sum of the lengths of the two other sides.} {\it Intuitive justification}. \InsertPSFigure 18.000 25.000 { gsave 0.4 setlinewidth 0 0 moveto 50 50 lineto 100 40 lineto closepath stroke 50 50 moveto 60.345 24.138 lineto stroke 60.345 24.138 moveto -1 2.5 rlineto 2.5 1 rlineto 1 -2.5 rlineto stroke 140 0 translate 0 0 moveto 120 60 lineto 100 40 lineto closepath stroke 120 60 moveto 124.138 49.655 lineto stroke 124.138 49.655 moveto -1 2.5 rlineto 2.5 1 rlineto 1 -2.5 rlineto stroke [3.5 1.5] 0 setdash 100 40 moveto 140 56 lineto stroke grestore} \LabelTeX -3 -3 $A$\ELTX \LabelTeX 16 19 $B$\ELTX \LabelTeX 36 14 $C$\ELTX \LabelTeX 21 5 $H$\ELTX \LabelTeX 46.5 -3 $A$\ELTX \LabelTeX 92 22 $B$\ELTX \LabelTeX 84 11 $C$\ELTX \LabelTeX 93 14.5 $H$\ELTX \EndFig \vskip3mm Let us draw the height of the triangle joining vertex $B$ to point $H$ on the opposite side~$(AC)$. If $H$ is located between $A$ and $C$, we get $AC = AH + HC$~; on the other hand, if the triangle is not flat (i.e.\ if $H\ne B$), we have $AH < AB$ and $HC < BC$ (since the hypotenuse is longer than the right-angle sides in a right-angle triangle -- this will be checked formally thanks to Pythagoras' theorem). If $H$ is located outside of the segment $[A, C]$, for instance beyond $C$, we already have $AC0$ is the set of points $M$ in the plane $\cP$ such that $d(A,M)=AM=R$. \item{\rm(h)} A circular arc is the intersection of a circle with an angular sector, the vertex of which is the center of the circle. \item{\rm(i)} The measure of an angle $($in degrees$)$ is proportional to the length of the circular arc that it intercepts on a circle whose center coincides with the vertex of the angle, in such a way that the full circle corresponds to $360^\degree$. A flat angle $($cut by a half-plane bounded by a diameter of the circle$)$ corresponds to an arc formed by a half-circle and has measure~$180^\degree$. A right angle is one half of a flat angle, that is, an angle corresponding to the quarter of a circle, in other words, an angle of measure equal to~$90^\degree$. \item{\rm(j)} Two half-lines with the same origin are said to be perpendicular if they form a right angle.{\parindent=0cm \note{The concepts of right and flat angles, as well as the notion of half angle are already primary school concerns. At this level, the best way to address these issues is probably to let pupils practice paper folding (the notion of horizontality and verticality are relative concepts, it is better to avoid them when introducing perpendicularity, so as to avoid any potential confusion).}} \vskip0pt}} The usual properties of parallel lines and of angles intercepted by such lines (``corresponding angles'' vs ``alternate angles'') easily leads to establishing the value of the sum of angles in a triangle (and, from there, in a quadrilateral). Definition (i) requires of course a few comments. The first and most obvious comment is that one needs to define what is the length of a circular arc, or more generally of a curvilign arc~: this is the limit (or the upper bound) of the lengths of polygonal line inscribed in the curve, when the curve is divided into smaller and smaller portions (cf.\ 2.2)\note{The definition and existence of limits are difficult issues that cannot be addressed before high school, but it seems appropriate to introduce this idea at least intuitively.}. The second one is that the measure of an angle is independent of the radius $R$ of the circle used to evaluate arc lengths; this follows from the fact that arc lengths are proportional to the radius~$R$, which itself follows from Thales' theorem (see below). Moreover, a proportionality argument yields the formula for the length of a circular arc located on a circle of radius~$R\;$: a full arc ($360^\degree$) has length $2\pi R$, hence the length of an arc of $1^\degree$ is 360 times smaller, that is $2\pi R/360 = \pi R/180$, and an arc of measure $a$ (in degrees) has length $$\ell=(\pi R/180)\times a = R\times a\times \pi/180.$$ \subsection{3.2. Construction with instruments and isometry criteria for triangles} As soon as they are introduced, it is extremely important to illustrate geometric concepts with figures and construction activities with instruments. Basic constructions with ruler and compasses, such as midpoints, medians, bissectors, are of an elementary level and should be already taught at primary school. The step that follows immediately next consists of constructing perpendiculars and parallel lines passing through a given point. At the beginning of junior high school, it becomes possible to consider conceptually more advanced matters, e.g. the problem of constructing a triangle ABC with a given base BC and two other elements, for instance : (3.2.1) the lengths of sides $AB$ and $AC$,\\ (3.2.2) the measures of angles $\oversalient{ABC}$ and $\oversalient{ACB}$,\\ (3.2.3) the length of $AB$ and the measure of angle $\oversalient{ABC}$. \InsertPSFigure 10.000 31.000 { gsave 0.6 setlinewidth 0 setgray 0 0 moveto 100 40 lineto stroke 0.4 setlinewidth 1 0 0 setrgbcolor /R 5000 sqrt def /ang1 25 def /ang2 67 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke /R 2600 sqrt def /ang1 145 def /ang2 190 def R ang1 cos mul 100 add R ang1 sin mul 40 add moveto 100 40 R ang1 ang2 arc stroke 0 0 1 setrgbcolor 0 0 moveto 50 50 lineto 100 40 lineto stroke %%%%%%%%%%%%%%%%% 110 0 translate 0.6 setlinewidth 0 setgray 0 0 moveto 100 40 lineto stroke 0.4 setlinewidth 1 0 0 setrgbcolor 0 0 moveto 60 72 lineto stroke /R 16 def /ang1 23 def /ang2 51 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0 1 setrgbcolor 100 40 moveto 40 64 lineto stroke /R 15 def /ang1 157 def /ang2 203 def R ang1 cos mul 100 add R ang1 sin mul 40 add moveto 100 40 R ang1 ang2 arc stroke 83.7 39.77 moveto 86.3 39.77 lineto stroke %%%%%%%%%%%%%%%% 110 0 translate 0.6 setlinewidth 0 setgray 0 0 moveto 100 40 lineto stroke 0.4 setlinewidth 0 0 1 setrgbcolor /R 4100 sqrt def /ang1 32 def /ang2 47 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 1 0 0 setrgbcolor 0 0 moveto 60 48 lineto stroke /R 15 def /ang1 23 def /ang2 39 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke /R 16.5 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0.7 0 setrgbcolor 100 40 moveto 50 40 lineto stroke grestore} \LabelTeX 13.2 16.5 $A$\ELTX \LabelTeX -3 -3 $B$\ELTX \LabelTeX 36 14 $C$\ELTX \LabelTeX 54.5 22.5 $A$\ELTX \LabelTeX 36 -3 $B$\ELTX \LabelTeX 75.2 14 $C$\ELTX \LabelTeX 94 15.9 $A$\ELTX \LabelTeX 75.4 -3 $B$\ELTX \LabelTeX 114 14 $C$\ELTX \EndFig \vskip3mm In the first case, the solution is obtained by constructing circles of centers $B$, $C$ and radii equal to the given lengths $AB$ and $AC$, in the second case a~protractor is used to draw two angular sectors with respective vertices $B$ and $C$, in the third case one draws an angular sector of vertex $B$ and a circle of center~$B$. In each case it can be seen that there are exactly two solutions, the second solution being obtained as a triangle $A'BC$ that is symmetric of $ABC$ with respect to line~$(BC)\;$: \InsertPSFigure 10.000 26.000 { gsave 0.6 setlinewidth 0 setgray 0 0 moveto 100 40 lineto stroke 0.4 setlinewidth 1 0 0 setrgbcolor /R 5000 sqrt def /ang1 -15 def /ang2 55 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke /R 2600 sqrt def /ang1 160 def /ang2 248 def R ang1 cos mul 100 add R ang1 sin mul 40 add moveto 100 40 R ang1 ang2 arc stroke 0 0 1 setrgbcolor 0 0 moveto 70.690 -1.724 lineto 100 40 lineto stroke %%%%%%%%%%%%%%%%% 110 0 translate 0.6 setlinewidth 0 setgray 0 0 moveto 100 40 lineto stroke 0.4 setlinewidth 1 0 0 setrgbcolor 0 0 moveto 77.586 1.2 mul -5.517 1.2 mul lineto stroke /R 16 def /ang1 -3 def /ang2 22 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0 1 setrgbcolor 100 40 moveto 77.586 1.2 mul 20 sub -5.517 1.2 mul 8 sub lineto stroke /R 15 def /ang1 203 def /ang2 244 def R ang1 cos mul 100 add R ang1 sin mul 40 add moveto 100 40 R ang1 ang2 arc stroke 87.8 28.7 moveto 90.4 31.0 lineto stroke %%%%%%%%%%%%%%%% 110 0 translate 0.6 setlinewidth 0 setgray 0 0 moveto 100 40 lineto stroke 0.4 setlinewidth 0 0 1 setrgbcolor /R 4100 sqrt def /ang1 -4 def /ang2 15 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 1 0 0 setrgbcolor 0 0 moveto 63.793 1.2 mul 5.517 1.2 mul lineto stroke /R 15 def /ang1 4.5 def /ang2 22 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke /R 16.5 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0.7 0 setrgbcolor 100 40 moveto 63.793 5.517 lineto stroke grestore} \LabelTeX 24.6 -5.5 $A'$\ELTX \LabelTeX -3 -3 $B$\ELTX \LabelTeX 36 14 $C$\ELTX \LabelTeX 65.5 -5.5 $A'$\ELTX \LabelTeX 36 -3 $B$\ELTX \LabelTeX 75.2 14 $C$\ELTX \LabelTeX 100.8 -1 $A'$\ELTX \LabelTeX 75.4 -3 $B$\ELTX \LabelTeX 114 14 $C$\ELTX \EndFig \vskip3mm One sees that the triangles $ABC$ and $A'BC$ have in each case sides with the same lengths. This leads to the important concept of {\it isometric figures}. \medskip {\bf 3.2.4. Definition.} {\it {\parindent=6mm \item{\rm (a)} One says that two triangles are isometric if the sides that are in correspondence have the same lengths, in such a way that if the first triangle has vertices $A$, $B$, $C$ and the corresponding vertices of the second one are $A'$, $B'$, $C'$, then $A'B'=AB$, $B'C'=BC$, $C'A'=CA$. \item{\rm (b)} More generally, one says that two figures in a plane or in space are isometric, the first one being defined by points $A_1$, $A_2$, $A_3$, $A_4\,\ldots$ and the second one by corresponding points $A'_1$, $A'_2$, $A'_3$, $A'_4\,\ldots$ if all mutual distances coincide.\vskip0pt}} \medskip The concept of isometric figures is related to the physical concept of {\it solid body}$\;$: a body is said to be a solid if the mutual distances of its constituents (molecules, atoms) do not vary while the object is moved; after such a mo ve, atoms which occupied certain positions $A_i$ occupy new positions $A'_i$ and we have $A'_iA'_j=A_iA_j$. This leads to a rigorous definition of solid displacements, that have a meaning from the viewpoints of mathematics and physics as well. \medskip {\bf 3.2.5. Definition.} {\it Given a geometric figure $($or a solid body in space$)$ defined by characteristic points $A_1$, $A_2$, $A_3$, $A_4\,\ldots$, a solid move is a continuous succession of positions $A_i(t)$ of these points with respect to the time~$t$, in such a way that all distances $A_i(t)A_j(t)$ are constant. If the points $A_i$ were the initial positions and the points $A'_i$ are the final positions, we say that the figure $(A'_1A'_2A'_3A'_4\,\ldots)$ is obtained by a displacement of figure $(A_1A_2A_3A_4\,\ldots)$.\note{The concept of continuity that we use is the standard continuity property for functions of one real variable - one can of course introuduce this only intuitively at the junior high school level. One can further show that an isometry between two figures or solids extends an affine isometry of the whole space, and that a solid move is represented by a positive affine isometry, see Section 10. The formal proof is not very hard, but certainly cannot be given before the end of high school (this would have been possible with the rather strong French curricula as they were 50 years ago in the grade 12 science class, but doing so would be nowadays completely impossible).}} Beyond displacements, another way of producing isometric figures is to use a reflection (with respect to a line in a plane, or with respect to a plane in space, as obtained by taking the image of an object through reflection in a mirror)\note{Conversely, an important theorem - which we will show later (see section 10) says that isometric figures can be deduced from each other either by a solid move or by a solid move preceded (or followed) by a reflection.}. This fact is already observed with triangles, the use of transparent graph paper is then a good way of visualizing isometric triangles that cannot be superimposed by a displacement without ``getting things out of the plane''~; in a similar way, it can be useful to construct elementary solid shapes (e.g.\ non regular tetrahedra) that cannot be superimposed by a solid move. \medskip {\bf 3.2.6. Exercise.} In order to ensure that two quadrilaterals $ABCD$ and $A'B'C'D'$ are isometric, it is not sufficient to check that the four sides $A'B'=AB$, $B'C'= BC$, $C'D'= CD$, $D'A' = DA$ possess equal lengths, one must also check that the two diagonals $A'C'=AC$ and $B'D'= BD$ be equal$\,$; equaling only one diagonal is not enough as shown by the following construction~: \InsertPSFigure 40.000 28.000 { gsave 0.6 setlinewidth 0 setgray 50 0 moveto 0 10 lineto 40 60 lineto stroke 0.4 setlinewidth 1 0 0 setrgbcolor /R 35 def /ang1 210 def /ang2 350 def R ang1 cos mul 40 add R ang1 sin mul 60 add moveto 40 60 R ang1 ang2 arc stroke /R 32 def /ang1 40 def /ang2 155 def R ang1 cos mul 50 add R ang1 sin mul moveto 50 0 R ang1 ang2 arc stroke 0.6 setlinewidth 0 0 1 setrgbcolor 50 0 moveto 31.6 26 lineto 40 60 lineto stroke 0 0.7 0 setrgbcolor 50 0 moveto 59 30.7 lineto 40 60 lineto stroke 0 setgray 0.4 setlinewidth [2 0.7] 0 setdash 50 0 moveto 40 60 lineto stroke grestore } \LabelTeX 16 -4 $A$\ELTX \LabelTeX -4 2 $B$\ELTX \LabelTeX 13 22.5 $C$\ELTX \LabelTeX 8.5 4.5 $D'$\ELTX \LabelTeX 20 12.5 $D$\ELTX \EndFig \bigskip The construction problems considered above for triangles lead us to state the following fundamental isometry criteria. {\bf 3.2.7. Isometry criteria for triangles\note{A rigorous formal proof of of these 3 isometry criteria will be given later, cf. Section 8.}.} {\it In order that two triangles be isometric, it is necessary and sufficient to check one of the following cases {\parindent=6mm \item{\rm(a)} that the three sides be respectively equal $($this is just the definition$)$, \RGBColor{0 0 1}{or} \item{\rm(b)} that they possess one angle with the same value and its adjacent sides equal, \RGBColor{0 0 1}{or} \item{\rm(c)} that they possess one side with the same length and its adjacent angles of equal values.\vskip0pt}} One should observe that conditions (b) and (c) are not sufficient if the adjacency specification is omitted - and it would be good to introduce (or to let pupils perform) constructions demonstrating this fact. A use of isometry criteria in conjunction with properties of alternate or corresponding angles leads to the various usual characterizations of quadrilaterals - parallelograms, lozenges, rectangles, squares$\;\ldots$ \subsection{3.3. Pythagoras' theorem} We first give the classical ``Chinese'' proof of Pythagoras' theorem, which is derived by a simple area argument based on moving four triangles (represented here in green, blue, yellow and light red). Its main advantage is to be visual and convincing\note{Again, in our context, the argument that will be described here is a justification rather than a formal proof. In fact, it would be needed to prove that the quadrilateral central figure on the right hand side is a square - this could certainly be checked with isometry properties of triangles - but one should not forget that they are not yet really proven at this stage. More seriously, the argument uses the concept of area, and it would be needed tp prove the existence of an area measure in the plane with all the desired properties : additivity by disjoint unions, translation invariance$\,\ldots$}. \InsertPSFigure 10 44 { gsave 1 mm unit initcoordinates %% Premier carre 0.1 setlinewidth 0.92 setgray 0 0 moveto 36 0 lineto 36 36 lineto 0 36 lineto closepath fill 0 setgray 0 0 moveto 36 0 lineto 36 36 lineto 0 36 lineto closepath stroke 0.8 1 0.8 setrgbcolor 24 36 moveto 24 12 lineto 36 36 lineto closepath fill 0 setgray 24 36 moveto 24 12 lineto 36 36 lineto closepath stroke 0.8 0.8 1 setrgbcolor 36 36 moveto 24 12 lineto 36 12 lineto closepath fill 0 setgray 36 36 moveto 24 12 lineto 36 12 lineto closepath stroke 1 1 0.5 setrgbcolor 0 12 moveto 24 12 lineto 24 0 lineto closepath fill 0 setgray 0 12 moveto 24 12 lineto 24 0 lineto closepath stroke 1 0.8 0.8 setrgbcolor 0 12 moveto 0 0 lineto 24 0 lineto closepath fill 0 setgray 0 12 moveto 0 0 lineto 24 0 lineto closepath stroke 0.3 setlinewidth 1 0 0 setrgbcolor 0 36 moveto 24 36 lineto 24 12 lineto 0 12 lineto closepath stroke 36 0 moveto 24 0 lineto 24 12 lineto 36 12 lineto closepath stroke 0.1 setlinewidth 15 23.5 moveto 31.5 -10 2.4 vector 33.5 7.5 moveto 23 22 2.4 vector %% Deuxieme carre 75 0 translate 0.1 setlinewidth 0.92 setgray 0 0 moveto 36 0 lineto 36 36 lineto 0 36 lineto closepath fill 0 setgray 0 0 moveto 36 0 lineto 36 36 lineto 0 36 lineto closepath stroke 0.8 1 0.8 setrgbcolor 0 36 moveto 0 12 lineto 12 36 lineto closepath fill 0 setgray 0 36 moveto 0 12 lineto 12 36 lineto closepath stroke 0.8 0.8 1 setrgbcolor 36 24 moveto 24 0 lineto 36 0 lineto closepath fill 0 setgray 36 24 moveto 24 0 lineto 36 0 lineto closepath stroke 1 1 0.5 setrgbcolor 12 36 moveto 36 36 lineto 36 24 lineto closepath fill 0 setgray 12 36 moveto 36 36 lineto 36 24 lineto closepath stroke 1 0.8 0.8 setrgbcolor 0 12 moveto 0 0 lineto 24 0 lineto closepath fill 0 setgray 0 12 moveto 0 0 lineto 24 0 lineto closepath stroke 0.3 setlinewidth 1 0 0 setrgbcolor 0 12 moveto 12 36 lineto 36 24 lineto 24 0 lineto closepath stroke 0.1 setlinewidth 16 18 moveto 25 180 2.4 vector grestore } %% Premier carre \LabelTeX -3 4 $b$\ELTX \LabelTeX -3 23 $a$\ELTX \LabelTeX 37.2 4 $b$\ELTX \LabelTeX 37.2 23 $a$\ELTX \LabelTeX 11 -3.5 $a$\ELTX \LabelTeX 29 -3.5 $b$\ELTX \LabelTeX 11 37.5 $a$\ELTX \LabelTeX 29 37.5 $b$\ELTX \LabelTeX 29.3 27 $c$\ELTX \LabelTeX 11 7.4 $c$\ELTX \LabelTeX 11 23 $\RGBColor{1 0 0}{a^2}$\ELTX \LabelTeX 29 5 $\RGBColor{1 0 0}{b^2}$\ELTX %% Deuxieme carre \LabelTeX 72 5 $b$\ELTX \LabelTeX 72 23 $a$\ELTX \LabelTeX 112.2 11 $a$\ELTX \LabelTeX 112.2 29 $b$\ELTX \LabelTeX 86 -3.5 $a$\ELTX \LabelTeX 104 -3.5 $b$\ELTX \LabelTeX 80 37.5 $b$\ELTX \LabelTeX 98 37.5 $a$\ELTX \LabelTeX 82 23 $c$\ELTX \LabelTeX 102.2 11 $c$\ELTX \LabelTeX 98 27.1 $c$\ELTX \LabelTeX 86 7.4 $c$\ELTX \LabelTeX 92 17 $\RGBColor{1 0 0}{c^2}$\ELTX \LabelTeX 46.5 17 $\RGBColor{1 0 0}{a^2+b^2=c^2}$\ELTX \EndFig \vskip5mm The point is to compare, in the left hand and right hand figures, the remaining grey area, which is the difference of the area of the square of side a + b with the area of the four rectangle triangles of sides $a$, $b$, $c$. The equality of the grey areas implies $a^2+b^2=c^2$. \medskip {\bf Complement.} {\it Let $(ABC)$ be a triangle and $a$, $b$, $c$ the lengths of the sides that are opposite to vertices $A$, $B$, $C$. {\parindent=6mm \item{\rm(i)} If the angle $\widehat C$ is smaller than a right angle, we have $c^2a^2+b^2$.\vskip0pt}} {\it Proof}. First consider the case where $(ABC)$ is rectangle$\,$: we have \RGBColor{0 0.7 0}{$c^2=a^2+b^2$} and the angle is equal to~$90^\degree$. \InsertPSFigure 10.000 22.000 { gsave 0.6 setlinewidth 0 setgray 0 0 moveto 100 0 lineto stroke 0.3 setlinewidth 1 0 0 setrgbcolor /R 1600 sqrt def /ang1 -90 def /ang2 90 def R ang1 cos mul 100 add R ang1 sin mul moveto 100 0 R ang1 ang2 arc stroke 0 0 1 setrgbcolor /R 1600 sqrt def /ang1 90 def /ang2 270 def R ang1 cos mul 100 add R ang1 sin mul moveto 100 0 R ang1 ang2 arc stroke 0 0.7 0 setrgbcolor /R 11600 sqrt def /ang1 -26 def /ang2 26 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0 moveto 100 40 lineto 100 0 lineto stroke /R 14 def /ang1 90 def /ang2 180 def R ang1 cos mul 100 add R ang1 sin mul moveto 100 0 R ang1 ang2 arc stroke 1 0 0 setrgbcolor /R 14600 sqrt def /ang1 -23 def /ang2 23 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0 moveto 114.9 37.1 lineto 100 0 lineto stroke /R 12 def /ang1 69 def /ang2 180 def R ang1 cos mul 100 add R ang1 sin mul moveto 100 0 R ang1 ang2 arc stroke 0 0 1 setrgbcolor /R 90 def /ang1 -30 def /ang2 30 def R ang1 cos mul R ang1 sin mul moveto 0 0 R ang1 ang2 arc stroke 0 0 moveto 82.4 35.9 lineto 100 0 lineto stroke /R 16 def /ang1 115.5 def /ang2 180 def R ang1 cos mul 100 add R ang1 sin mul moveto 100 0 R ang1 ang2 arc stroke grestore } \LabelTeX 13 -2.7 $a$\ELTX \LabelTeX 33.2 6.7 $\RGBColor{0 0.7 0}{b}$\ELTX \LabelTeX 16 8.5 $\RGBColor{0 0.7 0}{c}$\ELTX \LabelTeX 10 5.9 $\RGBColor{0 0 1}{c'}$\ELTX \LabelTeX 16 2.9 $\RGBColor{1 0 0}{c''}$\ELTX \LabelTeX -1 -4 $B$\ELTX \LabelTeX 35.3 14.8 $A$\ELTX \LabelTeX 28.9 14 $\RGBColor{0 0 1}{A'}$\ELTX \LabelTeX 40.8 13.7 $\RGBColor{1 0 0}{A''}$\ELTX \LabelTeX 34 -4 $C$\ELTX \LabelTeX 55 10 \RGBColor{0 0 1}{If angle $\widehat C$ is${}<90^\degree$, we have $c'90^\degree$, we have $c''>c$.}\ELTX \LabelTeX 0 -17 $~$\ELTX \EndFig \bigskip We argue by either increasing or decreasing the angle$\,$: if angle $\widehat C$ is${}<90^\degree$, we have $c'90^\degree$, we have $c''>c$. By this reasoning, we conclude~: {\bf Converse of Pythagoras' theorem.} {\it With the above notation, if $c^2=a^2+b^2$, then angle $\widehat C$ must be a right angle, hence the given triangle is rectangle in $C$.} \section{4. Cartesian coordinates in the plane} The next fundamental step of our approach is the introduction of cartesian coordinates and their use to give {\it formal proofs of properties that had previously been taken for granted} (or given with a partial justification only). This is done by working in orthonormal frames. \subsection{4.1. Expression of Euclidean distance} \InsertPSFigure 15.000 46.000 { gsave 0.8333 mm unit 20 10 translate 0.1 setlinewidth 1 0.8 0.6 setrgbcolor -5 1 35 { /k exch def -15 k moveto 75 k lineto stroke } for -15 1 75 { /k exch def k -5 moveto k 35 lineto stroke } for 0.3 setlinewidth 0 1 3 { /k exch 10 mul def -15 k moveto 75 k lineto stroke } for -1 1 7 { /k exch 10 mul def k -5 moveto k 35 lineto stroke } for 0.2 setlinewidth 0 setgray -20 0 moveto 100 0 2.4 vector 0 -10 moveto 50 90 2.4 vector 0 0 1 setrgbcolor 20 10 moveto 50 10 lineto 50 30 lineto closepath stroke 1 0 0 setrgbcolor 20 10 moveto 0.7 disk 50 30 moveto 0.7 disk 0 setgray 0.3 setlinewidth 20 0 moveto 0.7 0 indent 50 0 moveto 0.7 0 indent 0 10 moveto 0.7 90 indent 0 30 moveto 0.7 90 indent 0.15 setlinewidth 0 0.7 0 setrgbcolor [1.5 0.5] 0 setdash 20 1.2 moveto 20 9.3 lineto stroke 50 1.2 moveto 50 10 lineto stroke 1.2 10 moveto 19.3 10 lineto stroke 1.2 30 moveto 50 30 lineto stroke grestore } \LabelTeX 32 5.3 $x$\ELTX \LabelTeX 57.2 5.3 $x'$\ELTX \LabelTeX 13 16.3 $y$\ELTX \LabelTeX 13 32.3 $y'$\ELTX \LabelTeX 29 13.3 $M$\ELTX \LabelTeX 59.5 32.8 $M'$\ELTX \EndFig Pythagoras' theorem shows that the length $MM'$ of the hypotenuse is given by the formula $MM'^2=(x'-x)^2+(y'-y)^2$, as the two sides of the right angle are $x'-x$ and~$y'-y$ (up to sign). The distance from $M$ to $M'$ is therefore equal to $$ d(M,M')=MM'=\sqrt{(x'-x)^2+(y'-y)^2}.\leqno(4.1.1) $$ (It is of course advisable to first present the argument with simple numerical values). \subsection{4.2. Squares} Let us consider the figure formed by points $A\,(u\,;\,v)$, $B\,(-v\,;\,u)$, $C\,(-u\,;\,-v)$, $D\,(v\,;\,-u)$. \InsertPSFigure 15.000 60.000 { gsave 0.8333 mm unit 50 25 translate 0.1 setlinewidth 1 0.8 0.6 setrgbcolor -35 1 35 { /k exch def -45 k moveto 45 k lineto stroke } for -45 1 45 { /k exch def k -35 moveto k 35 lineto stroke } for 0.3 setlinewidth -3 1 3 { /k exch 10 mul def -45 k moveto 45 k lineto stroke } for -4 1 4 { /k exch 10 mul def k -35 moveto k 35 lineto stroke } for 0.2 setlinewidth 0 setgray -50 0 moveto 100 0 2.4 vector 0 -35 moveto 75 90 2.4 vector 0 0 1 setrgbcolor 30 20 moveto -20 30 lineto -30 -20 lineto 20 -30 lineto closepath stroke 0 0.7 0 setrgbcolor 30 20 moveto -30 -20 lineto stroke -20 30 moveto 20 -30 lineto stroke 1 0 0 setrgbcolor 30 20 moveto 0.7 disk -20 30 moveto 0.7 disk -30 -20 moveto 0.7 disk 20 -30 moveto 0.7 disk grestore } \LabelTeX 36.7 21.7 $O$\ELTX \LabelTeX 67.5 38.9 $A\;(u\,;\,v)$\ELTX \LabelTeX 18.3 48.1 $B\;(-v\,;\,u)$\ELTX \LabelTeX 5 -0.3 $C\;(-u\,;\,-v)$\ELTX \LabelTeX 59.2 0 $\smash{\raise-7.3mm\hbox{$D\;(v\,;\,-u)$}}$\ELTX \EndFig \vskip15mm Formula (4.1.1) yields $$ AB^2=BC^2=CD^2=DA^2=(u+v)^2+(u-v)^2=2(u^2+v^2), $$ hence the four sides have the same length, equal to $\sqrt{2}\sqrt{u^2+v^2}$. Similarly, we find $$ OA=OB=OC=OD=\sqrt{u^2+v^2}, $$ therefore the 4 isoceles triangles $OAB$, $OBC$, $OCD$ and $ODA$ are isometric, and as a consequence we have $\oversalient{OAB}=\oversalient{OBC}=\oversalient{OCD}=\oversalient{ODA}=90^\degree$ and the other angles are equal to~$45^\degree$. Hence $\oversalient{DAB}=\oversalient{ABC}=\oversalient{BCD}=\oversalient{CDA}=90^\degree$, and we have proved that our figure is a square. \subsection{4.3. ``Horizontal and vertical'' lines} The set $\cD$ of points $M(x\,;\,y)$ such that $y=c$ (where $c$ is a given numerical value) is a ``horizontal'' line. In fact, given any three points $M$, $M'$, $M''$ of abscissas $xax_1+b$, then this point is not aligned with $M_2$ and $M_3$, and similarly for $M_1''\;(x,y''_1)$ such that $y_1''0)$ if we have $A'_iA'_j/A_iA_j=k$ for all segments $[A_i,A_j]$ and $[A'_i,A'_j]$ that are in correspondance.} An important case where similar figures are obtained is by applying a homothety with a given center, say point $O\;$: if $O$ is chosen as the origin of coordinates and if to each point $M(x\,;\,y)$ we associate the point $M'(x'\,;\,y')$ such that $x'=kx$, $y'=ky$, then formula (4.1.1) shows that we indeed have $A'B'=|k|\,AB$, hence by assigning to each point $A_i$ the corresponding point $A'_i$ we obtain similar figures in the ratio $|k|\;$; this situation is described by saying that we have homothetic figures in the ratio~$k\;$; this ratio can be positive or negative (for instance, if $k=-1$, this is a central symmetry with respect to~$O$). The isometry criteria for triangles immediately extend into criteria for similarity. {\bf Similarity criteria for triangles.} {\it In order to conclude that two triangles are similar, ii is necessary and sufficient that one of the following conditions is met$\;:$ {\parindent=6mm \item{\rm(a)} the corresponding three sides are proportional in a certain ratio $k>0$ $($this is the definition$);$ \item{\rm(b)} the triangles have a corresponding equal angle and the adjacent sides are propor\-tional$\,;$ \item{\rm(c)} the triangles have two equal angles in correspondence.}} An interesting application of the similarity criteria consists in stating and proving the basic metric relations in rectangle triangles$\,$: if the triangle $ABC$ is rectangle in $A$ and if $H$ is the foot of the altitude drawn from vertex~$A$, we have the basic relations $$ AB^2=BH\cdot BC,~~AC^2=CH\cdot CB,~~AH^2=BH\cdot CH,~~AB\cdot AC=AH\cdot BC. $$ \InsertPSFigure 20 40 { gsave 0.6 setlinewidth 0 0 moveto 200 0 lineto 0 100 lineto closepath stroke 0 0 moveto 40 80 lineto stroke 0 0 18 0 2 1 atan arc stroke 0 100 18 270 dup 2 1 atan add arc stroke 0 0 22 90 1 2 atan sub 90 arc stroke 0 0 24 90 1 2 atan sub 90 arc stroke 200 0 22 180 1 2 atan sub 180 arc stroke 200 0 24 180 1 2 atan sub 180 arc stroke 0.3 setlinewidth 44 78 moveto 42 74 lineto 38 76 lineto stroke 6 0 moveto 6 6 lineto 0 6 lineto stroke grestore } \LabelTeX -3 -4 $A$\ELTX \LabelTeX 70 -4 $B$\ELTX \LabelTeX -3 36 $C$\ELTX \LabelTeX 13.7 30 $H$\ELTX \EndFig \bigskip In fact (for example) the similarity of rectangle triangles $ABH$ and $ABC$ leads to the equality of ratios $$ {AB\over BC}={BH\over AB}~~\Longrightarrow~~ AB^2=BH\cdot BC. $$ One is also led in a natural way to the definition of sine, cosine and tangent of an acute angle in a rectangle triangle. {\bf Definition.} {\it Consider a triangle $ABC$ that is rectangle in~$A$. One defines $$ \cos\oversalient{ABC}={AB\over BC},\qquad \sin\oversalient{ABC}={AC\over BC},\qquad \tan\oversalient{ABC}={AC\over AB}. $$} In fact, the ratios only depend on the angle $\oversalient{ABC}$ (which also determines uniquely the complementary angle $\oversalient{ACB}=90^\circ-\oversalient{ABC}$), since rectangle triangles that share a common angle else than their right angle are always similar by criterion~(c). Pythagoras' theorem then quickly leads to computing the values of cos, sin, tan for angles with ``remarkable~values'' $0^\degree$, $30^\degree$, $45^\degree$, $60^\degree$, $90^\degree$. \subsection{4.9. Computing areas and volumes} It is possible -- and therefore probably desirable -- to justify many basic formulas concerning areas and volumes of usual shapes and solid bodies (cylinders, pyramids, cones, spheres), just by using Thales and Pythagoras theorems, combined with elementary geometric arguments\note{We are using here the word ``justify'' rather than ``prove'' because the necessary theoretical foundations (e.g. measure theory) are missing -- and will probably be missing for 5-6 years or more. But in reality, one can see that these justifications can be made perfectly rigorous once the foundations considered here as intuitive are rigorously established. The concept of Hausdorff measure, as briefly explained in (11.3), can be used e.g.\ to give a rigorous definition of the $p$-dimensional measure of any object in a metric space, even when $p$ is not an integer.}. We give here some indication on such techniques, in the case of cones and spheres. The arguments are close to those developed by Archimedes more than two centuries BC (except that we take here the liberty of reformulating them in modern algebraic notations). \medskip \iflongertext {\bf Volume of a cone with an arbitrary planar base} Our goal is to evaluate the volume of a cone whose base $B$ is an arbitrary bounded measurable planar domain. Let $\cP$ be the plane containing the base domain $B$ and let $O$ be a point that does not belong to~$\cP$. \medskip {\bf Definition.} {\it The cone of vertex $O$ and of base $B$ is the union of all segments $[O,M]$ staring at $O$ and ending at a point $M\in B$. The altitude $h$ of the cone is the between $O$ and the plane $\cP$ that contains~$B$.} \medskip In the case where the base $B$ is a disk, we say that this is a {\it circular or elliptic cone} (straight or oblique). When the base is a triangle, the cone is actually a {\it tetrahedron}, and when the base is a square or a rectangle, the cone is a {\it pyramid} (straight or oblique). Here we make the hypothesis that the area $A$ of base $B$ is measurable, and we mean by this that by using a sufficiently fine grid, the approximate area obtained by multiplying the area of a square by the number of squares that are contained in $B$ tends to a limit when the side of squares tends to zero (the same limit being also attained when one takes all squares intersecting $B$, so that their union contains $B$). If we choose the vertex $O$ as the origin and coordinates $(x,y,z)$ such that the plane $\cP$ is horizontal and located ``below'' $O$, then $\cP$ can be expressed as a plane of equation $z=-h$ where $h$ is the altitude. Our first important observation is\?: \InsertPSFigure 27 39 {%%conditional gsave 1 mm unit initcoordinates %%premier cone 1 0.8 0.8 setrgbcolor -9.95 3 moveto 0 30 lineto 19.1 8.5 lineto closepath fill -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve 1 0.7 0.7 setrgbcolor fill gsave 0 30 moveto -9.95 3 lineto 19.1 8.5 lineto closepath clip newpath -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve 1 0 0 setrgbcolor [ 1.5 0.5 ] 0 setdash stroke grestore gsave -9.95 3 moveto 19.1 8.5 lineto 50 -20 lineto -50 -20 lineto closepath clip newpath 1 0 0 setrgbcolor -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve stroke grestore 1 0 0 setrgbcolor -9.95 3 moveto 0 30 lineto 19.1 8.5 lineto stroke 0 setgray 0 30 moveto 0.7 disk -15 5 moveto -20 -7 lineto 30 -7 lineto stroke 27 15 moveto 15 90 2.4 vector 27 15 moveto 11 -90 2.4 vector %%deuxieme cone 60 0 translate 0.8 0.8 1 setrgbcolor -8.5 5.5 moveto 12 30 lineto 19.7 7.4 lineto closepath fill -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve 0.7 0.7 1 setrgbcolor fill gsave -8.5 5.5 moveto 12 30 lineto 19.7 7.4 lineto closepath clip newpath -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve 0 0 1 setrgbcolor [ 1.5 0.5 ] 0 setdash stroke grestore gsave -8.5 5.5 moveto 19.7 7.4 lineto 50 -20 lineto -50 -20 lineto closepath clip newpath 0 0 1 setrgbcolor -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve stroke grestore 0 0 1 setrgbcolor -8.5 5.5 moveto 12 30 lineto 19.7 7.4 lineto stroke 0 setgray 12 30 moveto 0.7 disk -15 5 moveto -20 -7 lineto 30 -7 lineto stroke 27 15 moveto 15 90 2.4 vector 27 15 moveto 11 -90 2.4 vector grestore } \LabelTeX -17.5 -5.5 $\cP$\ELTX \LabelTeX 5 -4 $B$\ELTX \LabelTeX -1 32.6 $O$\ELTX \LabelTeX 24 17 $h$\ELTX \LabelTeX 42.5 -5.5 $\cP$\ELTX \LabelTeX 65 -4 $B$\ELTX \LabelTeX 71 32.6 $O$\ELTX \LabelTeX 84 17 $h$\ELTX \LabelTeX 30 -12 $~$\ELTX \EndFig (4.9.1) {\it The volume of the cone $V$ depends only on the base $B$ and on the altitude $h$, but not the relative position of $O$ and $B$ $($i.e., if $B$ is displaced horizontally with respect to $O$, the volume does not change$)$.\vskip0pt} To prove (4.9.1), we compare our cone and the deformed one by slicing both horizontally in very thin slices, as illustrated in the figure below\?: \InsertPSFigure 27 41 {%%conditional gsave 1 mm unit initcoordinates %%premier cone 1 0.8 0.8 setrgbcolor 0 30 moveto -12 0 lineto 22 0 lineto closepath fill 1 0 0 setrgbcolor /n 15 def /step 30 n div def 1 1 n { /k exch n div def /x1 k -12 mul def /x2 k 22 mul def /y 30 k 30 mul sub def x1 y step add moveto x2 y step add lineto x2 y lineto x1 y lineto closepath stroke } for 0 setgray 0 30 moveto 0.5 disk -15 21 moveto 9 90 2.4 vector -15 21 moveto 9 -90 2.4 vector 27 15 moveto 15 90 2.4 vector 27 15 moveto 15 -90 2.4 vector %%deuxieme cone 72 0 translate 0.8 0.8 1 setrgbcolor 0 30 moveto -24 0 lineto 10 0 lineto closepath fill 0 0 1 setrgbcolor /n 15 def /step 30 n div def 1 1 n { /k exch n div def /x1 k -24 mul def /x2 k 10 mul def /y 30 k 30 mul sub def x1 y step add moveto x2 y step add lineto x2 y lineto x1 y lineto closepath stroke } for 0 setgray 0 30 moveto 0.5 disk -25 21 moveto 9 90 2.4 vector -25 21 moveto 9 -90 2.4 vector 15 15 moveto 15 90 2.4 vector 15 15 moveto 15 -90 2.4 vector [ 1.5 0.5 ] 0 setdash 0 0.7 0 setrgbcolor -57 12 moveto -16 12 lineto stroke -90 12 moveto -80 12 lineto stroke grestore } \LabelTeX 5 -4 $B$\ELTX \LabelTeX -1 32.6 $O$\ELTX \LabelTeX -14.3 20 $|z|$\ELTX \LabelTeX 24 20 $h$\ELTX \LabelTeX 65 -4 $B$\ELTX \LabelTeX 71 32.6 $O$\ELTX \LabelTeX 48 20 $|z|$\ELTX \LabelTeX 84 20 $h$\ELTX \LabelTeX 30 -8 $~$\ELTX \EndFig If we replace the slices by cylinders with vertical lateral surface (and whose bases are homothetic to the base $B$ in the ratio $|z|/h$ as shown above), one gets a small error, since the calculated volumes become a little larger than those of the corresponding truncated cones$\,$, however the error becomes smaller and smaller when the number of slices increases. However, it is clear that the volumes of the two stacks of cylindrical slices are identical, as they have just been ``dragged'' horizontally against each other. \medskip The second observation is\?: {\parindent=11.8mm \item{(4.9.2)} {\it When the altitude $h$ is fixed, the volume $V$ of the cone is proportional to the area $A$ of base $B$.}} \InsertPSFigure 57 39 {%%conditional gsave 1 mm unit initcoordinates %%premier cone 1 0.8 0.8 setrgbcolor -9.95 3 moveto 0 30 lineto 19.1 8.5 lineto closepath fill -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve 1 0.7 0.7 setrgbcolor fill 0.6 0.9 0.6 setrgbcolor -2 1 11 { /k exch def /l k 2.4 div def -9.5 l add k moveto 18.5 l add k lineto stroke } for -5 1 8 { /k exch 2 mul def k -3.5 moveto k 6.75 add 12.7 lineto stroke } for 0.7 0.7 1 setrgbcolor 0 30 moveto -4 6 lineto -2 6 lineto closepath fill 0 30 moveto 5 4 lineto 7 4 lineto closepath fill 0.6 0.6 1 setrgbcolor 0 30 moveto -2 6 lineto -1.583 7 lineto closepath fill 0 30 moveto 7 4 lineto 7.417 5 lineto closepath fill gsave 0 30 moveto -9.95 3 lineto 19.1 8.5 lineto closepath clip newpath -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve 1 0 0 setrgbcolor [ 1.5 0.5 ] 0 setdash stroke grestore gsave -9.95 3 moveto 19.1 8.5 lineto 50 -20 lineto -50 -20 lineto closepath clip newpath 1 0 0 setrgbcolor -10 0 moveto [ -10 0 10 0 18 0 20 6 5 12 ] closedcurve stroke grestore 1 0 0 setrgbcolor -9.95 3 moveto 0 30 lineto 19.1 8.5 lineto stroke 0 setgray 0 30 moveto 0.7 disk -20 5 moveto -25 -7 lineto 30 -7 lineto stroke 35 15 moveto 15 90 2.4 vector 35 15 moveto 11 -90 2.4 vector grestore } \LabelTeX -22.5 -5.5 $\cP$\ELTX \LabelTeX 19.2 -4 base $B$ of area $A$\ELTX \LabelTeX -1 32.6 $O$\ELTX \LabelTeX 32 17 $h$\ELTX \LabelTeX 30 -12 $~$\ELTX \EndFig Indeed, we can calculate an approximate value of the area $A$ of the base by using a grid, and then it follows from (4.9.1) that the volumes of all oblique pyramids based on different grid squares are all identical. This shows that the total volume of these pyramids is proportional to the number of $n$ square grids included in $B$, and thus, in the limit, the volume of the cone $V$ is proportional to the area $A$ of the base. \medskip Our third observation is\?: {\parindent=11.8mm \item{(4.9.3)} {\it When the base $B$ is fixed, the volume $V$ of the cone is proportional to the altitude $h$.}} To prove this property, we consider two cones with the same base $B$ and different altitudes $h$, $h'$, and we compute their volumes through an approximation by a union of thin cylindrical slices\?: \InsertPSFigure 27 40 {%%conditional gsave 1 mm unit initcoordinates %%premier cone 1 0.8 0.8 setrgbcolor 0 30 moveto -12 0 lineto 22 0 lineto closepath fill 1 0 0 setrgbcolor /n 15 def /step 30 n div def 1 1 n { /k exch n div def /x1 k -12 mul def /x2 k 22 mul def /y 30 k 30 mul sub def x1 y step add moveto x2 y step add lineto x2 y lineto x1 y lineto closepath stroke } for 0 setgray 0 30 moveto 0.5 disk 27 15 moveto 15 90 2.4 vector 27 15 moveto 15 -90 2.4 vector %%deuxieme cone 62 -20 translate 0.8 0.8 1 setrgbcolor 0 50 moveto -12 0 lineto 22 0 lineto closepath fill 0 0 1 setrgbcolor /n 15 def /step 50 n div def 1 1 n { /k exch n div def /x1 k -12 mul def /x2 k 22 mul def /y 50 k 50 mul sub def x1 y step add moveto x2 y step add lineto x2 y lineto x1 y lineto closepath stroke } for 0 setgray 0 50 moveto 0.5 disk 27 25 moveto 25 90 2.4 vector 27 25 moveto 25 -90 2.4 vector grestore } \LabelTeX 5 -4 $B$\ELTX \LabelTeX -1 32.6 $O$\ELTX \LabelTeX 24 15 $h$\ELTX \LabelTeX 65 -24 $B$\ELTX \LabelTeX 65 -26 $~$\ELTX \LabelTeX 61 32.6 $O$\ELTX \LabelTeX 84.5 5 $h'$\ELTX \EndFig \medskip To obtain the second cone, each slice is scaled vertically in the ratio $h'/h $. As the volume of a cylinder is the product of the area of the base by the height, we see that the volume of the second stack is proportional to volume of the first, multiplied by the ratio $h'/h$. In the limit, when the slices become thinner and thinner, this clearly shows that the ratio of the volumes of our cones is $V'/V=h'/h$. The fourth and final step is to calculate the volume of a pyramid\?: we start with the case of a pyramid whose base is a square of side $c$ and whose altitude is also $h=c$. The picture below shows that one can fill the cube with side $c$ by 3 isometric pyramids of this kind (it would be useful to actually build them with paper and scissors and verify that they can be assembled into a cube\?!). \InsertPSFigure 5 42 {%%conditional gsave 1 mm unit initcoordinates %%Premier cube [ ] 0 setdash 1 0.8 0.8 setrgbcolor 0 0 moveto 13 35 lineto 25 0 lineto closepath fill 1 0.7 0.7 setrgbcolor 0 0 moveto 13 10 lineto 35 10 lineto 25 0 lineto closepath fill 1 0.6 0.6 setrgbcolor 25 0 moveto 13 35 lineto 38 10 lineto closepath fill 0 setgray 0 0 moveto 25 0 lineto 25 25 lineto 0 25 lineto closepath stroke 0 25 moveto 13 35 lineto 38 35 lineto 25 25 lineto stroke 38 35 moveto 38 10 lineto 25 0 lineto stroke 0.8 0 0 setrgbcolor [ 1.5 0.5 ] 0 setdash 0 0 moveto 13 10 lineto 38 10 lineto stroke 13 10 moveto 13 35 lineto stroke %%Deuxieme cube [ ] 0 setdash 42 0 translate 0.8 1 0.8 setrgbcolor 0 0 moveto 25 0 lineto 25 25 lineto 0 25 lineto closepath fill 0.6 1 0.6 setrgbcolor 0 25 moveto 13 35 lineto 25 25 lineto closepath fill 0 setgray 0 0 moveto 25 0 lineto 25 25 lineto 0 25 lineto closepath stroke 0 25 moveto 13 35 lineto 38 35 lineto 25 25 lineto stroke 38 35 moveto 38 10 lineto 25 0 lineto stroke [ 1.5 0.5 ] 0 setdash 0 0 moveto 13 10 lineto 38 10 lineto stroke 13 10 moveto 13 35 lineto stroke 0 0.8 0 setrgbcolor 0 0 moveto 13 35 lineto 25 0 lineto stroke %%Troisieme cube [ ] 0 setdash 42 0 translate 0.5 0.5 1 setrgbcolor 13 35 moveto 38 35 lineto 38 10 lineto closepath fill 0.7 0.7 1 setrgbcolor 13 35 moveto 38 35 lineto 25 25 lineto closepath fill 0.6 0.6 1 setrgbcolor 25 0 moveto 13 35 lineto 38 10 lineto closepath fill 0.8 0.8 1 setrgbcolor 13 35 moveto 25 25 lineto 25 0 lineto closepath fill 0 setgray 0 0 moveto 25 0 lineto 25 25 lineto 0 25 lineto closepath stroke 0 25 moveto 13 35 lineto 38 35 lineto 25 25 lineto stroke 38 35 moveto 38 10 lineto 25 0 lineto stroke [ 1.5 0.5 ] 0 setdash 0 0 moveto 13 10 lineto 38 10 lineto stroke 13 10 moveto 13 35 lineto stroke grestore } \LabelTeX -2.6 12 $c$\ELTX \LabelTeX 4 30.8 $c$\ELTX \LabelTeX 24 36.6 $c$\ELTX \EndFig \bigskip The volume of each of the three pyramids represented above is therefore one third of the volume of the cube, that is $V={1\over 3}c^3$. If we consider a pyramid with a square base $A=c^2 $ and an arbitrary altitude $h$, the volume must be multiplied by $h/c$ by (4.9.3), hence its volume is equal to $$ V={h\over c}\times{1\over 3}c^3={1\over 3}c^2h. $$ For an arbitrary base $B$, we can use a grid in the plane $\cP$ containing $B$, exactly as we did in the proof of (4.9.2). If $n$ is the number of squares of side $c$ contained in the base~$B$, the area $A$ of the base is approximately $A\approx n\,c^2$ and therefore we get $$ V\approx n\times{1\over 3}c^2h ={1\over 3}(n\,c^2)h \approx {1\over 3}Ah. $$ The approximation becomes better and better when the side $c$ tends to $0$, and in the limit, we see that the volume of an arbitrary cone is given by $$ V={1\over 3}Ah.\leqno(4.9.4) $$ \else {\bf Volume of a cone} The volume of a cone with an arbitrary plane base of area $A$ and altitude $h$ is given by $$ W={1\over 3}\,Ah \leqno(4.9.1) $$ One can indeed argue by a dilation argument that the volume $V$ is proportional to~$h$, and one also shows that is is proportional to $A$ by approximating the base with a union of small squares. The proof is then reduced to the case of an oblique pyramid (i.e.\ to the case when the base is a rectangle). The coefficient${1\over 3}$ is justified by observing that a cube can be divided in three identical oblique pyramids, whose summit is one of the vertices of the cube and the bases are the 3 adjacent opposite faces. The altitude of these pyramids is equal to the side of the cube, and their volume is thus ${1\over 3}$ of the volume of the cube. \fi \medskip {\bf Archimedes formula for the area of a sphere of radius $R$} Since any two spheres of the same radius are isometric, their area depends only on the radius~$R$. Let us take the center $O$ of the sphere as the origin, and consider the ``vertical'' cylinder of radius $R$ tangent to the sphere along the equator, and more precisely, the portion of cylinder located between the ``horizontal'' planes $z=-R$ and $z=R$. We use a ``projection'' of the sphere to the cylinder$\,$: for each point $M$ of the sphere, we consider the point $M'$ on the cylinder which is the intersection of the cylinder with the horizontal line $\cD_M$ passing by $M$ and intersecting the $Oz$~axis. This projection is actually one of the simplest possible cartographic representations of the Earth. After cutting the cylinder along a meridian (say the meridian of longitude~$180^\degree$), and unrolling the cylinder into a rectangle, we obtain the following cartographic map.\bigskip \InsertImage 1 32 0 30 1.0 0 {xrmap.png} \EndFig \vskip-34mm \InsertPSFigure 34.6 34 { gsave 1 mm unit initcoordinates 0 0.35 translate 0 15 moveto 15 90 2.4 vector 0 15 moveto 15 -90 2.4 vector 48.8 -1.6 moveto 15 3.1416 mul 0 2.4 vector 48.8 -1.6 moveto 15 3.1416 mul 180 2.4 vector -33.64 0 translate 0 0 moveto 0 30 lineto stroke 30 0 moveto 30 30 lineto stroke 0.2 setlinewidth 1 0.1 scale newpath 15 0 15 180 360 arc stroke newpath 15 300 15 0 360 arc stroke grestore } \LabelTeX 1.5 18 $2R$\ELTX \LabelTeX 45.7 -4.3 $2\pi R$\ELTX \EndFig \medskip We are going to check that the cylindrical projection preserves areas, hence that the area of the sphere is equal to that of the corresponding rectangular map of sides $2R$ and~$2\pi R$\?: $$ A=2R\times 2\pi R=4\pi R^2. \leqno\iflongertext(4.9.5)\else(4.9.2)\fi $$ In order to check that the areas are equal, we consider a ``rectangular field'' delimited by parallel and meridian lines, of very small size with respect to the sphere, in such a way that it can be seen as a planar surface, i.e. to a rectangle (for instance, on Earth, one certainly does not realize the rotundity of the globe when the size of the field does not exceed a few hundred meters). %%Here are different views of such a rectangular field\?: \InsertPSFigure 5 122 { gsave 1 mm unit initcoordinates 25 85 translate %% sph\`ere en perspective gsave 0.1 setlinewidth 0.9 0.9 1 setrgbcolor 0 21.75 moveto 16.27 15.17 lineto 12.52 14.65 lineto closepath fill 0 21.75 moveto 12.52 14.65 lineto 12.52 13.65 lineto 0 20.75 lineto closepath fill 0.75 0.75 1 setrgbcolor 9.52 16.35 moveto 12 16.86 lineto 12.7 15.57 lineto 9.97 15.10 lineto fill 0.5 0.5 1 setrgbcolor 0.04 setlinewidth 0 21.75 moveto 16.27 15.17 lineto stroke 0 21.75 moveto 12.52 14.65 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$4{\times}$\ELTX \EndFig \vskip20pt Let $a$, $b$ be the side lengths of our ``rectangular field'', respectively along parallel lines direction and meridian lines direction, and $a'$, $b'$ the side lengths of the corresponding rectangle projected on the tangent cylinder. In the view from above, Thales' theorem immediately implies $$ {a'\over a}={R\over r}. $$ In the lateral view, the two triangles represented in green are homothetic (they share a common angle, as the adjacent sides are perpendicular to each other). If we apply again Thales' theorem to the tangent triangle and more specifically to the sides adjacent to the common angle, we get $$ {b'\over b}={\hbox{adjacent small side}\over\hbox{hypotenuse}}={r\over R}. $$ The product of these equalities yields $$ {a'\times b'\over a\times b}={a'\over a}\times{b'\over b} ={R\over r}\times{r\over R}=1. $$ We conclude from there that the rectangle areas $a\times b$ and $a'\times b'$ are equal. This implies that the cylindrical projection preserves areas, and formula {\iflongertext(4.9.5)\else(4.9.2)\fi} follows. \medskip \iflongertext {\bf Volume of a ball of radius $R$} To obtain the volume of a ball\note{In mathematics, a ball is the portion of space delimited by a sphere, just as a disk is the plane domain bounded by a circle.} of radius $R$, we use a gridding of the sphere by meridians and parallels, and calculate the volume of pyramids based on the ``rectangular fields'' delimited in this way. When the grid is sufficiently fine, we can consider that the fields are actually almost planar and rectangular (note that this spatial argument is the exact analogue of the planar argument used to derive the area of a disc through angular sectors, see section~2.2). Let $A_1,\,A_2,\,A_3,\,\ldots$ be the areas of the rectangular fields and let $V_1,\,V_2,\,V_3,\,\ldots$ be the volumes of the corresponding pyramids (clearly, the altitude of these pyramids is $h=R$). We denote here $n=1$ or $2$ or $3\,\ldots$ \InsertPSFigure 40 57 {%%conditional gsave 1 mm unit initcoordinates 0 25 translate 0.1 setlinewidth 0.9 0.9 1 setrgbcolor 0 0 moveto 9.97 15.10 lineto 9.52 16.35 lineto closepath fill 0 0 moveto 9.97 15.10 lineto 12.7 15.57 lineto closepath fill 0.75 0.75 1 setrgbcolor 9.52 16.35 moveto 12 16.86 lineto 12.7 15.57 lineto 9.97 15.10 lineto fill 0.5 0.5 1 setrgbcolor 0.04 setlinewidth 0 0 moveto 9.97 15.10 lineto stroke 0 0 moveto 9.52 16.35 lineto stroke 0 0 moveto 12.7 15.57 lineto stroke gsave 1 0.2 scale 0 0 1 setrgbcolor newpath 0 95 16.248 180 360 arc stroke newpath 0 90 17.349 180 360 arc stroke grestore gsave 0 0 1 setrgbcolor 0.5 1 scale newpath 0 0 25 -90 90 arc stroke 1.3 1 scale newpath 0 0 25 -90 90 arc stroke grestore 0 setgray 0.1 setlinewidth newpath 0 0 25 0 360 arc stroke grestore } \LabelTeX -13 32 \RGBColor{0 0 1}{volume $V_n$}\ELTX \LabelTeX 13.9 38.3 \RGBColor{0 0 1}{area}\ELTX \LabelTeX 16 34.3 \RGBColor{0 0 1}{$A_n$}\ELTX \LabelTeX 40 35 Volume of pyramids\?:\ELTX \LabelTeX 45 28 $\displaystyle V_n={1\over 3}A_n\times R$\ELTX \EndFig \medskip If $A=4\pi R^2$ denotes the total area of the sphere, distributivity of multiplication withrespect to addition shows that the volume $V$ of the ball is given by $$ V=V_1+V_2+V_3+\ldots={1\over 3}(A_1+A_2+A_3+\ldots)\times R= {1\over 3}AR. $$ This gives the formula for the volume of the ball that we were aiming at\?: $$ V={1\over 3}AR={1\over 3}\times 4\pi R^2\times R={4\over 3}\pi R^3. \leqno(4.9.6) $$ \fi \section{5. An axiomatic approach to Euclidean geometry} Although we have been able to follow a deductive presentation when it is compared to some of the more traditional approaches -- almost all of the statements were ``proven'' from the definitions -- it should nevertheless be observed that some proofs relied merely on intuitive facts -- this was for instance the case of the ``proof'' of Pythagoras' Theorem. The only way to break the vicious circle is to take some of the facts that we feel necessary to use as "axioms", that is to say, to consider them as assumptions from which we first deduct all other properties by logical deduction~; a choice of other assumptions as our initial premises leads to non-Euclidean geometries (see section~\iflongertext 11\else 10\fi). As we shall see, the notion of a Euclidean plane can be defined using a single axiom, essentially equivalent to the conjunction of Pythagoras' Theorem - which was only partially justified - and the existence of Cartesian coordinates - which we had not discussed either. In case the idea of using an axiomatic approach would look frightening, we want to stress that this section may be omitted altogether -- provided pupils are in some way brought to the idea that the coordinate systems can be changed (translated, rotated, etc.) according to the needs. \subsection{5.1. The ``Pythagoras/Descartes'' model} In our vision, plane Euclidean geometry is based on the following ``axiomatic definition''. \vskip4mm \vbox{% {\bf Definition.} {\it What we will call a Euclidean plane is a set of points denoted $\cP$, for which mutual distances of points are supposed to be known, i.e.\ there is a predefined function $$ d:\cP\times \cP\longrightarrow\bR_+,\qquad (M,M')\longmapsto d(M,M')=MM'\ge 0, $$ and we assume that there exist ``orthonormal coordinate systems''\?: to each point one can assign a pair of coordinates, by means of a one-to-one correspon\-dence $M\mapsto (x\,;\,y)$ satisfying the axiom\note{As we will try to convince the reader in the sequel, this is a complete and perfectly rigorous description of Euclidean geometry, that is actually equivalent to the long compilation of axioms presented at more advanced university levels\?: vector space of dimension 2 over real numbers, affine plane associated to it, equipped with a positive definite symmetric bilinear form providing the Euclidean structure. At this point, the reader will probably realize how drastically simpler what we call the Pythagoras/Descartes axiom is\?! \iflongertext \else The same definition could be used to introduce higher dimensional Euclidean spaces, just by taking coordinates $(x_1,\ldots,x_n)$ instead.\fi Hyperbolic geometry, in the model of the Poincar\'e disk, would consist in assigning to every point $M$ of $\cP$ a complex number $z=x+iy$ in the unit disk (complex numbers of modulus $|z|<1$), equipped with the infinitesimal metric $|dz|/(1-|z|^2)$ and the resulting geodesic metric, see section \iflongertext 11\else 10\fi.} $$ d(M,M')=\sqrt{(x'-x)^2+(y'-y)^2}\leqno \hbox{\rm(Pythagoras/Descartes)} $$ for all points $M\;(x\,;y)$ and $M'\;(x'\,;\,y')$.}} \medskip It is certainly a good practice to represent the choice of an orthonomal coordinate system by using a transparent sheet of graph paper and placing it over the paper sheet that contains the working area of the Euclidean plane (here that area contains two triangles depicted in blue, above which the transparent sheet of graph paper has been placed). \InsertPSFigure 15.000 66.000 { gsave 0.8333 mm unit 50 25 translate 1 1 0.8 setrgbcolor -60 -40 moveto 60 -40 lineto 60 40 lineto -60 40 lineto closepath fill -20 rotate 0.1 setlinewidth 0.9 0.9 1 setrgbcolor 32 21 moveto -19 31 lineto -31 -9 lineto closepath fill 21 -11 moveto 31 -19 lineto 25 5 lineto closepath fill 1 0.8 0.6 setrgbcolor -35 1 35 { /k exch def -45 k moveto 45 k lineto stroke } for -45 1 45 { /k exch def k -35 moveto k 35 lineto stroke } for 0.3 setlinewidth -3 1 3 { /k exch 10 mul def -45 k moveto 45 k lineto stroke } for -4 1 4 { /k exch 10 mul def k -35 moveto k 35 lineto stroke } for 0.2 setlinewidth 0.6 setgray -50 0 moveto 100 0 2.4 vector 0 -35 moveto 75 90 2.4 vector 0.3 setlinewidth 0.6 0.6 1 setrgbcolor 32 21 moveto -19 31 lineto -31 -9 lineto closepath stroke 21 -11 moveto 31 -19 lineto 25 5 lineto closepath stroke 1 0.3 0.3 setrgbcolor 32 21 moveto 0.7 disk -19 31 moveto 0.7 disk -31 -9 moveto 0.7 disk 21 -11 moveto 0.7 disk 31 -19 moveto 0.7 disk 25 5 moveto 0.7 disk grestore } \LabelTeX 37.9 18.3 $O$\ELTX \LabelTeX 78.4 4 $x$\ELTX \LabelTeX 49.6 50.8 $y$\ELTX \LabelTeX -6 49 $\cP$\ELTX \LabelTeX 30 -23 $~$\ELTX \EndFig This already shows (at an intuitive level only at this point) that there is an infinite number of possible choices for the coordinate systems. We now investigate this in more detail. \medskip {\bf 5.1.1. Rotating the sheet of graph paper around $O$ by $180^\degree$} A rotation of $180^\degree$ of the graph paper around $O$ has the effect of just changing the orientation of axes. The new coordinates $(X\,;\,Y)$ are given with respect to the old ones by $$ X=-x,\qquad Y=-y. $$ Since $(-u)^2=u^2$ for every real number $u$, we see that the formula $$ d(M,M')=\sqrt{(X'-X)^2+(Y'-Y)^2}\leqno(*) $$ is still valid in the new coordinates, assuming it was valid in the original coordinates $(x\,;\,y)$. \medskip {\bf 5.1.2. Reversing the sheet of graph paper along one axis} If we reverse along $Ox$, we get $X=-x$, $Y=y$ and formula $(*)$ is still true. The argument is similar when reversing the sheet along $Oy$, we get the change of coordinates $X=x$, $Y=-y$ in that case. \medskip {\bf 5.1.3. Change of origin} Here we replace the origin $O$ by an arbitrary point $M_0\;(x_0\,;\,y_0)$. \InsertPSFigure 15.000 68.000 { gsave 0.8333 mm unit 50 25 translate 1 1 0.8 setrgbcolor -60 -40 moveto 60 -40 lineto 60 40 lineto -60 40 lineto closepath fill -20 rotate 0.1 setlinewidth 0.9 0.9 1 setrgbcolor 32 21 moveto -19 31 lineto -31 -9 lineto closepath fill 21 -11 moveto 31 -19 lineto 25 5 lineto closepath fill 1 0.8 0.6 setrgbcolor -35 1 35 { /k exch def -45 k moveto 45 k lineto stroke } for -45 1 45 { /k exch def k -35 moveto k 35 lineto stroke } for 0.3 setlinewidth -3 1 3 { /k exch 10 mul def -45 k moveto 45 k lineto stroke } for -4 1 4 { /k exch 10 mul def k -35 moveto k 35 lineto stroke } for 0.2 setlinewidth 0.6 setgray -50 0 moveto 100 0 2.4 vector 0 -35 moveto 75 90 2.4 vector 0 setgray -30 10 moveto 100 0 2.4 vector 20 -25 moveto 75 90 2.4 vector 0.3 setlinewidth 0.6 0.6 1 setrgbcolor 32 21 moveto -19 31 lineto -31 -9 lineto closepath stroke 21 -11 moveto 31 -19 lineto 25 5 lineto closepath stroke 1 0.3 0.3 setrgbcolor 32 21 moveto 0.7 disk -19 31 moveto 0.7 disk -31 -9 moveto 0.7 disk 21 -11 moveto 0.7 disk 31 -19 moveto 0.7 disk 25 5 moveto 0.7 disk 0 0.7 0 setrgbcolor 0.1 setlinewidth [ 1.5 0.5] 0 setdash -25 -25 moveto 90 0 rlineto 0 70 rlineto -90 0 rlineto closepath stroke grestore } \LabelTeX 37.9 18.3 $O$\ELTX \LabelTeX 78.4 4 $x$\ELTX \LabelTeX 49.6 50.8 $y$\ELTX \LabelTeX 52.6 13.7 $x_0$\ELTX \LabelTeX 40.6 26.6 $y_0$\ELTX \LabelTeX 96 5 $X$\ELTX \LabelTeX 68 53 $Y$\ELTX \LabelTeX 53.8 20.8 $M_0$\ELTX \LabelTeX 74 27.3 $M\;(x\,;\,y)$\ELTX \LabelTeX 30 -24 $~$\ELTX \EndFig The new coordinates of point $M\;(x\,;\,y)$ are given by $$X=x-x_0,\qquad Y=y-y_0.$$ For any two points $M$, $M'$, we get in this situation $$ X'-X=(x'-x_0)-(x-x_0)=x'-x,\qquad Y'-Y=(y'-y_0)-(y-y_0)=y'-y $$ and we see that formula $(*)$ is still unchanged. \medskip {\bf 5.1.4. Rotation of axes} We will show that when the origin $O$ is chosen, one can get the half-line $Ox$ to pass through an aribrary point $M_1\;(x_1\,;\,y_1)$ distinct from~$O$. This is intuitively obvious by ``rotating'' the sheet of graph paper around point~$O$, but requires a formal proof relying on our ``Pythagoras/Descartes'' axiom. This proof is substantially more involved than what we have done yet, and can probably be jumped over at first -- we give it here to show that there is no logical flaw in our approach. We start from the algebraic equality called {\it Lagrange's identity} $$ (au+bv)^2+(-bu+av)^2=a^2u^2+b^2v^2+b^2u^2+a^2v^2=(a^2+b^2)(u^2+v^2), $$ which is valid for all real numbers $a,\,b,\,u,\,v$. It can be obtained by developping the squares on the left and observing that the double products annihilate. As a consequence, if $a$ and $b$ satisfy $a^2+b^2=1$ (such an example is $a=3/5$, $b=4/5$) and if we perform the change of coordinates $$ X=ax+by,\qquad Y=-bx+ay $$ we get, for any two points $M$, $M'$ in the plane $$ \eqalign{ &X'-X=a(x'-x)+b(y'-y),\qquad Y'-Y=-b(x'-x)+a(y'-y),\cr &(X'-X)^2+(Y'-Y)^2=(x'-x)^2+(y'-y)^2\cr} $$ by Lagrange's identity with $u=x'-x$, $v=y'-y$. On the other hand, it is easy to check that $$ aX-bY=x,\qquad bX+aY=y, $$ hence the assignment $(x\,;\,y)\mapsto (X\,;\,Y)$ is one-to-one. We infer from there that in the sense of our definition, $(X\,;\,Y)$ is indeed an orthonormal coordinate system. If we now choose $a=kx_1$, $b=ky_1$, the coordinates of point $M_1\;(x_1\,;\,y_1)$ are transformed into $$ X_1=ax_1+by_1=k(x_1^2+y_1^2),\qquad \quad Y_1=-bx_1+ay_1=k(-y_1x_1+x_1y_1)=0, $$ and the condition $a^2+b^2=k^2(x_1^2+y_1^2)=1$ is satisfied by taking $k=1/\sqrt{x_1^2+y_1^2}$. Since $X_1=\smash{\sqrt{x_1^2+y_1^2}}>0$ and $Y_1=0$, the point $M_1$ is actually located on the half-line $OX$ in the new coordinate system. \subsection{5.2. Revisiting the triangular inequality} The proof given in 3.1.1, which relied on facts that were not entirely settled, can now be made completely rigorous. \InsertPSFigure 11.000 27.000 { gsave 0.4 setlinewidth 0 0.7 0 setrgbcolor -15 -6 moveto 160 21.801 7 vector 0 0 moveto 40 111.801 7 vector 0 setgray 0 0 moveto 50 50 lineto 100 40 lineto closepath stroke 50 50 moveto 60.345 24.138 lineto stroke 60.345 24.138 moveto -1 2.5 rlineto 2.5 1 rlineto 1 -2.5 rlineto stroke 170 0 translate 0 0.7 0 setrgbcolor -15 -6 moveto 190 21.801 7 vector 0 0 moveto 40 111.801 7 vector 0 setgray 0 0 moveto 120 60 lineto 100 40 lineto closepath stroke 120 60 moveto 124.138 49.655 lineto stroke 124.138 49.655 moveto -1 2.5 rlineto 2.5 1 rlineto 1 -2.5 rlineto stroke grestore} \LabelTeX -2 -3.3 $A=O$\ELTX \LabelTeX 16 19 $B\;(u\,;\,v)$\ELTX \LabelTeX 34 10.7 $C\;(c\,;\,0)$\ELTX \LabelTeX 45 15 $x$\ELTX \LabelTeX -8 11 $y$\ELTX \LabelTeX 21 5 $H$\ELTX \LabelTeX 58 -3.3 $A=O$\ELTX \LabelTeX 102.5 22 $B$\ELTX \LabelTeX 94.5 11 $C$\ELTX \LabelTeX 103.5 14.5 $H$\ELTX \LabelTeX 115 18.9 $x$\ELTX \LabelTeX 52 11 $y$\ELTX \EndFig \vskip3mm Given three distinct points $A$, $B$, $C$ distincts, we select $O=A$ as the origin and the half line $[A,C)$ as the $Ox$ axis. Our three points then have coordinates $$ A\;(0\,;\,0),\qquad B\;(u\,;\,v),\qquad C\;(c\,;\,0),\qquad c>0, $$ and the foot $H$ of the altitude starting at $B$ is $H\;(u\,;\,0)$. We find $AC=c$ and $$ AB=\sqrt{u^2+v^2}\ge AH=|u|\ge u,\quad BC=\sqrt{(c-u)^2+v^2}\ge HC=|c-u|\ge c-u. $$ Therefore $AC=c=u+(c-u)\le AB+BC$ in all cases. The equality only holds when we have at the same time $v=0$, $u\ge 0$ and $c-u\ge 0$, i.e.\ $u\in[0,c]$ and $v=0$, in other words when $B$~is located on the segment $[A,C]$ of the $Ox$~axis. \subsection{5.3. Axioms of higher dimensional affine spaces} The approach that we have described is also appropriate for the introduction of Euclidean geometry in any dimension, especially in dimension~$3$. The starting point is the calculation of the diagonal $\delta$ of a rectangular parallelepiped with sides $a$, $b$, $c\,$: \InsertPSFigure 40.000 38.000 { gsave 0.5 setlinewidth 0 0 moveto 100 0 lineto 130 30 lineto stroke 0 60 moveto 30 90 lineto 130 90 lineto 100 60 lineto closepath stroke 0 0 moveto 0 60 lineto stroke 100 0 moveto 100 60 lineto stroke 130 30 moveto 130 90 lineto stroke [4 1] 0 setdash 0 0 moveto 30 30 lineto 130 30 lineto stroke 30 30 moveto 30 90 lineto stroke [] 0 setdash 1 0 0 setrgbcolor 0 0 moveto 130 30 lineto 130 90 lineto stroke 0 0 1 setrgbcolor 0 0 moveto 130 90 lineto stroke grestore } \LabelTeX -2 -3.5 $A$\ELTX \LabelTeX 17 -3 $a$\ELTX \LabelTeX 33.5 -3.5 $B$\ELTX \LabelTeX 42 3.2 $b$\ELTX \LabelTeX 47 10 $C$\ELTX \LabelTeX 47 20.5 $c$\ELTX \LabelTeX 47 31 $D$\ELTX \LabelTeX 21 16.8 $\delta$\ELTX \EndFig \bigskip As the triangles $ACD$ and $ABC$ are rectangle in $C$ and $B$ respectively, we have $$ AD^2=AC^2+CD^2\quad\hbox{and}\quad AC^2=AB^2+BC^2 $$ hence the ``great diagonal'' of our rectangle parallelepiped is given by $$ \delta^2=AD^2=AB^2+BC^2+CD^2=a^2+b^2+c^2~~\Rightarrow~~ \delta=\sqrt{a^2+b^2+c^2}. $$ \iflongertext This leads to the following definition \medskip {\bf Definition.} {\it A Euclidean space of dimension $3$ is a set of points denoted~$\cE$, equipped with a distance $d$, namely a mapping $$ d:\cE\times \cE\longrightarrow\bR_+,\qquad (M,M')\longmapsto d(M,M')=MM'\ge 0, $$ such that one can find ``orthonormal coordinate systems'' in the form of a one-to-one correspondence $\cE\ni M\longmapsto(x\,;\,y\,;\,z)\in\bR^3$ satisfying the axiom $$ d(M,M')=\sqrt{(x'-x)^2+(y'-y)^2+(z'-z)^2}\leqno \hbox{\rm(Pythagore/Descartes)} $$ for all points $M\;(x\,;\,y\,;\,z)$ and $M'\;(x'\,;\,y'\,;\,z')$ in $\cE$.} \medskip One could of course give a similar definition in any dimension $n$ by taking coordinate systems $(x_1\,,\,\ldots\,,\,,x_n)\in\bR^n$. The triangle inequality can be proved with almost no changed\?: given three points $A$, $B$, $C$, we choose $A$ as the origin, the half-line $[A,C)$ as the axis $Ox$. This reduces the calculation to the case where $B=(u\,;\,v\,;\,w)$ and $C=(c\,;\,0\,;\,0)$. For this, one must first verify that it is possible to find an orthonormal coordinate system pointing the $Ox$ axis in the direction of an arbitrary point $M_1\;(x_1\,;\,y_1\,;\,z_1)$. One begins by getting $z_1$ to vanish by a change of variables $Y=ay+bz$, $Z=by-az$ in the last two coordinates\?; this brings $M_1$ in the ``horizontal'' plane $Z=0$\?; then one makes $y_1$ vanish with the same method, using a change of variables in $x$, $y$ only. Idem in higher dimensions. \else This leads to the expression of the distance function in dimension 3 $$ d(M,M')=\sqrt{(x'-x)^2+(y'-y)^2+(z'-z)^2} $$ and we can just adopt the latter formula as the 3-dimensional Pythagoras/Descartes axiom. \fi \section{6. Foundations of vector calculus} We will work here in the plane to simplify the exposition, but the only change in higher dimension would be the appearance of additional coordinates. \subsection{6.1. Median formula} Consider points $A$, $B$ with coordinates $(x_A\,;\,y_A)$, $(x_B\,;\,y_B)$ in an orthonormal frame~$Oxy$.The point $I$ of coordinates $$ x_I={x_A+x_B\over 2},\qquad y_I={y_A+y_B\over 2} $$ satisfies $IA=IB={1\over 2}AB$\?: this is the midpoint of segment $[A,B]$. \medskip {\bf Median formula.} {\it For every point $M\;(x\,;\,y)$, one has $$ MA^2+MB^2= 2\,MI^2+{1\over 2}AB^2 = 2\,MI^2+2\,IA^2. $$} \InsertPSFigure 30.000 22.000 { gsave 0.5 setlinewidth 0 0 moveto 40 65 lineto 130 30 lineto closepath stroke 1 0 0 setrgbcolor 40 65 moveto 65 15 lineto stroke grestore } \LabelTeX -2 -3.5 $A$\ELTX \LabelTeX 44 7 $B$\ELTX \LabelTeX 22.3 1.9 $I$\ELTX \LabelTeX 12 24.5 $M$\ELTX \EndFig \bigskip {\it Proof.} In fact, by expanding the squares, we get $$ (x-x_A)^2+(x-x_B)^2= 2x^2-2(x_A+x_B)x+x_A^2+x_B^2, $$ while $$ \eqalign{ 2(x-x_I)^2+{1\over 2}(x_B-x_A)^2 &=2(x^2-2x_Ix+x_I^2)+{1\over 2}(x_B-x_A)^2\cr &=2\Big(x^2-(x_A+x_B)x+{1\over 4}(x_A+x_B)^2\Big)+{1\over 2}(x_B-x_A)^2\cr &=2x^2-2(x_A+x_B)x+x_A^2+x_B^2.\cr} $$ Therefor we get $$ (x-x_A)^2+(x-x_B)^2=2(x-x_I)^2+{1\over 2}(x_B-x_A)^2. $$ The median formula is obtained by adding the analogous equality for coordinates y and applying Pythagoras' theorem.\qed \medskip It follows from the median formula that there is a unique point $M$ such that $MA=MB={1\over 2}\,AB$, in fact we then find $MI^2=0$, hence $M=I$. The coordinate formulas that we initially gave to define midpoints are therefore independent of the choice of coordinates. \subsection{6.2. Parallelograms} A quadrilateral $ABCD$ is a parallelogram if and only if its diagonals $[A,C]$ and $[B,D]$ intersect at their midpoint\?: \InsertPSFigure 30.000 20.000 { gsave /rotmem {currentpoint /cpty exch def /cptx exch def currentpoint translate dup rotate /angmem exch def} def /rotrev {angmem neg rotate cptx neg cpty neg translate cptx cpty moveto} def /vector {dup abs /arlgth exch def 3 div /arlthd exch def rotmem currentlinewidth 1.5811 mul sub dup 0 ge {0 rlineto currentpoint stroke moveto} {0 rmoveto} ifelse currentdash [] 0 setdash arlgth neg arlthd rmoveto arlgth arlthd neg rlineto arlgth neg arlthd neg rlineto arlthd 0 ge {stroke} {closepath gsave fill grestore stroke} ifelse setdash rotrev} def 0.5 setlinewidth 0 0 moveto 50 40 lineto 130 30 lineto 80 -10 lineto closepath stroke 1 0 0 setrgbcolor 0 0 moveto 130 30 lineto stroke 50 40 moveto 80 -10 lineto stroke 0 0 1 setrgbcolor 0.75 setlinewidth 0 0 moveto 80.622 -7.125 7.5 vector 50 40 moveto 80.622 -7.125 7.5 vector grestore } \LabelTeX -4 -1.5 $A$\ELTX \LabelTeX 26.8 -6.9 $B$\ELTX \LabelTeX 46.5 9 $C$\ELTX \LabelTeX 16 15.6 $D$\ELTX \LabelTeX 21 1.5 $I$\ELTX \EndFig \bigskip In this way, we find the necessary and sufficient condition $$ x_I={1\over 2}(x_B+x_D)={1\over 2}(x_A+x_C),\qquad y_I={1\over 2}(y_B+y_D)={1\over 2}(y_A+y_C), $$ which is equivalent to $$ x_B+x_D=x_A+x_C,\qquad y_B+y_D=y_A+y_C $$ or, alternatively, t $$ x_B-x_A=x_C-x_D,\qquad y_B-y_A=y_C-y_D, $$ in other words, the variation of coordinates involved in getting from $A$ to $B$ is the same as the one involved in getting from $D$ to $C$. \subsection{6.3. Vectors} A {\it bipoint} is an ordered pair $(A,B)$ of points$\,$; we say that $A$ is the {\it origin} and that $B$ is the {\it extremity} of the bipoint. The bipoints $(A,B)$ and $(A',B')$ are said to be {\it equipollent} if the quadrilateral $ABB'A'$ is a parallelogram (which can possibly be a ``flat'' parallelogram in case the four points are aligned). \InsertPSFigure 30.000 20.000 { gsave /rotmem {currentpoint /cpty exch def /cptx exch def currentpoint translate dup rotate /angmem exch def} def /rotrev {angmem neg rotate cptx neg cpty neg translate cptx cpty moveto} def /vector {dup abs /arlgth exch def 3 div /arlthd exch def rotmem currentlinewidth 1.5811 mul sub dup 0 ge {0 rlineto currentpoint stroke moveto} {0 rmoveto} ifelse currentdash [] 0 setdash arlgth neg arlthd rmoveto arlgth arlthd neg rlineto arlgth neg arlthd neg rlineto arlthd 0 ge {stroke} {closepath gsave fill grestore stroke} ifelse setdash rotrev} def 0.5 setlinewidth 0 0 moveto 50 40 lineto 130 30 lineto 80 -10 lineto closepath stroke 1 0 0 setrgbcolor 0 0 moveto 130 30 lineto stroke 50 40 moveto 80 -10 lineto stroke 0 0 1 setrgbcolor 0.75 setlinewidth 0 0 moveto 80.622 -7.125 7.5 vector 50 40 moveto 80.622 -7.125 7.5 vector grestore } \LabelTeX -4 -1.5 $A$\ELTX \LabelTeX 26.8 -6.9 $B$\ELTX \LabelTeX 46.5 9 $B'$\ELTX \LabelTeX 16 15.6 $A'$\ELTX \LabelTeX 21 1.5 $I$\ELTX \EndFig \bigskip {\bf Definition.} {\it Given two points $A$, $B$, the vector $\overrightarrow{AB}$ is the ``variation of position'' needed to get from $A$ to $B$. Given a coordinate frame $Oxy$, this ``variation of position'' is expressed along the $Ox$ axis by $x_B-x_A$ and along the $Oy$ axis by $y_B-y_A$. If the bipoints $(A,B)$ and $(A',B')$ are equipollent, the vectors $\overrightarrow{AB}$ and $\overrightarrow{A'B'}$ are equal since the variations $x_{B'}-x_{A'}=x_B-x_A$ and $y_{B'}-y_{A'}=y_B-y_A$ are the same $($this is true in any coordinate system$)$.} The ``component'' of vector $\overrightarrow{AB}$ in the coordinate system $Oxy$ are the numbers denoted in the form of an ordered pair $(x_B-x_A\,;\,y_B-y_A)$. The components $(s\,;\,t)$ of a vector $\overrightarrow{V}$ depend of course on the choice of the coordinate frame $Oxy\;$: to a given vector $\overrightarrow{V}$ one assigns different components $(s\,;\,t)$, $(s'\,;\,t')$ in different coordinate frames $Oxy$, $Ox'y'$. \InsertPSFigure 12.000 42.000 { gsave 0.8333 mm unit 15 8 translate 10 rotate 0.1 setlinewidth 1 0.8 0.6 setrgbcolor -8 1 29 { /k exch def -15 k moveto 35 k lineto stroke } for -15 1 35 { /k exch def k -8 moveto k 29 lineto stroke } for 0.3 setlinewidth 0 1 2 { /k exch 10 mul def -15 k moveto 35 k lineto stroke } for -1 1 3 { /k exch 10 mul def k -8 moveto k 29 lineto stroke } for 0.15 setlinewidth 0 setgray -20 0 moveto 60 0 2.4 vector 0 -13 moveto 46 90 2.4 vector 0.3 setlinewidth 1 0 0 setrgbcolor 11.3 0 moveto 33 0 lineto stroke 0 7.2 moveto 0 11 lineto stroke 0 0.7 0 setrgbcolor 0.2 setlinewidth [1.5 0.5] 0 setdash 11.3 0 moveto 11.3 7.2 lineto stroke 33 0 moveto 33 11 lineto stroke 0 7.2 moveto 11.3 7.2 lineto stroke 0 11 moveto 33 11 lineto stroke 0.3 setlinewidth 0 0 1 setrgbcolor [] 0 setdash -10 rotate 10 9 moveto 22 20 2.4 vector 0.1 setlinewidth [1.5 0.5] 0 setdash 10 9 moveto 87 9 lineto stroke 10 22 20 cos mul add 22 20 sin mul 9 add moveto 77 0 rlineto stroke 77 0 translate -20 rotate [] 0 setdash 0.1 setlinewidth 1 0.8 0.6 setrgbcolor -8 1 29 { /k exch def -15 k moveto 35 k lineto stroke } for -15 1 35 { /k exch def k -8 moveto k 29 lineto stroke } for 0.3 setlinewidth 0 1 2 { /k exch 10 mul def -15 k moveto 35 k lineto stroke } for -1 1 3 { /k exch 10 mul def k -8 moveto k 29 lineto stroke } for 0.15 setlinewidth 0 setgray -20 0 moveto 60 0 2.4 vector 0 -13 moveto 46 90 2.4 vector 0.3 setlinewidth 1 0 0 setrgbcolor 6.5 0 moveto 23 0 lineto stroke 0 11.8 moveto 0 26 lineto stroke 0 0.7 0 setrgbcolor 0.2 setlinewidth [1.5 0.5] 0 setdash 6.5 0 moveto 6.5 11.8 lineto stroke 23 0 moveto 23 26 lineto stroke 0 11.8 moveto 6.5 11.8 lineto stroke 0 26 moveto 23 26 lineto stroke 0.3 setlinewidth 0 0 1 setrgbcolor [] 0 setdash 0.3 setlinewidth 20 rotate 10 9 moveto 22 20 2.4 vector grestore } \LabelTeX 9.5 3.3 $O$\ELTX \LabelTeX 43.3 9.3 $x$\ELTX \LabelTeX 5 32.5 $y$\ELTX \LabelTeX 72.5 4 $O$\ELTX \LabelTeX 106 -8 $x'$\ELTX \LabelTeX 82.2 31.4 $y'$\ELTX \LabelTeX 31.8 15 $\RGBColor{0 0 1}{\overrightarrow{V}}$\ELTX \LabelTeX 91.7 13.2 $\RGBColor{0 0 1}{\overrightarrow{V}}$\ELTX \LabelTeX 29.8 7.2 $\RGBColor{1 0 0}{s}$\ELTX \LabelTeX 86.5 -0.9 $\RGBColor{1 0 0}{s'}$\ELTX \LabelTeX 8.2 13.2 $\RGBColor{1 0 0}{t}$\ELTX \LabelTeX 80.2 22.9 $\RGBColor{1 0 0}{t'}$\ELTX \EndFig \subsection{6.4. Addition of vectors} \InsertPSFigure 30.000 21.000 { gsave /rotmem {currentpoint /cpty exch def /cptx exch def currentpoint translate dup rotate /angmem exch def} def /rotrev {angmem neg rotate cptx neg cpty neg translate cptx cpty moveto} def /vector {dup abs /arlgth exch def 3 div /arlthd exch def rotmem currentlinewidth 1.5811 mul sub dup 0 ge {0 rlineto currentpoint stroke moveto} {0 rmoveto} ifelse currentdash [] 0 setdash arlgth neg arlthd rmoveto arlgth arlthd neg rlineto arlgth neg arlthd neg rlineto arlthd 0 ge {stroke} {closepath gsave fill grestore stroke} ifelse setdash rotrev} def 0.5 setlinewidth 0 0 moveto 50 40 lineto 130 30 lineto 80 -10 lineto closepath stroke 0 0 1 setrgbcolor 0.75 setlinewidth 80 -10 moveto 64.031 38.660 7.5 vector 0 0 moveto 80.622 -7.125 7.5 vector 1 0 0 setrgbcolor 0 0 moveto 133.417 12.995 7.5 vector 0 0.7 0 setrgbcolor 0 0 moveto 64.031 38.660 7.5 vector grestore } \LabelTeX -4 -1.5 $A$\ELTX \LabelTeX 26.8 -6.9 $B$\ELTX \LabelTeX 46.5 9 $C$\ELTX \LabelTeX 16 15.6 $D$\ELTX \EndFig \medskip The addition of vectors is defined by means of {\it Chasles' relation} $$ \RGBColor{0 0 1}{\overrightarrow{AB}+\overrightarrow{BC}}= \RGBColor{1 0 0}{\overrightarrow{AC}}\leqno(6.4.1) $$ for any three points $A$, $B$, $C\;$: when one takes the sum of the variation of position required to get from $A$ to $B$, and then from $B$ to $C$, one finds the variation of position to get from $A$ to $C\;$; actually, we have for instance $$ (x_B-x_A)+(x_C-x_B)=x_C-x_A $$ for the component along the $Ox$ axis. Equivalently, if $ABCD$ is a parallelogram, one can also put $$ \RGBColor{0 0 1}{\overrightarrow{AB}}+\RGBColor{0 0.7 0}{\overrightarrow{AD}}= \RGBColor{1 0 0}{\overrightarrow{AC}}.\leqno(6.4.2) $$ That (6.4.1) and (6.4.2) are equivalent follows from the fact that $\overrightarrow{AD}=\overrightarrow{BC}$ in parallelogram $ABCD$. For any choice of coordinae frame $Oxy$, the sum of vectors of components $(s\,;\,t)$, $(s'\,;\,t')$ has components $(s+s'\,;\,t+t')$. For every point A, the vector $\overrightarrow{AA}$ has zero componaents\?: it will be denoted simply $\overrightarrow{0}$. Obviously, we have $\overrightarrow{V}+\overrightarrow{0}=\overrightarrow{0}+\overrightarrow{V}=\overrightarrow{V}$ for every vector~$\overrightarrow{V}$. On the other hand, Chasles' relation yields $$ \overrightarrow{AB}+\overrightarrow{BA}=\overrightarrow{AA}=\overrightarrow{0} $$ for all points $A$, $B$. Therefore we define $$ -\overrightarrow{AB}=\overrightarrow{BA}, $$ in other words, the opposite of a vector is obtained by exchanging the origin and extremity of any corresponding bipoint. \subsection{6.5. Multiplication of a vector by a real number} Given a vector $\overrightarrow{V}$ of components $(s\,;\,t)$ in a coordinate frame $Oxy$ and an arbitrary real number~$\lambda$, we define $\lambda\overrightarrow{V}$ as the vector of components $(\lambda s\,;\,\lambda t)$. This definition is actually independent of the coordinate frame~$Oxy$. In fact if $\overrightarrow{V}=\overrightarrow{AB}\ne \overrightarrow{0}$ and $\lambda\ge 0$, we have $\lambda\overrightarrow{AB}=\overrightarrow{AC}$ where $C$ is the unique point located on the half-line $[A,B)$ such that $AC=\lambda\,AB$. On the other hand, if $\lambda\le 0$, we have $-\lambda\ge 0$ and $$ \lambda\overrightarrow{AB}= (-\lambda)(-\overrightarrow{AB})=(-\lambda)\overrightarrow{BA}. $$ Finally, it is clear that $\lambda\overrightarrow{0}=\overrightarrow{0}$. Multiplication of vectors by a number is distributive with respect to the addition of vectors (this is a consequence of the distributivity of multiplication with respect to addition in the set of real numbers). \section{7. Cartesian equation of circles and trigonometric functions} By Pythagoras' thorem, the circle of center $A\;(a,b)$ and radius $R$ in the plane is the set of points $M$ satisfying the equation $$ AM=R~~\Leftrightarrow~~AM^2=R^2~~\Leftrightarrow~~(x-a)^2+(y-b)=R^2, $$ which can also be put in the form $x^2+y^2-2ax-2by+c=0$ with $c=a^2+b^2-R^2$. Conversely, the set of solutions of such an equation defines a circle of center $A\;(a\,;\,b)$ and of radius $R=\sqrt{a^2+b^2-c}$ if $ca^2+b^2$. The {\it trigonometric circle} $\cC$ is defined to be the unit circle centered at the origin in an orthormal coordinate system $Oxy$, that is, the of points $M\;(x\,;\,y)$ such that $x^2+y^2=1$. Let $U$ be the point of coordinates $(1\,;\,0)$ and $V$ the point of coordinates $(0\,;\,1)$. The usual trigonometric functions cos, sin and tan are then defined for arbitrary angle arguments as shown on the above figure\note{It seems essential at this stage that functions cos, sin, tan have already been introduced as the ad hoc ratios of sides in a right triangle, i.e.\ at least for the case of acute angles, and that their values for the remarkable angle values $0^\degree$, $30^\degree$, $45^\degree$, $60^\degree$, $90^\degree$ are known.}$\,$: \InsertPSFigure 42.000 55.000 { /ellipsearc {/cmtr0 matrix currentmatrix def currentpoint translate 4 2 roll scale 0 0 1 5 -2 roll newpath arc cmtr0 setmatrix} def gsave 1 mm unit 20.000 20.000 moveto 20 circle stroke 20.000 20.000 moveto 35.000 0.000 2.4 vector 20.000 20.000 moveto 30.000 90.000 2.4 vector 40.000 5.000 moveto 40.000 45.000 lineto stroke 40.000 5.000 moveto 40.000 50.000 lineto stroke 20.000 20.000 moveto 40.000 45.000 lineto stroke 20.000 20.000 moveto 6 6 0 51 ellipsearc stroke 23.700 24.700 moveto 0 135 2.4 vector [ 1.000 0.500 ] 0 setdash 0.549 0.137 1.0 setrgbcolor 20.000 35.600 moveto 32.500 35.600 lineto stroke 0.0 0.0 1.0 setrgbcolor 32.500 20.000 moveto 32.500 35.600 lineto stroke 1.0 0.0 0.0 setrgbcolor 32.500 35.600 moveto 40.000 20.000 lineto stroke [ ] 0 setdash 0.3 setlinewidth 20.000 20.000 moveto 19.850 19.850 0 51 ellipsearc stroke 0.0 0.0 1.0 setrgbcolor 20.000 20.000 moveto 20.000 35.600 lineto stroke 0.549 0.137 1.0 setrgbcolor 20.000 20.000 moveto 32.500 20.000 lineto stroke 0.682 0.0 0.0 setrgbcolor 40.150 20.000 moveto 40.150 44.900 lineto stroke grestore } \LabelTeX 26.600 21.500 $\theta$ \ELTX \LabelTeX 21.600 16.500 $\RGBColor{0.549 0.137 1}{x=\cos(\theta)}$ \ELTX \LabelTeX 4.000 28.500 $\RGBColor{0 0 1}{y=\sin(\theta)}$ \ELTX \LabelTeX 41.300 32.500 $\RGBColor{0.682 0 0}{\displaystyle{y\over x}=\tan(\theta)}$ \ELTX \LabelTeX 16.300 16.500 $O$ \ELTX \LabelTeX 41.000 16.500 $U$ \ELTX \LabelTeX 16.300 41.500 $V$ \ELTX \LabelTeX 30.500 38.000 $M$ \ELTX \LabelTeX 41.000 45.500 $T$ \ELTX \EndFig \medskip The equation of the circle implies the relation $(\cos\theta)^2+(\sin\theta)^2=1$ for every $\theta$. \section{8. Intersection of lines and circles} Let us begin by intersecting a circle $\cC$ of center $A$ and radius $R$ with an arbitrary line~$\cD$. In order to simplify the calculation, we take $A=O$ as the origin and we take the axis $Ox$ to be perpendicular to the line~$\cD$. The line $\cD$ is then ``vertical'' in the coordinate frame $Oxy$. (We start here right away with the most general case, but, once again, it would be desirable to approach the question by treating first simple numerical examples$\;\ldots$). \InsertPSFigure 52.000 63.000 { gsave 1 mm unit 0.000 30.000 translate 20 rotate 0.15 setlinewidth -30.000 0.000 moveto 60.000 0.000 2.4 vector 0.000 -30.000 moveto 60.000 90.000 2.4 vector 0.25 setlinewidth 0 0 1 setrgbcolor 0 0 moveto 20 circle stroke 1 0 0 setrgbcolor 15 -25 moveto 15 25 lineto stroke grestore } \LabelTeX -21.2 39 $\RGBColor{0 0 1}{\cC}$ \ELTX \LabelTeX -2.7 26 $O$ \ELTX \LabelTeX 11.1 31.5 $x_0$ \ELTX \LabelTeX 20.1 34.3 $R$ \ELTX \LabelTeX 27.1 36.7 $x$ \ELTX \LabelTeX -13.1 56.7 $y$ \ELTX \LabelTeX 23.1 13 $\RGBColor{1 0 0}{D}$ \ELTX \EndFig This leads to equation $$ \cC:x^2+y^2=R^2,\qquad D:x=x_0, $$ hence $$ y^2=R^2-x_0^2. $$ As a consequence, if $|x_0|0$ and there are two solutions $y=\sqrt{R^2-x_0^2}$ and $y=-\sqrt{R^2-x_0^2}$, corresponding to two intersection points$(x_0,\sqrt{R^2-x_0^2})$ and $(x_0,-\sqrt{R^2-x_0^2})$ that are symmetric with respect to the $Ox$ axis. If $|x_0|=R$, we find a single solution $y=0\;$: the line $\cD:x=x_0$ is tangent to circle $\cC$ at point $(x_0\,;\,0)$. If $|x_0|>R$, the equation $y^2=R^2-x_0^2<0$ has no solution$\,$; the line $\cD$ does not intersect the circle. Consider now the intersection of a circle $\cC$ of center $A$ and radius $R$ with a circle $\cC'$ of center $A'$ and radius~$R'$. Let $d=AA'$ be the distance between their centers. If $d=0$ the circles are concentric and the discussion is easy (the circles coincide if $R=R'$, and are disjoint if $R\ne R'$). We will therefore assume that $A\ne A'$, i.e.\ $d>0$. By selecting $O=A$ as the origin and $Ox=[A,A')$ as the positive $x$ axis, we are reduces to the case where $A\;(0\,;\,0)$ and $A'\;(d\,;\,0)$. We then get equations $$ \cC:x^2+y^2=R^2,\qquad \cC':(x-d)^2+y^2=R^{\prime2}\Longleftrightarrow x^2+y^2=2dx+R^{\prime2}-d^2. $$ For any point $M$ in the intersection $\cC\cap\cC'$, we thus get $2dx+R^{\prime2}-d^2=R^2$, hence $$ x=x_0={1\over 2d}(d^2+R^2-R^{\prime2}). $$ This shows that the intersection $\cC\cap\cC'$ is contained in the intersection $\cC\cap D$ of $\cC$ with the line $\cD:x=x_0$. Conversely, one sees that if $x^2+y^2=R^2$ and $x=x_0$, then $(x\,;\,y)$ also satisfies the equation $$ x^2+y^2-2dx=R^2-2dx_0=R^2-(d^2+R^2-R^{\prime2})=R^{\prime2}-d^2 $$ which is the equation of $\cC'$, hence $\cC\cap D\subset \cC\cap\cC'$ and finally $\cC\cap\cC'=\cC\cap D$. \InsertPSFigure 45.000 65.000 { gsave 1 mm unit 0.000 30.000 translate 20 rotate 0.15 setlinewidth -30.000 0.000 moveto 85.000 0.000 2.4 vector 0.000 -30.000 moveto 60.000 90.000 2.4 vector 0.25 setlinewidth 0 0 1 setrgbcolor 0 0 moveto 20 circle stroke 0 0.7 0 setrgbcolor 22 0 moveto 15 circle stroke 22 -0.7 moveto 22 0.7 lineto stroke 1 0 0 setrgbcolor 15 -25 moveto 15 25 lineto stroke grestore } \LabelTeX -2.7 26 $A$ \ELTX \LabelTeX 11.1 31.5 $x_0$ \ELTX \LabelTeX 21.1 34.6 $A'$ \ELTX \LabelTeX 49.8 44.9 $x$ \ELTX \LabelTeX -13.1 56.7 $y$ \ELTX \LabelTeX 23.1 13 $\RGBColor{1 0 0}{D}$ \ELTX \LabelTeX -21.2 39 $\RGBColor{0 0 1}{\cC}$ \ELTX \LabelTeX 28 52 $\RGBColor{0 0.7 0}{\cC'}$ \ELTX \EndFig The intersection points are thus given by $y=\pm\sqrt{R^2-x_0^2}$. As a consequene, we have exactly two solutions that are symmetric with respect to the line $(AA')$ as soon as $-RR^{\prime 2}~~\hbox{et}~~(d-R)^2R',~~d-R-R',\cr} $$ i.e.\ $|R-R'|0 $ is a small real number. Finally, let $I$ be the midpoint of $A$, $B$ with respect to the geodesic distance. \InsertPSFigure 30.000 24.000 {%%conditional gsave 0.6 setlinewidth 0 0 moveto 40 32 5 vector stroke 0 0 moveto 50 -18 5 vector stroke 0.3 setlinewidth 0 0 moveto 40 25 80 33 119 40 curveto stroke 0 0 moveto 50 -15 100 -20 149 -23 curveto stroke 119 40 moveto 130 20 140 0 149 -23 curveto stroke 1 0 0 setrgbcolor 0 0 moveto 45 5 90 7 135 9 curveto stroke grestore } \LabelTeX -3.5 -2.5 $O$\ELTX \LabelTeX 49 3 $I$\ELTX \LabelTeX 43 14.5 $A$\ELTX \LabelTeX 53.5 -9.5 $B$\ELTX \LabelTeX 10 9 $u$\ELTX \LabelTeX 14 -8.5 $v$\ELTX \EndFig \vskip5mm Then infinitesimally small geodesic triangles $OAB$ satisfy in the limit $$\leqalignno{ &\lim_{\varepsilon\to 0}{OA^2+OB^2-2OI^2-2AI^2 \over OA^2\;OB^2} =-{1 \over 6}{\langle R(u,v)u,v\rangle_g \over \Vert u\Vert_g^2\;\Vert v\Vert_g^2},&(13.1)\cr &\lim_{\varepsilon\to 0}{OA^2+OB^2-2OI^2-2AI^2 \over OA^2\;OB^2-(OI^2-AI^2)^2} =-{1 \over 6}{\langle R(u,v)u,v\rangle_g \over \Vert u\wedge v\Vert_g^2},&(13.2)\cr} $$ where $R$ is the Riemannian curvature tensor. The numerator $OA^2+OB^2-2OI^2-2AI^2$ vanishes for Euclidean geometry (median theorem\?!), and the above formulas tell us that the deviation with respect to the Euclidean situation is essentially described by the sectional curvature.\\ {\it Proof}~: this is left as a very interesting -- and non trivial -- exercise to readers~! \medskip If $X$ and $Y$ are two compact metric spaces, one defines their {\it Gromov-Hausdorff distance} $d_{\rm GH}(X,Y)$ to be the infimum of all Hausdorff distances $d_{\rm H}(f(X),f(Y))$ for all possible isometric embeddings $f:X\to\cE$, $g:Y\to\cE$ of $X$ and $Y$ in another compact metric space~$\cE$. This provides a crucial tool to study deformations and degenerations of Riemannian manifolds. \else \section{10. More advanced material} At this point, we have all the necessary foundations, and the succession of concepts to be introduced becomes much more flexible -- much of what we discuss below only concerns high school level and beyond. One can for example study further properties of triangles and circles, and gradually introduce the main geometric transformations (in the plane to start with)~: translations, homotheties, affinities, axial symmetries, projections, rotations with respect to a point~; and in space, symmetries with respect to a point, a line or a plane, orthogonal projections on a plane or on a line, rotation around an axis. Available tools allow making either intrinsic geometric reasonings (with angles, distances, similarity ratios,~$\ldots$), or calculations in Cartesian coordinates. It is actually desirable that these techniques remain intimately connected, as this is common practice in contemporary mathematics (the period that we describe as ``contemporary'' actually going back to several centuries for mathematicians, engineers, physicists~$\ldots$) It is then time to investigate the phenomenon of linearity, independently of any distance consideration. This leads to the concepts of linear combinations of vectors, linear dependence and independence, non orthonormal frames, etc,~in relation with the resolution of systems of linear equations. One is quickly led to determinants $2\times 2$, $3\times 3$, to equations of lines, planes, etc. The general concept of vector space provides an intrinsic vision of linear algebra, and one can introduce general affine spaces, bilinear symmetric forms, Euclidean and Hermitian geometry in arbitrary dimension. What we have done before can be deepened in various ways, especially by studying the general concept of isometry. \medskip {\bf 10.1. Definition.} {\it Let $\cE$ and $\cF$ be two Euclidean spaces and let $s:\cE\to\cF$ be an arbitrary map between these. We say that $s$ is an isometry from $\cE$ to $\cF$ if for every pair of points $(M,N)$ of~$\cE$, we have $d(s(M),s(N))=d(M,N)$.} \medskip Isometries are closely tied to inner product via the following fundamental theorem.\medskip {\bf 10.2. Theorem.} {\it If $s:\cE\to\cF$ is an isometry, then $s$ is an affine transformation, and its associated linear map $\sigma:\overrightarrow{\cE}\to\overrightarrow{\cF}$ is an orthogonal transform of Euclidean vector spaces, namely a linear map preserving orthogonality and inner products~: $$ \sigma(\overrightarrow{V})\cdot\sigma(\overrightarrow{W})= \overrightarrow{V}\cdot\overrightarrow{W}\leqno(10.3) $$ for all vectors $\overrightarrow{V},\overrightarrow{W}\in \overrightarrow{E}$.} In the same vein, one can prove the following result, which provides a rigorous mathematical justification to all definitions and physical considerations appeared in section~3.2. \medskip {\bf 10.4. Theorem.} {\it Let $(A_1A_2A_3A_4\,\ldots)$ and $(A'_1A'_2A'_3A'_4\,\ldots)$ be two isometric figures formed by points $A_i$, $A'_i$ of a Euclidean space~$\cE$. Then there exists an isometry $s$ of the entire Euclidean space $\cE$ such that $A'_i=s(A_i)$ for all~$i$.} \bigskip \InsertImage 43 60 0 50 1.0 0 {Riemann0.jpg} \LabelTeX -43 55 {\twelvebf Non Euclidean geometries.}\ELTX \LabelTeX -43 -5 \centerline{\it Bernhard Riemann $($1826--1866$\,)$}\ELTX \EndFig \medskip In contemporary mathematics, non Euclidean geometries are best seen as a special instance of Riemannian geometry, so called in reference to Bernhard Riemann, one of the founders of modern complex analysis and differential geometry~[Rie]. A Riemannian manifold is by definition a differential manifold $M$, namely a topological space that admits local differentiable systems of coordinates $x=(x_1\,;\,x_2\,;\,\ldots\,;\,x_n)$, equipped with an infinitesimal metric $g$ of the form $$ ds^2=g(x)=\sum_{1\le i,j\le n}a_{ij}(x)\,dx_idx_j.\leqno(10.5) $$ By integrating the infinitesimal metric along paths, one obtains the geodesic distance which is used as a substitute of the Euclidean distance (in physics, general relativity also arises in a similar way by considering Lorentz-like metrics of the form $ds^2=dx_1^2+dx_2^2+dx_3^2-c^2\,dt^2$). On the unit disk $\bD=\{z\in\bC\,;\,|z|<1\}$ in the complex plane, denoting $z=x+iy$, one considers the so-called Poincar\'e metric (named after Henri Poincar\'e, 1854-1912, see [Poi]) $$ ds={|dz|\over 1-|z|^2} ~~\Leftrightarrow~~ ds^2={dx^2+dy^2\over(1-(x^2+y^2))^2}.\leqno(10.6) $$ The associated geodesic distance can be computed to be $$ d_{\rm P}(a,b)={1\over 2}\ln{1+{|b-a|\over |1-\ovl a b|}\over 1-{|b-a|\over |1-\ovl a b|}}. \leqno(10.7) $$ When substituting this distance to the Pythagoras/Descartes axiom, one actually obtains a non Euclidean geometry, which is a model of the hyperbolic geometry discovered by Nikolai Lobachevski (1793-1856). In this geometry, there are actually infinitely many parallel lines to a given line $\cD$ through a given point $p$ exterior to~$\cD$, so that Euclid's fifth postulate fails~! \InsertPSFigure 30.000 65.000 { 1 mm unit initcoordinates gsave 30 34 translate /hom 27 def hom dup scale 0.085 hom div setlinewidth 2 setlinejoin /expit { /x exch def /r x x mul 1 add def x 2 mul r div def 2 r sub r div def } def /doarc { /yp exch def /xp exch def /y exch def /x exch def /z x xp add 0.5 mul def /w y yp add 0.5 mul def /r z z mul w w mul add sqrt def /z z r div def /w w r div def /r xp x sub def /rp yp y sub def /rp r r mul rp rp mul add sqrt 0.5 mul def /r rp 1 rp rp mul sub sqrt div def /rp 1 r r mul add sqrt def /z z rp mul def /w w rp mul def /a1 y x atan def /a2 yp xp atan def a1 a2 180 sub lt { /a1 a1 360 add def } if a1 a2 180 add gt { /a1 a1 360 sub def } if a1 a2 gt { z w r a1 90 add a2 270 add arc } { z w r a1 270 add a2 90 add arcn } ifelse } def 1 0 moveto /x1 /y1 -9 expit /x2 /y2 -4 expit /x3 /y3 -0.5 expit /x4 /y4 0.1 expit 0.9 setgray x1 y1 moveto x1 y1 x3 y3 doarc 0 0 1 y3 x3 atan y4 x4 atan arc x4 y4 x2 y2 doarc 0 0 1 y2 x2 atan y1 x1 atan arcn fill 0 setgray x1 y1 x3 y3 doarc stroke x2 y2 x3 y3 doarc stroke x2 y2 x4 y4 doarc stroke 0 0 1 0 360 arc stroke 0.250 hom div setlinewidth /x5 /y5 -5.4 expit /x6 /y6 -0.2 expit x5 y5 x6 y6 doarc stroke -0.335 -0.21 moveto 0.6 hom div disk grestore } \LabelTeX 22.000 22.000 $\cD$ \ELTX \LabelTeX 20.000 31.000 $p$ \ELTX \LabelTeX 41.000 30.000 $\Delta$ \ELTX \LabelTeX -30 0 \centerline{\it Lobachevski's hyperbolic geometry and the failure of Euclid's fifth postulate}\ELTX \EndFig \bigskip {\twelvebf On some ideas of Felix Hausdorff and Mikhail Gromov.} We first describe a few important ideas due to Felix Hausdorff (1868~-~1942), one of the founders of modern topology~[Hau]. The first one is that the Lebesgue measure of $\bR^n$ can be generalized without any reference to the vector space structure, but just by using the metric. If $(\cE,d)$ is an arbitrary metric space, one defines the $p$-dimensional Hausdorff measure of a subset $A$ of $\cE$ as $$ \cH_p(A)=\lim_{\varepsilon\to 0} \cH_{p,\varepsilon}(A),\qquad \cH_{p,\varepsilon}(A)=\inf_{\diam A_i\le\varepsilon}\sum_i(\diam A_i)^p \leqno(10.8) $$ where $\cH_{p,\varepsilon}(A)$ is the least upper bound of sums $\sum_i(\diam A_i)^p$ running over all countable partitions $A=\bigcup A_i$ with~$\dim A_i\le\varepsilon$. For $p=1$ (resp.\ $p=2$, $p=3$) one recovers the usual concepts of length, area, volume, and the definition even works when $p$ is not an integer (it is then extremely useful to define the dimension of fractal sets). Another important idea of Hausdorff is the existence of a natural metric structure on the set of compact subsets of a given metric space $(\cE,d)$. If $K,\,L$ are two compact subsets of~$\cE$, the {\it Hausdorff distance} of $K$ and $L$ is defined to be $$ d_{\rm H}(K,L)=\max\big\{\max_{x\in K}\min_{y\in L}d(x,y), \max_{y\in L}\min_{x\in K}d(x,y)\big\}.\leqno(10.9) $$ The study of metric structures has become today one of the most active domains in mathematics. We should mention here the work of Mikhail Gromov (Abel prize 2009) on length spaces and ``moduli spaces'' of Riemannian manifolds. If $X$ and $Y$ are two compact metric spaces, one defines their {\it Gromov-Hausdorff distance} $d_{\rm GH}(X,Y)$ to be the infimum of all Hausdorff distances $d_{\rm H}(f(X),f(Y))$ for all possible isometric embeddings $f:X\to\cE$, $g:Y\to\cE$ of $X$ and $Y$ in another compact metric space~$\cE$. This provides a crucial tool to study deformations and degenerations of Riemannian manifolds. \vfill\eject \fi \vskip6mm {\twelvebf References} \medskip {\eightpoint \bibitem[Art]&Emil Artin&Geometric Algebra&Interscience publishers Inc., New York, 1957, download at\\ {\tt http://ia700300.us.archive.org/24/items/geometricalgebra033556mbp/\\ geometricalgebra033556mbp.pdf}& \bibitem[Des]&Ren\'e Descartes&La G\'eom\'etrie, {\rm appendix of} Discours de la m\'ethode&available from Project Gutenberg {\tt http://www.gutenberg.org/files/26400/26400-pdf.pdf}& \bibitem[Die]&Jean Dieudonn\'e&Alg\`ebre lin\'eaire et g\'eom\'etrie \'el\'ementaire&Hermann, 1968, 240 pages& \bibitem[Euc]&Euclid&Euclid's elements of geometry&English translation from Greek by Richard Fitzpatrick, {\tt http://farside.ph.utexas.edu/euclid/Elements.pdf}& \bibitem[Gro]&Mikhail Gromov&Structures m\'etriques pour les vari\'et\'es riemanniennes&edited by J.~Lafontaine and P.~Pansu, Textes Math\'ematiques, no.~1, CEDIC, Paris, 1981, translated by S.M. Bates, {\it Metric Structures for Riemannian and Non-Riemannian Spaces}, Birkh\"auser, 1999& \bibitem[Hau]&Felix Hausdorff&Dimension und \"ausseres Mass&Mathematische Annalen 79, 1-2 (1919) 157-179, and {\it Grundz\"uge der Mengenlehre}, Leipzig, 1914. viii + 476 pages, English translation {\it Set Theory}, American Mathematical Soc., 1957, 352 pages& \bibitem[Hil]&David Hilbert&Grundlagen der Geometrie&Teubner, Leipzig, 1899, English translation {\it The Foundations of Geometry}, by E.J. Townsend (2005), available from project Gutenberg,\\ {\tt http://www.gutenberg.org/files/17384/17384-pdf.pdf}& \bibitem[Lob]&Nikolai Lobachevski&Geometrical researches on the theory of parallels&\\ {\tt http://quod.lib.umich.edu/u/umhistmath/AAN2339.0001.001/1?rgn=full+text;view=pdf}& \bibitem[Pas]&Moritz Pasch&Vorlesungen \"uber neuere Geometrie&Teubner, Leipzig, 1882, download at\\ {\tt https://ia700308.us.archive.org/32/items/vorlesungenbern00pascgoog/\\ vorlesungenbern00pascgoog.pdf}& \bibitem[Poi]&Henri Poincar\'e&Œuvres&tome 2, 1916 - Fonctions fuchsiennes, et {\it Les g\'eom\'etries non euclidiennes} {\tt http://www.univ-nancy2.fr/poincare/bhp/}& \bibitem[Pon]&Jean-Victor Poncelet&Trait\'e des propri\'et\'es projectives des figures&Gauthier-Villars, Paris, 1866, download at {\tt http://gallica.bnf.fr/ark:/12148/bpt6k5484980j}& \bibitem[Rie]&Bernhard Riemann&Grundlagen f\"ur eine allgemeine Theorie der Funktionen einer ver\"ander\-lichen complexen Gr\"osse&Inauguraldissertation, G\"ottingen, 1851, and {\it Ueber die Hypothesen, welche der Geometrie zu Grunde liegen}, Habilitationsschrift, 1854, Abhandlungen der K\"oniglichen Gesellschaft der Wissenschaften zu G\"ottingen, 13 (1868), English translation {\it On the Hypotheses which lie at the Bases of Geometry}, {\tt http://www.emis.de/classics/Riemann/}& } \iflongertext \message{^^J!!! 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