The Dude Minds…

 

On considère \(n\) points rouges et \(n\) points bleus dispersés dans le plan, trois points quelconques n’étant pas colinéaires. Est-il toujours possible de relier les \(n\) points rouges aux \(n\) points bleus avec des segments de telle sorte qu’aucun segment n’en croise un autre ?

En essayant quelques cas à la main avec un \(n\) petit, on se doute que la réponse sera affirmative mais avec un \(n\) grand, la démarche qui nous mène à une solution n’est pas claire. Comment choisir les points et les relier ? Comment éviter systématiquement les croisements ?

 

Si deux segments se croisent, il est effectivement possible et même facile de remplacer ces segments par d’autres qui ne partagent aucun point d’intersection.

Considérons le quadrilatère convexe formé par les quatre points aux extrémités des segments qui se croisent et qui possède les deux segments comme diagonales.

Deux des côtés (opposés) relient évidemment deux points de la même couleur ; cependant les deux autres côtés (opposés aussi) relient un point rouge à un point bleu. On propose donc de remplacer les segments initiaux par ces deux côtés opposés du polygone.

Et hop ! Les segments ont été remplacés par d’autres qui ne se croisent plus.

 

L’approche à emprunter qui nous permettrait de relier les \(n\) points bleus aux \(n\) points rouges en évitant les croisements étant nébuleuse, on adopte plutôt avec une approche radicalement différente : on relie les points rouges aux points bleus sans faire attention, de manière arbitraire et on essaie de les démêler ensuite. Notons au passage qu’il y a \(n!\) façons de relier les \(n\) points bleus aux \(n\) points rouges.

Est-il possible alors de décroiser tous les segments avec la technique décrite ci-haut ?

En effet, il suffit d’identifier des segments qui se croisent puis de les remplacer par d’autres qui ne se croisent pas. Le nombre d’intersections diminue ainsi de \(1\) chaque fois qu’on fait appel à la technique précédente. Si on considère le nombre d’intersections total, en procédant de cette manière, à répétition, à chaque étape le nombre d’intersections diminuant de \(1\), éventuellement, on devrait atteindre le nombre désiré de \(0\) intersection, non ?

Vraiment ?

Hélas, ce n’est malheureusement pas si simple. Il est fort possible que les côtés du quadrilatère choisis croisent à leur tour d’autres segments et que le nombre d’intersections reste le même ou pire, augmente. Oups.

On identifie le quadrilatère.

devient

Dans ce cas-ci le nombre d’intersections augmente.

Cependant, tout n’est pas perdu !

 

L’idée de remplacer les deux segments qui forment les diagonales par les côtés du quadrilatère est la bonne. Mais ce n’est pas le nombre d’intersections qui doit nous intéresser : c’est plutôt longueur totale des segments. Une observation importante à faire est celle-ci : la somme des mesures des deux segments qui se croisent est supérieure à la somme des mesures des côtés opposés du quadrilatère qui les remplacent.

Considérons que les segments \(\textcolor{Red}{A}\textcolor{Blue}{C}\) et \(\textcolor{Blue}{B}\textcolor{Red}{D}\) se croisent en \(E\). Avec l’inégalité du triangle, on sait que \[m\overline{AE} + m\overline{EB} > m\overline{AB}\]et que \[m\overline{DE} + m\overline{EC} > m\overline{DC}\]La somme de ces expressions correspond à \[m\overline{AE} + m\overline{EB}  + m\overline{DE} + m\overline{EC} > m\overline{AB} + m\overline{DC}\]ce qui fait \[m\overline{AE} +m\overline{EC}+  m\overline{DE} + m\overline{EB}  > m\overline{AB} + m\overline{DC}\]ou plus simplement \[m\overline{AC} + m\overline{DB}> m\overline{AB} + m\overline{DC}\]ce qui est le résultat recherché.

Il y a \(n!\) façons différentes de lier les \(n\) points bleus et les \(n\) points rouges avec \(n\) segments. Considérons la façon de les relier qui minimise la somme des mesures des \(n\) segments. On prétend qu’il n’y a à ce moment aucuns segments qui se croisent. En effet, s’il y a avait un croisement, on pourrait appliquer la procédure décrite ci-dessus et remplacer les segments par les côtés du quadrilatère correspondant pour diminuer la somme des mesures des \(n\) segments. C’est la contradiction recherchée car on avait considéré la façon de relier les segments qui minimisait la longueur totale \(n\) segments et on en a trouvé une autre dans laquelle la longueur totale était encore plus petite !

Il est donc toujours possible…

de relier les \(n\) points rouges…

aux \(n\) points bleus…

sans que deux segments ne se croisent.

Notons que notre réflexion repose sur le fait qu’il existe (au moins) une façon minimale de relier les points mais elle n’exclut pas qu’il y ait, d’abord, plus qu’une façon minimale (d’ailleurs, est-ce possible ?) et ensuite plusieurs autres façons valides de relier les points (donc sans aucun croisement) mais dans lesquelles la longueur totale des segments n’est pas minimale.

Dans The History of Mathematics: A Source-Based Approach: Volume 1, au début du chapitre concernant les mathématiques indiennes, chinoises et arabes (jusqu’au 17^e siècle), tout en subtilité :

[…]

This complicated and extensive history raises problems for historians, and for us as students. Our sources are few, and their transmission was varied (copies of copies, etc.). Even when studied by those comfortable with the language they can be difficult to read, and they may be entirely silent on why the work was done at all, or why in this particular way. We may at best glimpse matters that are otherwise lost. We may see examples of profound discoveries but be unable to see how they were made, or how and whether) they were proved to be correct. The temptation to sweep the few things we have into tidy generalisations about Indian, Chinese, or Islamic mathematics can be hard to resist, and we must remember that even when the surviving documents are richly informative much has been lost. This problem is particularly acute when the earliest periods of Indian and Chinese cultures are concerned, when sources are few and especially when great claims are made for the antiquity and profundity of what some writers would have us believe was there.

A further difficulty attends the study of these cultures for those coming from the West. It is hard for us not to see them as alien or exotic, even if we have tried our best not to see them as inferior or hostile. We must remember that our view (as students who cannot read the original sources) is brought to us by historians and writers with a variety of agendas. They include explorers and colonial administrators as well as scholars from several centuries. It is possible to detect several distinct ways in which these cultures have been analysed, and the evidence selected accordingly. Some writers have looked at them to see what they had that ‘we’ (Europeans and, later, Americans) have, and to see who had it first. This is a very Western-centric approach, in which mathematics is defined, often tacitly, as the mathematics we do, and the other culture scrutinised for what is has in common with ours. Evidence of difference is suppressed.

Other writers emphasise the exotic and the different, whether or not they present it as better or as weird. This can be a genuine effort to see the other culture more on its own terms, but it can pander to stereotypes of that culture that are often pernicious, and it can be blind to similarities in our ‘own’, Western, culture, past or present. Finally, because much of the writing about these cultures was done originally by Western scholars with their own agendas, Indian, Chinese, and Islamic writers have sometimes fought back too strongly, over-stating the originality of their own cultures. Maoist China was not the only place to make exaggerated claims for its ancient ways.

The greatest problem facing us as students, however, is the neglect of the history of mathematics in these cultures. To be intelligible, sources have to be accessible, collected, catalogued, edited, and perhaps transcribed. Only then can they speak to us, and the sad truth is that for much of the time, and in most of the territories we are studying, this has simply not been the case. Many losses are due to the ravages of time: documents are fragile and easily destroyed. The Indian climate is particularly unsuited to the survival of paper. Other losses are the consequences of wars and fires, some (as in China) the result of deliberate destruction. But not until the 1950s were there any attempts to collect, preserve, catalogue, and make available the documents that do survive, and the results are still patchy. There are undoubtedly major discoveries still to be made in libraries across Asia and elsewhere.

[…]

Our study of how this mathematics was done leads us to a consideration of whether and why it was accepted. A rough and ready distinction can be made in the history of ‘European’ mathematics between Euclidean geometry and algebra, according to which geometry was concerned with proof and algebra with calculation, perhaps according to a set of rules that formed an algorithm. Indeed, mathematicians did sometimes turn to Euclidean geometry to give their methods the security of a decent proof, and the shift at the start of the 17th century to regard algebra as itself being capable of proof was a momentous one in the history of ‘European’ mathematics. We shall see that in India, China, and the Islamic world there was always a greater emphasis on number, calculation, and algorithms than was the case in the West, and this has given rise to a controversy as to the extent to which those cultures possessed a concept of proof.

It is evident that we are plunged into issues of historians’ presumptions. Those who assert that mathematics began in Greece and have proof at its core are likely to think that other cultures must have lacked a proper concept of proof, and to that extent be deficient. Those looking for the exotic may agree that proof was lacking, but elevate some other virtue instead. Yet other historians may believe that some concepts, such as science or mathematics, are not at all culturally dependent, while allowing for any amount of regional variation: they may expect to find proofs in the mathematics of India, the Islamic world, and China, but not Euclidean-style geometrical proofs. So we have to ask what we take proof to be, and this leads to an interesting debate among historians as to whether, and to what extent, an understanding of algorithms implies the existence of a proof.

[…]

Barrow-Green, June, Jeremy Gray & Robin Wilson, The History of Mathematics: A Source-Based Approach: Volume 1, 2019, AMS/MAA Textbooks

On considère deux angles non nuls \(x\) et \(y\) tels que \[0< x + y<\pi\]

\[(\text{Aire du }  \triangle ABC) = (\text{Aire du } \triangle BCD) + (\text{Aire du } \triangle ACD)\]

En utilisant la formule trigonométrique de l’aire du triangle, on obtient \[\tfrac{1}{2} a  b \sin\left(x+y\right) = \tfrac{1}{2}a h\sin\left(x\right) + \tfrac{1}{2}  b h \sin\left(y\right)\]

Dans le triangle \(BCD\), on a aussi \[\cos(x) = \frac{h}{a}\]ou \[h = a \cos(x)\]Dans le triangle \(ACD\), on a \[\cos(y) = \frac{h}{b}\]ou \[h = b \cos(y)\]

L’astuce est de remplacer \(h\) dans\[\tfrac{1}{2} a  b\sin\left(x+y\right) = \tfrac{1}{2}a h\sin\left(x\right) + \tfrac{1}{2}  b h  \sin\left(y\right)\]tantôt par \(b \cos(y)\), tantôt par \(a \cos(x)\).

\begin{align*}\tfrac{1}{2} a b  \sin\left(x+y\right) &= \tfrac{1}{2}a b \cos(y) \sin\left(x\right) + \tfrac{1}{2} b  a \cos(x) \sin\left(y\right) \\ \\ &=\tfrac{1}{2}ab\left(\sin(x)\cos(y) + \sin(y)\cos(x)\right) \end{align*}

Il ne reste qu’à diviser les deux côtés par \(\frac{1}{2} ab\) pour obtenir \[\sin(x+y) = \sin(x)\cos(y) + \sin(y)\cos(x)\]