géométrie projective, exemple de calculs(2)

\bullet Soit D une droite affine d’équation f et soient a, b deux points distincts du plan affine. On suppose que D coupe (ab) en m. On a les propriétés suivantes :

1) On a m =\dfrac{f(b)a - f(a)b}{f(b) - f(a)}.

2) On a la formule :\dfrac{\overline{ma}}{\overline{mb}}=\dfrac{f(a)T(b)}{f(b)T(a)}

\bullet \underline{\text{Théorème}\; \text{de}\; \text{Thalès}}

Soient A,B,C trois droites parallèles. Deux droites D,D' coupent respectivement A,B,C en a, b, c ; a', b', c'.

On a la formule : \dfrac{\overline{ab}}{\overline{ac}}=\dfrac{\overline{a'b'}}{\overline{a'c'}}

En effet, soit f une équation de A, on a:
\dfrac{\overline{ab}}{\overline{ac}}=\dfrac{f(b)T(c)}{f(c)T(b)} et \dfrac{\overline{a'b'}}{\overline{a'c'}} =\dfrac{f(b')T(c')}{f(c')T(b')} . Mais, commeB = (bb') est parallèle à A on a f(b)T(b') =f(b')T(b) et de même, f(c)T(c') = f(c')T(b).
Le résultat s’ensuit.

Exemple, situation 1, (T=x+y+z).

Rendered by QuickLaTeX.com

A : 3x+y-z=0

B : 2x-y-4z=0

C : x-y-3z=0

Commençons par montrer que A et B sont bien parallèles, en calculant T(f \wedge g) = 0, avec f et g équations respectives de A et B.

T(f \wedge g) =

T(1\times(-4)-(-1)\times(-1),-(3\times(-4)-2\times(-1)), 3\times(-1)-2\times1)

=T(-5,10,-5)=T(1,-2,1) (ici on normalise les coordonnées du point obtenu, z=1)

=1-2-1=0 !

Idem pour les droites A et C.

Les points a , b , a' et c' ont pour coordonnées respectives (0,1,1) , (2,0,1) , (-1,4,1) , (4,1,1)

On obtient les coordonnée de c et b' grâce à la formule c =\dfrac{f(b)a - f(a)b}{f(b) - f(a)} , où f est ici l’équation de C.

Soit c =\dfrac{-a +4b}{3}= (\frac{8}{3},-\frac{1}{3},1), on obtiendra de même, b'= (\frac{37}{13},-\frac{22}{13},1)

On obtient alors bien, avec f équation de A,

\dfrac{\overline{ab}}{\overline{ac}}=\dfrac{f(b)T(c)}{f(c)T(b)}=\dfrac{5\times \frac{10}{3}}{\frac{20}{3}\times 3}=\dfrac{5}{6}

et \dfrac{\overline{a'b'}}{\overline{a'c'}}=\dfrac{f(b')T(c')}{f(c')T(b')}=\dfrac{\frac{120}{13}\times 6}{12\times \frac{72}{13}}=\dfrac{5}{6} !

\bullet On suppose E_{\infty} muni d’une base e_1 , e_2. Soit abc un triangle de X. La mesure de l’aire algébrique du triangle abc relativement à cette base est la moitié du déterminant des vecteurs \overrightarrow{ab} et \overrightarrow{ac} sur la base e_1 , e_2 ou plus simplement:

\matbb{A}(abc) =\dfrac{[a,b,c]}{2 T(a)T(b)T(c) }

Exemple, situation1 , (T: x+y+z=0)

Soient a(0,1,1) , b(2,0,1) et c(3,3,1)

\overrightarrow{ab}=\dfrac{b}{T(b)}- \dfrac{a}{T(a)} =(\dfrac{2}{3},\dfrac{-1}{2},\dfrac{-1}{6})

\overrightarrow{ac}=\dfrac{c}{T(c)}- \dfrac{a}{T(a)} =(\dfrac{3}{7},\dfrac{-1}{14},\dfrac{-5}{14})

Prenons e_1(1,0,-1) et e_2(0,1,-1) comme base de E_{\infty}, les vecteurs ont alors pour coordonnées respectives (  \dfrac{2}{3}  ,-\dfrac{1}{2} ) et (  \dfrac{3}{7}  ,-\dfrac{1}{14} ) .

Avec la première formule : \matbb{A}(abc) =\dfrac{\begin{bmatrix}  \frac{2}{3}  &  \frac{3}{7} \\ -\frac{1}{2} &  -\frac{1}{14} \end{bmatrix} }{2}=\dfrac{7}{84}

et la deuxième : \matbb{A}(abc) =\dfrac{\begin{bmatrix}  0  &  2&3 \\ 1&0&3  \\ 1&1&1 \end{bmatrix} }{2\times 2\times 3\times 7}=\dfrac{7}{84} !

D’autres formules fondamentales :

f\wedge (a \wedge  b) =f(b)a-f(a)b

a \wedge (f \wedge g) =g(a)f-f(a)g

[a,b,f \wedge g] =f(a)g(b)-f(b)g(a) = [f,g,a \wedge b]

[a  \wedge a',b  \wedge b',c  \wedge c'] = [a,a',c][b,b',c']-[a,a',c'][b,b',c]

…ou encore [a,b,c]d = [b,c,d]a + [c,a,d]b + [a,b,d]c

\bulletSi maintenant on munit E^* d’une forme quadratique dégénérée telle que T est une base du noyau de cette forme. Cela
permet définir sur le plan affine X,
et sans aucune donnée supplémentaire, les notions de parallélisme et d’orthogonalitée, ainsi que la forme q sur E_{\infty} en posant q(n_f)=q^*(f) et la forme polaire \phi de q appelée produit scalaire.
On définit alors avec des points normalisés q(\overrightarrow{ab})=q(b-a)
et \phi(\overrightarrow{ab},\overrightarrow{ac})= \phi(b-a,c-a)

On parlera de \underline{ \text{plan}\; \text{affine}\; \text{euclidien}} lorsqu’on est dans cette situation.

\sqrt{q(\overrightarrow{ab})} définit alors une distance.

Soient \overline{A} et \overline{B} deux droites affines, l’invariant I^*(\overline{A},\overline{B})=\dfrac{\phi^*(A,B)^2}{q^*(A)q^*(B)} permet de définir la notion d’\underline{\text{angle}\; \text{non} \;\text{orienté}\; \text{de}\; \text{droites}}.

Si A, A’ sont parallèles et si B est sécante à A et A’, on a I^*(\overline{A},\overline{B}) =I^*(\overline{A'},\overline{B}). C’est l’égalité des angles « alternes internes »

Un autre invariant de deux droites \dfrac{[A,B,l]^2}{q^*(A)q^*(B)} correspond lui au sinus dans le cas réel et la formule de Lagrange: q^*(A)q^*(B)=\phi^*(A,B)^2+\delta(q^*)[A,B,l]^2 est la traduction de \cos^2\theta+\sin^2\theta=1.

La formule des aires :b \wedge c + c \wedge a + a \wedge b = [a, b, c]l permet alors d’obtenir la formule d’Al Kashi:
q(\overrightarrow{bc})=q(\overrightarrow{ab})+q(\overrightarrow{ac})-2\phi(\overrightarrow{ab},\overrightarrow{ac}), dont découle le théorème de Pythagore.

La quantité I^+([ab),[ac))=\dfrac{\phi(\overrightarrow{ab},\overrightarrow{ac})}{\sqrt{q(\overrightarrow{ab})q(\overrightarrow{ac})}} permet elle de définir la notion d’\underline{\text{angle}\; \text{non}\; \text{orienté}\; \text{de}\; \text{demi-droites}} : \widehat{bac}=ArccosI^+([ab),[ac))

On retrouve alors la formule familière: \phi(\overrightarrow{ab},\overrightarrow{ac})=ab\; ac\; cos\;\widehat{bac}

L’aire algébrique de abc, \dfrac{1}{2}\dfrac{[a,b,c]}{l(a)l(b)l(c)}, s’écrit alors: \dfrac{1}{2}\;ab\;ac\;\sin\;\widehat{bac}.

Angles orientés de vecteurs:

Rappelons que le groupe O^+(q) est isomorphe au groupe \mathbb{R} / 2 \pi \mathbb{Z} par l’application qui à \overline{\theta} associe la rotation vectorielle d’angle \overline{\theta}, notée \rho(\overline{\theta}).

Soient v et w deux vecteurs unitaires de E_{\infty}. On appelle angle des vecteurs v et w et on note
(v,w) l’unique élément \overline{\theta} \in \mathbb{R} / 2 \pi \mathbb{Z} tel que \rho(\overline{\theta})(v) = w.

On a alors les propriétés suivantes :

1) (u,w) = (u, v) + (v,w) (relation de Chasles).

2) (u,v) = (u', v') \Leftrightarrow (u, u') = (v, v') (règle du parallélogramme).

La relation de Chasles vaut pour des angles de sommets différents et cette propriété est constitutive de la géométrie euclidienne.

On appelle angle orienté des droites (ab) et (ac) et on note ((ab),(ac)) la classe modulo \pi de l’angle de vecteurs (\overrightarrow{ab},\overrightarrow{ac}).

Soient A,B deux équations de droites normalisées et soient a =l\wedge A et b =l\wedge B les vecteurs unitaires associés. On a les formules cos(a, b) =\phi^*(A,B) et sin(a, b) =[A,B, l].

la formule suivante:

[A;B; l][A;C; l] =\left|\begin{array}{cl}q^*(A)& \phi^*(A,C)\\\phi^*(B,A)&\phi^*(B,C)\\\end{array}\right|.

correspond alors à la relation: \cos(\beta +\gamma) = \cos \beta \cos\gamma - \sin \beta \sin \gamma

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    %3′ lük Etil Askorbik Asit kullan?lm??t?r.
    Lipozomal ta??y?c? sistemler, hedef noktaya aktif maddeyi ta??mak için geli?tirilen üst düzey teknolojidir.
    Ürün içindeki aktif madde (C V?TAM?N?) cilt alt?na indikten sonra çözülür ve daha k?sa sürede maksimum etki gösterir.

    Bu sayede hem uzun raf ömrü boyunca stabilitesini kaybetmeden etkisini sürdürür, hem de etkisi di?er formlara göre çok daha ba?ar?l?d?r.

    c vitamini serumu

  25. ???? ???????? ???? ?????????. ??????? ???????? ????????? ??????, ??? ?????? ????? ??????, ????? ???? …

    ?????
    30.09.2019 ?.
    ??????? ?? ????, ??? ?????????? ? ????? ?????? ?? ???????? ???????? ?? ??? 100%

    ??????
    29.09.2019 ?.
    ??? ????? ???????????: ???????, ??????, ????? ?????????? ??????????, ??????? ?? ????? ??? ??????? + ??????? …

    ???????? ???????, ??-? ???????????, 135
    21.04.2019 ?.

    elfa ??????? ???????? ????????

  26. YA?LANMA BEL?RT?LER?NE KAR?I ETK?S? KANITLANMI? ?ÇER?K

    HC Age Defense Krem, cildi en yayg?n ya?lanma belirtilerine kar?? korur ve
    zaman?n ciltteki y?prat?c? etkisini görünür ?ekilde azalt?r.
    Cildin parlakl???n?, s?k?l???n? ve esnekli?ini art?r?rken uzun süren nemlendirme sa?lamay? ve cilt bariyerini güçlendirmeyi destekler.
    K?r???kl?k kar??t? ve yenileyici peptitler, antioksidanlar, vitaminler ve bitki özleri ile formüle edilmi?tir.

    Yeniden dirili? bitkisi, tek bir damla su olmadan uzun süre hayatta kalabilmek için e?siz ve büyüleyici bir kabiliyete sahiptir.
    Bu süre Sahra Çölü’nde 100 y?l? bulabilir.

    Bu süre zarf?nda dirili? bitkisi tamamen kurur ve ya?ad???na dair herhangi bir belirti göremezsiniz.
    Bitkinin yapraklar? küçülür, solmu? ve ölü görünür.
    Ancak görünü?ler aldat?c?d?r; her seferinde ya?murlar ba?lar ba?lamaz bitki ad?na yak???r ?ekilde
    yeniden canlan?r. Ya?murlar durduktan sonra, tekrar kurur ve bir sonraki dirili? mucizesini bekler.

    HC Ya?lanma kar??t? krem ürünlerde kullan?lan Glycoin® natural,
    yeniden dirili? bitkisinin büyüleyici varl???n? ve hayatta kalma stratejisini
    sa?layan yap? ta??d?r. Tam kuruma s?ras?nda, Glycoin® natural
    bitkinin yap?s?n? hasara kar?? korur.
    Ya?murlar geldi?inde bir enerji yükseltici görevi görür ve
    bitkinin yeniden ye?illenmesini etkinle?tirir.

    ya?lanma kar??t? krem