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<h1 class="firstHeading">Utilisateur:Rungaldier/monobook.css</h1>
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<h3 id="siteSub">Un article de Wikipédia, l'encyclopédie libre.</h3>
<div id="contentSub"><span class="subpages">< <a href="../../../../articles/r/u/n/Utilisateur%7ERungaldier_121d.html" title="Utilisateur:Rungaldier">Utilisateur:Rungaldier</a></span></div>
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</p><div dir="ltr"><pre class="source-css">Commençons par remarquer qu<span class="st0">'étant donné trois droites issues d'</span>un point,
il n<span class="st0">'existe qu'</span>une seule droite formant avec celles-là un faisceau
harmonique.
Supposons en effet les trois droites issues de <math>O</math>; une sécante <math>\mathcal D</math>
quelconque coupe ces droites en trois points <math>M_1,M_2,M_3</math>. Il existe sur
<math>\mathcal D</math> un unique point <math>M<span class="st0">'</math> tel que <math>M_1,M_2,M'</span>,M_4</math> soit en division
harmonique. La droite <math><span class="br0">(</span>OM<span class="st0">')</math> est alors la seule droite qui convient.
Notons <math>[O|A_1,A_2,A_3,A_4]</math> le faisceau des droites
<math>(OA_1),(OA_2),(OA_3),(OA_4)</math> (Les points <math>A_1,A_2,A_3,A_4</math> n'</span>étant pas
forcément alignés<span class="br0">)</span>.
<table><td>
Soit <math>I</math> le point d<span class="st0">'intersection des diagonales <math>(AC)</math> et <math>(BD)</math>. Sur la
droite <math>(OI)</math>, soit <math>M</math> l'</span>unique point tel que le faisceau
<math><span class="br0">[</span>O<span class="st0">'|O,C,M,D]</math> soit harmonique. Posons <math>E=(O'</span>M<span class="br0">)</span>\cap<span class="br0">(</span>CD<span class="br0">)</span></math> et
<math>E<span class="st0">'=(O'</span>M<span class="br0">)</span>\cap<span class="br0">(</span>AB<span class="br0">)</span></math>.
On a <math><span class="br0">[</span>O<span class="st0">'|O,C,M,D]=[O'</span>|O,C,E,D<span class="br0">]</span></math>, de sorte que le
faisceau <math><span class="br0">[</span>I|O,C,E,D<span class="br0">]</span></math> est harmonique.
Pour une raison analogue, il en
est de même de <math><span class="br0">[</span>I|O,B,E<span class="st0">',A]</math>.
Mais comme <math>(IA)=(IC)</math> et que <math>(IB)=(ID)</math>,
on a <math>[I|O,C,E,D]=[I|O,A,E,B]</math> de sorte que les deux faisceaux
<math>[I|O,B,E'</span>,A<span class="br0">]</span></math> et <math><span class="br0">[</span>I|O,A,E,B<span class="br0">]</span></math> sont tous deux harmoniques et possèdent
trois droites communes. En vertu de la propriété d<span class="st0">'unicité, ces deux
faisceaux sont identiques et par conséquent <math>(IE)=(IE'</span><span class="br0">)</span></math>.
Ainsi
<math>I=<span class="br0">(</span>EE<span class="st0">')<span class="es0">\c</span>ap(OI)=M</math> par définition de <math>M</math>.</td>
<td>[[Image:FaisceauHarmonique.svg]]</td>
</table>
Soit <math>Y=<span class="es0">\l</span>ambda X</math>, <math>Y=<span class="es0">\m</span>u X</math> deux droites issues de <math>O</math>. <math>A=(a,0)</math> un point
de l'</span>axe des <math>x</math>; <math>Y=\alpha<span class="br0">(</span>X-a<span class="br0">)</span></math> et <math>Y=\beta<span class="br0">(</span>X-a<span class="br0">)</span></math> deux droites issues de <math>A</math>.
On note <math>M_i<span class="br0">(</span>x_i,y_i<span class="br0">)</span></math> les quatre points d<span class="st0">'intersection.
On calcule facilement <math> x_1=<span class="es0">\f</span>rac{a<span class="es0">\a</span>lpha}{ <span class="es0">\a</span>lpha-<span class="es0">\l</span>ambda}</math> d'</span>où l<span class="st0">'on tire par
permutation :<math><span class="es0">\q</span>uad x_2=<span class="es0">\f</span>rac{a<span class="es0">\b</span>eta}{ <span class="es0">\b</span>eta-<span class="es0">\l</span>ambda},<span class="es0">\q</span>uad x_3=<span class="es0">\f</span>rac{a<span class="es0">\b</span>eta}{ <span class="es0">\b</span>eta-<span class="es0">\m</span>u},<span class="es0">\q</span>uad
x_4=<span class="es0">\f</span>rac{a<span class="es0">\a</span>lpha}{<span class="es0">\a</span>lpha-<span class="es0">\m</span>u}.</math>
La droite <math>(M_1M_3)</math> a pour équation :
<math><span class="es0">\q</span>uad<span class="es0">\b</span>egin{vmatrix}X&x_1&x_3<span class="es0">\\</span>Y&<span class="es0">\l</span>ambda x_1&<span class="es0">\m</span>u x_3<span class="es0">\\</span>1&1&1<span class="es0">\e</span>nd{vmatrix}=
<span class="es0">\b</span>egin{vmatrix}X&a<span class="es0">\a</span>lpha&a<span class="es0">\b</span>eta<span class="es0">\\</span>Y&a<span class="es0">\l</span>ambda<span class="es0">\a</span>lpha&a<span class="es0">\m</span>u<span class="es0">\b</span>eta<span class="es0">\\</span>1&<span class="es0">\a</span>lpha-<span class="es0">\l</span>ambda&<span class="es0">\b</span>eta-<span class="es0">\m</span>u<span class="es0">\e</span>nd{vmatrix}=0.</math>
On en tire l'</span>abcisse <math>\omega</math> du point d<span class="st0">'intersection avec l'</span>axe <math>Ox</math> :
<math>\quad\omega=\frac<span class="br0">{</span>a^<span class="nu0">2</span>\alpha\beta<span class="br0">(</span>\lambda-\mu<span class="br0">)</span><span class="br0">}</span><span class="br0">{</span>\begin<span class="br0">{</span>vmatrix<span class="br0">}</span>a\lambda\alpha&a\mu\beta\\\alpha-\lambda&\beta-\mu\end<span class="br0">{</span>vmatrix<span class="br0">}</span><span class="br0">}</span>.</math>
Par permutation on déduit celle <math>\omega<span class="st0">'</math> de <math>(M_2M_4)<span class="es0">\c</span>ap(Ox)</math> :
<math><span class="es0">\q</span>uad<span class="es0">\o</span>mega'</span>=\frac<span class="br0">{</span>a^<span class="nu0">2</span>\alpha\beta<span class="br0">(</span>\lambda-\mu<span class="br0">)</span><span class="br0">}</span><span class="br0">{</span>\begin<span class="br0">{</span>vmatrix<span class="br0">}</span>a\lambda\beta&a\mu\alpha\\\beta-\lambda&\alpha-\mu\end<span class="br0">{</span>vmatrix<span class="br0">}</span><span class="br0">}</span>.</math>
Il en résulte
<math>\frac1\omega+\frac1<span class="br0">{</span>\omega<span class="st0">'}=<span class="es0">\f</span>rac1{a^2<span class="es0">\a</span>lpha<span class="es0">\b</span>eta(<span class="es0">\l</span>ambda-<span class="es0">\m</span>u)}<span class="es0">\l</span>eft(
<span class="es0">\b</span>egin{vmatrix}a<span class="es0">\l</span>ambda<span class="es0">\a</span>lpha&a<span class="es0">\m</span>u<span class="es0">\b</span>eta<span class="es0">\\</span><span class="es0">\a</span>lpha-<span class="es0">\l</span>ambda&<span class="es0">\b</span>eta-<span class="es0">\m</span>u<span class="es0">\e</span>nd{vmatrix}+
<span class="es0">\b</span>egin{vmatrix}a<span class="es0">\l</span>ambda<span class="es0">\b</span>eta&a<span class="es0">\m</span>u<span class="es0">\a</span>lpha<span class="es0">\\</span><span class="es0">\b</span>eta-<span class="es0">\l</span>ambda&<span class="es0">\a</span>lpha-<span class="es0">\m</span>u<span class="es0">\e</span>nd{vmatrix}<span class="es0">\r</span>ight)=<span class="es0">\f</span>rac2a</math>
après développement des déterminants.
Remarque : on aurait pu prendre <math>a=1</math> mais la moyenne harmonique aurait
été moins visible.</span></pre></div><div class="printfooter">
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