{"id":561,"date":"2018-09-02T15:09:13","date_gmt":"2018-09-02T13:09:13","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=561"},"modified":"2021-10-26T17:17:55","modified_gmt":"2021-10-26T15:17:55","slug":"le-theoreme-de-viviani","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/02\/le-theoreme-de-viviani\/","title":{"rendered":"Le th\u00e9or\u00e8me de Viviani"},"content":{"rendered":"\n<p>Le th\u00e9or\u00e8me de Viviani stipule que : &#8220;<em>dans un triangle \u00e9quilat\u00e9ral, la somme des distances d&#8217;un point int\u00e9rieur quelconque aux trois c\u00f4t\u00e9s est constante<\/em>.&#8221;<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-01.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"277\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-01-300x277.png\" alt=\"th\u00e9or\u00e8me de Viviani\" class=\"wp-image-562\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-01-300x277.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-01.png 396w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Autrement dit,&nbsp;quelle que soit la position du point M dans le triangle ABC, \\[ \\text{MS}+\\text{MQ}+\\text{MO} = \\text{constante}.\\]<\/p>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_83 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/02\/le-theoreme-de-viviani\/#Demonstration_du_theoreme_de_Viviani_avec_les_aires\" >D\u00e9monstration du th\u00e9or\u00e8me de Viviani avec les aires<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/02\/le-theoreme-de-viviani\/#Demonstration_avec_lanalyse\" >D\u00e9monstration avec l&#8217;analyse<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Demonstration_du_theoreme_de_Viviani_avec_les_aires\"><\/span>D\u00e9monstration du th\u00e9or\u00e8me de Viviani avec les aires<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-02.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"262\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-02-300x262.png\" alt=\"th\u00e9or\u00e8me de Viviani\" class=\"wp-image-563\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-02-300x262.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/viviani-02.png 396w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Posons \\(A_1\\) l&#8217;aire du triangle ABM ; alors,&nbsp;\\[ A_1=\\frac{1}{2}\\text{AB}\\times\\text{MO}.\\]<\/p>\n\n\n\n<p>Posons \\(A_2\\) l&#8217;aire du triangle ACM ; alors,&nbsp;\\[ A_2=\\frac{1}{2}\\text{AC}\\times\\text{MQ}.\\]<\/p>\n\n\n\n<p>Posons \\(A_3\\) l&#8217;aire du triangle CBM ; alors,&nbsp;\\[ A_3=\\frac{1}{2}\\text{BC}\\times\\text{MS}.\\]<\/p>\n\n\n\n<p>Posons \\(A_0\\) l&#8217;aire du triangle ABC ; alors,&nbsp;\\[ A_0=\\frac{1}{2}\\text{AC}\\times h\\]<\/p>\n\n\n\n<p>o\u00f9&nbsp;<em>h<\/em> est la hauteur de ABC issue de B. \u00c0 l&#8217;aide du th\u00e9or\u00e8me de Pythagore, on peut d\u00e9montrer que :&nbsp;\\[ h = \\frac{\\sqrt{3}}{2}\\text{AC}.\\]<\/p>\n\n\n\n<p>On en d\u00e9duit que :&nbsp;\\[ A_0=\\frac{\\sqrt{3}}{4}\\text{AC}^2.\\]<\/p>\n\n\n\n<p>Or,&nbsp;\\[ A_0=A_1+A_2+A_3\\]&nbsp; et\\[ \\text{AC} = \\text{AB} = \\text{BC}. \\]<\/p>\n\n\n\n<p>Donc :&nbsp;\\[ \\frac{\\sqrt{3}}{4}\\text{AC}^2=\\frac{1}{2}\\text{AC}\\left(\\text{MO}+\\text{MS}+\\text{MQ}\\right) \\]<\/p>\n\n\n\n<p>soit :&nbsp;\\[ \\text{MO}+\\text{MS}+\\text{MQ} = \\frac{\\sqrt{3}}{2}\\text{AC}.\\]<\/p>\n\n\n\n<p>On voit alors que la somme des 3 distances ne d\u00e9pend pas de la position de M.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Demonstration_avec_lanalyse\"><\/span>D\u00e9monstration avec l&#8217;analyse<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p>On rapporte le plan au rep\u00e8re \\(\\left(\\text{A}~;\\vec{\\text{AC}},\\vec{j}\\right)\\), o\u00f9 \\(\\vec{\\text{AC}}\\cdot\\vec{j}=0\\) et \\(\\|\\vec{\\text{AC}}\\|=\\|\\vec{j}\\|\\). Alors,&nbsp;\\[ \\text{A}(0~;0)\\quad ;\\quad \\text{C}(1~;0)\\quad ;\\quad \\text{B}\\left(\\frac{1}{2}~;\\frac{\\sqrt{3}}{2}\\right).\\]<\/p>\n\n\n\n<p>Ainsi,&nbsp;\\[ \\vec{\\text{AB}}\\left(\\begin{array}{c}\\frac{1}{2}\\\\ \\frac{\\sqrt{3}}{2}\\end{array}\\right),\\]<\/p>\n\n\n\n<p>d&#8217;o\u00f9 : \\[ (\\text{AB})\\;:\\; \\frac{\\sqrt{3}}{2}x-\\frac{1}{2}y=0,\\]<\/p>\n\n\n\n<p>Ou encore :&nbsp;\\[ (\\text{AB})\\;:\\; \\sqrt{3}x-y=0.\\]<\/p>\n\n\n\n<p>De plus, on a :&nbsp;\\[ \\vec{\\text{BC}}\\left(\\begin{array}{c}\\frac{1}{2}\\\\-\\frac{\\sqrt{3}}{2}\\end{array} \\right) ,\\]<\/p>\n\n\n\n<p>d&#8217;o\u00f9 :&nbsp;\\[ (\\text{BC})\\;:\\; -\\sqrt{3}x-y+y_0=0.\\]<\/p>\n\n\n\n<p>\\[\\text{C}\\in(\\text{BC}) \\Rightarrow y_0=\\sqrt{3},\\]<\/p>\n\n\n\n<p>donc :&nbsp;\\[ (\\text{BC})\\;:\\; -\\sqrt{3}x-y+\\sqrt{3}=0.\\]<\/p>\n\n\n\n<p>Posons alors \\(M(\\alpha~;\\beta)\\) dans ce rep\u00e8re. On sait que la distance du point M \u00e0 une droite (<em>d<\/em>) d&#8217;\u00e9quation <em>ax+by+c=0<\/em>&nbsp;est :&nbsp;\\[ \\text{d}(\\text{M};(d))=\\frac{\\vert a\\alpha + b\\beta + c\\vert}{\\sqrt{a^2+b^2}}.\\]<\/p>\n\n\n\n<p>D&#8217;o\u00f9 :&nbsp;\\[ \\text{d}(\\text{M};(\\text{AB}))=\\frac{\\vert \\alpha\\sqrt{3} &#8211; \\beta \\vert}{\\sqrt{\\sqrt{3}^2+(-1)^2}}=\\frac{\\vert \\alpha\\sqrt{3}-\\beta\\vert}{2} \\]&nbsp;et&nbsp;\\[ \\text{d}(\\text{M};(\\text{BC}))=\\frac{\\vert -\\alpha\\sqrt{3} &#8211; \\beta+\\sqrt{3} \\vert}{\\sqrt{(-\\sqrt{3}^2+(-1)^2}}=\\frac{\\vert -\\alpha\\sqrt{3} &#8211; \\beta+\\sqrt{3}\\vert}{2}. \\]<\/p>\n\n\n\n<p>De plus,&nbsp;\\[ \\text{d}(\\text{M};(\\text{AC}))=\\beta.\\]<\/p>\n\n\n\n<p>De plus, M est toujours au-dessous de (AB), d&#8217;\u00e9quation r\u00e9duite :&nbsp;\\[ y=x\\sqrt{3}.\\]<\/p>\n\n\n\n<p>Donc :&nbsp;\\[ \\beta \\leq\\alpha\\sqrt{3},\\]<\/p>\n\n\n\n<p>ce qui signifie que :&nbsp;\\[ \\alpha\\sqrt{3}-\\beta\\geq 0\\]<\/p>\n\n\n\n<p>et donc :&nbsp;\\[ \\text{d}(\\text{M};(\\text{AB}))=\\frac{\\alpha\\sqrt{3}-\\beta}{2}. \\]<\/p>\n\n\n\n<p>De m\u00eame, M est toujours au-dessous de (BC), d&#8217;\u00e9quation r\u00e9duite :&nbsp;\\[ y=-x\\sqrt{3}+\\sqrt{3},\\]<\/p>\n\n\n\n<p>d&#8217;o\u00f9 : \\[ \\beta\\leq-\\alpha\\sqrt{3}+\\sqrt{3},\\]<\/p>\n\n\n\n<p>ce qui signifie que : \\[ -\\alpha\\sqrt{3}-\\beta+\\sqrt{3}\\geq 0\\]<\/p>\n\n\n\n<p>et donc :<\/p>\n\n\n\n<p>\\[ \\text{d}(\\text{M};(\\text{BC}))=\\frac{-\\alpha\\sqrt{3} &#8211; \\beta+\\sqrt{3}}{2}. \\]<\/p>\n\n\n\n<p>Ainsi,<\/p>\n\n\n\n<p>\\[ \\text{d}(\\text{M};(\\text{BC})) + \\text{d}(\\text{M};(\\text{AC})) + \\text{d}(\\text{M};(\\text{AB}))=\\frac{\\alpha\\sqrt{3}-\\beta-\\alpha\\sqrt{3}-\\beta+\\sqrt{3}}{2}+\\beta,\\]<\/p>\n\n\n\n<p>soit :<\/p>\n\n\n\n<p>\\[ \\text{d}(\\text{M};(\\text{BC})) + \\text{d}(\\text{M};(\\text{AC})) + \\text{d}(\\text{M};(\\text{AB}))=\\frac{\\sqrt{3}}{2}.\\]<\/p>\n\n\n\n<p>La somme des 3 distances ne d\u00e9pend donc pas de la position du point M.<\/p>\n\n\n\n<div class=\"wp-block-file aligncenter um_article\"><a  href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Th\u00e9or\u00e8me-de-Viviani.pdf\" data-fancybox data-type=\"iframe\" data-width=\"90%\" data-height=\"100%\" data-preload=\"false\">Th\u00e9or\u00e8me de Viviani<\/a><a  href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Th\u00e9or\u00e8me-de-Viviani.pdf\" class=\"wp-block-file__button\" download data-fancybox data-type=\"iframe\" data-width=\"90%\" data-height=\"100%\" data-preload=\"false\">T\u00e9l\u00e9charger<\/a><\/div>\n\n\n\n<p>Voir les fichiers sources \\(\\LaTeX\\) du document PDF:<\/p>\n\n\n\n<div class=\"wp-block-file aligncenter um_article\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Th\u00e9or\u00e8me-de-Viviani.zip\">Th\u00e9or\u00e8me de Viviani<\/a><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Th\u00e9or\u00e8me-de-Viviani.zip\" class=\"wp-block-file__button\" download>T\u00e9l\u00e9charger<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Le th\u00e9or\u00e8me de Viviani stipule que : &#8220;dans un triangle \u00e9quilat\u00e9ral, la somme des distances d&#8217;un point int\u00e9rieur quelconque aux trois c\u00f4t\u00e9s est constante.&#8221; Autrement dit,&nbsp;quelle que soit la position du point M dans le triangle ABC, \\[ \\text{MS}+\\text{MQ}+\\text{MO} = \\text{constante}.\\]<\/p>\n","protected":false},"author":1,"featured_media":3702,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[25,52,53,46],"class_list":["post-561","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","tag-demonstration","tag-geometrie","tag-theoreme","tag-triangle"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.6 - 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