{"id":614,"date":"2018-09-04T14:40:07","date_gmt":"2018-09-04T12:40:07","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=614"},"modified":"2021-10-26T17:15:27","modified_gmt":"2021-10-26T15:15:27","slug":"triangle-orthique-et-probleme-de-fagnano","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/04\/triangle-orthique-et-probleme-de-fagnano\/","title":{"rendered":"Triangle orthique et probl\u00e8me de Fagnano"},"content":{"rendered":"\n<p>Giulio Fagnano \u00e9tait un math\u00e9maticien italien de la fin du XVIIe si\u00e8cle.<\/p>\n\n\n\n<p>Il a probablement \u00e9t\u00e9 le premier \u00e0 s&#8217;\u00eatre int\u00e9ress\u00e9 \u00e0 la th\u00e9orie des int\u00e9grales elliptiques, mais ce n&#8217;est pas l&#8217;objet de cet article.<\/p>\n\n\n\n<p>Le probl\u00e8me connu sous le nom de <em>probl\u00e8me de Fagnano<\/em>&nbsp;est le suivant :<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\"><p>Peut-on inscrire un triangle de p\u00e9rim\u00e8tre minimal dans un triangle acutangle ?<\/p><\/blockquote>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_82_2 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-1'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/04\/triangle-orthique-et-probleme-de-fagnano\/#Le_triangle_orthique\" >Le triangle orthique<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-1'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/04\/triangle-orthique-et-probleme-de-fagnano\/#La_solution_au_probleme_de_Fagnano\" >La solution au probl\u00e8me de Fagnano<\/a><\/li><\/ul><\/nav><\/div>\n<h1 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Le_triangle_orthique\"><\/span>Le triangle orthique<span class=\"ez-toc-section-end\"><\/span><\/h1>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-01.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"269\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-01-300x269.png\" alt=\"\" class=\"wp-image-615\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-01-300x269.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-01.png 510w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Le triangle orthique du triangle ABC est le triangle A&#8217;B&#8217;C&#8217;, dont les sommets sont les pieds des hauteurs de ABC.<\/p>\n\n\n\n<p>Maintenant, vous allez sans doute me demander le rapport entre le triangle orthique et le probl\u00e8me de Fagnano, n&#8217;est-ce pas ?<\/p>\n\n\n\n<p>Et bien, on peut d\u00e9montrer que le triangle solution au probl\u00e8me de Fagnano est le triangle orthique du triangle donn\u00e9.<\/p>\n\n\n\n<p>Voyons cela dans la section suivante&#8230;<\/p>\n\n\n\n<h1 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"La_solution_au_probleme_de_Fagnano\"><\/span>La solution au probl\u00e8me de Fagnano<span class=\"ez-toc-section-end\"><\/span><\/h1>\n\n\n\n<p>Consid\u00e9rons donc un triangle acutangle ABC.<\/p>\n\n\n\n<p><strong>Dans un premier temps, on fixe A&#8217;<\/strong> sur (BC), puis nous allons trouver les points B&#8217; et C&#8217;, respectivement sur (AC) et (AB) de sorte que le p\u00e9rim\u00e8tre de A&#8217;B&#8217;C&#8217; soit la plus petite possible.<\/p>\n\n\n\n<p>On va construire alors :<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>\\(A_1\\) le sym\u00e9trique de A&#8217; par rapport \u00e0 (AB) ;<\/li><li>\\(A_2\\) le sym\u00e9trique de A&#8217; par rapport \u00e0 (AC).<\/li><\/ul>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-02.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"181\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-02-300x181.png\" alt=\"\" class=\"wp-image-616\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-02-300x181.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-02-600x363.png 600w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-02.png 645w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>On a alors :&nbsp;\\[ AA_2=AA&#8217;=AA_1 \\qquad\\text{et}\\qquad \\left\\lbrace\\begin{array}{l}&nbsp;\\widehat{A&#8217;AB}=\\widehat{A_1AB}\\\\\\widehat{A&#8217;AC}=\\widehat{A_2AC}\\end{array}\\right.\\]<\/p>\n\n\n\n<p>car les triangles \\(AA&#8217;A_2\\) et \\(AA&#8217;A_1\\) sont isoc\u00e8les en A.<\/p>\n\n\n\n<p>Notons :&nbsp;\\[ \\gamma = \\widehat{BAC}\\quad\\text{exprim\u00e9 en degr\u00e9s}.\\]<\/p>\n\n\n\n<p>Alors,&nbsp;\\[ \\widehat{A_1AA_2}=2\\gamma.\\]<\/p>\n\n\n\n<p>Le triangle ABC \u00e9tant acutangle, \\(0&lt;\\gamma&lt;90\\) donc \\(0&lt;2\\gamma&lt;180\\). Ainsi, \\((A_1A_2)\\) coupe (AC) et (AB). Notons M et N les points d&#8217;intersection comme indiqu\u00e9s sur la figure ci-dessous :<\/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\/triangle-orthique-03.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"198\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-03-300x198.png\" alt=\"\" class=\"wp-image-617\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-03-300x198.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-03-600x396.png 600w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-03.png 634w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Les triangles \\(A&#8217;NA_2\\) et \\(A&#8217;MA_1\\) sont isoc\u00e8les respectivement en <em>N<\/em> et <em>M<\/em>. Donc,&nbsp;\\[ NA_2=NA&#8217;\\qquad\\text{et}\\qquad MA_1=MA&#8217;.\\]<\/p>\n\n\n\n<p>Ainsi, le p\u00e9rim\u00e8tre de <em>MNA&#8217;<\/em>&nbsp;est \u00e9gal \u00e0 \\(A_1A_2\\).<\/p>\n\n\n\n<p>Si on consid\u00e8re <em>M&#8217;<\/em>&nbsp;sur (AB) et <em>N&#8217;<\/em>&nbsp;sur (AC), respectivement distincts de <em>M<\/em>&nbsp;et <em>N<\/em>, alors le p\u00e9rim\u00e8tre de <em>M&#8217;A&#8217;N&#8217;<\/em>&nbsp;est \u00e9gal \u00e0 la longueur de la ligne bris\u00e9e \\(A_2N&#8217;A&#8217;M&#8217;A_1\\) pour les m\u00eames raisons que pr\u00e9c\u00e9demment.<\/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\/triangle-orthique-04.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"180\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-04-300x180.png\" alt=\"\" class=\"wp-image-618\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-04-300x180.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-04-600x359.png 600w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-04.png 755w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>C&#8217;est bien connu : le chemin le plus court est la ligne droite donc la ligne bris\u00e9e (bleue) est n\u00e9cessairement de longueur sup\u00e9rieure \u00e0 la longueur de \\([A_1A_2]\\).<\/p>\n\n\n\n<p><strong>Dans un deuxi\u00e8me temps, on cherche la position de <em>A&#8217;<\/em><\/strong>&nbsp;sur (BC).<\/p>\n\n\n\n<p>Pour minimiser le p\u00e9rim\u00e8tre du triangle <em>A&#8217;MN<\/em>, on doit donc trouver la position de <em>A&#8217;<\/em>&nbsp;sur (BC).<\/p>\n\n\n\n<p>Comme nous l&#8217;avons vu, ce p\u00e9rim\u00e8tre vaut \\(A_1A_2\\), o\u00f9 \\([A_1A_2]\\) est la base du triangle isoc\u00e8le \\(AA_1A_2\\) dont l&#8217;angle au sommet principal est constant et vaut toujours \\(2\\gamma\\).<\/p>\n\n\n\n<p>Pour minimiser \\(A_1A_2\\), il faut minimiser \\(AA_1\\), soit <em>AA&#8217;<\/em>&nbsp;car ces deux longueurs sont \u00e9gales.<\/p>\n\n\n\n<p>Or, la distance la plus courte entre un point (ici, <em>A<\/em>) et une droite (ici, (BC)) est la longueur du segment perpendiculaire \u00e0 la droite et passant par le point, donc ici la hauteur issue de <em>A<\/em>.<\/p>\n\n\n\n<p>Ainsi, <em>A&#8217;<\/em>&nbsp;est le pied de la hauteur du triangle <em>ABC<\/em> issue de <em>A<\/em>.<\/p>\n\n\n\n<p><strong>Dans un troisi\u00e8me temps, on montre que <em>M&nbsp;<\/em>= <em>C&#8217;<\/em> et <em>N&nbsp;<\/em>= <em>B&#8217;<\/em>,<\/strong>&nbsp;o\u00f9 <em>B&#8217;<\/em>&nbsp;et <em>C&#8217;<\/em>&nbsp;sont respectivement les pieds des hauteurs de <em>ABC<\/em>&nbsp;issues de <em>B<\/em>&nbsp;et <em>C<\/em>.<\/p>\n\n\n\n<p>Cette d\u00e9monstration repose sur le fait que : \\[&nbsp;\\widehat{BC&#8217;A&#8217;}=\\widehat{BCA}\\qquad\\text{et}\\qquad\\widehat{BCA}=\\widehat{AC&#8217;B&#8217;}.\\qquad(1) \\]<\/p>\n\n\n\n<p>Alors, comme \\(A&#8217;C&#8217;A_1\\) est isoc\u00e8le en <em>C&#8217;<\/em>,&nbsp;\\[ \\widehat{A_1C&#8217;B}=\\widehat{BC&#8217;A&#8217;}=\\widehat{BCA}=\\widehat{AC&#8217;B&#8217;}\\;,\\]<\/p>\n\n\n\n<p>ce qui signifie que <em>B&#8217;<\/em>, <em>C&#8217;<\/em>&nbsp;et \\(A_1\\) sont align\u00e9s.<\/p>\n\n\n\n<p>De la m\u00eame fa\u00e7on, on montre que <em>C&#8217;<\/em>, <em>B&#8217;<\/em>&nbsp;et \\(A_2\\) sont aussi align\u00e9s, ce qui finalement d\u00e9montre que <em>B&#8217;<\/em>&nbsp;et <em>C&#8217;<\/em>&nbsp;sont sur la droite \\(A_1A_2\\) et donc que <em>M&nbsp;<\/em>= <em>C&#8217;<\/em>&nbsp;et <em>N&nbsp;<\/em>= <em>B&#8217;<\/em>.<\/p>\n\n\n\n<p><strong>N.B.<\/strong>&nbsp;Les \u00e9galit\u00e9s (1) ne sont pas \u00e9videntes ; il faut donc les justifier.<\/p>\n\n\n\n<p>Les triangles <em>A&#8217;CA<\/em>&nbsp;et <em>C&#8217;CA<\/em>&nbsp;sont rectangles d&#8217;hypot\u00e9nuse [<em>AC<\/em>] ; donc [<em>AC<\/em>] est un diam\u00e8tre du cercle \\(\\mathcal{C}\\) passant par <em>A<\/em>, <em>C<\/em>, <em>A&#8217;<\/em>&nbsp;et <em>C&#8217;<\/em>&nbsp;comme l&#8217;illustre le sch\u00e9ma suivant :<\/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\/triangle-orthique-05.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"284\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-05-300x284.png\" alt=\"\" class=\"wp-image-619\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-05-300x284.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/triangle-orthique-05.png 543w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Les angles \\(\\widehat{CAA&#8217;}\\) et \\(\\widehat{CC&#8217;A&#8217;}\\) interceptent le m\u00eame arc \\(\\overparen{CA&#8217;}\\) et sont inscrits dans le m\u00eame cercle donc ils ont la m\u00eame mesure : \\[&nbsp;\\widehat{A&#8217;CA}=\\widehat{CC&#8217;A&#8217;}\\qquad(2)\\]<\/p>\n\n\n\n<p>De plus, le triangle <em>A&#8217;CA<\/em>&nbsp;est rectangle en <em>A&#8217;<\/em>&nbsp;donc :&nbsp;\\[ \\widehat{CAA&#8217;}+\\widehat{A&#8217;CA}=90^\\circ.\\]<\/p>\n\n\n\n<p>Le triangle <em>BCC&#8217;<\/em>&nbsp;est rectangle en <em>C&#8217;<\/em>&nbsp;donc :&nbsp;\\[ \\widehat{CC&#8217;A&#8217;}+\\widehat{A&#8217;C&#8217;B}=90^\\circ.\\]<\/p>\n\n\n\n<p>Ainsi,&nbsp;\\[ \\widehat{A&#8217;C&#8217;B}+\\widehat{CC&#8217;A&#8217;}=\\widehat{CAA&#8217;}+\\widehat{A&#8217;CA}\\]<\/p>\n\n\n\n<p>et donc, d&#8217;apr\u00e8s l&#8217;\u00e9galit\u00e9 (2)&nbsp; :&nbsp;\\[&nbsp;\\widehat{A&#8217;C&#8217;B}=\\widehat{A&#8217;CA} \\;,\\]<\/p>\n\n\n\n<p>ce qui d\u00e9montre la premi\u00e8re \u00e9galit\u00e9 des \u00e9galit\u00e9s (1).<\/p>\n\n\n\n<p>L&#8217;angle \\(\\widehat{CA&#8217;C&#8217;}\\) intercepte le petit arc \\(\\overparen{CC&#8217;}\\) et l&#8217;angle \\(\\widehat{CAC&#8217;}=\\widehat{BAC}\\) intercepte le grand angle \\(\\widehat{CC&#8217;}\\) ; ils sont donc suppl\u00e9mentaires.<\/p>\n\n\n\n<p>De mani\u00e8re analogue, on d\u00e9montre que \\(\\widehat{BA&#8217;B&#8217;}\\) et \\(\\widehat{BAB&#8217;}=\\widehat{BAC}\\) sont aussi suppl\u00e9mentaires (en consid\u00e9rant les triangles rectangles <em>AA&#8217;B<\/em>&nbsp;et <em>AB&#8217;B<\/em>).<\/p>\n\n\n\n<p>On obtient alors :&nbsp;\\[ \\widehat{BA&#8217;B&#8217;}=\\widehat{CA&#8217;C&#8217;}\\;,\\]<\/p>\n\n\n\n<p>ce qui montre que [<em>A&#8217;A<\/em>) est une bissectrice de l&#8217;angle \\(\\widehat{B&#8217;A&#8217;C&#8217;}\\).<\/p>\n\n\n\n<p>On peut d\u00e9montrer de la m\u00eame fa\u00e7on que [<em>C&#8217;C<\/em>) est une bissectrice de \\(\\widehat{A&#8217;C&#8217;B&#8217;}\\), et donc que :&nbsp;\\[ \\widehat{AC&#8217;B&#8217;}=\\widehat{BC&#8217;A&#8217;}\\;,\\]<\/p>\n\n\n\n<p>soit :&nbsp;\\[ \\widehat{BCA}=\\widehat{AC&#8217;B&#8217;}\\;,\\]<\/p>\n\n\n\n<p>ce qui d\u00e9montre la seconde \u00e9galit\u00e9 des \u00e9galit\u00e9s (1).<\/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\/Triangle-orthique-et-probleme-de-Fagnano.pdf\" data-fancybox data-type=\"iframe\" data-width=\"90%\" data-height=\"100%\" data-preload=\"false\">Triangle orthique et probleme de Fagnano<\/a><a  href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Triangle-orthique-et-probleme-de-Fagnano.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>Obtenir les 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\/Triangle-orthique-et-probl\u00e8me-de-Fagnano.zip\">Triangle orthique et probl\u00e8me de Fagnano<\/a><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Triangle-orthique-et-probl\u00e8me-de-Fagnano.zip\" class=\"wp-block-file__button\" download>T\u00e9l\u00e9charger<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Giulio Fagnano \u00e9tait un math\u00e9maticien italien de la fin du XVIIe si\u00e8cle. Il a probablement \u00e9t\u00e9 le premier \u00e0 s&#8217;\u00eatre int\u00e9ress\u00e9 \u00e0 la th\u00e9orie des int\u00e9grales elliptiques, mais ce n&#8217;est pas l&#8217;objet de cet article. Le probl\u00e8me connu sous le nom de probl\u00e8me de Fagnano&nbsp;est le suivant : Peut-on inscrire [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[59,25,46],"class_list":["post-614","post","type-post","status-publish","format-standard","hentry","category-mathematiques","tag-angles","tag-demonstration","tag-triangle"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.5 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Triangle orthique et probl\u00e8me de Fagnano - Mathweb.fr<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/04\/triangle-orthique-et-probleme-de-fagnano\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Triangle orthique et probl\u00e8me de Fagnano - Mathweb.fr\" \/>\n<meta property=\"og:description\" content=\"Giulio Fagnano \u00e9tait un math\u00e9maticien italien de la fin du XVIIe si\u00e8cle. Il a probablement \u00e9t\u00e9 le premier \u00e0 s&#8217;\u00eatre int\u00e9ress\u00e9 \u00e0 la th\u00e9orie des int\u00e9grales elliptiques, mais ce n&#8217;est pas l&#8217;objet de cet article. 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