{"id":631,"date":"2018-09-06T09:25:52","date_gmt":"2018-09-06T07:25:52","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=631"},"modified":"2021-02-04T17:51:26","modified_gmt":"2021-02-04T16:51:26","slug":"aire-entre-trois-cercles-tangents","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/06\/aire-entre-trois-cercles-tangents\/","title":{"rendered":"Aire entre trois cercles tangents"},"content":{"rendered":"\n<p>Cet article a pour objectifs de construire trois cercles tangents de rayons diff\u00e9rents et de calculer l&#8217;aire du domaine compris entre ces trois cercles.<\/p>\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\/06\/aire-entre-trois-cercles-tangents\/#Construction\" >Construction<\/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\/06\/aire-entre-trois-cercles-tangents\/#Aire_du_domaine_entre_les_trois_cercles\" >Aire du domaine entre les trois cercles<\/a><ul class='ez-toc-list-level-2' ><li class='ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/06\/aire-entre-trois-cercles-tangents\/#Aire_de_O_1O_2O_3\" >Aire de \\(O_1O_2O_3\\)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/06\/aire-entre-trois-cercles-tangents\/#Aire_du_secteur_angulaire_de_centre_O_1\" >Aire du secteur angulaire de centre \\(O_1\\)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/06\/aire-entre-trois-cercles-tangents\/#Aire_des_secteurs_angulaires_de_centres_O_2_et_O_3\" >Aire des secteurs angulaires de centres \\(O_2\\) et \\(O_3\\)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/06\/aire-entre-trois-cercles-tangents\/#Aire_du_secteur_hachure\" >Aire du secteur hachur\u00e9<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-1'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.mathweb.fr\/euclide\/2018\/09\/06\/aire-entre-trois-cercles-tangents\/#Application\" >Application<\/a><\/li><\/ul><\/nav><\/div>\n<h1 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Construction\"><\/span>Construction<span class=\"ez-toc-section-end\"><\/span><\/h1>\n\n\n\n<p>Rappelons que deux cercles tangents sont deux cercles qui se coupent en un unique point. Ainsi, trois cercles \\(\\mathcal{C}_1,\\ \\mathcal{C}_2\\) et \\(\\mathcal{C}_3\\) tangents sont tels que \\(\\mathcal{C}_i\\cap\\mathcal{C}_j=\\{M_i\\}\\), avec \\(i\\neq j\\), <em>i<\/em> = 1, 2 ou 3 et <em>j<\/em> = 1, 2 ou 3.<\/p>\n\n\n\n<p>Notons respectivement \\(r_i\\) et \\(O_i\\)&nbsp; le rayon et le centre de \\(\\mathcal{C}_i\\),&nbsp;<em>i<\/em> = 1, 2 ou 3.<\/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\/cercles-tangents-01.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"211\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-01-300x211.png\" alt=\"\" class=\"wp-image-632\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-01-300x211.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-01.png 389w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Construire \\(\\mathcal{C}_1\\) et \\(\\mathcal{C}_2\\) est ais\u00e9 : il suffit de tracer le segment \\([O_1O_2]\\) de longueur \\(r_1+r_2\\).Nous cherchons donc \u00e0 construire \\(\\mathcal{C}_3\\), de rayon \\(r_3\\) de sorte qu&#8217;il soit tangent aux deux cercles. On doit alors avoir \\(O_1O_3=r_1+r_3\\) et \\(O_2O_3=r_2+r_3\\).<\/p>\n\n\n\n<p>On peut ainsi construire le cercle de centre \\(O_1\\) et de rayon \\(r_1+r_3\\), puis le cercle de centre \\(O_2\\) de rayon \\(r_2+r_3\\). Ces deux cercles sont s\u00e9cants en un point qui est \\(O_3\\).<\/p>\n\n\n\n<p>Pour construire \\(\\mathcal{C}_3\\), il suffit de tracer le cercle de centre \\(O_3\\) passant par le point d&#8217;intersection de \\([O_1O_3]\\) et de \\(\\mathcal{C}_1\\) (par exemple).<\/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\/cercles-tangents-02.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"264\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-02-264x300.png\" alt=\"\" class=\"wp-image-633\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-02-264x300.png 264w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-02-300x341.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-02.png 524w\" sizes=\"auto, (max-width: 264px) 100vw, 264px\" \/><\/a><\/figure><\/div>\n\n\n\n<h1 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aire_du_domaine_entre_les_trois_cercles\"><\/span>Aire du domaine entre les trois cercles<span class=\"ez-toc-section-end\"><\/span><\/h1>\n\n\n\n<p>Ce qui nous int\u00e9resse est l&#8217;aire du domaine hachur\u00e9 repr\u00e9sent\u00e9 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\/cercles-tangents-03.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"269\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-03-269x300.png\" alt=\"\" class=\"wp-image-634\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-03-269x300.png 269w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-03-300x335.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/cercles-tangents-03.png 529w\" sizes=\"auto, (max-width: 269px) 100vw, 269px\" \/><\/a><\/figure><\/div>\n\n\n\n<p>Th\u00e9oriquement, son aire est obtenue en soustrayant \u00e0 l&#8217;aire du triangle \\(O_1O_2O_3\\) la somme des aires des secteurs angulaires de centres \\(O_1,\\ O_2\\) et \\(O_3\\).<\/p>\n\n\n\n<p>\u00c0 ce stade, nous ne connaissons que \\(O_1O_2,\\ O_2O_3\\) et \\(O_1O_3\\).<\/p>\n\n\n\n<p>\u00c0 l&#8217;aide de la formule de H\u00e9ron, on peut d\u00e9terminer l&#8217;aire du triangle \\(O_1O_2O_3\\) et \u00e0 l&#8217;aide de la formule d&#8217;Al-Kashi, nous pouvons trouver la mesure des angles \\(\\widehat{O_2O_1O_3},\\ \\widehat{O_1O_3O_2}\\) et \\(\\widehat{O_3O_2O_1}\\).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aire_de_O_1O_2O_3\"><\/span><strong>Aire de<\/strong> \\(O_1O_2O_3\\)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p>\\(\\mathcal{A}_T=\\sqrt{p(p-a)(p-b)(p-c)}\\), avec \\(a=O_1O_2,\\ b=O_2O_3,\\ c=O_3O_1,\\ p=\\frac{1}{2}(a+b+c)\\).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aire_du_secteur_angulaire_de_centre_O_1\"><\/span>Aire du secteur angulaire de centre \\(O_1\\)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p>Je note \\(\\alpha_1=\\widehat{O_1}\\) dans le triangle \\(O_1O_2O_3\\).<\/p>\n\n\n\n<p>Avec les notations pr\u00e9c\u00e9dentes, on a :&nbsp;\\[&nbsp;b^2=a^2+c^2-2ac\\cos\\alpha_1\\;,&nbsp;\\]<\/p>\n\n\n\n<p>soit :&nbsp;\\[&nbsp;\\alpha_1=\\arccos\\left(\\frac{a^2+c^2-b^2}{2ac}\\right).&nbsp;\\]<\/p>\n\n\n\n<p>Ainsi, l&#8217;aire du secteur angulaire correspondant est \u00e9gale \u00e0 :&nbsp;\\[&nbsp;\\mathcal{A}_1=\\frac{1}{2}\\alpha_1r_1^2=\\frac{1}{2}\\arccos\\left(\\frac{a^2+c^2-b^2}{2ac}\\right)r_1^2.&nbsp;\\]<\/p>\n\n\n\n<p><em>(Je conviens ici d&#8217;exprimer les angles en radians)<\/em><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aire_des_secteurs_angulaires_de_centres_O_2_et_O_3\"><\/span>Aire des secteurs angulaires de centres \\(O_2\\) et \\(O_3\\)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p>En notant \\(\\alpha_2=\\widehat{O_2}\\) et \\(\\alpha_3=\\widehat{O_3}\\) dans le triangle \\(O_1O_2O_3\\), et en utilisant ce que nous venons de faire pr\u00e9c\u00e9demment, on a :&nbsp;\\[\\alpha_2=\\arccos\\left(\\frac{a^2+b^2-c^2}{2ab}\\right)\\qquad\\text{et}\\qquad\\alpha_3=\\arccos\\left(\\frac{b^2+c^2-a^2}{2bc}\\right).\\]<\/p>\n\n\n\n<p>Ainsi, l&#8217;aire des secteurs angulaires correspondant respectivement \u00e0 \\(\\alpha_2\\) et \\(\\alpha_3\\) sont \u00e9gales \u00e0 :&nbsp;\\[&nbsp;\\mathcal{A}_2=\\frac{1}{2}\\arccos\\left(\\frac{a^2+b^2-c^2}{2ab}\\right)r_2^2\\quad\\text{et}\\quad\\mathcal{A}_3=\\frac{1}{2}\\arccos\\left(\\frac{b^2+c^2-a^2}{2bc}\\right)r_3^2.&nbsp;\\]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aire_du_secteur_hachure\"><\/span>Aire du secteur hachur\u00e9<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p>\\[&nbsp;\\begin{align*}&nbsp;\\begin{split}&nbsp;\\mathcal{A} &amp; = \\mathcal{A}_T-\\mathcal{A}_1-\\mathcal{A}_2-\\mathcal{A}_3\\\\&nbsp;&amp; = \\sqrt{p(p-a)(p-b)(p-c)}-\\frac{1}{2}\\left[\\arccos\\left(\\frac{a^2+c^2-b^2}{2ac}\\right)r_1^2\\right.\\\\ &amp; \\left.+\\arccos\\left(\\frac{a^2+b^2-c^2}{2ab}\\right)r_2^2+\\arccos\\left(\\frac{b^2+c^2-a^2}{2bc}\\right)r_3^2\\right].&nbsp;\\end{split}\\end{align*} \\]<\/p>\n\n\n\n<h1 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Application\"><\/span>Application<span class=\"ez-toc-section-end\"><\/span><\/h1>\n\n\n\n<p>Sur le sch\u00e9ma fait en d\u00e9but de section 3, j&#8217;ai pris \\(r_k=k\\) cm. On a alors : \\[ \\begin{align*} a&amp;=1+2=3\\\\ b&amp;=2+3=5\\\\ c&amp;=1+3=4\\\\p&amp;=\\frac{1}{2} (3+5+4)=6\\end{align*}\\]<\/p>\n\n\n\n<p>ce qui donne : \\[&nbsp;\\begin{align*}&nbsp;\\mathcal{A}&amp;=\\sqrt{6\\times3\\times1\\times2}-\\frac{1}{2}\\left(1^2\\times\\arccos\\frac{9+16-25}{2\\times3\\times4}\\right.\\\\ &amp; \\left.\\qquad+2^2\\times\\arccos\\frac{9+25-16}{2\\times3\\times5}+3^2\\times\\arccos\\frac{25+16-9}{2\\times5\\times4}\\right)\\\\&nbsp;\\mathcal{A}&amp; = 6-\\frac{1}{2}\\left(\\arccos 0+4\\arccos\\frac{3}{5}+9\\arccos\\frac{4}{5}\\right)\\\\&nbsp;\\mathcal{A} &amp; \\approx 0,464~\\text{cm}^2.&nbsp;\\end{align*} \\]<\/p>\n\n\n\n<div class=\"wp-block-file aligncenter\"><a  href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Aire-entre-trois-cercles-tangents.pdf\" data-fancybox data-type=\"iframe\" data-width=\"90%\" data-height=\"100%\" data-preload=\"false\">Aire entre trois cercles tangents<\/a><a  href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Aire-entre-trois-cercles-tangents.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 le fichier source \\(\\LaTeX\\) du document PDF:<\/p>\n\n\n\n<div class=\"wp-block-file aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Aire-entre-3-cercles-tangents.zip\">Aire entre 3 cercles tangents<\/a><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2018\/09\/Aire-entre-3-cercles-tangents.zip\" class=\"wp-block-file__button\" download>T\u00e9l\u00e9charger<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Cet article a pour objectifs de construire trois cercles tangents de rayons diff\u00e9rents et de calculer l&#8217;aire du domaine compris entre ces trois cercles.<\/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":[54,49,60,61],"class_list":["post-631","post","type-post","status-publish","format-standard","hentry","category-mathematiques","tag-aire","tag-al-kashi","tag-cercle","tag-formule-de-heron"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.5 - 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