{"id":11192,"date":"2025-08-04T11:27:30","date_gmt":"2025-08-04T09:27:30","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=11192"},"modified":"2025-08-10T11:09:05","modified_gmt":"2025-08-10T09:09:05","slug":"arbre-fractal-de-pythagore","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2025\/08\/04\/arbre-fractal-de-pythagore\/","title":{"rendered":"Arbre fractal de Pythagore"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Un arbre fractal de Pythagore est une figure plut\u00f4t esth\u00e9tique dans certaines conditions.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Nous allons voir ce que c&rsquo;est&#8230;<\/p>\n\n\n\n<!--more-->\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Je pr\u00e9cise que cet article est r\u00e9dig\u00e9 en s&rsquo;appuyant sur la vid\u00e9o de Micka\u00ebl Launay visible sur <a href=\"https:\/\/www.youtube.com\/watch?v=BCR7IhRFNDo\" target=\"_blank\" rel=\"noreferrer noopener\">Youtube<\/a>.<\/p>\n<\/blockquote>\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_86 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\/2025\/08\/04\/arbre-fractal-de-pythagore\/#Notion_de_similitude\" >Notion de similitude<\/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\/2025\/08\/04\/arbre-fractal-de-pythagore\/#Introduction_a_larbre_de_Pythagore\" >Introduction \u00e0 l&rsquo;arbre de Pythagore<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2025\/08\/04\/arbre-fractal-de-pythagore\/#Construction_de_larbre_de_Pythagore\" >Construction de l&rsquo;arbre de Pythagore<\/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\/2025\/08\/04\/arbre-fractal-de-pythagore\/#Larbre_est-il_%C2%AB_infini_%C2%BB\" >L&rsquo;arbre est-il \u00ab\u00a0infini\u00a0\u00bb ?<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Notion_de_similitude\"><\/span><strong>Notion de similitude<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Consid\u00e9rons:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>un point \\(\\Omega\\) du plan,<\/li>\n\n\n\n<li>un r\u00e9el \\(k\\),<\/li>\n\n\n\n<li>un ange $\\alpha$.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">\u00c0 tout point $A$ du plan, on peut consid\u00e9rer successivement:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>son image $A&rsquo;$ par la rotation de centre $\\Omega$ et d&rsquo;angle $\\alpha$,<\/li>\n\n\n\n<li>puis $A\u00a0\u00bb$ l&rsquo;image de $A&rsquo;$ par l&rsquo;homoth\u00e9tie de centre $\\Omega$ et de rapport $k$.<\/li>\n<\/ul>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/similitude.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"180\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/similitude-300x180.webp\" alt=\"\" class=\"wp-image-11196\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/similitude-300x180.webp 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/similitude.webp 425w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">\u00c0 l&rsquo;aide de la trigonom\u00e9trie, on d\u00e9montre que: $$\\left.\\begin{array}{l}\\Omega(x_\\Omega;y_\\Omega)\\\\A(a;b)\\end{array}\\right\\rbrace \\Longrightarrow\\begin{cases}A'(x_\\Omega+(a-x_\\Omega)\\cos\\alpha-(b-y_\\Omega)\\sin\\alpha; y_\\Omega+(a-x_\\Omega)\\sin\\alpha+(b-y_\\Omega)\\cos\\alpha)\\\\A\u00a0\u00bb(x_\\Omega + k(a-x_\\Omega)\\cos\\alpha-k(b-y_\\Omega)\\sin\\alpha;y_\\Omega+k(a-x_\\Omega)\\sin\\alpha+k(b-y_\\Omega)\\cos\\alpha)\\end{cases}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On dit ici que $A\u00a0\u00bb$ est obtenu \u00e0 partir du point $A$ par similitude de centre $\\Omega$, d&rsquo;angle $\\alpha$ et de rapport $k$.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Introduction_a_larbre_de_Pythagore\"><\/span><strong>Introduction \u00e0 l&rsquo;arbre de Pythagore<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Partons d&rsquo;un carr\u00e9 $ABCD$ de c\u00f4t\u00e9 $L$, puis tra\u00e7ons un triangle rectangle $ECD$ tel que $\\widehat{CDE}=\\theta$, sur lequel nous tra\u00e7ons des carr\u00e9s:<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre_pythagore_1.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"266\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre_pythagore_1-266x300.webp\" alt=\"arbre fractal de Pythagore: \u00e9tape 1\" class=\"wp-image-11199\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre_pythagore_1-266x300.webp 266w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre_pythagore_1.webp 739w\" sizes=\"auto, (max-width: 266px) 100vw, 266px\" \/><\/a><\/figure>\n<\/div>\n\n\n<ul class=\"wp-block-list\">\n<li>$E$ est obtenu par similitude de centre $D$, d&rsquo;angle $\\theta$ et de rapport $k=L\\cos\\theta$ car $\\dfrac{DE}{DC}=\\cos\\theta$.<\/li>\n\n\n\n<li>$F$ est obtenu \u00e0 partir de $D$ par rotation de centre $E$ et d&rsquo;angle $-\\frac{\\pi}{2}$.<\/li>\n\n\n\n<li>$G$ est obtenu \u00e0 partir de $E$ par rotation de centre $D$ et d&rsquo;angle $\\frac{\\pi}{2}$.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">On peut construire de m\u00eame le carr\u00e9 sur $|EC]$.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Construction_de_larbre_de_Pythagore\"><\/span><strong>Construction de l&rsquo;arbre de Pythagore<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">L&rsquo;id\u00e9e est de reproduire ce qui vient d&rsquo;\u00eatre expliqu\u00e9 sur les carr\u00e9s obtenus, et sur ceux qui vont l&rsquo;\u00eatre par la suite. Nous allons faire cela \u00e0 l&rsquo;aide de Python:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from math import pi, cos, sin\nfrom PIL import Image, ImageDraw \n\nwidth = 1748\nheight = 1240\n\ntree = Image.new('RGB', (width,height), (255,255,255))\ndraw = ImageDraw.Draw(tree,'RGBA')\n\ntheta = pi\/180 * 40 # 40 degr\u00e9s\nL = 200 # longueur de l'ar\u00eate du premier carr\u00e9\nA = (width\/2 - L\/2 , height-250)\nB = (width\/2 + L\/2 , height-250)\n\ndef sim(centre, point, angle, k):\n    # angle devient \"-angle\" dans les sinus \u00e0 cause de la logique graphique de PIL\n    x = centre[0] + k*(point[0]-centre[0])*cos(angle)-k*(point[1]-centre[1])*sin(-angle)\n    y = centre[1] + k*(point[0]-centre[0])*sin(-angle)+k*(point[1]-centre[1])*cos(angle)\n    \n    return x,y\n\ndef arbre(A,B,angle,n):\n    if n==0: return \n    \n    C = sim(B,A,-pi\/2,1)\n    D = sim(A,B,pi\/2,1)\n    E = sim(D,C,angle,cos(angle))\n    \n    draw.polygon([A,B,C,D], fill=(0,0,0,125))\n    \n    arbre(D, E, angle, n-1)\n    arbre(E, C, angle, n-1)\n    \n\narbre(A,B,theta,10)\ntree.show()<\/pre>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-2.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"207\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-2-300x207.webp\" alt=\"arbre fractal de Pythagore: version 1\" class=\"wp-image-11200\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-2-300x207.webp 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-2-1024x706.webp 1024w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-2-768x530.webp 768w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-2.webp 1269w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">Si l&rsquo;on souhaite mettre un peu de couleur, on peut utiliser le script suivant:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from math import pi, cos, sin\nfrom PIL import Image, ImageDraw \nfrom random import randint\n\nwidth = 1748\nheight = 1240\n\ntree = Image.new('RGB', (width,height), (255,255,255))\ndraw = ImageDraw.Draw(tree,'RGBA')\n\ntheta = pi\/180 * 40 # 40 degr\u00e9s\nL = 200 # longueur de l'ar\u00eate du premier carr\u00e9\nA = (width\/2 - L\/2 , height-250)\nB = (width\/2 + L\/2 , height-250)\n\ndef sim(centre, point, angle, k):\n    # angle devient \"-angle\" dans les sinus \u00e0 cause de la logique graphique de PIL\n    x = centre[0] + k*(point[0]-centre[0])*cos(angle)-k*(point[1]-centre[1])*sin(-angle)\n    y = centre[1] + k*(point[0]-centre[0])*sin(-angle)+k*(point[1]-centre[1])*cos(angle)\n    \n    return x,y\n\ndef arbre(A,B,angle,n):\n    if n==0: return \n    \n    C = sim(B,A,-pi\/2,1)\n    D = sim(A,B,pi\/2,1)\n    E = sim(D,C,angle,cos(angle))\n    \n    draw.polygon([A,B,C,D], fill=(randint(0,255),randint(0,255),randint(0,255),125))\n    \n    arbre(D, E, angle, n-1)\n    arbre(E, C, angle, n-1)\n    \n\narbre(A,B,theta,10)\ntree.show()<\/pre>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-3.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"208\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-3-300x208.webp\" alt=\"arbre fractal de Pythagore en couleurs\" class=\"wp-image-11201\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-3-300x208.webp 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-3-1024x710.webp 1024w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-3-768x533.webp 768w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-3.webp 1240w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div>\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Larbre_est-il_%C2%AB_infini_%C2%BB\"><\/span><strong>L&rsquo;arbre est-il \u00ab\u00a0infini\u00a0\u00bb ?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Nous allons noter $C_n$ le centre des carr\u00e9s allant vers la gauche.<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-4.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"290\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-4-290x300.webp\" alt=\"\" class=\"wp-image-11202\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-4-290x300.webp 290w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-4.webp 621w\" sizes=\"auto, (max-width: 290px) 100vw, 290px\" \/><\/a><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">On peut observer que $$\\frac{C_2C_1}{C_1C_0}=\\cos\\theta.$$ En effet, $DEFG$ est l&rsquo;image de $ABCD$ ($DC$ se transforme en $DG$) par la similitude de centre $D$, d&rsquo;angle $\\theta+\\frac{\\pi}{2}$ et de rapport $k=\\cos\\theta$ (par construction). De m\u00eame, le carr\u00e9 vert est l&rsquo;image du carr\u00e9 rouge par la similitude de centre $G$ et de m\u00eames angles et rapport.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Toutes les longueurs sont donc multipli\u00e9es par $\\cos\\theta$ pour passer d&rsquo;un carr\u00e9 au suivant.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">La suite $(C_{n-1}C_n)$ est donc g\u00e9om\u00e9trique de raison $\\cos\\theta$.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculons alors longueur:$$\\begin{align*}L_n &amp; = \\sum_{k=0}^{n-1} C_kC_{k+1}= C_0C_1 + C_1C_2 + \\cdots + C_{n-1}C_{n}\\\\&amp; = C_0C_1 \\times (1 + \\cos\\theta + \\cos^2\\theta + \\cdots + \\cos^n\\theta)\\\\&amp; = C_0C_1\\times\\dfrac{1-\\cos^{n+1}\\theta}{1-\\cos\\theta}.\\end{align*}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Comme $0 \\leqslant \\cos\\theta &lt; 1$, la limite de $L_n$ quand $n$ tend vers $+\\infty$ est finie et vaut $\\frac{C_0C_1}{1-\\cos\\theta}$.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">L&rsquo;arbre de Pythagore (not\u00e9 $\\mathcal{A}$) ne peut donc pas \u00eatre \\og infini \\fg{} (dans le sens o\u00f9 la distance entre $C_0$ et $C_\\infty$ ne peut pas \u00eatre infinie).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Je vais appeler <em>plus grand rayon<\/em> (PGR), et je vais le noter $\\mathcal{PGR}(\\mathcal{A})$ le maximum de la distance entre $C_0$ et $C_n$, pour tout entier naturel $n$: $$\\mathcal{PGR}(\\mathcal{A})=\\max\\{C_0C_n,\\ n\\geqslant1\\}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On peut ainsi modifier le script pr\u00e9c\u00e9dent pour d\u00e9terminer une valeur approch\u00e9e du PGR (en pixels) de l&rsquo;arbre:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from math import pi, cos, sin\nfrom PIL import Image, ImageDraw, ImageFont\nfrom random import randint\n\nwidth = 1748\nheight = 1240\n\ntree = Image.new('RGB', (width,height), (255,255,255))\ndraw = ImageDraw.Draw(tree,'RGBA')\n\ntheta = pi\/180 * 40 # 40 degr\u00e9s\nL = 200 # longueur de l'ar\u00eate du premier carr\u00e9\nA = (width\/2 - L\/2 , height-250)\nB = (width\/2 + L\/2 , height-250)\n\ncentres = []\n\ndef sim(centre, point, angle, k):\n    # angle devient \"-angle\" dans les sinus \u00e0 cause de la logique graphique de PIL\n    x = centre[0] + k*(point[0]-centre[0])*cos(angle)-k*(point[1]-centre[1])*sin(-angle)\n    y = centre[1] + k*(point[0]-centre[0])*sin(-angle)+k*(point[1]-centre[1])*cos(angle)\n    \n    return x,y\n \ndef arbre(A,B,angle,n):\n    if n==0: return \n    \n    global centres\n    \n    C = sim(B,A,-pi\/2,1)\n    D = sim(A,B,pi\/2,1)\n    E = sim(D,C,angle,cos(angle))\n    \n    centres.append(((A[0]+C[0])\/2,(A[1]+C[1])\/2))\n    \n    draw.polygon([A,B,C,D], fill=(randint(0,255),randint(0,255),randint(0,255),125))\n    \n    arbre(D, E, angle, n-1)\n    arbre(E, C, angle, n-1)\n    \ndef pgr():\n    global centres\n    origine = centres[0] # premier centre\n    d = 0\n    \n    for c in centres:\n        if c != origine:\n            distance = (origine[0]-c[0])**2 + (origine[1]-c[1])**2\n            if distance > d:\n                d = distance\n                faraway = c\n                \n    return d**0.5, faraway\n\narbre(A,B,theta,15)\npgr = f\"PGR de l'arbre: {pgr()[0]}\"\ndraw.text((10,10), pgr, font=ImageFont.truetype(\"arial.ttf\", size=30), fill=(0, 0, 0, 255))\n# on trace le PGR\ndraw.line([centres[0],pgr()[1]],fill=(100,100,100),width=1) \n\ntree.show()<\/pre>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-7387b849 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\"><div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"213\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5-300x213.webp\" alt=\"\" class=\"wp-image-11205\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5-300x213.webp 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5-1024x726.webp 1024w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5-768x545.webp 768w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5-1536x1090.webp 1536w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-5.webp 1748w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div><\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\"><div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-6.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"279\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-6-300x279.webp\" alt=\"\" class=\"wp-image-11206\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-6-300x279.webp 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-6-1024x953.webp 1024w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-6-768x715.webp 768w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-6.webp 1332w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div><\/div>\n<\/div>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Lors de la r\u00e9daction de cet article, j&rsquo;ai appel\u00e9 \u00ab\u00a0diam\u00e8tre\u00a0\u00bb ce que j&rsquo;appelle actuellement \u00ab\u00a0PGR\u00a0\u00bb. Cette terminologie me semblant trompeuse, je l&rsquo;ai chang\u00e9 pour \u00ab\u00a0PGR\u00a0\u00bb, mais n&rsquo;ai pas pris le temps de refaire les images. C&rsquo;est pourquoi vous pouvez lire \u00ab\u00a0Diam\u00e8tre\u00a0\u00bb et non \u00ab\u00a0PGR\u00a0\u00bb. Il en sera de m\u00eame pour les images suivantes.<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">Ce dernier exemple est int\u00e9ressant (j&rsquo;ai pris $n=20$ et un angle initial de 22.5 degr\u00e9s): on aimerait que le nombre d&rsquo;it\u00e9rations soit plus grand afin que la \u00ab\u00a0spirale\u00a0\u00bb continue davantage&#8230; Nous allons donc l\u00e9g\u00e8rement modifier le script afin de le permettre car avec ce que nous avons, pour $n=25$, \u00e7a plante&#8230;<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from math import pi, cos, sin\nfrom PIL import Image, ImageDraw, ImageFont\nfrom random import randint\n\nwidth = 1748\nheight = 1240\n\ntree = Image.new('RGB', (width,height), (255,255,255))\ndraw = ImageDraw.Draw(tree,'RGBA')\n\ntheta = pi\/180 * 22.5\nL = 150 # longueur de l'ar\u00eate du premier carr\u00e9\nA = (width\/2 - L\/2 +100, height-250)\nB = (width\/2 + L\/2 +100, height-250)\n\ncentres = []\n\ndef sim(centre, point, angle, k):\n    # angle devient \"-angle\" dans les sinus \u00e0 cause de la logique graphique de PIL\n    x = centre[0] + k*(point[0]-centre[0])*cos(angle)-k*(point[1]-centre[1])*sin(-angle)\n    y = centre[1] + k*(point[0]-centre[0])*sin(-angle)+k*(point[1]-centre[1])*cos(angle)\n    \n    return x,y\n \ndef arbre(A,B,angle,n):\n    if (n==0) or (length(A,B)&lt;8): return \n    \n    global centres\n    \n    C = sim(B,A,-pi\/2,1)\n    D = sim(A,B,pi\/2,1)\n    E = sim(D,C,angle,cos(angle))\n    \n    centres.append(((A[0]+C[0])\/2,(A[1]+C[1])\/2))\n    \n    draw.polygon([A,B,C,D], fill=(randint(0,255),randint(0,255),randint(0,255),125))\n    \n    arbre(D, E, angle, n-1)\n    arbre(E, C, angle, n-1)\n\ndef length(A,B):\n    return (A[0]-B[0])**2 + (A[1]-B[1])**2\n    \ndef pgr():\n    global centres\n    origine = centres[0] # premier centre\n    d = 0\n    \n    for c in centres:\n        if c != origine:\n            distance = length(origine,c) #(origine[0]-c[0])**2 + (origine[1]-c[1])**2\n            if distance > d:\n                d = distance\n                faraway = c\n                \n    return d**0.5, faraway\n\narbre(A,B,theta,50)\npgr = f\"PGR de l'arbre: {pgr()[0]}\"\ndraw.text((10,10), pgr, font=ImageFont.truetype(\"arial.ttf\", size=30), fill=(0, 0, 0, 255))\ndraw.line([centres[0],pgr()[1]],fill=(100,100,100),width=1) \n\ntree.show()<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Nous avons introduit:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>une nouvelle fonction \u00ab\u00a0<strong>length<\/strong>\u00a0\u00bb qui permet de renvoyer le carr\u00e9 de la distance entre deux points;<\/li>\n\n\n\n<li>un test dans la fonction \u00ab\u00a0<strong>arbre<\/strong>\u00ab\u00a0: on n&rsquo;ex\u00e9cute la fonction que si $AB^2 &gt; 8$ (par exemple). En effet, inutile d&rsquo;aller plus loin si l&rsquo;on ne voit plus les points.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Cette modification donne:<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-7.webp\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"286\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-7-300x286.webp\" alt=\"\" class=\"wp-image-11207\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-7-300x286.webp 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-7-1024x976.webp 1024w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-7-768x732.webp 768w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2025\/08\/arbre-pythagore-7.webp 1301w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>Un arbre fractal de Pythagore est une figure plut\u00f4t esth\u00e9tique dans certaines conditions. Nous allons voir ce que c&rsquo;est&#8230;<\/p>\n","protected":false},"author":1,"featured_media":11209,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,5],"tags":[],"class_list":["post-11192","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-python"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Arbre fractal de Pythagore - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"Un arbre fractal de Pythagore est obtenu \u00e0 partir d&#039;un carr\u00e9 sur lequel on met un triangle rectangle. 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