{"id":202175,"date":"2023-10-06T01:08:10","date_gmt":"2023-10-06T04:08:10","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=202175"},"modified":"2023-10-06T01:08:10","modified_gmt":"2023-10-06T04:08:10","slug":"a-problema-de-gincana-um-triangulo-dividido","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/a-problema-de-gincana-um-triangulo-dividido\/","title":{"rendered":"(A) Problema de Gincana: Um tri\u00e2ngulo dividido"},"content":{"rendered":"<p class=\"ti3\"  style=\"text-align:center;\">Clique no bot\u00e3o abaixo para visualizar o problema.<\/p>\n<div style=\"text-align:center;margin-top:10px;\">\n<input class=\"bo1 c1 cb13\" style=\"width:150px;\" value=\"Problema\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Problema';}\" \/>  <\/p>\n<div style=\"display: none;\">\n<p style=\"margin:20px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nNo tri\u00e2ngulo ret\u00e2ngulo [tex]ABC[\/tex] mostrado abaixo, [tex]BD = CD[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/08\/f034.png\"  style=\"border: 0px;\"  alt=\"\" width=\"306\" height=\"250\" class=\"aligncenter size-full wp-image-199287\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/08\/f034.png 306w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/08\/f034-300x245.png 300w\" sizes=\"auto, (max-width: 306px) 100vw, 306px\" \/><br \/>\nSabendo que [tex]AB = 3[\/tex] e [tex]AC = 4[\/tex], determine o valor de [tex]CD[\/tex].<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o 1<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nDenotemos por [tex]x[\/tex] a medida [tex]CD[\/tex], ent\u00e3o [tex]BD = x[\/tex] e, como [tex]AC = 4[\/tex], temos [tex]AD = AC-CD = 4-x[\/tex], conforme indicado na figura abaixo.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/09\/f034_1.png\" style=\"border: 0px;\"  alt=\"\" width=\"320\" height=\"331\" class=\"aligncenter size-full wp-image-201231\" \/><br \/>\nAplicado o Teorema de Pit\u00e1goras ao tri\u00e2ngulo [tex]ABD[\/tex], temos:<br \/>\n[tex]\\qquad x^2 = 3^2+(4-x)^2[\/tex]<br \/>\n[tex]\\qquad x^2 = 3^2+4^2-8x+x^2[\/tex]<br \/>\n[tex]\\qquad 8x = 9+16[\/tex]<br \/>\n[tex]\\qquad 8x = 25[\/tex]<br \/>\n[tex]\\qquad x = \\dfrac{25}{8} = 3,125.[\/tex]<br \/>\nPortanto, [tex]\\fcolorbox{black}{#eee0e5}{$CD=3,125$}\\,.[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o 2<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPrimeiramente, focando no tri\u00e2ngulo ret\u00e2ngulo [tex]ABD[\/tex], podemos aplicar o teorema de Pit\u00e1goras para obtermos o seguinte:<br \/>\n[tex]\\qquad (BD)^2 = (AB)^2 + (AD)^2.\\qquad \\textcolor{#800000}{(i)}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Analisando o tri\u00e2ngulo [tex]ABC[\/tex], observamos que [tex]\\boxed{AD = AC-DC}.[\/tex]<br \/>\nSubstituindo essa igualdade em [tex]\\textcolor{#800000}{(i)}[\/tex] e desenvolvendo, ficamos com:<br \/>\n[tex]\\qquad (BD)^2 = (AB)^2 + (AC &#8211; DC)^2\\\\<br \/>\n\\qquad (BD)^2 = (AB)^2 + (AC)^2-2 \\cdot (AC) \\cdot (DC) + (DC)^2.\\qquad \\textcolor{#800000}{(ii)}\\\\<br \/>\n~~[\/tex]<br \/>\nComo [tex]BD = DC[\/tex], ent\u00e3o [tex]BD^2 = DC^2[\/tex]; assim, podemos simplificar estes termos em [tex]\\textcolor{#800000}{(ii)}[\/tex]:<br \/>\n[tex]\\qquad 0 = (AB)^2 + (AC)^2 &#8211; 2 \\cdot (AC) \\cdot (DC)\\\\<br \/>\n~~[\/tex]<br \/>\nAgora, aplicando os valores num\u00e9ricos para [tex]AB[\/tex] e [tex]AC[\/tex], obtemos:<br \/>\n[tex]\\qquad 0 = 3^2 + 4^2-2 \\cdot 4 \\cdot (DC)\\\\<br \/>\n\\qquad 0 = 9 + 16-8 \\cdot (DC)\\\\<br \/>\n\\qquad 0 = 25-8(DC)\\\\<br \/>\n\\qquad 8(DC) = 25\\\\<br \/>\n\\qquad DC = \\dfrac{25}{8}.[\/tex]<br \/>\nAssim, conclu\u00edmos que o valor de [tex]CD[\/tex] \u00e9 [tex]\\,\\fcolorbox{black}{#eee0e5}{$\\dfrac{25}{8}=3,125$}\\,.[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelo <strong>COM Pot\u00eancias de Euler<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:20px\">\n<div class=\"g3\" style=\"width:490px;\">\nParticipou da discuss\u00e3o o Clube <strong> Pot\u00eancias de Euler<\/strong>.<\/div>\n<p style=\"margin:10px\">\n<p style=\"margin:30px\">\n<div class=\"gg\">\nPrimeira Gincana de 2023 &#8211; Clubes de Matem\u00e1tica da OBMEP<br \/>\nN\u00edvel A &#8211; Quest\u00e3o Mediana<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Clique no bot\u00e3o abaixo para visualizar o problema. Problema No tri\u00e2ngulo ret\u00e2ngulo [tex]ABC[\/tex] mostrado abaixo, [tex]BD = CD[\/tex]. Sabendo que [tex]AB = 3[\/tex] e [tex]AC = 4[\/tex], determine o valor de [tex]CD[\/tex]. Solu\u00e7\u00e3o 1 Denotemos por [tex]x[\/tex] a medida [tex]CD[\/tex], ent\u00e3o [tex]BD = x[\/tex] e, como [tex]AC = 4[\/tex], temos [tex]AD = AC-CD = 4-x[\/tex], &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/a-problema-de-gincana-um-triangulo-dividido\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-202175","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/202175","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=202175"}],"version-history":[{"count":1,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/202175\/revisions"}],"predecessor-version":[{"id":202176,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/202175\/revisions\/202176"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=202175"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}