{"id":88435,"date":"2017-11-12T14:19:55","date_gmt":"2017-11-12T16:19:55","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=88435"},"modified":"2023-04-14T02:29:35","modified_gmt":"2023-04-14T05:29:35","slug":"problema-perpendiculares","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-perpendiculares\/","title":{"rendered":".Problema: Perpendiculares"},"content":{"rendered":"<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nNum tri\u00e2ngulo equil\u00e1tero [tex]ABC[\/tex], os pontos [tex]M[\/tex], [tex]N[\/tex] e [tex]P[\/tex] s\u00e3o tomados em [tex]\\overline{BC}[\/tex], [tex]\\overline{AC}[\/tex] e [tex]\\overline{AB}[\/tex], respectivamente, de modo que [tex]\\overline{AM} \\perp \\overline{BC}[\/tex], [tex]\\overline{MN} \\perp \\overline{AC}[\/tex] e [tex]\\overline{NP} \\perp \\overline{AB}[\/tex]. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/01\/Perpendiculares.png\" style=\"border:none;\" alt=\"\" width=\"355\" height=\"312\" class=\"aligncenter size-full wp-image-71339\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/01\/Perpendiculares.png 355w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/01\/Perpendiculares-300x264.png 300w\" sizes=\"auto, (max-width: 355px) 100vw, 355px\" \/><\/p>\n<p>Mostre que a interse\u00e7\u00e3o de [tex]\\overline{NP}[\/tex] e [tex]\\overline{AM}[\/tex] \u00e9 ponto m\u00e9dio de [tex]\\overline{AM}[\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"cb36 g1n\" style=\"margin:30px;line-height:30px; width:70%; margin-left:auto; margin-right:auto; \">\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" style=\"border:none\"  alt=\"explicador_p\" width=\"80\" height=\"96\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Lembretes <\/p>\n<p style=\"margin:10px\">\n<p>Alguns resultados sobre \u00e2ngulos internos de um tri\u00e2ngulo:<br \/>\n[tex] \\color{#800000}{(*)}[\/tex] A soma das medidas dos \u00e2ngulos internos de um tri\u00e2ngulo \u00e9 [tex]180^{\\circ}[\/tex].<br \/>\n[tex] \\color{#800000}{(**)}[\/tex] Todo tri\u00e2ngulo equil\u00e1tero tem \u00e2ngulos internos de [tex]60^{\\circ}[\/tex].<br \/>\n[tex] \\color{#800000}{(***)}[\/tex] \u00c2ngulos opostos pelo v\u00e9rtice (OPV) t\u00eam a mesma medida.\n<\/p><\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\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\"\/>\nNote que, a partir das proposi\u00e7\u00f5es [tex] \\color{#800000}{(*)}[\/tex], [tex] \\color{#800000}{(**)}[\/tex]  e [tex] \\color{#800000}{(***)}[\/tex] do Lembrete, podemos completar as medidas de alguns \u00e2ngulos da figura inicial, de acordo com a figura a seguir.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_08.png\" style=\"border:none\" alt=\"\" width=\"339\" height=\"323\" class=\"aligncenter size-full wp-image-88561\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_08.png 339w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_08-300x286.png 300w\" sizes=\"auto, (max-width: 339px) 100vw, 339px\" \/><\/p>\n<p style=\"margin:10px\">\n<p>Agora, observe que:<\/p>\n<ul>\n<li> o tri\u00e2ngulo [tex]IMN[\/tex] \u00e9 equil\u00e1tero, logo [tex]IM=MN=NI[\/tex],<\/li>\n<li> o tri\u00e2ngulo [tex]AIN[\/tex] \u00e9 is\u00f3sceles, logo [tex]NI=IA[\/tex].<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_09.png\" style=\"border:none\" alt=\"\" width=\"343\" height=\"320\" class=\"aligncenter size-full wp-image-88564\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_09.png 343w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_09-300x280.png 300w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><\/p>\n<p style=\"margin:10px\">\n<p>Dessa forma, [tex]IM=IA[\/tex] e, com isso, temos que [tex]I[\/tex] \u00e9 ponto m\u00e9dio de [tex]\\overline{AM}[\/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\" style=\"line-height:35px;\">\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\"\/>\nPara aqueles que gostam de ver para crer, vamos apresentar uma segunda solu\u00e7\u00e3o, na qual mostraremos explicitamente que o comprimento de [tex]\\overline{AI}[\/tex] \u00e9 a metade do comprimento de  [tex]\\overline{AM}[\/tex], sendo [tex]I[\/tex] o ponto de interse\u00e7\u00e3o de [tex]\\overline{NP}[\/tex] e [tex]\\overline{AM}[\/tex].<br \/>\nInicialmente, observe que, a partir duas proposi\u00e7\u00f5es [tex] \\color{#800000}{(*)}[\/tex]  e [tex] \\color{#800000}{(**)}[\/tex] do Lembrete, podemos completar as medidas de alguns \u00e2ngulos da figura inicial, como mostramos a seguir.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_06.png\" style=\"border:none\" alt=\"\" width=\"333\" height=\"305\" class=\"aligncenter size-full wp-image-88381\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_06.png 333w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2017\/11\/Eduardo_06-300x275.png 300w\" sizes=\"auto, (max-width: 333px) 100vw, 333px\" \/><\/p>\n<p>Seja [tex]a[\/tex] a medida do lado do tri\u00e2ngulo equil\u00e1tero [tex]ABC[\/tex]. <\/p>\n<ul>\n<li>No tri\u00e2ngulo [tex]AMC[\/tex], temos:<br \/>\n[tex]\\qquad \\cos 30^{\\circ} = \\dfrac{AM}{a}\\\\ [\/tex]<br \/>\nMas, em um tri\u00e2ngulo equil\u00e1tero uma altura \u00e9 tamb\u00e9m mediana; assim, [tex]M[\/tex] \u00e9 ponto m\u00e9dio de [tex]\\overline{AM}[\/tex]. Logo, segue da igualdade anterior que:<br \/>\n[tex]\\qquad  \\dfrac{\\sqrt{3}}{2} = \\dfrac{AM}{a} [\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\qquad  \\fcolorbox{black}{#eee0e5}{$AM = \\dfrac{a\\sqrt{3}}{2}$} \\, .\\qquad   \\color{#800000}{(i)}[\/tex]<\/li>\n<\/ul>\n<ul>\n<li>No tri\u00e2ngulo [tex]MNC[\/tex], temos<br \/>\n[tex]\\qquad\\text{sen}\\, 30^{\\circ} = \\dfrac{NC}{MC} \\\\<br \/>\n[\/tex]<br \/>\n[tex]\\qquad\\text{sen}\\, 30^{\\circ} = \\dfrac{NC}{\\frac{a}{2}}\\\\<br \/>\n [\/tex]<br \/>\n[tex]\\qquad \\dfrac{1}{2} = \\dfrac{NC}{\\frac{a}{2}} [\/tex]<br \/>\n[tex]\\qquad \\boxed{NC = \\dfrac{a}{4}}.\\qquad   \\color{#800000}{(ii)}[\/tex]<\/li>\n<\/ul>\n<ul>\n<li> [tex]AN + NC = AC [\/tex]<br \/>\n[tex]AN \\stackrel{\\textcolor{#800000}{(ii)}}{=} a-\\dfrac{a}{4}[\/tex].<br \/>\n[tex]\\boxed{AN = \\dfrac{3a}{4}}. \\qquad   \\color{#800000}{(iii)}[\/tex]<\/li>\n<\/ul>\n<ul>\n<li> No tri\u00e2ngulo [tex]NPA[\/tex], temos:<br \/>\n[tex]\\qquad \\text{sen}\\, 30^{\\circ} =\\dfrac{AP}{AN}\\\\<br \/>\n[\/tex]<br \/>\n[tex]\\qquad \\text{sen}\\, 30^{\\circ} \\stackrel{\\textcolor{#800000}{(iii)}}{=} \\dfrac{AP}{\\dfrac{3a}{4}}[\/tex]<br \/>\n[tex]\\qquad \\dfrac{1}{2} = \\dfrac{AP}{\\frac{3a}{4}}[\/tex]<br \/>\n[tex]\\qquad \\boxed{AP = \\dfrac{3a}{8}}.\\qquad   \\color{#800000}{(iv)}[\/tex]<\/li>\n<\/ul>\n<ul>\n<li>No tri\u00e2ngulo [tex]API[\/tex], temos:<br \/>\n[tex]\\qquad \\cos 30^{\\circ} = \\dfrac{AP}{AI} \\\\<br \/>\n[\/tex]<br \/>\n[tex]\\qquad \\cos 30^{\\circ} \\stackrel{\\textcolor{#800000}{(iv)}}{=} \\dfrac{\\frac{3a}{8}}{AI} [\/tex]<br \/>\n[tex]\\qquad  \\dfrac{\\sqrt{3}}{2} = \\dfrac{\\frac{3a}{8}}{AI} [\/tex]<br \/>\n[tex]\\qquad \\fcolorbox{black}{#eee0e5}{$AI = \\dfrac{a\\sqrt{3}}{4}$} \\, .\\qquad   \\color{#800000}{(v)}[\/tex]<\/p>\n<p>De [tex] \\color{#800000}{(i)}[\/tex] e [tex] \\color{#800000}{(v)}[\/tex], segue que [tex]I[\/tex] \u00e9 ponto m\u00e9dio de [tex]\\overline{AM}[\/tex].<\/li>\n<\/ul>\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","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) Num tri\u00e2ngulo equil\u00e1tero [tex]ABC[\/tex], os pontos [tex]M[\/tex], [tex]N[\/tex] e [tex]P[\/tex] s\u00e3o tomados em [tex]\\overline{BC}[\/tex], [tex]\\overline{AC}[\/tex] e [tex]\\overline{AB}[\/tex], respectivamente, de modo que [tex]\\overline{AM} \\perp \\overline{BC}[\/tex], [tex]\\overline{MN} \\perp \\overline{AC}[\/tex] e [tex]\\overline{NP} \\perp \\overline{AB}[\/tex]. Mostre que a interse\u00e7\u00e3o de [tex]\\overline{NP}[\/tex] e [tex]\\overline{AM}[\/tex] \u00e9 ponto m\u00e9dio de [tex]\\overline{AM}[\/tex]. Lembretes &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-perpendiculares\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":179,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-88435","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/88435","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\/179"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=88435"}],"version-history":[{"count":19,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/88435\/revisions"}],"predecessor-version":[{"id":186891,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/88435\/revisions\/186891"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=88435"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}