{"id":163063,"date":"2020-08-30T21:24:15","date_gmt":"2020-08-31T00:24:15","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=163063"},"modified":"2020-11-15T01:07:17","modified_gmt":"2020-11-15T03:07:17","slug":"problemao-relacao-entre-areas-em-um-tetraedro","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-relacao-entre-areas-em-um-tetraedro\/","title":{"rendered":".Problem\u00e3o: Rela\u00e7\u00e3o entre \u00e1reas em um tetraedro"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:25px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"margin:10px\">\n<p>A figura mostra um tetraedro [tex]ABCD[\/tex] cujos v\u00e9rtices [tex]A[\/tex], [tex]B[\/tex], [tex]C[\/tex] e [tex]D[\/tex] s\u00e3o v\u00e9rtices de um paralelep\u00edpedo retangular de dimens\u00f5es com medidas [tex]x[\/tex], [tex]y[\/tex] e [tex]z[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra9.png\" alt=\"\" width=\"600\" height=\"325\" class=\"aligncenter size-full wp-image-163343\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra9.png 600w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra9-300x163.png 300w\" sizes=\"auto, (max-width: 600px) 100vw, 600px\" \/><\/p>\n<p>Se [tex]F_A[\/tex], [tex]F_B[\/tex], [tex]F_C[\/tex] e [tex]F_D[\/tex] s\u00e3o as \u00e1reas das faces do tetraedro opostas, respectivamente, aos v\u00e9rtices [tex]A[\/tex], [tex]B[\/tex], [tex]C[\/tex] e [tex]D[\/tex], <strong>prove que [tex]\\boxed{F^2_A + F^2_C= F^2_B + F^2_D}[\/tex]<\/strong>.<\/p>\n<p style=\"font-size:12px;margin:1px,line-height:10px;text-align:right;\">(Extra\u00eddo de <strong>Maclaurin Olympiad<\/strong>, 2003-2009)<\/p>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 8px; margin-bottom: 20px;border: 0 none\"\/>\n<table style=\" height:auto; margin-left: auto;margin-right:auto; width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"explicador_p\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"90\" height=\"108\" \/><\/td>\n<td class=\"bf \">\n<p class=\"fala bg1 cb33\" style=\"width:400px; line-height: 20px; font-size:14px\" > Se voc\u00eas precisarem de uma ajuda para visualizar melhor o tetraedro e o paralelep\u00edpedo, cliquem no bot\u00e3o abaixo. <\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\"text-align:center;\">\n<input class=\"bo1 c3 cba3\" style=\"width:150px;\" value=\"Applet\" 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 = 'Applet';}\" \/>    <\/p>\n<div  style=\"display: none;\">\n<div class=\"g1n\" style=\"border: 2px dashed #800000; text-align:justify; padding: 10px; width:85%; margin-left: auto; margin-right: auto;\" >\n<p class=\"c3 ti3\"style=\"line-height:25px;margin:10px\" > Um applet para ajudar <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"margin:10px\">\n<p><strong>Instru\u00e7\u00f5es<\/strong>:<br \/>\n<strong>(1)<\/strong> Espere o applet carregar. <span style=\"font-size:12px\">(O aplicativo abrir\u00e1 uma janela 3D; assim, ele pode demorar um pouquinho para carregar.)<\/span><br \/>\n<strong>(2)<\/strong> Para movimentar o paralelep\u00edpedo e, consequentemente, o tetraedro, clique em qualquer lugar da janela do aplicativo, mantenha o mouse pressionado e fa\u00e7a o movimento.<br \/>\n<strong>(3)<\/strong> Se quiser voltar para a posi\u00e7\u00e3o inicial, clique nas setinhas circulares que aparecem no canto superior direito do aplicativo.<\/p>\n<p style=\"text-align:center;\">\n<iframe loading=\"lazy\" scrolling=\"no\" title=\"Tetraedro no paralelep\u00edpedo\" src=\"https:\/\/www.geogebra.org\/material\/iframe\/id\/btvk7wxm\/width\/726\/height\/722\/border\/800000\/sfsb\/false\/smb\/false\/stb\/false\/stbh\/false\/ai\/false\/asb\/false\/sri\/true\/rc\/false\/ld\/false\/sdz\/true\/ctl\/false\" width=\"726px\" height=\"722px\" style=\"border:0px;\"> <\/iframe><br \/>\n <span style=\"font-size: 14px;\">OBMEP_ srdg, criado com o <a href=\"http:\/\/www.geogebra.org\/\" target=\"_blank\">GeoGebra<\/a><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cb36 g1n\" style=\"margin:40px;line-height:26px; width:80%; 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=\"63\" height=\"76\" class=\"alignright size-full wp-image-1001\" \/><\/p>\n<p class=\"c3 ti3\" style=\"line-height:30px;margin:10px\"> AJUDA<\/p>\n<p><span style=\"color:#800000;\">&#9658;<\/span><strong> Defini\u00e7\u00e3o:<\/strong><br \/>\nUma reta e um plano s\u00e3o ditos <strong>perpendiculares <\/strong>se, e somente se, eles t\u00eam um ponto comum e a reta for perpendicular a todas as retas do plano que passam por esse ponto comum. Neste caso, dizemos tamb\u00e9m que <strong>a reta \u00e9 perpendicular ao plano<\/strong> e que <strong>o plano \u00e9 perpendicular \u00e0 reta<\/strong>.<br \/>\nNa figura abaixo, a reta [tex]a[\/tex] \u00e9 perpendicular ao plano [tex]\\alpha[\/tex] ([tex]a \\perp \\alpha[\/tex]) e o plano [tex]\\alpha[\/tex] \u00e9 perpendicular \u00e0 reta [tex]a[\/tex] ([tex]\\alpha \\perp a[\/tex]).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra1.png\" style=\"border:none\" alt=\"\" width=\"640\" height=\"266\" class=\"aligncenter size-full wp-image-163338\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra1.png 640w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra1-300x125.png 300w\" sizes=\"auto, (max-width: 640px) 100vw, 640px\" \/><\/p>\n<p>Se uma reta [tex]a[\/tex] \u00e9 perpendicular a um plano [tex]\\alpha[\/tex] (ou o plano [tex]\\alpha[\/tex] \u00e9 perpendicular \u00e0 reta [tex]a[\/tex]), a intersec\u00e7\u00e3o de [tex]a[\/tex] em [tex]\\alpha[\/tex] \u00e9 chamada &quot;p\u00e9 da perpendicular&quot;. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra2.png\" style=\"border:none\"alt=\"\" width=\"640\" height=\"244\" class=\"aligncenter size-full wp-image-163339\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra2.png 640w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra2-300x114.png 300w\" sizes=\"auto, (max-width: 640px) 100vw, 640px\" \/><br \/>\nUma reta e um plano que apresentem um \u00fanico ponto em comum, mas n\u00e3o forem perpendiculares, s\u00e3o ditos <strong>obl\u00edquos<\/strong>. Na figura a seguir, [tex]r[\/tex] e [tex]\\alpha[\/tex] s\u00e3o perpendiculares, mas [tex]s[\/tex] e [tex]\\alpha[\/tex] s\u00e3o obl\u00edquos.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra4.png\" style=\"border:none\" alt=\"\" width=\"640\" height=\"266\" class=\"aligncenter size-full wp-image-163340\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra4.png 640w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra4-300x125.png 300w\" sizes=\"auto, (max-width: 640px) 100vw, 640px\" \/><\/p>\n<p><span style=\"color:#800000;\">&#9658;<\/span> <strong>Consequ\u00eancia da Defini\u00e7\u00e3o:<\/strong><br \/>\nSe uma reta [tex]p[\/tex] \u00e9 perpendicular a um plano [tex]\\alpha[\/tex], ent\u00e3o qualquer reta contida em [tex]\\alpha[\/tex] \u00e9 dita ortogonal a [tex]p[\/tex]. Particularmente, se as duas retas ortogonais intersectam-se, elas s\u00e3o ditas perpendiculares.<br \/>\nAssim, se  [tex]p[\/tex] \u00e9 perpendicular a [tex]\\alpha\\;[\/tex] e [tex]\\;q[\/tex] \u00e9 uma reta contida em [tex]\\alpha[\/tex], ent\u00e3o duas situa\u00e7\u00f5es podem ocorrer::<br \/>\n&#8211; [tex]p[\/tex] e [tex]q[\/tex] intersectam-se no ponto em que [tex]p[\/tex] &quot;fura&quot; o plano, e neste caso, [tex]p[\/tex]  e [tex]q[\/tex] s\u00e3o <strong>perpendiculares<\/strong>;<br \/>\n&#8211; [tex]p[\/tex] e [tex]q[\/tex] n\u00e3o se intersectam ([tex]p \\cap q=\\emptyset[\/tex]), e neste caso [tex]p[\/tex]  e [tex]q[\/tex] s\u00e3o <strong>ortogonais<\/strong>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra10.png\" style=\"border:none\" alt=\"\" width=\"640\" height=\"249\" class=\"aligncenter size-full wp-image-163354\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra10.png 640w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra10-300x117.png 300w\" sizes=\"auto, (max-width: 640px) 100vw, 640px\" \/><\/p>\n<p><span style=\"color:#800000;\">&#9658;<\/span><strong> Teorema:<\/strong><br \/>\nSe uma reta [tex]a[\/tex] \u00e9 perpendicular a duas retas [tex]b[\/tex] e [tex]c[\/tex] concorrentes de um plano [tex]\\alpha[\/tex], ent\u00e3o [tex]a[\/tex] \u00e9 perpendicular ao plano [tex]\\alpha[\/tex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra11.png\"  style=\"border:none\" alt=\"\" width=\"640\" height=\"213\" class=\"aligncenter size-full wp-image-163355\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra11.png 640w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra11-300x100.png 300w\" sizes=\"auto, (max-width: 640px) 100vw, 640px\" \/>\n<\/div>\n<div class=\"g1n\" style=\"line-height:32px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o  <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 30px;border: 0 none\"\/>\n<span class=\"c3\"><strong>Nota\u00e7\u00f5es:<\/strong> <\/span>Denotaremos o segmento definido por dois pontos, digamos [tex]X[\/tex] e  [tex]Y[\/tex], por [tex]\\overline{XY}[\/tex] e o seu comprimento por [tex]XY[\/tex].<br \/>\nNa nossa solu\u00e7\u00e3o, vamos considerar mais dois v\u00e9rtices do paralelep\u00edpedo: [tex]E\\,[\/tex] e [tex]\\,F\\,.[\/tex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra14.png\"  style=\"border:none\" alt=\"\" width=\"403\" height=\"288\" class=\"aligncenter size-full wp-image-165155\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra14.png 403w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/tetra14-300x214.png 300w\" sizes=\"auto, (max-width: 403px) 100vw, 403px\" \/><\/p>\n<ul>\n<li>Inicialmente, calculemos a \u00e1rea da face do tetraedro oposta ao v\u00e9rtice [tex]A[\/tex]:<br \/>\n[tex]\\qquad F_A=\\dfrac{BC \\cdot DC}{2}=\\dfrac{y \\cdot z}{2}[\/tex].<\/li>\n<li>Agora, calculemos a \u00e1rea da face do tetraedro oposta ao v\u00e9rtice [tex]D[\/tex]:<br \/>\n[tex]\\qquad F_D=\\dfrac{AB \\cdot CB}{2}=\\dfrac{x \\cdot y}{2}[\/tex].<\/li>\n<li>Para calcularmos [tex]F_B[\/tex], precisamos notar que, como [tex]\\overline{DC}[\/tex] \u00e9 perpendicular a [tex]\\overline{BC}[\/tex] e a [tex]\\overline{EC}[\/tex], ent\u00e3o, pelo <strong>Teorema da Ajuda<\/strong>, [tex]\\overline{DC}[\/tex] \u00e9 perpendicular ao plano que passa por [tex]ABCE[\/tex]. Assim, [tex]\\overline{DC}[\/tex] \u00e9 perpendicular a [tex]\\overline{AC}[\/tex] (<strong>Consequ\u00eancia da Defini\u00e7\u00e3o<\/strong>). Com isso<br \/>\n[tex]\\qquad F_B=\\dfrac{AC \\cdot DC}{2}[\/tex].<br \/>\nAplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo [tex]ACB[\/tex], obtemos: [tex]AC^2=x^2+y^2[\/tex]. Logo,<br \/>\n[tex]\\qquad F_B=\\dfrac{\\sqrt{x^2+y^2\\,} \\cdot z}{2}[\/tex].<\/li>\n<li>De forma an\u00e1loga, para calcularmos [tex]F_C[\/tex], precisamos perceber que como [tex]\\overline{AB}[\/tex] \u00e9 perpendicular a [tex]\\overline{FB}[\/tex] e  a [tex]\\overline{CB}[\/tex], ent\u00e3o, pelo <strong>Teorema da Ajuda<\/strong>, [tex]\\overline{AB}[\/tex] \u00e9 perpendicular ao plano que passa por [tex]BCDF[\/tex]. Assim, [tex]\\overline{AB}[\/tex] \u00e9 perpendicular a [tex]\\overline{BD}[\/tex] (<strong>Consequ\u00eancia da Defini\u00e7\u00e3o<\/strong>), e:<br \/>\n[tex]\\qquad F_C=\\dfrac{AB \\cdot BD}{2}[\/tex].<br \/>\nAplicando, agora, o Teorema de Pit\u00e1goras no tri\u00e2ngulo [tex]BDC[\/tex], temos que [tex]BD^2=y^2+z^2[\/tex] e, portanto:<br \/>\n[tex]\\qquad F_C=\\dfrac{x \\cdot \\sqrt{y^2+z^2}}{2}[\/tex].<\/li>\n<\/ul>\n<p>Agora, segue que:<br \/>\n[tex]\\qquad F^2_A + F^2_C=\\left(\\dfrac{y \\cdot z}{2}\\right)^2+\\left(\\dfrac{x \\cdot \\sqrt{y^2+z^2}}{2}\\right)^2\\\\<br \/>\n\\qquad F^2_A + F^2_C=\\dfrac{y^2 \\cdot z^2+x^2 \\cdot (y^2+z^2)}{4}\\\\<br \/>\n\\qquad \\boxed{F^2_A + F^2_C=\\dfrac{y^2 \\cdot z^2+x^2 \\cdot y^2 + x^2 \\cdot z^2}{4}}\\,.[\/tex]<br \/>\nPor outro lado,<br \/>\n[tex]\\qquad F^2_B + F^2_D=\\left(\\dfrac{\\sqrt{x^2+y^2} \\cdot z}{2}\\right)^2+\\left(\\dfrac{x \\cdot y}{2}\\right)^2\\\\<br \/>\n\\qquad F^2_B + F^2_D=\\dfrac{(x^2+y^2) \\cdot z^2+x^2 \\cdot y^2}{4}\\\\<br \/>\n\\qquad F^2_B + F^2_D=\\dfrac{x^2 \\cdot z^2 +y^2  \\cdot z^2+x^2 \\cdot y^2}{4}\\\\<br \/>\n\\qquad \\boxed{F^2_B + F^2_D=\\dfrac{y^2  \\cdot z^2+x^2 \\cdot y^2+x^2 \\cdot z^2}{4}}\\,.[\/tex]<br \/>\nLogo, de fato, [tex]\\, \\fcolorbox{black}{#eee0e5}{$F^2_A + F^2_C= F^2_B + F^2_D$}\\,.[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p class=\"g2\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) A figura mostra um tetraedro [tex]ABCD[\/tex] cujos v\u00e9rtices [tex]A[\/tex], [tex]B[\/tex], [tex]C[\/tex] e [tex]D[\/tex] s\u00e3o v\u00e9rtices de um paralelep\u00edpedo retangular de dimens\u00f5es com medidas [tex]x[\/tex], [tex]y[\/tex] e [tex]z[\/tex]. Se [tex]F_A[\/tex], [tex]F_B[\/tex], [tex]F_C[\/tex] e [tex]F_D[\/tex] s\u00e3o as \u00e1reas das faces do tetraedro opostas, respectivamente, aos v\u00e9rtices [tex]A[\/tex], &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-relacao-entre-areas-em-um-tetraedro\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-163063","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/163063","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\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=163063"}],"version-history":[{"count":22,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/163063\/revisions"}],"predecessor-version":[{"id":170954,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/163063\/revisions\/170954"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=163063"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}