{"id":59325,"date":"2016-04-18T14:52:31","date_gmt":"2016-04-18T17:52:31","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=59325"},"modified":"2023-03-31T20:50:47","modified_gmt":"2023-03-31T23:50:47","slug":"problema-saindo-inteiro-pela-tangente","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-saindo-inteiro-pela-tangente\/","title":{"rendered":".Problema: Saindo inteiro pela tangente"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\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\"\/>\nAs tangentes dos \u00e2ngulos internos de um tri\u00e2ngulo s\u00e3o n\u00fameros inteiros positivos. Quais s\u00e3o os poss\u00edveis valores dessas tangentes?<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:40px;\">\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: 20px;border: 0 none\"\/>\nSejam [tex] \\,  \\, \\alpha \\leq \\beta \\leq \\gamma \\,  \\, [\/tex] as medidas dos \u00e2ngulos do tri\u00e2ngulo cujas tangentes s\u00e3o  inteiros positivos. Dessa forma, temos<br \/>\n[tex]\\qquad\\qquad \\alpha \\leq \\beta \\leq \\gamma \\lt \\dfrac{\\pi}{2}[\/tex].[tex]\\qquad\\qquad (i)[\/tex]<br \/>\nPerceba que<br \/>\n[tex]\\qquad\\qquad 0 \\lt \\alpha \\leq \\dfrac{\\pi}{3}[\/tex], [tex]\\qquad\\qquad (ii)[\/tex]<br \/>\ncaso contr\u00e1rio ter\u00edamos [tex]\\alpha + \\beta + \\gamma \\gt \\pi[\/tex].<br \/>\nNo intervalo [tex] (ii)[\/tex] a fun\u00e7\u00e3o tangente \u00e9 crescente, portanto, [tex] \\,  \\,  0 \\lt tg\\,{\\alpha} \\leq tg\\,\\dfrac{\\pi}{3} = \\sqrt{3}[\/tex].<br \/>\nComo [tex] \\,  \\, tg\\,{\\alpha} \\,  \\, [\/tex] \u00e9 um valor inteiro positivo, ent\u00e3o [tex] \\,  \\, tg\\,{\\alpha} = 1 \\,  \\, [\/tex], logo [tex] \\,  \\, \\alpha = \\dfrac{\\pi}{4} \\,  \\, [\/tex] e, portanto, [tex] \\,  \\, \\beta + \\gamma = \\dfrac{3 \\pi}{4} \\,  \\, [\/tex], j\u00e1 que  [tex]\\alpha + \\beta + \\gamma=\\pi[\/tex].<br \/>\nAssim,<br \/>\n[tex]\\qquad tg\\,(\\beta + \\gamma) = tg\\,\\dfrac{3 \\pi}{4}[\/tex]<br \/>\n[tex]\\qquad \\Leftrightarrow \\dfrac{tg\\,\\beta + tg\\,\\gamma}{1 &#8211; tg\\,\\beta \\cdot tg\\,\\gamma} = -1[\/tex]<br \/>\n[tex]\\qquad \\Leftrightarrow tg\\,\\beta \\cdot tg\\,\\gamma-1 = tg\\,\\beta + tg\\,\\gamma[\/tex]<br \/>\n[tex]\\qquad \\Leftrightarrow tg\\,\\beta \\cdot tg\\,\\gamma-tg\\,\\beta-tg\\,\\gamma + 1 = 2[\/tex]<br \/>\n[tex]\\qquad \\Leftrightarrow (tg\\,\\beta-1) \\cdot (tg\\,\\gamma-1) = 2[\/tex]<br \/>\nComo as tangentes s\u00e3o inteiros positivos, temos que um dos fatores \u00e9 [tex]2[\/tex] e outro [tex]1[\/tex]. Por outro lado, a fun\u00e7\u00e3o tangente crescente no intervalo considerado [tex] (i)[\/tex], portanto:<br \/>\n[tex]\\qquad tg\\,\\beta-1 = 1\\,[\/tex] e [tex]\\,tg\\,\\gamma-1 = 2[\/tex],<br \/>\nou seja,<br \/>\n[tex]\\qquad  tg\\,\\beta = 2\\,[\/tex] e [tex]\\,tg\\,\\gamma = 3[\/tex].<br \/>\nAssim, [tex] \\,  \\,  \\boxed{tg\\,{\\alpha} = 1; \\,  \\,  \\,  \\,  tg\\,\\beta = 2; \\,  \\,  \\,  \\,  tg\\,\\gamma = 3}[\/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:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) As tangentes dos \u00e2ngulos internos de um tri\u00e2ngulo s\u00e3o n\u00fameros inteiros positivos. Quais s\u00e3o os poss\u00edveis valores dessas tangentes? Solu\u00e7\u00e3o Sejam [tex] \\, \\, \\alpha \\leq \\beta \\leq \\gamma \\, \\, [\/tex] as medidas dos \u00e2ngulos do tri\u00e2ngulo cujas tangentes s\u00e3o inteiros positivos. Dessa forma, &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-saindo-inteiro-pela-tangente\/\">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-59325","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/59325","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=59325"}],"version-history":[{"count":32,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/59325\/revisions"}],"predecessor-version":[{"id":185983,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/59325\/revisions\/185983"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=59325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}