{"id":58157,"date":"2016-03-23T18:56:32","date_gmt":"2016-03-23T21:56:32","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=58157"},"modified":"2023-03-31T20:40:31","modified_gmt":"2023-03-31T23:40:31","slug":"problema-de-olimpiada-controlando-beta","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-olimpiada-controlando-beta\/","title":{"rendered":".Problema de Olimp\u00edada: Controlando Beta&#8230;"},"content":{"rendered":"<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSejam [tex]\\alpha[\/tex] e [tex]\\beta[\/tex] n\u00fameros inteiros positivos tais que<br \/>\n[tex]\\qquad \\qquad \\dfrac{43}{197}\\lt\\dfrac{\\alpha}{\\beta}\\lt\\dfrac{17}{77}[\/tex].<br \/>\nEncontre o menor valor poss\u00edvel para [tex]\\beta[\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" >\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\"\/>\nComo [tex]\\dfrac{43}{197}\\lt\\dfrac{\\alpha}{\\beta}\\lt\\dfrac{17}{77}[\/tex] e [tex]\\alpha[\/tex] e [tex]\\beta[\/tex] s\u00e3o n\u00fameros positivos, segue que [tex]\\dfrac{77}{17}\\lt\\dfrac{\\beta}{\\alpha}\\lt\\dfrac{197}{43}[\/tex], portanto<\/p>\n<p style=\"margin:15px\">\n<p>[tex]\\qquad 4+\\dfrac{9}{17}\\lt\\dfrac{\\beta}{\\alpha}\\lt 4+\\dfrac{25}{43}.\\qquad\\qquad \\qquad (i)[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Assim, podemos concluir que [tex]4\\lt\\dfrac{\\beta}{\\alpha}\\lt 5[\/tex] e, com isso,<br \/>\n[tex]\\qquad \\beta=4\\alpha+z[\/tex], com [tex]0\\lt z \\lt \\alpha.\\qquad\\qquad (ii)[\/tex]<br \/>\nDe [tex](i)[\/tex] e [tex](ii)[\/tex], segue que<br \/>\n[tex]\\qquad 4+\\dfrac{9}{17}\\lt 4+\\dfrac{z}{\\alpha}\\lt 4+\\dfrac{25}{43}[\/tex],<\/p>\n<p style=\"margin:15px\">\n<p>portanto, [tex]\\dfrac{9}{17}\\lt \\dfrac{z}{\\alpha}\\lt  \\dfrac{25}{43}[\/tex] e, consequentemente, <\/p>\n<p style=\"margin:15px\">\n<p>[tex]\\qquad \\dfrac{43z}{25}\\lt \\alpha \\lt  \\dfrac{17z}{9}.\\qquad\\qquad \\quad (iii)[\/tex]<br \/>\nMas qual \u00e9 o menor valor inteiro de [tex]z[\/tex] para o qual as desigualdades [tex](iii)[\/tex] definem um inteiro positivo [tex]\\alpha[\/tex]?<br \/>\nVejamos.<\/p>\n<ul>\n<li>[tex]z=1 \\Rightarrow  \\dfrac{43}{25}\\lt \\alpha \\lt  \\dfrac{17}{9} \\Rightarrow  1+\\dfrac{18}{25}\\lt \\alpha \\lt  1+\\dfrac{8}{9} \\Rightarrow 1\\lt\\alpha\\lt 2[\/tex];<br \/>\nmas n\u00e3o existe um n\u00famero inteiro entre [tex]1[\/tex] e [tex]2[\/tex].<\/li>\n<li>[tex]z=2 \\Rightarrow  \\dfrac{86}{25}\\lt \\alpha \\lt  \\dfrac{34}{9}  \\Rightarrow   3+\\dfrac{11}{25}\\lt \\alpha \\lt  3+\\dfrac{7}{9} \\Rightarrow 3\\lt\\alpha\\lt 4[\/tex];<br \/>\nmas n\u00e3o existe um n\u00famero inteiro entre [tex]3[\/tex] e [tex]4[\/tex].<\/li>\n<li>[tex]z=3 \\Rightarrow  \\dfrac{129}{25}\\lt \\alpha \\lt  \\dfrac{51}{9} \\Rightarrow  5+\\dfrac{4}{25}\\lt \\alpha \\lt  5+\\dfrac{2}{3} \\Rightarrow 5\\lt\\alpha\\lt 6[\/tex];<br \/>\nmas n\u00e3o existe um n\u00famero inteiro entre [tex]5[\/tex] e [tex]6[\/tex].<\/li>\n<li>[tex]z=4 \\Rightarrow  \\dfrac{172}{25}\\lt \\alpha \\lt  \\dfrac{68}{9} \\Rightarrow  6+\\dfrac{22}{25}\\lt \\alpha \\lt  7+\\dfrac{5}{9} \\Rightarrow 6\\lt\\alpha\\lt 8[\/tex];<br \/>\nassim, temos um valor inteiro para [tex]\\alpha[\/tex]: [tex]\\alpha=7[\/tex]. Neste caso, por [tex](ii)[\/tex], [tex]\\beta=4\\alpha + z=4\\times 7+4=32[\/tex].<\/li>\n<\/ul>\n<p>Observe que se [tex]z\\ge 5[\/tex], ent\u00e3o [tex]\\alpha \\gt  \\dfrac{43z}{25} \\ge\\dfrac{43}{5}\\gt 8[\/tex] e com isso, por [tex](ii)[\/tex], ter\u00edamos [tex]\\beta\\gt 37[\/tex].<\/p>\n<p style=\"margin:10px\">\n<p>Portanto, de fato, o valor m\u00ednimo para [tex]\\beta[\/tex] \u00e9 [tex]\\fbox{$\\beta=32$}[\/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=\"gg\">\nOlimp\u00edada: <strong>INMO &#8211; 2005<\/strong> (\u00cdndia)\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema Sejam [tex]\\alpha[\/tex] e [tex]\\beta[\/tex] n\u00fameros inteiros positivos tais que [tex]\\qquad \\qquad \\dfrac{43}{197}\\lt\\dfrac{\\alpha}{\\beta}\\lt\\dfrac{17}{77}[\/tex]. Encontre o menor valor poss\u00edvel para [tex]\\beta[\/tex]. Solu\u00e7\u00e3o Como [tex]\\dfrac{43}{197}\\lt\\dfrac{\\alpha}{\\beta}\\lt\\dfrac{17}{77}[\/tex] e [tex]\\alpha[\/tex] e [tex]\\beta[\/tex] s\u00e3o n\u00fameros positivos, segue que [tex]\\dfrac{77}{17}\\lt\\dfrac{\\beta}{\\alpha}\\lt\\dfrac{197}{43}[\/tex], portanto [tex]\\qquad 4+\\dfrac{9}{17}\\lt\\dfrac{\\beta}{\\alpha}\\lt 4+\\dfrac{25}{43}.\\qquad\\qquad \\qquad (i)[\/tex] Assim, podemos concluir que [tex]4\\lt\\dfrac{\\beta}{\\alpha}\\lt 5[\/tex] e, com isso, [tex]\\qquad \\beta=4\\alpha+z[\/tex], com [tex]0\\lt z \\lt \\alpha.\\qquad\\qquad &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-de-olimpiada-controlando-beta\/\">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-58157","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/58157","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=58157"}],"version-history":[{"count":34,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/58157\/revisions"}],"predecessor-version":[{"id":185969,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/58157\/revisions\/185969"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=58157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}