{"id":149413,"date":"2020-03-18T21:51:17","date_gmt":"2020-03-19T00:51:17","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=149413"},"modified":"2023-06-11T01:06:26","modified_gmt":"2023-06-11T04:06:26","slug":"problemao-um-problema-antigo","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-um-problema-antigo\/","title":{"rendered":".Problem\u00e3o: Um problema antigo da OBM"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n

\n

(OBM<\/strong>)<\/span> Sejam [tex]a, b, c[\/tex] e [tex]d[\/tex] n\u00fameros reais distintos tais que [tex]a[\/tex] e [tex]b[\/tex] s\u00e3o ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-3cx-8d=0[\/tex] e [tex]c[\/tex] e [tex]d[\/tex] s\u00e3o ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-3ax-8b=0[\/tex]. Calcule a soma [tex]a+b+c+d[\/tex].<\/p>\n

\n<\/div>\n

\"explicador_p\" <\/p>\n

AJUDA<\/p>\n

\n

Voc\u00ea se lembra das Rela\u00e7\u00f5es de Girard<\/em><\/strong>?
\nCaso n\u00e3o, visite esta Sala<\/strong><\/a> do nosso Blog, pois essas rela\u00e7\u00f5es ajudar\u00e3o a resolver este problem\u00e3o.\n<\/div>\n

\n

Solu\u00e7\u00e3o<\/p>\n

\n

Utilizando as Rela\u00e7\u00f5es de Girard<\/em><\/strong> nas equa\u00e7\u00f5es [tex]\\textcolor{red}{x^2-3ax-8b=0}\\,[\/tex] e [tex]\\,\\textcolor{blue}{x^2-3cx-8d=0}[\/tex], \u00e9 f\u00e1cil perceber que [tex]\\textcolor{blue}{a+b=3c}\\,[\/tex] e que [tex]\\,\\textcolor{red}{c+d=3a}\\,.[\/tex] Agora, somando e subtraindo membro a membro essas duas igualdades obtidas, temos que:<\/p>\n\n\n
\n[tex](a+b)+(c+d)=3c+3a\\\\
\nb+d=3c+3a-a-c\\\\
\nb+d=2c+2a\\\\
\n \\boxed{b+d=2(a+c)}.\\qquad \\qquad \\textcolor{#800000}{(I)}[\/tex] <\/td>\n		\n[tex]\\qquad (a+b)-(c+d)=3c-3a\\\\
\n\\qquad b-d=3c-3a-a+c\\\\
\n\\qquad b-d=4c-4a\\\\
\n\\qquad \\boxed{b-d=4(c-a)}\\,.\\qquad \\qquad \\textcolor{#800000}{(II)} [\/tex]<\/td>\n<\/tr>\n<\/table>\n

Por outro lado, como [tex]a[\/tex] \u00e9 raiz de [tex]x^2-3cx-8d=0[\/tex] e [tex]c[\/tex] \u00e9 raiz da equa\u00e7\u00e3o [tex]x^2-3ax-8b=0[\/tex], obtemos respectivamente:
\n [tex]\\qquad a^2-3ca-8d=0\t\\qquad \\textcolor{#800000}{(III)}[\/tex]
\ne
\n [tex]\\qquad c^2-3ca-8b=0\\,.\t\\qquad \\textcolor{#800000}{(IV)}[\/tex]<\/p>\n

Subtraindo as igualdades [tex]\\textcolor{#800000}{(III)}[\/tex] e [tex]\\textcolor{#800000}{(IV)}[\/tex] e utilizando a igualdade [tex]\\textcolor{#800000}{(II)}[\/tex], segue que:
\n[tex]\\quad a^2-c^2-8d+8b=0 \\\\
\n\\quad a^2-c^2-8(d-b)=0 \\\\
\n\\quad a^2-c^2=8(d-b) \\\\
\n\\quad a^2-c^2\\stackrel{\\textcolor{#800000}{(II)}}{=}8\\cdot 4(a-c) \\\\
\n\\quad (a-c)(a+c)=32(a-c).\\qquad \\textcolor{#800000}{(V)}[\/tex]
\nMas [tex]a\\,[\/tex] e [tex]\\, c[\/tex] s\u00e3o n\u00fameros distintos; logo, [tex]a-c \\ne 0[\/tex] e podemos dividir a equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(V)}[\/tex] por [tex]a-c[\/tex], concluindo que [tex]a+c=32[\/tex].
\nComo [tex]a+c=32[\/tex] e, por [tex]\\textcolor{#800000}{(I)}[\/tex], [tex]b+d=2(a+c)[\/tex], temos que [tex]b+d=2\\cdot 32=64[\/tex].
\nPortanto, [tex](a+c)+(b+d)=32+64[\/tex], ou seja, [tex] \\fcolorbox{black}{#eee0e5}{$a+b+c+d=96$}\\,.[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 9\u00ba ano do E. F.) (OBM) Sejam [tex]a, b, c[\/tex] e [tex]d[\/tex] n\u00fameros reais distintos tais que [tex]a[\/tex] e [tex]b[\/tex] s\u00e3o ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-3cx-8d=0[\/tex] e [tex]c[\/tex] e [tex]d[\/tex] s\u00e3o ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-3ax-8b=0[\/tex]. Calcule a soma [tex]a+b+c+d[\/tex]. AJUDA Voc\u00ea se lembra das Rela\u00e7\u00f5es de Girard? Caso n\u00e3o, visite esta … <\/p>\n

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-149413","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/149413","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=149413"}],"version-history":[{"count":11,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/149413\/revisions"}],"predecessor-version":[{"id":194287,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/149413\/revisions\/194287"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=149413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}