{"id":13634,"date":"2014-02-02T23:27:02","date_gmt":"2014-02-03T01:27:02","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=13634"},"modified":"2023-01-19T01:45:56","modified_gmt":"2023-01-19T04:45:56","slug":"problema-olimpico-nivel-b-triangulos-e-angulos","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-olimpico-nivel-b-triangulos-e-angulos\/","title":{"rendered":".Problema Ol\u00edmpico – N\u00edvel B: Tri\u00e2ngulos e \u00c2ngulos"},"content":{"rendered":"
\n

Problema<\/strong><\/p>\n

\n

Na figura, [tex]CA=CB[\/tex] e [tex]CD=CE[\/tex].<\/p>\n

\"Figura4\"<\/p>\n

Calcule, em graus, a medida [tex]x[\/tex] do \u00e2ngulo [tex]EDB[\/tex]. <\/p>\n

<\/span><\/div>\n

 <\/p>\n

\n

Solu\u00e7\u00e3o 1<\/strong><\/p>\n

\nVamos considerar [tex]\\measuredangle{BAC} = \\alpha[\/tex] para obter alguns resultados preliminares que nos indicar\u00e3o [tex] \\measuredangle{DCE}[\/tex].
\n\"\"<\/p>\n
    \n
  • Como [tex]CA = CB[\/tex], o tri\u00e2ngulo [tex]ABC[\/tex] \u00e9 is\u00f3sceles de base [tex]AB[\/tex], logo segue que [tex]\\fbox{$\\measuredangle{CBA} = \\alpha$}[\/tex].<\/li>\n

    \"\"<\/p>\n

  • Como a soma dos \u00e2ngulos internos de um tri\u00e2ngulo \u00e9 [tex]180^{\\circ}[\/tex], temos que [tex]\\measuredangle{ADC} = 180^{\\circ}-40^{\\circ}-\\alpha [\/tex], ou seja [tex]\\fbox{$\\measuredangle{ADC}=140^{\\circ}-\\alpha$}[\/tex].<\/li>\n
  • Tamb\u00e9m temos que
    \n[tex]\\qquad \\measuredangle{CDE} = 180^{\\circ}-\\measuredangle{ADC}-x=180^{\\circ}-(140^{\\circ}-\\alpha)-x[\/tex]
    \n e com isso [tex]\\fbox{$\\measuredangle{CDE} = 40^{\\circ} + \\alpha-x $}[\/tex].<\/li>\n
  • Visto que [tex]\\triangle{CDE}[\/tex] \u00e9 is\u00f3sceles de base [tex]DE[\/tex], conclu\u00edmos que [tex]\\fbox{$\\measuredangle{CED}=\\measuredangle{CDE} = 40^{\\circ} + \\alpha-x$}[\/tex].<\/li>\n
  • E, finalmente,
    \n [tex]\\qquad \\measuredangle{DCE} = 180^{\\circ}-\\measuredangle{CDE}-\\measuredangle{CED}=180^{\\circ}-2\\cdot(40^{\\circ} + \\alpha-x)[\/tex],
    \ndonde [tex]\\fbox{$\\measuredangle{DCE} = 100^{\\circ}-2\\alpha + 2x$} [\/tex].<\/li>\n<\/ul>\n

    Agora, observe que [tex]\\hat{A} + \\hat{B} + \\hat{C} = 180^{\\circ}[\/tex], ou seja, [tex]\\alpha + \\alpha +(40^{\\circ} + 100^{\\circ} – 2\\alpha + 2x) = 180^{\\circ}[\/tex].
    \nSimplificando essa igualdade, obtemos [tex]140^{\\circ} + 2x = 180^{\\circ}[\/tex] e, consequentemente, a medida solicitada: [tex]\\fcolorbox{black}{yellow}{$x= 20^{\\circ}$}\\,[\/tex].<\/p>\n

    <\/span><\/p>\n

    \n

    Solu\u00e7\u00e3o elaborada pelo aluno do PIC-OBMEP Angelo Donizeti Lorenconi Junior<\/strong>, com contribui\u00e7\u00f5es dos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

     <\/p>\n

    \n

    Solu\u00e7\u00e3o 2 <\/strong><\/p>\n

    \nPelo enunciado temos a seguinte figura:<\/p>\n

    \"gincana\"<\/p>\n

    Suponhamos [tex]C\\hat{A}B=\\beta [\/tex] e [tex] C\\hat{D}E=\\alpha[\/tex].<\/p>\n