{"id":43283,"date":"2015-07-31T15:31:19","date_gmt":"2015-07-31T18:31:19","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=43283"},"modified":"2023-03-22T23:27:52","modified_gmt":"2023-03-23T02:27:52","slug":"problema-de-gincana-mais-um-problema-de-distancia","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-mais-um-problema-de-distancia\/","title":{"rendered":".Problema de Gincana: Mais um problema de  dist\u00e2ncia"},"content":{"rendered":"<p style=\"margin:20px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nNa figura, temos os pontos [tex]A = (0 , 9)[\/tex], [tex]B = (9 , 9)[\/tex], [tex]C = (9 , 0)[\/tex], [tex]D = (0 , 0)[\/tex] e os pontos [tex]P[\/tex] e [tex]Q[\/tex] dividem [tex]\\overline{DC}[\/tex] em tr\u00eas segmentos congruentes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2022\/08\/f244_2.png\" style=\"border:0px;\"  alt=\"\" width=\"530\" height=\"599\" class=\"aligncenter size-full wp-image-181655\" \/><\/p>\n<p>Sabendo que [tex]G[\/tex] \u00e9 o baricentro do tri\u00e2ngulo [tex]BPQ[\/tex], <strong>determine a dist\u00e2ncia [tex]d[\/tex] entre os pontos [tex]A[\/tex] e [tex]G[\/tex]<\/strong>.<\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:43px;\">\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\"\/>\nSendo [tex]B = (9 , 9)[\/tex], [tex]P = (3 , 0)[\/tex], [tex]Q = (6 , 0)[\/tex] e [tex]G = (x, y)[\/tex] o baricentro do tri\u00e2ngulo [tex]PBQ[\/tex], temos<br \/>\n[tex]\\qquad x = \\dfrac{9 + 3 + 6}{3} = 6 \\qquad [\/tex]  e  [tex]\\qquad y = \\dfrac{0 + 0 + 9}{3} = 3[\/tex].<br \/>\nA partir desses dados, podemos calcular a dist\u00e2ncia [tex]d[\/tex] usando as coordenadas dos pontos [tex]A = (0 , 9)\\,[\/tex] e [tex]\\,G = (6 , 3)[\/tex]. Vejamos:<br \/>\n[tex]\\qquad d = \\sqrt{(0-6)^2 + (9-3)^{2^\\textcolor{white}{1}}} = 6\\sqrt{2^\\textcolor{white}{(}}[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2022\/08\/f244_3.png\" style=\"border:0px;\"  alt=\"\" width=\"599\" height=\"600\" class=\"aligncenter size-full wp-image-181656\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2022\/08\/f244_3.png 599w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2022\/08\/f244_3-150x150.png 150w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2022\/08\/f244_3-300x300.png 300w\" sizes=\"auto, (max-width: 599px) 100vw, 599px\" \/><\/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\">\nPrimeira Gincana de 2015 &#8211; Clubes de Matem\u00e1tica da OBMEP<br \/>\nN\u00edvel B &#8211; Quest\u00e3o M\u00e9dia<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema Na figura, temos os pontos [tex]A = (0 , 9)[\/tex], [tex]B = (9 , 9)[\/tex], [tex]C = (9 , 0)[\/tex], [tex]D = (0 , 0)[\/tex] e os pontos [tex]P[\/tex] e [tex]Q[\/tex] dividem [tex]\\overline{DC}[\/tex] em tr\u00eas segmentos congruentes. Sabendo que [tex]G[\/tex] \u00e9 o baricentro do tri\u00e2ngulo [tex]BPQ[\/tex], determine a dist\u00e2ncia [tex]d[\/tex] entre os pontos [tex]A[\/tex] &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-mais-um-problema-de-distancia\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-43283","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/43283","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=43283"}],"version-history":[{"count":53,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/43283\/revisions"}],"predecessor-version":[{"id":185133,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/43283\/revisions\/185133"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=43283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}