{"id":3257,"date":"2024-04-13T16:54:20","date_gmt":"2024-04-13T16:54:20","guid":{"rendered":"https:\/\/blogs.ua.es\/dimates\/?p=3257"},"modified":"2024-04-21T16:11:01","modified_gmt":"2024-04-21T16:11:01","slug":"de-hexagono-a-triangulo","status":"publish","type":"post","link":"https:\/\/blogs.ua.es\/dimates\/2024\/04\/13\/de-hexagono-a-triangulo\/","title":{"rendered":"De hex\u00e1gono a tri\u00e1ngulo"},"content":{"rendered":"<pre>Problema 2 del concurso Marat\u00f3 de problemes 2024\nSe dirige a una edad de: 14-15 a\u00f1os<\/pre>\n<p>El hex\u00e1gono de la izquierda tiene todos los \u00e1ngulos rectos, y las longitudes de dos lados son a y b, tal y como est\u00e1 indicado en la figura.<\/p>\n<p>Si dividimos el hex\u00e1gono en las tres partes que muestra la figura y las giramos adecuadamente, podemos componer un tri\u00e1ngulo.<\/p>\n<p>Expresa la longitud x, que corresponde al lado izquierdo del pol\u00edgono inicial, en funci\u00f3n de a y b.<\/p>\n<p>(Por si es demasiado peque\u00f1o, aclaro que a y b son las longitudes de los dos lados horizontales superiores del hex\u00e1gono.)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/blogs.ua.es\/dimates\/files\/2024\/04\/345poligonos.png\" alt=\"\" width=\"300\" height=\"300\" class=\"alignnone size-full wp-image-3258\" \/><\/p>\n<p>Soluci\u00f3n: <a href=\"https:\/\/blogs.ua.es\/dimates\/2024\/04\/21\/solucion-a-de-hexagono-a-triangulo\/\">Aqu\u00ed<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problema 2 del concurso Marat\u00f3 de problemes 2024 Se dirige a una edad de: 14-15 a\u00f1os El hex\u00e1gono de la izquierda tiene todos los \u00e1ngulos rectos, y las longitudes de dos lados son a y b, tal y como est\u00e1 indicado en la figura. Si dividimos el hex\u00e1gono en las tres partes que muestra la [&hellip;]<\/p>\n","protected":false},"author":4267,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2242030,2242026,1738,2849],"tags":[],"class_list":["post-3257","post","type-post","status-publish","format-standard","hentry","category-marato","category-marato-de-problemes","category-olimpiadas","category-problemas"],"_links":{"self":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/3257","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/users\/4267"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/comments?post=3257"}],"version-history":[{"count":3,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/3257\/revisions"}],"predecessor-version":[{"id":3269,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/3257\/revisions\/3269"}],"wp:attachment":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/media?parent=3257"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/categories?post=3257"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/tags?post=3257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}