{"id":3082,"date":"2023-02-18T07:16:18","date_gmt":"2023-02-18T12:16:18","guid":{"rendered":"https:\/\/josmfs.net\/?p=3082"},"modified":"2023-06-09T12:51:12","modified_gmt":"2023-06-09T16:51:12","slug":"linked-triangles-problem","status":"publish","type":"post","link":"https:\/\/josmfs.net\/wordpress\/2023\/02\/18\/linked-triangles-problem\/","title":{"rendered":"Linked Triangles Problem"},"content":{"rendered":"<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft  wp-image-3081\" src=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Prob-Fig2.jpg\" alt=\"\" width=\"200\" height=\"153\" srcset=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Prob-Fig2.jpg 997w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Prob-Fig2-300x230.jpg 300w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Prob-Fig2-768x588.jpg 768w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Prob-Fig2-624x478.jpg 624w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/>I found this problem from the 1981 Canadian Math Society\u2019s magazine, <em>Crux Mathematicorum<\/em>, to be quite challenging.<\/p>\n<p>\u201c<em>Proposed by Kaidy Tan, Fukien Teachers\u2019 University, Foochow, Fukien, China.<\/em><\/p>\n<p>An isosceles triangle has vertex A and base BC. Through a point F on AB, a perpendicular to AB is drawn to meet AC in E and BC produced in D. Prove synthetically that<\/p>\n<p style=\"text-align: center;\">Area of AFE = 2 Area of CDE \u00a0\u00a0if and only if\u00a0 AF = CD.\u201d<\/p>\n<p>See the <a href=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Trangles-Problem-221001.pdf\">Linked Triangles Problem<\/a><\/p>\n<p><strong>(Update 2\/22\/2023, 6\/9\/2023) Alternative Solutions<\/strong><\/p>\n<p><!--more--><\/p>\n<p><strong>(Update 2\/22\/2023)<\/strong><\/p>\n<p>The indefatigable Oscar Rojas has come up with an alternative solution to the problem. It takes a bit of study to verify.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-full wp-image-3088\" src=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Rojas.jpg\" alt=\"\" width=\"825\" height=\"552\" srcset=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Rojas.jpg 825w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Rojas-300x201.jpg 300w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Rojas-768x514.jpg 768w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/02\/Linked-Triangles-Rojas-624x418.jpg 624w\" sizes=\"auto, (max-width: 825px) 100vw, 825px\" \/><\/p>\n<p><strong>(Update 6\/9\/2023)<\/strong><\/p>\n<p>Srhiri Noe from Morocco, a former high school teacher and teacher trainer, has provided another solution to the problem.\u00a0 I have edited the solution slightly for clarity.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-large wp-image-3199\" src=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/06\/Linked-Triangles-Srhiri-1024x841.jpg\" alt=\"\" width=\"625\" height=\"513\" srcset=\"https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/06\/Linked-Triangles-Srhiri-1024x841.jpg 1024w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/06\/Linked-Triangles-Srhiri-300x246.jpg 300w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/06\/Linked-Triangles-Srhiri-768x630.jpg 768w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/06\/Linked-Triangles-Srhiri-624x512.jpg 624w, https:\/\/josmfs.net\/wordpress\/wp-content\/uploads\/2023\/06\/Linked-Triangles-Srhiri.jpg 1200w\" sizes=\"auto, (max-width: 625px) 100vw, 625px\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>I found this problem from the 1981 Canadian Math Society\u2019s magazine, Crux Mathematicorum, to be quite challenging. \u201cProposed by Kaidy Tan, Fukien Teachers\u2019 University, Foochow, Fukien, China. An isosceles triangle has vertex A and base BC. Through a point F on AB, a perpendicular to AB is drawn to meet AC in E and BC [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[212,13],"class_list":["post-3082","post","type-post","status-publish","format-standard","hentry","category-puzzles-and-problems","tag-crux-mathematicorum","tag-plane-geometry"],"_links":{"self":[{"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/posts\/3082","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/comments?post=3082"}],"version-history":[{"count":4,"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/posts\/3082\/revisions"}],"predecessor-version":[{"id":3200,"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/posts\/3082\/revisions\/3200"}],"wp:attachment":[{"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/media?parent=3082"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/categories?post=3082"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/josmfs.net\/wordpress\/wp-json\/wp\/v2\/tags?post=3082"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}