{"id":3059,"date":"2021-05-29T10:00:32","date_gmt":"2021-05-29T15:00:32","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=3059"},"modified":"2021-05-29T10:08:36","modified_gmt":"2021-05-29T15:08:36","slug":"tetrahedron-optimization","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/tetrahedron-optimization\/","title":{"rendered":"Tetrahedron optimization"},"content":{"rendered":"<p>This week&#8217;s <a href=\"https:\/\/fivethirtyeight.com\/features\/can-you-crack-the-case-of-the-crystal-key\/\">Riddler Classic<\/a> is a short problem 3D geometry. Here we go! (I paraphrased the question)<\/p>\n<blockquote><p>\nA polyhedron has six edges. Five of the edges have length $1$. What is the largest possible volume?\n<\/p><\/blockquote>\n<p>Here is my solution<br \/>\n<a href=\"javascript:Solution('soln_tetravol','toggle_tetravol')\" id=\"toggle_tetravol\">[Show Solution]<\/a><\/p>\n<div id=\"soln_tetravol\" style=\"display: none\">\n<p>The only polyhedron with six edges is a <a href=\"https:\/\/en.wikipedia.org\/wiki\/Tetrahedron\">tetrahedron<\/a>, which is a pyramid with a triangular base. Two of the faces will be equilateral triangles that share a common edge. This accounts for the five edges of length 1. The length of the sixth edge is determined by the angle between the faces, which we will call $\\theta$. Here is an animation showing the different tetrahedra you get as you vary $\\theta$:<\/p>\n<p><a href=\"https:\/\/laurentlessard.com\/bookproofs\/wp-content\/uploads\/2021\/05\/tetra_anim_reduced.gif\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/laurentlessard.com\/bookproofs\/wp-content\/uploads\/2021\/05\/tetra_anim_reduced.gif\" alt=\"\" width=\"1104\" height=\"748\" class=\"aligncenter size-full wp-image-3060\" \/><\/a><\/p>\n<p>In this diagram, $AB=BC=AC=AD=BD=1$ and $OD=OC=\\frac{\\sqrt{3}}{2}$.<\/p>\n<p>The volume is equal to<br \/>\n\\begin{align}<br \/>\nV&#038;=\\frac{1}{3}(\\text{Area of base})\\cdot(\\text{altitude}) \\\\<br \/>\n&#038;= \\frac{1}{3}(\\text{Area ABC})\\cdot(DG) \\\\<br \/>\n&#038;= \\frac{1}{3}\\left( \\frac{1}{2} (AB)(OC) \\right) \\cdot \\left( (OD) \\sin\\theta \\right) \\\\<br \/>\n&#038;= \\frac{1}{3} \\cdot \\frac{1}{2}\\cdot 1 \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\sin\\theta \\\\<br \/>\n&#038;= \\frac{1}{8}\\cdot\\sin\\theta<br \/>\n\\end{align}Therefore, the maximum volume is $\\frac{1}{8}$ and it occurs when $\\theta=90^\\circ$. This is intuitive because the area of the base is fixed, so the largest volume occurs when the altitude is as large as possible.\n<\/p><\/div>\n","protected":false},"excerpt":{"rendered":"<p>This week&#8217;s Riddler Classic is a short problem 3D geometry. Here we go! (I paraphrased the question) A polyhedron has six edges. Five of the edges have length $1$. What is the largest possible volume? Here is my solution [Show Solution] The only polyhedron with six edges is a tetrahedron, which is a pyramid with &hellip; <a href=\"https:\/\/laurentlessard.com\/bookproofs\/tetrahedron-optimization\/\" class=\"more-link\">Continue reading<span class=\"screen-reader-text\"> &#8220;Tetrahedron optimization&#8221;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":3060,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[7],"tags":[10,26],"class_list":["post-3059","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-geometry","tag-optimization"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3059","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/comments?post=3059"}],"version-history":[{"count":9,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3059\/revisions"}],"predecessor-version":[{"id":3070,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3059\/revisions\/3070"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/3060"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=3059"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=3059"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=3059"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}