{"id":2083,"date":"2017-08-27T23:00:17","date_gmt":"2017-08-28T04:00:17","guid":{"rendered":"http:\/\/www.laurentlessard.com\/bookproofs\/?p=2083"},"modified":"2017-08-26T17:19:44","modified_gmt":"2017-08-26T22:19:44","slug":"a-tetrahedron-puzzle","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/a-tetrahedron-puzzle\/","title":{"rendered":"A tetrahedron puzzle"},"content":{"rendered":"

This post is about a 3D geometry Riddler puzzle<\/a> involving spheres and tetrahedra! Here is the problem:<\/p>\n

\nWe want to create a new gift for fall, and we have a lot of spheres, of radius 1, left over from last year\u2019s fidget sphere craze, and we\u2019d like to sell them in sets of four. We also have a lot of extra tetrahedral packaging from last month\u2019s Pyramid Fest. What\u2019s the smallest tetrahedron into which we can pack four spheres?\n<\/p><\/blockquote>\n

Here is my solution:
\n[Show Solution]<\/a><\/p>\n

\n

We’ll solve this problem using old school (read: high school) geometry. Rather than starting with four spheres and looking for the smallest tetrahedron that contains them, we’ll solve the equivalent problem of starting with a fixed tetrahedron and looking for the largest spheres we can pack inside. We’ll use a tetrahedron of unit sidelength. Place one vertex at the origin $\\vec O(0,0,0)$, one vertex on the $x$-axis $\\vec P(1,0,0)$, and one face in the $xy$-plane. This leads to a vertex at $\\vec Q(\\frac{1}{2},\\frac{\\sqrt{3}}{2},0)$ because $OPQ$ is an equilateral triangle. To find the final vertex $R$, we know by symmetry that its $x$ and $y$ coordinates are the same as those of the centroid of $OPQ$, so $x=\\frac{1}{2}$ and $y=\\frac{\\sqrt{3}}{6}$ (it’s at one third the altitude!). To find the final coordinate, we look for $z$ such that $x^2+y^2+z^2=1$, since all sides have length $1$. This leads us to the final vertex of the tetrahedron: $\\vec R(\\frac{1}{2},\\frac{\\sqrt{3}}{6},\\frac{\\sqrt{6}}{3})$.<\/p>\n

The center of the tetrahedron can be found by looking for a point of the form $\\alpha(\\vec P+ \\vec Q+\\vec R)$ (again by symmetry) whose $x$-coordinate is $\\frac{1}{2}$. Doing this leads us to the center: $\\vec C(\\frac{1}{2},\\frac{\\sqrt{3}}{6},\\frac{\\sqrt{6}}{12})$. Next, let’s look for the center of the sphere closest to the origin. By symmetry, it must be of the form $\\vec C_1 = \\beta \\vec C$. We want to find $\\beta$ such that it’s $x$-distance from the point $x=\\frac{1}{2}$ is $r$ (so it just touches the sphere closest to $\\vec P$), and we also want its $z$ coordinate to be $r$ (so it’s a distance $r$ from the $xy$ plane, because it’s a tangent point of the sphere). For a reference, see the figure below. I’m saying that $C_1$ (center of the sphere) should be on the line $OC$ and we should have $|C_1M| = |C_1A_1| = r$.<\/p>\n

\"\"<\/a><\/p>\n

The equations just described are:
\n\\begin{align}
\n\\frac{1}{2} \\beta &= \\frac{1}{2}-r\\\\
\n\\frac{\\sqrt{6}}{12} \\beta &= r
\n\\end{align}
\nSolving this system of equations yields $\\beta=\\frac{6-\\sqrt{6}}{5}$ and $r=\\frac{\\sqrt{6}-1}{10}\\approx 0.145$. So this is the radius of the largest sphere we can use when the tetrahedron has sidelength $1$. If the radius is instead $1$, then we must scale the tetrahedron accordingly. The resulting sidelength will be:
\n\\[
\ns = \\frac{10}{\\sqrt{6}-1} = 2\\sqrt{6}+2 \\approx 6.89898
\n\\]Here is a scale diagram of the entire pyramid with the four spheres inside:<\/p>\n

\"\"<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

This post is about a 3D geometry Riddler puzzle involving spheres and tetrahedra! Here is the problem: We want to create a new gift for fall, and we have a lot of spheres, of radius 1, left over from last year\u2019s fidget sphere craze, and we\u2019d like to sell them in sets of four. We … Continue reading “A tetrahedron puzzle”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":2086,"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":[10,2,14],"class_list":["post-2083","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-geometry","tag-riddler","tag-symmetry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2083","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=2083"}],"version-history":[{"count":4,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2083\/revisions"}],"predecessor-version":[{"id":2089,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2083\/revisions\/2089"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/2086"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=2083"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=2083"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=2083"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}