{"id":476,"date":"2014-07-23T08:18:39","date_gmt":"2014-07-23T15:18:39","guid":{"rendered":"http:\/\/www.imaginary-institute.com\/blog\/?p=476"},"modified":"2014-07-26T11:12:22","modified_gmt":"2014-07-26T18:12:22","slug":"a-little-geometry-problem","status":"publish","type":"post","link":"https:\/\/www.imaginary-institute.com\/blog\/2014\/07\/23\/a-little-geometry-problem\/","title":{"rendered":"Circle Packing In A Square"},"content":{"rendered":"

\"circleInBox\"<\/a>Given a circle of radius r<\/em>, and around it a box of side 2r<\/em>. We want to put another circle into the gap between the circle and a corner of the box. Find a formula for the center and radius of the largest circle that can be fit into that space. This might be a little trickier than you first think.<\/p>\n

The solution’s after the break, but don’t give up too soon!\u00a0 <\/p>\n

\"circleInBox3a\"<\/a><\/p>\n

They key thing is to not<\/em> assume the circle is centered in the “middle” of the gap.<\/p>\n

Symmetry says the circle must be on the diagonal. The radius we seek, which I’ll call a<\/em>, makes the inner circle touch both edges of the box and the big circle; clearly no circle can be bigger. The diagonal of the square is r\u221a2<\/em>, so the piece of the diagonal in the gap is g = r (\u221a2-1)<\/em>.<\/p>\n

\"circleInBox3b\"<\/a><\/p>\n

 <\/p>\n

Focus on the little region in the bottom left. The line from the corner to the circle has length g<\/em>. The upper part of the line has length a<\/em>. The lower part is the diagonal of a box with side a, so that lower part has length a\u221a2<\/em>. Thus g = a + a\u221a2<\/em>, or a = g\/(1+\u221a2<\/em>).<\/p>\n

When I first worked this out I had all sorts of triangles and things floating around. I was definitely over-thinking it! The answer is nice and easy.<\/p>\n

Note that this is basically an Appolonian gasket, but because we’re tangent to a circle and two perpendicular lines (rather than three circles) our result is a simple special case.<\/p>\n","protected":false},"excerpt":{"rendered":"

Given a circle of radius r, and around it a box of side 2r. We want to put another circle into the gap between the circle and a corner of the box. Find a formula for the center and radius of the largest circle that can be fit into that space. This might be a […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-476","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/posts\/476","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/comments?post=476"}],"version-history":[{"count":19,"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/posts\/476\/revisions"}],"predecessor-version":[{"id":507,"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/posts\/476\/revisions\/507"}],"wp:attachment":[{"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/media?parent=476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/categories?post=476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.imaginary-institute.com\/blog\/wp-json\/wp\/v2\/tags?post=476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}