{"id":1329,"date":"2016-09-26T21:59:45","date_gmt":"2016-09-27T02:59:45","guid":{"rendered":"http:\/\/www.laurentlessard.com\/bookproofs\/?p=1329"},"modified":"2016-09-27T01:25:19","modified_gmt":"2016-09-27T06:25:19","slug":"cutting-a-circular-table","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/cutting-a-circular-table\/","title":{"rendered":"Cutting a circular table"},"content":{"rendered":"

This Riddler Classic<\/a> puzzle is about cutting circles out of rectangles!<\/p>\n

You\u2019re on a DIY kick and want to build a circular dining table which can be split in half so leaves can be added when entertaining guests. As luck would have it, on your last trip to the lumber yard, you came across the most pristine piece of exotic wood that would be perfect for the circular table top. Trouble is, the piece is rectangular. You are happy to have the leaves fashioned from one of the slightly-less-than-perfect pieces underneath it, but there\u2019s still the issue of the main circle. You devise a plan: cut two congruent semicircles from the perfect 4-by-8-foot piece and reassemble them to form the circular top of your table. What is the radius of the largest possible circular table you can make?\n<\/p><\/blockquote>\n

Here is my solution to the case of a general rectangular table. The result may surprise you!
\n[Show Solution]<\/a><\/p>\n

\n

We’ll solve the more general problem in which we vary the dimensions of the rectangle. Specifically, we’ll suppose the rectangle has a height of $1$ and a width of $x \\ge 1$. After some thought, you can convince yourself that there are two arrangements that lead to large areas. The first arrangement is to use semicircles with flat sides aligned with the top and bottom sides of the table. Here is a diagram:
\n\"circular_table2\"<\/a>
\nIn this case, when the semicircle is as large as possible, it will either touch the other semicircle, or it will touch the opposite side of the rectangle, whichever comes first. The top side enforces the restriction $r \\le 1$. Meanwhile, being tangent to the other semicircle can be characterized by calculating the width and height in terms of $r$. Specifically,
\n\\begin{align*}
\n2r \\sin \\theta &= 1 \\\\
\n2r + 2r \\cos \\theta &= x
\n\\end{align*}Rearranging and using the fact that $\\sin^2\\theta + \\cos^2\\theta = 1$ for any $\\theta$, we conclude that:
\n\\[
\n\\left(\\frac{1}{2r}\\right)^2 + \\left(\\frac{x}{2r}-1\\right)^2 = 1
\n\\]Solving this equation for $x$ yields, and together with $r\\le 1$ found before, we conclude that:
\n\\[
\nr = \\textrm{min}\\left( \\frac{x^2+1}{4x}, 1 \\right)
\n\\]<\/p>\n

The other competitive arrangement of semicircles involves a slanted arrangement depicted in the figure below. Here, both semicircles are tangent to two sides, touch a third side, and also touch each other along the flat edge.
\n\"circular_table1\"<\/a>
\nThe algebra in this case is more complicated, but it can be resolved in a similar fashion. Once again, we calculate the height and width as a function of $r$ and $x$, and we obtain:
\n\\begin{align*}
\nr + r\\sin\\theta &= 1 \\\\
\n2r + r\\cot\\theta-r\\cos\\theta &= x
\n\\end{align*}We’ll work to eliminate $\\theta$ as we did in the simpler case. Solving for $\\sin\\theta$ in the first equation and substituting into the second, we obtain:
\n\\begin{align*}
\nr + r\\sin\\theta &= 1 \\\\
\n2r + r\\cos\\theta\\left(\\frac{2r-1}{1-r}\\right) &= x
\n\\end{align*}Once again, use the fact that $\\sin^2\\theta + \\cos^2\\theta = 1$ for any $\\theta$, and we find:
\n\\[
\n\\left(\\frac{1-r}{r}\\right)^2 + \\left(\\frac{1-r}{2r-1}\\right)^2\\left(\\frac{x}{r}-2\\right)^2 = 1
\n\\]Unfortunately, this equation isn’t as easy to solve as the previous one. Clearing all fractions, we are left with a fourth degree polynomial in $r$:
\n\\[
\n4 r^4 -4(x+4)r^3+(x+4)^2r^2-(6+4x+2x^2)r+(1 + x^2) = 0
\n\\]For any fixed $r$, this can be solved numerically. There is a closed-form expression for $r$ in terms of $x$, but it’s quite messy and not particularly instructive.<\/p>\n

The interesting result here is that different schemes are optimal for different $x$ values. In the animation below, I plot $\\frac{\\pi r^2}{x}$ as a function of $x$. This is the area ratio of the circular area to the rectangular area. The transition occurs at $x\\approx 2.47243$.
\n\"circular_table\"<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

This Riddler Classic puzzle is about cutting circles out of rectangles! You\u2019re on a DIY kick and want to build a circular dining table which can be split in half so leaves can be added when entertaining guests. As luck would have it, on your last trip to the lumber yard, you came across the … Continue reading “Cutting a circular table”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":1338,"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],"class_list":["post-1329","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-geometry","tag-riddler"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/1329","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=1329"}],"version-history":[{"count":11,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/1329\/revisions"}],"predecessor-version":[{"id":1343,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/1329\/revisions\/1343"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/1338"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=1329"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=1329"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=1329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}