$n$<\/th>\n	$k_n$<\/th>\n	Range of $k$ for this $n$<\/th>\n<\/tr>\n<\/thead>\n
$2$<\/td>\n	$0.333333$<\/td>\n	$0.333333 \\leq k$<\/td>\n<\/tr>\n
$3$<\/td>\n	$0.089642$<\/td>\n	$0.089642 \\leq k \\leq 0.333333$<\/td>\n<\/tr>\n
$4$<\/td>\n	$0.039573$<\/td>\n	$0.039573 \\leq k \\leq 0.089642$<\/td>\n<\/tr>\n
$5$<\/td>\n	$0.021016$<\/td>\n	$0.021016 \\leq k \\leq 0.039573$<\/td>\n<\/tr>\n
$6$<\/td>\n	$0.012511$<\/td>\n	$0.012511 \\leq k \\leq 0.021016$<\/td>\n<\/tr>\n
$7$<\/td>\n	$0.008056$<\/td>\n	$0.008056 \\leq k \\leq 0.012511$<\/td>\n<\/tr>\n
$8$<\/td>\n	$0.005494$<\/td>\n	$0.005494 \\leq k \\leq 0.008056$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n<\/body>","protected":false},"excerpt":{"rendered":" his week\u2019s Riddler Classic involves designing maximum-area polygons with a fixed budget on the length of the perimeter and the number of vertices. The original problem involved designing enclosures for hamsters, but I have paraphrased the problem to make it more concise. You want to build a polygonal enclosure consisting of posts connected by walls. … Continue reading “Polygons with perimeter and vertex budgets”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":2936,"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":[10,26,2],"class_list":["post-2906","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-geometry","tag-optimization","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t

\n
For a fixed perimeter, the maximum-area polygon with $n$ sides is a regular<\/em> polygon. Since we are interested in maximizing area, we should therefore always use regular polygons as our shape, and we should always use our entire mass budget of $1$ kg.<\/p>\n
Let\u2019s assume the weight of the posts $k$ is given and fixed. If the perimeter is $p$ and the number of sides is $n$, the formula for the area of a regular $n$-gon is as follows (explanation here<\/a>).
\n\\[
\nA = \\frac{p^2}{4n \\tan\\frac{\\pi}{n}}
\n\\]Our total budget constraint is that $p + kn \\leq 1$. Since our goal is to maximize area we should always use our entire budget. So $p = 1-kn$. This means that $n \\leq \\frac{1}{k}$ in order to ensure we respect the weight budget. So for a fixed $k$, the maximum area of an $n$-gon enclosure is given by the formula<\/p>\n
$\\displaystyle
\nA = \\frac{(1-kn)^2}{4n\\tan\\frac{\\pi}{n}}\\qquad\\text{for: }2\\leq n \\leq \\frac{1}{k}
\n$<\/span><\/p>\n
Let\u2019s talk about limiting cases for a moment.<\/p>\n
\n
If $n=2$, we have $\\tan(\\pi\/n) = \\infty$ so $A=0$. This makes sense; an enclosure with only two sides will be flat and have zero area.\n<\/li>\n
If $k\\to 0$ we can take $n\\to\\infty$. A polygon with infinitely many sides but a fixed perimeter is simply a circle, so let\u2019s see if this checks out. Using the fact that $\\lim_{n\\to\\infty} n \\tan\\tfrac{\\pi}{n} = \\pi$, we obtain a total area of $A = \\tfrac{1}{4\\pi}$. This is precisely the area of a circle with perimeter $1$! Indeed, if $2\\pi r = 1$, then we obtain $r = \\frac{1}{2\\pi}$ and therefore $A = \\pi r^2 = \\frac{1}{4\\pi}$.\n<\/li>\n<\/ul>\n
So for each $k$, we should choose the $n \\in \\{2,3,4,\\dots,\\lfloor \\tfrac{1}{k}\\rfloor\\}$ that maximizes $A$. Here is a plot showing plots of $A$ for each $n$ and $k$. When $k$ is large (posts weigh a lot), we try to use as few of them as possible. So at first, the triangle ($n=3$) is best. As we decrease $k$, it eventually becomes more efficient to use a square ($n=4$), and so on. As $k$ goes to zero, we obtain the solution with $n\\to\\infty$, which is the circular enclosure discussed above.<\/p>\n
$\"\"$ <\/a><\/p>\n
To find out where the transition occurs between $n$ and $n+1$, we should look for the value of $k$ at which the areas match (indicated by the black dots on the plot). So we want to find $k$ such that:
\n\\[
\n\\frac{(1-kn)^2}{4n\\tan\\frac{\\pi}{n}} = \\frac{(1-k(n+1))^2}{4(n+1)\\tan \\frac{\\pi}{n+1}}
\n\\]Solving this equation for $k$, we obtain:<\/p>\n
$\\displaystyle
\nk_n = \\frac{ \\sqrt{n \\tan \\frac{\\pi}{n}}-\\sqrt{(n+1) \\tan \\frac{\\pi}{n+1}} }{ (n+1)\\sqrt{n \\tan \\frac{\\pi}{n}}-n\\sqrt{(n+1) \\tan \\frac{\\pi}{n+1}} }
\n$<\/span><\/p>\n
We can interpret this formula as follows: $k_n$ is the smallest value of $k$ for which we should use $n$ posts in our enclosure. So if we make $k$ smaller, it becomes more efficient to use $(n+1)$ posts instead. Here are the first few numerical values of $k_n$:<\/p>\n\n\n\n\n\n <\/colgroup>\n\n\n\n\n\n\n\n\n\n\n
$n$<\/th>\n $k_n$<\/th>\n Range of $k$ for this $n$<\/th>\n<\/tr>\n<\/thead>\n
$2$<\/td>\n $0.333333$<\/td>\n $0.333333 \\leq k$<\/td>\n<\/tr>\n
$3$<\/td>\n $0.089642$<\/td>\n $0.089642 \\leq k \\leq 0.333333$<\/td>\n<\/tr>\n
$4$<\/td>\n $0.039573$<\/td>\n $0.039573 \\leq k \\leq 0.089642$<\/td>\n<\/tr>\n
$5$<\/td>\n $0.021016$<\/td>\n $0.021016 \\leq k \\leq 0.039573$<\/td>\n<\/tr>\n
$6$<\/td>\n $0.012511$<\/td>\n $0.012511 \\leq k \\leq 0.021016$<\/td>\n<\/tr>\n
$7$<\/td>\n $0.008056$<\/td>\n $0.008056 \\leq k \\leq 0.012511$<\/td>\n<\/tr>\n
$8$<\/td>\n $0.005494$<\/td>\n $0.005494 \\leq k \\leq 0.008056$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
<\/p>\n<\/body>","protected":false},"excerpt":{"rendered":"
his week\u2019s Riddler Classic involves designing maximum-area polygons with a fixed budget on the length of the perimeter and the number of vertices. The original problem involved designing enclosures for hamsters, but I have paraphrased the problem to make it more concise. You want to build a polygonal enclosure consisting of posts connected by walls. … Continue reading “Polygons with perimeter and vertex budgets”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":2936,"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":[10,26,2],"class_list":["post-2906","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-geometry","tag-optimization","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t