{"id":2906,"date":"2020-08-21T16:23:10","date_gmt":"2020-08-21T21:23:10","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=2906"},"modified":"2020-08-21T16:55:37","modified_gmt":"2020-08-21T21:55:37","slug":"polygons-with-perimeter-and-vertex-budgets","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/polygons-with-perimeter-and-vertex-budgets\/","title":{"rendered":"Polygons with perimeter and vertex budgets"},"content":{"rendered":"

his week’s Riddler Classic<\/a> 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.<\/p>\n

\nYou want to build a polygonal enclosure consisting of posts connected by walls. Each post weighs $k$ kg. The walls weigh $1$ kg per meter. You are allowed a maximum budget of $1$ kg for the posts and walls.<\/p>\n

What\u2019s the greatest value of $k$ for which you should use four posts rather than three?<\/p>\n

Extra credit:<\/em> For which values of $k$ should you use five posts, six posts, seven posts, and so on?\n<\/p><\/blockquote>\n

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

\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’s 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’s talk about limiting cases for a moment.<\/p>\n