{"id":3571,"date":"2022-11-11T09:12:15","date_gmt":"2022-11-11T15:12:15","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=3571"},"modified":"2022-11-11T23:33:33","modified_gmt":"2022-11-12T05:33:33","slug":"randomly-cutting-a-sandwich","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/randomly-cutting-a-sandwich\/","title":{"rendered":"Randomly cutting a sandwich"},"content":{"rendered":"

This week’s Riddler Classic<\/a> is geometry puzzle about randomly slicing a square sandwich.<\/p>\n

I have made a square sandwich, and now it’s time to slice it. But rather than making a standard horizontal or diagonal cut, I instead pick two random points along the perimeter of the sandwich and make a straight cut from one point to the other. (These points can be on the same side.)<\/p>\n

What is the probability that the smaller resulting piece has an area that is at least one-quarter of the whole area?<\/p><\/blockquote>\n

My solution:
\n[Show Solution]<\/a><\/p>\n

\n

Suppose the sandwich has side length 1. Let $t \\in [0,1]$ be location of the first point (measured from a corner). We have three cases to consider:<\/p>\n

    \n
  1. If the second point is on the same side (probability $\\frac{1}{4}$), the area of the resulting smallest piece will be zero.\n
  2. If the second point is on an adjacent side (probability $\\frac{1}{2}$), the area of the resulting smallest piece will be $\\frac{1}{2}tr$, where $r \\in [0,1]$ is the location of the second point on the adjacent side.\n
  3. If the second point is on the opposite side (probability $\\frac{1}{4}$), the area of the resulting piece will be the smaller of the two resulting quadrilaterals. Specifically, it will be $\\min(\\frac{t+r}{2},1-\\frac{t+r}{2})$, where $r \\in [0,1]$ is the location of the second point on the opposite side.\n<\/ol>\n

    We will solve the general case, which is to find the probability that the smallest slice will have a fraction at least $p$ of the total area. the problem asks for $p=\\frac{1}{4}$. Since the total area of the square is $1$, we are effectively asking: “what is the probability that the smallest area is at least $p$?”.<\/p>\n

    Case 1.<\/strong> If we are in the first case (points on the same side), the probability that the area is greater than $p$ is zero, since the area is always zero. So we conclude that:
    \n\\[
    \na_1 = 0
    \n\\]<\/p>\n

    Case 2.<\/strong> If we are in the second case (points on an adjacent sides), we want the probability that $\\frac{1}{2}tr > p$, where $r$ and $t$ are chosen uniformly at random in $[0,1]$. This amounts to computing:
    \n\\begin{align}
    \na_2 &= \\int_0^1 \\int_0^1 \\mathbf{1}\\bigl( \\tfrac{1}{2}tr \\gt p \\bigr)\\,\\mathrm{d}r\\,\\mathrm{d}t \\\\
    \n&= \\int_{2p}^1 \\int_0^1 \\mathbf{1}\\bigl(r \\gt \\tfrac{2p}{t} \\bigr)\\,\\mathrm{d}r\\,\\mathrm{d}t \\\\
    \n&= \\int_{2p}^1 \\left( 1-\\frac{2p}{t}\\right)\\,\\mathrm{d}t \\\\
    \n&= 1-2p+2p\\log(2p)
    \n\\end{align}Note that we had to change the lower bound of the integral for $t$ because when $t \\lt 2p$, we have $\\tfrac{2p}{t}\\gt 1$, so the integrand is zero. Geometrically, this is the fraction of the region $(t,r)\\in[0,1]^2$ where $tr\\gt p^2$, sketched below.<\/p>\n

    \"\"<\/p>\n

    Case 3.<\/strong> If we are in the third case (points on opposite sides), we perform a similar computation with the area of the quadrilateral:
    \n\\begin{align}
    \na_3 &= \\int_0^1 \\int_0^1 \\mathbf{1}\\bigl( \\min(\\tfrac{t+r}{2},1-\\tfrac{t+r}{2}) \\gt p \\bigr)\\,\\mathrm{d}r\\,\\mathrm{d}t \\\\
    \n&= \\int_0^1 \\int_0^1 \\mathbf{1}\\bigl( \\tfrac{t+r}{2}\\gt p\\text{ and } 1-\\tfrac{t+r}{2} \\gt p \\bigr) \\,\\mathrm{d}r\\,\\mathrm{d}t \\\\
    \n&= \\int_0^1 \\int_0^1 \\mathbf{1}\\bigl( p \\lt \\tfrac{t+r}{2} \\lt 1-p\\bigr) \\,\\mathrm{d}r\\,\\mathrm{d}t \\\\
    \n&= \\int_0^1 \\int_0^1 \\mathbf{1}\\bigl( 2p \\lt t+r \\lt 2-2p\\bigr) \\,\\mathrm{d}r\\,\\mathrm{d}t \\\\
    \n&= 1-4p^2
    \n\\end{align}A trick to evaluating the above integral is to do it geometrically. Sketching the region for which $2p\\lt t+r\\lt 2-2p$, we see that it is the square of area $1$ with the two corners cut off, whose area is $(2p)^2$, so the result is $1-4p^2$. See below for an illustration.<\/p>\n

    \"\"<\/p>\n

    Putting everything together, the probability that the area of the smallest piece is at least $p$ is given by sum of the probabilities we already calculated, each weighted by their likelihood of occurrence, so $f(p) = \\frac{1}{4}a_1 + \\frac{1}{2}a_2 + \\frac{1}{4}a_3$. Plugging in the results from above and simplifying, we obtain our final answer:<\/p>\n

    $\\displaystyle
    \nf(p) = \\frac{3}{4}-p-p^2+p \\log (2 p)
    \n$<\/span><\/p>\n

    Here is what the function looks like:<\/p>\n

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

    Naturally, the plot only extends to $p=\\frac{1}{2}$ because the area of the smaller piece cannot be more than half of the total area. When $p=\\frac{1}{4}$ (what the original problem asked about), we obtain $f(\\frac{1}{4}) = \\frac{7}{16}-\\frac{\\log (2)}{4}\\approx 0.264213$. So the probability that the smaller piece is at least one quarter of the whole area is about 26.4%.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    This week’s Riddler Classic is geometry puzzle about randomly slicing a square sandwich. I have made a square sandwich, and now it’s time to slice it. But rather than making a standard horizontal or diagonal cut, I instead pick two random points along the perimeter of the sandwich and make a straight cut from one … Continue reading “Randomly cutting a sandwich”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":3578,"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":[28,10,8,2],"class_list":["post-3571","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-calculus","tag-geometry","tag-probability","tag-riddler"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3571","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=3571"}],"version-history":[{"count":11,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3571\/revisions"}],"predecessor-version":[{"id":3587,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3571\/revisions\/3587"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/3578"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=3571"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=3571"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=3571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}