{"id":2710,"date":"2019-12-14T00:19:06","date_gmt":"2019-12-14T06:19:06","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=2710"},"modified":"2019-12-17T11:19:34","modified_gmt":"2019-12-17T17:19:34","slug":"prismatic-puzzle","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/prismatic-puzzle\/","title":{"rendered":"Prismatic puzzle"},"content":{"rendered":"<p>This week&#8217;s <a href=\"https:\/\/fivethirtyeight.com\/features\/can-you-solve-a-particularly-prismatic-puzzle\/\">Riddler Classic<\/a> is simple to state, but tricky to figure out:<\/p>\n<blockquote><p>\nWhat whole number dimensions can rectangular prisms have so that their volume (in cubic units) is the same as their surface area (in square units)?\n<\/p><\/blockquote>\n<p>Here is my solution:<br \/>\n<a href=\"javascript:Solution('soln_prism','toggle_prism')\" id=\"toggle_prism\">[Show Solution]<\/a><\/p>\n<div id=\"soln_prism\" style=\"display: none\">\n<p>If the side lengths of the prism are $a,b,c$, then the volume being equal to the surface area implies that<br \/>\n\\[<br \/>\nabc = 2ab+2ac+2bc.<br \/>\n\\]Let&#8217;s assume the side lengths are ordered, so $a \\ge b \\ge c$. We therefore have: $ab \\ge ac \\ge bc$. So $abc = 2ab+2ac+2bc \\leq 6ab$. Simplifying, we obtain $c \\leq 6$. We can now consider each case separately:<\/p>\n<ul>\n<li> If $c=6$, our equation reduces to $ab=3a+3b$. Rearranging and factoring, this is $(a-3)(b-3)=9$. Since $c=6$ was assumed to be the smallest side length, each of the factors $(a-3)$ and $(b-3)$ is at least $3$. So the only solution is $(6,6,6)$.\n<li> If $c=5$, our equation reduces to $3ab=10a+10b \\le 20a$. Therefore, $3b \\le 20$ and so $b \\le 6$. Since $c\\le b$, this means we only have to check $b=5$ and $b=6$. Checking each case, we find the only solution is $(10,5,5)$.\n<li> If $c=4$, our equation reduces to $ab=4a+4b$. Rearranging and factoring, this is $(a-4)(b-4)=16$. Each way of factoring $16$ as a product of two factors yields a different solution. These are: $16\\times 1$, $8\\times 2$, and $4\\times 4$. The resulting $(a,b,c)$ triples are: $(20,5,4)$, $(12,6,4)$, $(8,8,4)$.\n<li> If $c=3$, our equation reduces to $ab=6a+6b$. Rearranging and factoring, this is $(a-6)(b-6)=36$. Each way of factoring $36$ as a product of two factors yields a different solution. These are: $36\\times 1$, $18\\times 2$, $12\\times 3$, $9\\times 4$, $6\\times 6$. The resulting triples are: $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$.\n<li> If $c=2$, our equation reduces to $0=4a+4b$, which has no solutions.\n<li> If $c=1$, our equation reduces to $0=ab+2a+2b$, which also has no solutions.\n<\/ul>\n<p>And that&#8217;s it! Overall, the problem has exactly 10 solutions. They are:<br \/>\n$(6,6,6)$, $(10,5,5)$, $(20,5,4)$, $(12,6,4)$, $(8,8,4)$, $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$.<\/p>\n<p>Note: I assumed that &#8220;whole number&#8221; in this context means a positive integer. If we allow for side lengths of zero, then any degenerate prism with side lengths $(a,0,0)$ will have both area and volume equal to zero, so there are infinitely many solutions.<\/p>\n<p>Another note: If there were some way to upper-bound $a$, then we could do an exhaustive numerical search to find all solutions because there would only be finitely many triples $(a,b,c)$ to search. Alas, I have not been able to find an upper bound on $a$. I suspect it might not be possible, since if we set $a\\to\\infty$, the problem reduces to $bc=2b+2c$, which is the 2D version of the problem (a rectangle whose area equals its perimeter)!\n<\/p><\/div>\n","protected":false},"excerpt":{"rendered":"<p>This week&#8217;s Riddler Classic is simple to state, but tricky to figure out: What whole number dimensions can rectangular prisms have so that their volume (in cubic units) is the same as their surface area (in square units)? Here is my solution: [Show Solution] If the side lengths of the prism are $a,b,c$, then the &hellip; <a href=\"https:\/\/laurentlessard.com\/bookproofs\/prismatic-puzzle\/\" class=\"more-link\">Continue reading<span class=\"screen-reader-text\"> &#8220;Prismatic puzzle&#8221;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"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":[36,2],"class_list":["post-2710","post","type-post","status-publish","format-standard","hentry","category-riddler","tag-number-theory","tag-riddler"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2710","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=2710"}],"version-history":[{"count":6,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2710\/revisions"}],"predecessor-version":[{"id":2714,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2710\/revisions\/2714"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=2710"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=2710"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=2710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}