{"id":2348,"date":"2018-07-13T15:47:25","date_gmt":"2018-07-13T20:47:25","guid":{"rendered":"http:\/\/www.laurentlessard.com\/bookproofs\/?p=2348"},"modified":"2018-07-19T00:53:31","modified_gmt":"2018-07-19T05:53:31","slug":"l-bisector","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/l-bisector\/","title":{"rendered":"L-bisector"},"content":{"rendered":"

This post is about a geometry Riddler puzzle<\/a> involving bisecting a shape using only a straightedge and a pencil. Here is the problem:<\/p>\n

\nSay you have an “L” shape formed by two rectangles touching each other. These two rectangles could have any dimensions and they don’t have to be equal to each other in any way. (A few examples are shown below.)<\/p>\n

Using only a straightedge and a pencil (no rulers, protractors or compasses), how can you draw a single straight line that cuts the L into two halves of exactly equal area, no matter what the dimensions of the L are? You can draw as many lines as you want to get to the solution, but the bisector itself can only be one single straight line.<\/p>\n

\"\"\n<\/p><\/blockquote>\n

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

\n

This is a wonderful little puzzle with a very elegant solution. This solution is based on three observations:<\/p>\n

    \n
  1. You can find the center of a rectangle using only a straightedge and pencil. This can be done by drawing a straight lines connecting opposite corners of the rectangle. The center of the rectangle is where the two lines intersect.\n
  2. Any line passing through the center of a rectangle bisects its area. This follows because of symmetry. The two halves of the rectangle thus created are actually rotated versions of one another (180 deg. about the center).\n
  3. The “L” shape is simply two smaller rectangles. Alternatively, you can think of it as a large rectangle with a smaller rectangle subtracted from it. This is evident if you look at the shapes in the problem statement.\n<\/ol>\n

    And now, the solution is simple. Extend side lengths of the “L” shape to mark all the corners of the rectangles involved. Then find the centers of those rectangles as described in item 1 above, and join the centers using a straight line. This line bisects the L-shape because it bisects each of the rectangles. Depending on how you partition your L-shape, you can get two different bisectors:<\/p>\n

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

    Alternatively, you can think of the L-shape as the difference<\/em> of two rectangles. The logic is similar: since the line bisects each rectangle, it must also bisect their difference! Here is an illustration of this construction:<\/p>\n

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

    This strategy works for any<\/em> shape that is the sum or difference of two rectangles! So it’s not restricted to L-shapes. More examples:<\/p>\n

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

    Finally, the construction also works if the rectangles are replaced by parallelograms, or any other shapes with the two properties that we can construct the center easily and all lines through the center bisect the area. For example…<\/p>\n

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

    But why stop there? You can also extend this construction to 3D! Any plane passing through the center of a rectangular prism bisects its volume. Therefore, if you made a shape by adding and\/or subtracting three rectangular prisms, the plane passing through the three centers would bisect the volume of the entire shape!<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    This post is about a geometry Riddler puzzle involving bisecting a shape using only a straightedge and a pencil. Here is the problem: Say you have an “L” shape formed by two rectangles touching each other. These two rectangles could have any dimensions and they don’t have to be equal to each other in any … Continue reading “L-bisector”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":2359,"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-2348","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\/2348","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=2348"}],"version-history":[{"count":8,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2348\/revisions"}],"predecessor-version":[{"id":2365,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2348\/revisions\/2365"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/2359"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=2348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=2348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=2348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}