{"id":745,"date":"2016-07-09T21:02:25","date_gmt":"2016-07-10T02:02:25","guid":{"rendered":"http:\/\/www.laurentlessard.com\/bookproofs\/?p=745"},"modified":"2016-07-20T21:52:39","modified_gmt":"2016-07-21T02:52:39","slug":"cutting-polygons-in-half","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/cutting-polygons-in-half\/","title":{"rendered":"Cutting polygons in half"},"content":{"rendered":"

This Riddler puzzle<\/a> is about cutting polygons in half. Here is the problem:<\/p>\n

\nThe archvillain Laser Larry threatens to imminently zap Riddler Headquarters (which, seen from above, is shaped like a regular pentagon with no courtyard or other funny business). He plans to do it with a high-powered, vertical planar ray that will slice the building exactly in half by area, as seen from above. The building is quickly evacuated, but not before in-house mathematicians move the most sensitive riddling equipment out of the places in the building that have an extra high risk of getting zapped.<\/p>\n

Where are those places, and how much riskier are they than the safest spots? (It\u2019s fine to describe those places qualitatively.)<\/p>\n

Extra credit: Get quantitative! Seen from above, how many high-risk points are there? If there are infinitely many, what is their total area?\n<\/p><\/blockquote>\n

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

\n

A note on interpretations<\/h3>\n

Because the problem doesn’t explain how to define “risk”, there are two main ways to interpret the problem.<\/p>\n

    \n
  1. We can assume the laser cut will happen randomly, and the “risk” of a point is related to the likelihood that the cut will hit it. If the point is a geometrical point and the cut is infinitely thin, then all points have zero probability. If assume the point has a finite size, then the problem can make sense, but we still need to worry about how the cuts are distributed<\/em>. For example:\n