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Question: There is a castle shaped like an equilateral triangle. The length of each side of the castle is $200$ meters. Five guards are placed around the perimeter of the castle spaced as far apart as possible. Using the pigeonhole principle, prove that there are always two guards within $100$ meters of each other. You may assume the following theorems are true:

• The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

• The distance between two points inside an equilateral triangle is less than the side of the triangle.

My attempt: To place guards around the castle, I would start by placing three at the vertices. Then because they are spaced as far apart as possible, the next two would be placed at the midpoints of two sides. Then the guards on the vertices will be 100 meters from the guards on the midpoints and the guards on the midpoints will be 100 meters apart because of the first theorem. Where does the pigeonhole principle come in however?

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You can subdivide the triangle into four regions by connecting the midpoints of each of the segments. Note that only three of these regions share an edge with the triangle so these are the ones where guards may be placed. Thus there are 3 regions and 4 guards so by PHP one of the regions contains at least 2 guards. By the theorems cited above these 2 guards are within 100m of each other.

As Misha points out below, placing 4 guards suffices to prove the result.

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    $\begingroup$ This seems to answer a different question in which the guards are placed inside the triangle, not around the perimeter. (Granted, I'm not sure why the perimeter question needs 5 guards; for that version, 4 are enough.) $\endgroup$ Nov 17, 2021 at 3:53
  • $\begingroup$ Ah, I did misread the question. Thanks for catching that. Fixing the answer now. $\endgroup$
    – raka
    Nov 17, 2021 at 4:00

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