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I am working on a bit of a silly problem in my introductory discrete mathematics course. I have five pandas that I need to place in a pen, and I have a pen that is the shape of an equilateral triangle with sides of length 2. I want to prove that If I were to place these five pandas in the pen, there must be at least two pandas that are of less than or equal to unit distance from each other. I definitely think this is a pigeonhole principle problem, although I am not sure what should be the "boxes'' in this situation. Any assistance would be greatly appreciated.

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    $\begingroup$ Think about triforce. $\endgroup$ – T. Bongers Jan 15 '16 at 1:05
  • $\begingroup$ Yes, pigeonhole principle is your friend here. For a more difficult challenge, determine the maximum of minimum distances between pandas (across all possible placements). $\endgroup$ – hardmath Jan 15 '16 at 3:04
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In this diagram, the large triangle has sides of $2$ while the four smaller ones have sides of $1$.

enter image description here

So place five pandas into those four small triangles. The maximum distance between any two points in any small triangle is clearly just $1$. Use the pigeonhole principle and finish from here.

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Hint: you can cut the pen into four triangles with side $1$

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Connect the mid-points of the sides of the triangle with line segments.This cuts the triangle into 4 pieces. At least 2 of 5 pandas are in the same piece. Each piece is a 1x1x1 equilateral triangle. So two pandas in the same piece cannot be at a distance of more than 1 from each other.

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