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Suppose 5 points are chosen at random inside an equilateral triangle with sides of length 1. Show that there is at least one pair of these points that are separated by a distance of at most 1/2.

I can't figure this out, this applies to pigeon hole principle I believe but I dont get how this problem can be approached

Roy Kesserwani
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  • You are right: you need to apply the pigeonhole principle. You have $5$ points which is a hint that you need to divide the area of the triangle into $4$ sub-areas. – Hoda Feb 05 '14 at 07:19
  • i divided the area of the rectangle by 4 sub areas which makes it an area of 1/8 each piece and then what? – Roy Kesserwani Feb 05 '14 at 07:24
  • How did you do that? You need to be more specific. Not every partitioning works. It should be done in a way that if two points fall in the same region, then you can say their distance is at most $1/2$. – Hoda Feb 05 '14 at 07:27

1 Answers1

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Draw the equilateral triangle whose vertices are the midpoints of the larger triangle. This creates four equilateral triangles inside the larger one. The side of these triangles is $1/2$.

Now, observe that the maximum separation of two points inside one of the smaller triangles approaches $1/2$. The separation equals $1/2$ if the two points lie on two of the vertices.

Since there are four such equilateral triangles, and five points, two of the points must lie in the same small equilateral triangle. And the separation of these two points is at most $1/2$.

John
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  • Hey John, have you learned this before or was this answer based on your logic? its a weird question but my question is how did you know the answer? – Roy Kesserwani Feb 05 '14 at 07:32
  • I hadn't heard it before. I first thought about stuffing three of the points in the corners of the big triangle, and noted that their separation was $1$. Then I saw that if I stuck a fourth point on the midpoint of one of the sides, the separation was $1/2$. It took me a few minutes to recognize that the side of the four smaller triangles was $1/2$, and that the same argument applied for the corners of those triangles. But anyway ... I just played with it. Mathematics is a lot about playing with things. – John Feb 05 '14 at 07:36