How many?

Probability Level 3

Consider an equilateral triangle in which each side has length 1 centimetre. What is the smallest number of points that must be chosen from the region enclosed by the triangle (including the boundary) so that at least two of these points have distance of at most 1 2 \dfrac{1}{2} centimetre between them.

This is a part of the Set .


The answer is 5.

Khang Nguyen Thanh by Khang Nguyen Thanh , 27, Vietnam

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1 solution

If we choose the 4 points ( A , B , C , D A,B,C,D as shown in Figure 1), we have: A B = B C = C A = 1 , A D = B D = C D = 3 3 AB=BC=CA=1, AD=BD=CD=\dfrac{\sqrt3}{3} , so any two of these points have distance of at most 1 2 \dfrac{1}{2} .

Consider the case we choose 5 points from the region enclosed by the triangle. We divide this triangle into 4 smaller equilateral triangle (by midpoints of the sides). From pigeonhole principle, there exist at least two points belong to the same smaller triangle. The distance of them do not exceed 1 2 \dfrac{1}{2} .

So, the smallest number of points we need is 5 5 .

1 Helpful 1 Interesting 0 Brilliant 0 Confused

As always with "minimum", you have to show that

  1. We have a lower bound.
  2. No smaller number is a lower bound.

What you need to do, is to show why 4 isn't a lower bound, in order to conclude that "the smallest number of points is 5".

Calvin Lin Staff - 4 years ago

I found a generalisation. If n²+1 points are chosen in an equilateral triangle with side length 1, there are 2 points whose distance is atmost 1/n

Prayas Rautray - 4 years ago

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