Question

Consider an equilateral triangle of side equal to 1. Choose 7 points inside the triangle, arbitrarly. Show that three of them will form a triangle of area less that you Hint: consider the center of the triangle and connect it to each of the vertices.
I first discovered that the area of the equilateral triangle is $\frac{\sqrt{3}}{4}$ . I then considered a point in the center of a triangle, and then connected it to each of the vertices. From there I gathered that each of the 3 triangles formed by these line segments gave an area of $\frac{\sqrt{3}}{12}$ . Please use the pigeonhole principle to explain.

Answer 1

Pigeon hole principal: If total kn+1 pigeons are distributed among the n holes then there is at least one hole containing at least k+1 pigeons.

Now till you have done is: The total area is subdivided into three triangular areas where each of them has area
V3/12
. Now for any three points lying inside one of these triangular area will form a triangle of area less than
V3/12
. So now only thing that we have to show is that, among these three areas there is at least one area which contains at least three of the seven points. And the area of the triangle made by those three points will be less than
V3/12
, which will prove the required.

Now use the pigeon hole principal, put n= number of triangular areas = 3 and nk+1= total number of points = 7 (Which gives k=2). These seven points are distributed among n=3 triangular arears. So from the principal there is atleast one area contains atleast k+1=3 points​​​​​​. Hence we are done.