Pigeonhole principle for a triangle Consider a equilateral triangle of total area 1. Suppose 7 points are chosen inside. Show that some 3 points form a triangle of area $\leq\frac 14$.
 A: Choose one point $p$ and draw a line from $p$ to each of the other six points.
We can order the six points $a,b,c,d,e,f$ going clockwise around point $p$ and draw lines from $a$ to $b$, from $b$ to $c$, from $c$ to $d$, from $d$ to $e$ and from $e$ to $f$. 
This gives five disjoint triangles which fit inside the triangle of area $1$.
Therefore the total area of the five triangles is less than or equal to one and at least one triangle has area less than or equal to $\frac 15$. 
A: I have seen answers with 9 points (in comment link above) and 5 points (above). 
However, not with 7 points. Please see below:
Drawlines from midpoints of all 3 sides to each other
Now we have 6 points (3 vertices and 3 mid points)
we have 4 triangles of equal area 1/4. (equilateral triangle)
To add the 7th point will create a triangle of area less than 1/4.
QED
A: 
Divide the triangle into 4 equal size areas.
As the diagram above,
there are only 6 vertices the 4 triangles share such that the area of each group of 3 vertices is $\frac{1}{4}$, with any additional vertex attached will it fall into the center of any group of 3 vertices. Hence if 7 points are chosen, we can always see the area of 3 of them is less than $\frac{1}{4}$.
