# Squares and Pigeonhole Principle

• Jul 7th 2013, 11:31 AM
I-Think
Squares and Pigeonhole Principle
Let S be a square region of side length 2. Show that among any 9 points in the square, there are 3 which form a triangle of area $\displaystyle \leq{\frac{1}{2}$.

I subdivided S into 8 triangles of equal area, $\displaystyle \frac{1}{2}$.
Now I divide into 2 cases
Case A: Where 3 points are in one triangle.
Then the area of said triangle $\displaystyle \leq{\frac{1}{2}$.

Case B: A max of 2 points are in any triangle
I'm having trouble with this case. Help please?
• Jul 7th 2013, 12:34 PM
Plato
Re: Squares and Pigeonhole Principle
Quote:

Originally Posted by I-Think
Let S be a square region of side length 2. Show that among any 9 points in the square, there are 3 which form a triangle of area $\displaystyle \leq{\frac{1}{2}$.

Consider a unit square, a square of side length one and area of one square unit.
Any three points from that unit square that do form a triangle,
The area of such a triangle is $\displaystyle \le0.5$ square units.
Can you show that?

Now divide your given square into four sub-squares each of one square unit each.

Can you finish?