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 .

I subdivided S into 8 triangles of equal area, .

Now I divide into 2 cases

Case A: Where 3 points are in one triangle.

Then the area of said triangle .

Case B: A max of 2 points are in any triangle

I'm having trouble with this case. Help please?

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

.

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 square units.

**Can you show that?**

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

Can you finish?