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?

Re: Squares and Pigeonhole Principle

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**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?