Problem 1221

1. Divide the Square into Smaller Regions:
Consider a unit square with side length 1. Divide this square into 4 smaller squares, each with side length $\frac{1}{2}$. Each of these smaller squares has an area of $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

2. Pigeonhole Principle:
By the pigeonhole principle, if we place 9 points inside the unit square, at least one of the smaller squares must contain at least 3 of these points. This is because there are only 4 smaller squares and 9 points, so at least one square must contain $\left\lceil \frac{9}{4} \right\rceil = 3$ points.

3. Area of Triangle in Smaller Square:
Consider one of these smaller squares with side length $\frac{1}{2}$. Any triangle formed by 3 points within this smaller square will have an area less than or equal to the area of the smaller square. The maximum area of a triangle within this smaller square is achieved when the triangle is right-angled and its legs are the sides of the smaller square. The area of such a triangle is:
$$\text{Area} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

4. Conclusion:
Therefore, any triangle formed by 3 points within this smaller square will have an area less than or equal to $\frac{1}{8}$. Since we have shown that at least one of the smaller squares must contain at least 3 of the 9 points, we can conclude that there always exists a triangle with an area less than or equal to $\frac{1}{8}$.

$$\boxed{\text{True}}$$