Problem 1410

1. Understanding the Problem:
We are given a circle with a diameter of 1 and 65 points inside it, with no three points being collinear. We need to prove that there exist three points among these 65 that form a triangle with an area less than or equal to $\frac{1}{72}$.

2. Convex Hull and Triangulation:
For $n$ points in the plane, where no three points are collinear, the convex hull of these points is a polygon with $k$ vertices, where $3 \leq k \leq n$. A complete triangulation of this convex hull results in $t = 2n - k - 2$ triangles. The minimum number of triangles, $t = n - 2$, occurs when $k = n$.

3. Area of Convex Hull:
Since the points are within a circle of diameter 1, the area of the convex hull is less than or equal to the area of the circle, which is $\pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}$.

4. Average Area of Triangles:
If we have $n$ points, the convex hull can be triangulated into $n - 2$ triangles. Therefore, the average area of each triangle in the triangulation is less than or equal to $\frac{\pi/4}{n-2}$.

5. Finding the Bound:
We need to find $n$ such that $\frac{\pi/4}{n-2} \leq \frac{1}{72}$. Solving this inequality:
$$\frac{\pi}{4(n-2)} \leq \frac{1}{72}$$
$$\pi \leq \frac{4(n-2)}{72}$$
$$\pi \leq \frac{n-2}{18}$$
$$18\pi \leq n - 2$$
$$n \geq 18\pi + 2$$
Using $\pi \approx 3.14159$:
$$n \geq 18 \times 3.14159 + 2 \approx 56.54862 + 2 \approx 58.54862$$
Since $n$ must be an integer, we round up to get $n \geq 59$.

6. Conclusion:
For $n = 65$, which is greater than 59, there must exist a triangle with an area less than or equal to $\frac{1}{72}$.