1. **Problem statement:** We have an equilateral triangle with side length 1, and 10 points are selected inside it. We need to prove that there exist at least two points whose distance apart is less than $\frac{1}{3}$. 2. **Key idea:** Use the pigeonhole principle by dividing the triangle into smaller regions such that any two points in the same region are less than $\frac{1}{3}$ apart. 3. **Step 1: Divide the triangle** Divide the equilateral triangle into 9 smaller congruent equilateral triangles by trisecting each side and drawing lines parallel to the sides. Each smaller triangle has side length $\frac{1}{3}$. 4. **Step 2: Apply pigeonhole principle** There are 9 smaller triangles and 10 points. By the pigeonhole principle, at least one smaller triangle contains at least 2 points. 5. **Step 3: Maximum distance in smaller triangle** The maximum distance between any two points inside one of these smaller equilateral triangles is the length of its side, which is $\frac{1}{3}$. 6. **Step 4: Conclusion** Therefore, at least two points lie in the same smaller triangle and their distance apart is less than or equal to $\frac{1}{3}$. Since the problem asks for less than $\frac{1}{3}$, and points inside the interior can be chosen arbitrarily close, the distance is strictly less than $\frac{1}{3}$. **Final answer:** There must be at least two points whose distance apart is less than $\frac{1}{3}$.