3. Can a certain number of circles, the sum of whose radii is 1999, be placed in a circle of radius 1 such that no two of them have common interior points?
Solution. Inside the circle, we can construct a square with side 1. This square is divided into $n^{2}$ small squares with side $\frac{1}{n}$. In each of these small squares, we inscribe a circle with radius $\frac{1}{2 n}$. Then the sum of the radii of these circles is $n^{2} \cdot \frac{1}{2 n}=\frac{n}{2}$. From $\frac{n}{2}=1999$ we get $n=3998$. Therefore, in a circle with radius 1, 3998 non-overlapping circles can be placed, the sum of whose radii is 1999.