3. For each pair of different natural numbers a and b, not greater than 20, Petya drew on the board the line $y=a x+b$ (that is, he drew the lines $y=x+2, \ldots, y=x+20, y=2 x+1, y=2 x+3, \ldots$, $y=2 x+20, \ldots, y=3 x+1, y=3 x+2, y=3 x+4, \ldots, y=3 x+20, \ldots, y=20 x+1, \ldots, y=20 x+19)$. Vasya drew on the same board a circle of radius 1 centered at the origin. How many of Petya's lines intersect Vasya's circle? (I. Rubanov)
Answer: 190. First solution. The graph $y=a x+b$ intersects the coordinate axes at points $A(-b / a, 0)$ and $B(0, b)$. If $ba$. Then points $K(-1,0)$ and $N(0,1)$ lie on the legs of triangle $O A B$, where $O$ is the origin, and point $M(-1,1)$ lies inside this triangle, since $a \cdot(-1)+b=b-a \geq 1$. Thus, the angle $K M N$ is entirely below the line $y=a x+b$, and Vasya's circle does not intersect this line, as it is entirely contained within the angle $K M N$.

Therefore, Petya's line intersects Vasya's circle if and only if $bb$. We then reason as at the beginning of the first solution.