1998 AIME 第 2 题

All four constraints together confine $(x, y)$ to the square $1 \le x, y \le 30$

Inside that square, subtract the pairs with $y \gt 2x$ and those with $x \gt 2y;$ the two bad sets are disjoint and have equal size

The chains unpack into four conditions: $x \le 2y,$ $2y \le 60,$ $y \le 2x,$ and $2x \le 60.$ So $(x, y)$ lies in the square $1 \le x, y \le 30,$ and within it we must avoid $x \gt 2y$ and $y \gt 2x,$ which cannot both happen.

Pairs with $y \gt 2x:$ for each $x$ from $1$ to $14$ the values $y = 2x + 1, \ldots, 30$ work, giving $\sum_{x=1}^{14} (30 - 2x)$ $= 420 - 210$ $= 210$ pairs. By the symmetry swapping $x$ and $y,$ there are also $210$ pairs with $x \gt 2y.$

The answer is $30 \cdot 30 - 210 - 210 = 480.$