HMMT 二月 2004 · 几何 · 第 5 题

英文解析

5. Find the area of the region of the xy -plane defined by the inequality | x | + | y | + | x + y | ≤ 1.
   Solution: 3 / 4
   To graph this region we divide the xy -plane into six sectors depending on which ofx, y, x + y are ≥ 0, or ≤ 0. The inequality simplifies in each case:
   Sector Inequality Simplified inequalityx ≥ 0 , y ≥ 0 , x + y ≥ 0 x + y + ( x + y ) ≤ 1 x + y ≤ 1 / 2
   x ≥ 0 , y ≤ 0 , x + y ≥ 0 x − y + ( x + y ) ≤ 1 x ≤ 1 / 2
   x ≥ 0 , y ≤ 0 , x + y ≤ 0 x − y − ( x + y ) ≤ 1 y ≥ − 1 / 2
   x ≤ 0 , y ≥ 0 , x + y ≥ 0 − x + y + ( x + y ) ≤ 1 y ≤ 1 / 2
   x ≤ 0 , y ≥ 0 , x + y ≤ 0 − x + y − ( x + y ) ≤ 1 x ≥ − 1 / 2
   x ≤ 0 , y ≤ 0 , x + y ≤ 0 − x − y − ( x + y ) ≤ 1 x + y ≥ − 1 / 2
   We then draw the region; we get a hexagon as shown. The hexagon intersects eachregion in an isosceles right triangle of area 1 / 8, so the total area is 6 · 1 / 8 = 3 / 4.