HMMT 十一月 2010 · 冲刺赛 · 第 2 题

英文解析

2. [ 5 ] A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integercoordinate vertices intersect the interior of this circle?
   Answer: 132 By symmetry, the answer is four times the number of squares in the first quadrant.
   Let’s identify each square by its coordinates at the bottom-left corner, ( x, y ). When x = 0, we canhave y = 0 . . . 5, so there are 6 squares. (Letting y = 6 is not allowed because that square intersectsonly the boundary of the circle.) When x = 1, how many squares are there? The equation of the circle
   √ √
   2 2
   is y = 36 − x = 36 − 1 is between 5 and 6, so we can again have y = 0 . . . 5. Likewise for x = 2
   √
   and x = 3. When x = 4 we have y = 20 which is between 4 and 5, so there are 5 squares, and when
   √
   x = 5 we have y = 11 which is between 3 and 4, so there are 4 squares. Finally, when x = 6, we havey = 0, and no squares intersect the interior of the circle. This gives 6 + 6 + 6 + 6 + 5 + 4 = 33. Sincethis is the number in the first quadrant, we multiply by four to get 132.