2020 AIME II Problem 12

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12.

Let mm and nn be odd integers greater than 1.1. An m×nm \times n rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers 11 through n,n, those in the second row are numbered left to right with the integers n+1n + 1 through 2n,2n, and so on. Square 200200 is in the top row, and square 20002000 is in the bottom row. Find the number of ordered pairs (m,n)(m, n) of odd integers greater than 11 with the property that, in the m×nm \times n rectangle, the line through the centers of squares 200200 and 20002000 intersects the interior of square 1099.1099.

Answer: 248
Concepts:coordinate geometryslopemidpointcasework
Difficulty rating: 3160
Solution:

Use coordinates (column,row).(\text{column}, \text{row}). Square 200200 is in the top row, so n200n \ge 200 (hence n201n \ge 201 as nn is odd) and its center is (200,1).(200, 1). Square 20002000 is in the bottom row, so its column is b=2000(m1)nb = 2000 - (m-1)n with 1bn,1 \le b \le n, i.e. (m1)n<2000mn,(m-1)n \lt 2000 \le mn, and its center is (b,m).(b, m). Since mm and nn are odd, bb is even, so the midpoint (200+b2,m+12)\left(\frac{200 + b}{2}, \frac{m+1}{2}\right) of the two centers has integer coordinates; its square number is (m1)n+200+b2=22002=1100.\frac{(m-1)n + 200 + b}{2} = \frac{2200}{2} = 1100. Its column 100+b2100 + \frac{b}{2} lies between 101101 and n,n, so the line passes through the center of square 1100,1100, and square 10991099 sits immediately to its left in the same row.

Square 10991099 lies only in that row, and the line crosses that row's horizontal strip in a segment centered (by symmetry) at the center of square 1100,1100, extending 12s\frac{1}{2|s|} to each side, where s=m1b200s = \frac{m-1}{b - 200} is the slope. So the line meets the interior of square 10991099 exactly when 12s>12,\frac{1}{2|s|} \gt \frac{1}{2}, that is s<1,|s| \lt 1, i.e. m1<b200m - 1 \lt |b - 200| =1800(m1)n= |1800 - (m-1)n| (a vertical line, b=200,b = 200, fails).

Since n201n \ge 201 and (m1)n<2000,(m-1)n \lt 2000, we need m1<10,m - 1 \lt 10, so m{3,5,7,9}.m \in \{3, 5, 7, 9\}. For m=3:m = 3: odd n[667,999],n \in [667, 999], 167167 values, excluding n9001|n - 900| \le 1 (odd cases 899,901899, 901) leaves 165.165. For m=5:m = 5: odd n[401,499],n \in [401, 499], 5050 values, excluding 449,451449, 451 leaves 48.48. For m=7:m = 7: odd n[287,333],n \in [287, 333], 2424 values, excluding 299,301299, 301 leaves 22.22. For m=9:m = 9: odd n[223,249],n \in [223, 249], 1414 values, excluding 225225 leaves 13.13. The total is 165+48+22+13=248.165 + 48 + 22 + 13 = 248.

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