2016 AIME I Problem 14

Below is the professionally curated solution for Problem 14 of the 2016 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AIME I solutions, or check the answer key.

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Concepts:lattice pointgreatest common divisordistance formula

Difficulty rating: 3370

14.

Centered at each lattice point in the coordinate plane are a circle radius 110\frac{1}{10} and a square with sides of length 15\frac{1}{5} whose sides are parallel to the coordinate axes. The line segment from (0,0)(0, 0) to (1001,429)(1001, 429) intersects mm of the squares and nn of the circles. Find m+n.m + n.

Solution:

Since gcd(1001,429)=143,\gcd(1001, 429) = 143, the segment passes through the lattice points (7k,3k)(7k, 3k) for k=0,,143k = 0, \ldots, 143 and consists of 143143 translated copies of the segment from (0,0)(0, 0) to (7,3).(7, 3). The line is y=37x.y = \frac{3}{7}x. It meets the square centered at (m,n)(m, n) exactly when its height passes within 110\frac{1}{10} of nn for some xx within 110\frac{1}{10} of m,m, that is when 3m7n110+37110=17,\left|\frac{3m}{7} - n\right| \le \frac{1}{10} + \frac{3}{7} \cdot \frac{1}{10} = \frac{1}{7}, or equivalently 3m7n1.|3m - 7n| \le 1.

For 0m70 \le m \le 7 the solutions are (0,0)(0, 0) and (7,3)(7, 3) with 3m7n=0,3m - 7n = 0, and (2,1)(2, 1) and (5,2)(5, 2) with 3m7n=1.3m - 7n = \mp 1. In the first two the line passes through the center, so it meets the circle as well. In the other two, equality means the line passes exactly through a corner of the square (for (2,1),(2, 1), the corner (2.1,0.9)(2.1, 0.9)), while its distance to the center is 132+72=158>110,\frac{1}{\sqrt{3^2 + 7^2}} = \frac{1}{\sqrt{58}} \gt \frac{1}{10}, so it misses the circle. Thus each copy of the segment meets 44 squares and 22 circles.

The 142142 interior lattice points (7k,3k)(7k, 3k) are each shared by two consecutive copies, so m=4143142=430m = 4 \cdot 143 - 142 = 430 and n=2143142=144,n = 2 \cdot 143 - 142 = 144, giving m+n=574.m + n = 574.

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