Problem:
Centered at each lattice point in the coordinate plane are a circle radius 101​ and a square with sides of length 51​ whose sides are parallel to the coordinate axes. The line segment from (0,0) to (1001,429) intersects m of the squares and n of the circles. Find m+n.
Solution:
Let A=(0,0) and B=(1001,429)=(143â‹…7,143â‹…3). Between the points (0,0) and (7,3),AB can intersect the square at lattice point (m,n) only if it passes between the upper-left and lower-right corners of the square. That is,
n+101​≥73​(m−101​) and n−101​≤73​(m+101​)
This implies 3m−1≤7n≤3m+1. The only lattice points (m,n) with 0≤m≤7 which satisfy this requirement are (0,0),(2,1),(5,2), and (7,3). In the cases of (0,0) and (7,3),AB passes through the center of the square centered at that point, so it also intersects the circle centered at that point. In the cases of (2,1) and (5,2),AB passes through the lower-right and upper-left corners of the square centered at that point, respectively, so it does not interest the circle centered at that point. Altogether the segment joining (0,0) to (7,3) intersects 4 squares and 2 circles. The same conclusion applies to the segment joining (7k−7,3k−3) to (7k,3k) for each k with 1≤k≤143. Because the points (7k,3k ) belong to two of these segments for 1≤k≤142,AB intersects 4⋅143−142=430 of the squares and 2⋅143−142=144 of the circles. The requested sum is 430+144=574​.
The problems on this page are the property of the MAA's American Mathematics Competitions