2016 AIME I Problems/Problem 14

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Problem

Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ 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

First note that $1001 = 143 \cdot 7$ and $429 = 143 \cdot 3$ so every point of the form $(7k, 3k)$ is on the line. Then consider the line $l$ from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$. Translate the line $l$ so that $(7k, 3k)$ is now the origin. There is one square and one circle that intersect the line around $(0,0)$. Then the points on $l$ with an integral $x$-coordinate are, since $l$ has the equation $y = \frac{3x}{7}$:

\[(0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3).\]

We claim that the lower right vertex of the square centered at $(2,1)$ lies on $l$. Since the square has side length $\frac{1}{5}$, the lower right vertex of this square has coordinates $(2 + \frac{1}{10},  1 - \frac{1}{10}) = (\frac{21}{10}, \frac{9}{10})$. Because $\frac{9}{10} = \frac{3}{7} \cdot \frac{21}{10}$, $(\frac{21}{10}, \frac{9}{10})$ lies on $l$. Since the circle centered at $(2,1)$ is contained inside the square, this circle does not intersect $l$. Similarly the upper left vertex of the square centered at $(5,2)$ is on $l$. Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between $(0,0)$ and $(7,3)$ that intersect $l$. Since there are $\frac{1001}{7} = \frac{429}{3} = 143$ segments from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$, the above count is yields $143 \cdot 2 = 286$ circles. Since every lattice point on $l$ is of the form $(3k, 7k)$ where $0 \le k \le 143$, there are $144$ lattice points on $l$. Centered at each lattice point, there is one square and one circle, hence this counts $288$ squares and circles. Thus $m + n = 286 + 288 = \boxed{574}$.

(Solution by gundraja) [asy]size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted); for(int i=0;i<8;++i)for(int j=0;j<4;++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));} [/asy]

See also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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