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Difference between revisions of "2016 AIME I Problems/Problem 14"

(Created page with "== Problem == Centered at each lattice point in the coordinate plane are a circle radius <math>\frac{1}{10}</math> and a square with sides of length <math>\frac{1}{5}</math> ...")
 
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== Solution ==
 
== Solution ==
  
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First note that <math>1001 = 143 \cdot 7</math> and <math>429 = 143 \cdot 3</math> so every point of the form <math>(7k, 3k)</math> is on the line.  Then consider the line <math>l</math> from <math>(7k, 3k)</math> to <math>(7(k + 1), 3(k + 1))</math>.  Translate the line <math>l</math> so that <math>(7k, 3k)</math> is now the origin.
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First note that <math>1001 = 143 \cdot 7</math> and <math>429 = 143 \cdot 3</math> so every point of the form <math>(7k, 3k)</math> is on the line.  Then consider the line <math>l</math> from <math>(7k, 3k)</math> to <math>(7(k + 1), 3(k + 1))</math>.  Translate the line <math>l</math> so that <math>(7k, 3k)</math> is now the origin.   There is one square and one circle that intersect the line around <math>(0,0)</math>.  Then the points on <math>l</math> with an integral <math>x</math>-coordinate are, since <math>l</math> has the equation <math>y = \frac{3x}{7}</math>:
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<cmath>  (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). </cmath>
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We claim that the lower right vertex of the square centered at <math>(2,1)</math> lies on <math>l</math>.  Since the square has side length <math>\frac{1}{5}</math>, the lower right vertex of this square has coordinates $(2 + \frac{1}{10},

Revision as of 17:25, 4 March 2016

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},

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