Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎See also)
Line 2: Line 2:
  
 
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> whose sides are parallel to the coordinate axes. The line segment from <math>(0,0)</math> to <math>(1001, 429)</math> intersects <math>m</math> of the squares and <math>n</math> of the circles. Find <math>m + n</math>.
 
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> whose sides are parallel to the coordinate axes. The line segment from <math>(0,0)</math> to <math>(1001, 429)</math> intersects <math>m</math> of the squares and <math>n</math> of the circles. Find <math>m + n</math>.
−
 
−
== See also ==
 
−
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
 
−
{{MAA Notice}}
 
  
 
== Solution ==
 
== Solution ==

Revision as of 16:51, 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 $(2 + \frac{1}{10}, 1 - \frac{1}{10}) = (\frac{21}{10}, \frac{9}{10})$. 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_14&oldid=77290"