2016 AIME I Problems/Problem 14
Problem
Centered at each lattice point in the coordinate plane are a circle radius 
and a square with sides of length 
whose sides are parallel to the coordinate axes. The line segment from 
to 
intersects 
of the squares and 
of the circles. Find 
.
Solution
First note that 
and 
so every point of the form 
is on the line. Then consider the line 
from to 
. Translate the line so that is now the origin. There is one square and one circle that intersect the line around . Then the points on with an integral 
-coordinate are, since has the equation 
:
![\[(0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3).\]](http://latex.artofproblemsolving.com/8/9/1/891edbf0029fe08e770b58312fe80a4e1f1f0371.png)
We claim that the lower right vertex of the square centered at 
lies on . Since the square has side length , the lower right vertex of this square has coordinates 
. Because 
, 
lies on . Since the circle centered at is contained inside the square, this circle does not intersect . Similarly the upper left vertex of the square centered at 
is on . 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 and 
that intersect . Since there are 
segments from to , the above count is yields 
circles. Since every lattice point on is of the form 
where 
, there are 
lattice points on . Centered at each lattice point, there is one square and one circle, hence this counts 
squares and circles. Thus 
.
Solution by gundraja
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| 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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