Difference between revisions of "2016 AIME I Problems/Problem 14"
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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> | <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> | ||
− | 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 | + | 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 <math>(2 + \frac{1}{10})</math>, |
Revision as of 17:34, 4 March 2016
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
:
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
,