Difference between revisions of "2016 AIME I Problems/Problem 14"
(→See also) |
(→Solution) |
||
Line 12: | Line 12: | ||
Solution by gundraja | Solution by gundraja | ||
+ | |||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=I|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Revision as of 16:51, 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
. 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 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.