Difference between revisions of "2012 AMC 12A Problems/Problem 23"
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By periodicity, this probability is the same if <math>v = (x,y)</math>, where <math>0 \le x \le 2012</math> and <math>0 \le y \le 2012</math>. Therefore, the answer is <math>0.16</math>. <math>\boxed{C}</math>. | By periodicity, this probability is the same if <math>v = (x,y)</math>, where <math>0 \le x \le 2012</math> and <math>0 \le y \le 2012</math>. Therefore, the answer is <math>0.16</math>. <math>\boxed{C}</math>. | ||
− | Note: the | + | Note: the range of <math>x</math> and <math>y</math> in the problem is arbitrary. |
Revision as of 19:57, 18 February 2012
Problem
Let be the square one of whose diagonals has endpoints
and
. A point
is chosen uniformly at random over all pairs of real numbers
and
such that
and
. Let
be a translated copy of
centered at
. What is the probability that the square region determined by
contains exactly two points with integer coefficients in its interior?
Solution
We first notice that for the translated square to contain two points with integer coordinates (lattice points) in its interior, these two points must be adjacent. This can be shown by considering the diagonal of . The diagonal is
, which is the length of the diagonal of a unit square. Because
square is not parallel to the axis, square
cannot two points that are not adjacent.
Because we have showed that the two lattice points contained in must be adjacent, let us consider the unit square
with vertices
and
. Let us first consider only two vertices,
and
. We want to find the area of the region within
that the point
will create the translation of
,
such that it covers both
and
. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
For to contain the point
,
must be inside square
. Similarly, for
to contain the point
,
must be inside a translated square
with center at
, which we will call
. Therefore, the area we seek is Area
.
To calculate the area, we notice that Area Area
by symmetry. Let
. Let
be the midpoint of
, and
along the line
. Let
be the intersection of
and
within
, and
be the intersection of
and
outside
. Therefore, the area we seek is
Area
. Because
all have
coordinate
, they are collinear. Noting that the side length of
and
is
(as shown above), we also see that
, so
. If follows that
and
. Therefore, the area is
Area
.
Because there are three other regions in the unit square that we need to count, the total area that
contains two adjacent lattice points within
is
.
By periodicity, this probability is the same if , where
and
. Therefore, the answer is
.
.
Note: the range of and
in the problem is arbitrary.