Difference between revisions of "2012 AMC 12A Problems/Problem 23"
Sugar rush (talk | contribs) |
|||
(7 intermediate revisions by 2 users not shown) | |||
Line 3: | Line 3: | ||
Let <math>S</math> be the square one of whose diagonals has endpoints <math>(1/10,7/10)</math> and <math>(-1/10,-7/10)</math>. A point <math>v=(x,y)</math> is chosen uniformly at random over all pairs of real numbers <math>x</math> and <math>y</math> such that <math>0 \le x \le 2012</math> and <math>0\le y\le 2012</math>. Let <math>T(v)</math> be a translated copy of <math>S</math> centered at <math>v</math>. What is the probability that the square region determined by <math>T(v)</math> contains exactly two points with integer coefficients in its interior? | Let <math>S</math> be the square one of whose diagonals has endpoints <math>(1/10,7/10)</math> and <math>(-1/10,-7/10)</math>. A point <math>v=(x,y)</math> is chosen uniformly at random over all pairs of real numbers <math>x</math> and <math>y</math> such that <math>0 \le x \le 2012</math> and <math>0\le y\le 2012</math>. Let <math>T(v)</math> be a translated copy of <math>S</math> centered at <math>v</math>. What is the probability that the square region determined by <math>T(v)</math> contains exactly two points with integer coefficients in its interior? | ||
− | <math> \textbf{(A)}\ | + | <math> \textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25} </math> |
== Solution == | == Solution == | ||
Line 30: | Line 30: | ||
Because there are three other regions in the unit square <math>U</math> that we need to count, the total area of <math>v</math> within <math>U</math> such that <math>T(v)</math> contains two adjacent lattice points is <math>0.04 \cdot 4 = 0.16</math>. | Because there are three other regions in the unit square <math>U</math> that we need to count, the total area of <math>v</math> within <math>U</math> such that <math>T(v)</math> contains two adjacent lattice points is <math>0.04 \cdot 4 = 0.16</math>. | ||
− | By periodicity, this probability is the same for all <math>0 \le x \le 2012</math> and <math>0 \le y \le 2012</math>. Therefore, the answer is <math>\boxed{\textbf{(C)} | + | By periodicity, this probability is the same for all <math>0 \le x \le 2012</math> and <math>0 \le y \le 2012</math>. Therefore, the answer is <math> 0.16 = \boxed{\frac{4}{25} \textbf{(C)} }</math> |
− | |||
==Video Solution by Richard Rusczyk== | ==Video Solution by Richard Rusczyk== |
Latest revision as of 19:37, 19 January 2021
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
![[asy] pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1); draw (A--B--C--D--A); draw(A--C); draw(B--D); draw(W--X); draw(Y--Z); label("\((0.1,0.7)\)",A,NE); label("\((-0.1,-0.7)\)",C,SW); label("\(x\)",X,NW); label("\(y\)",Y,NE); [/asy]](http://latex.artofproblemsolving.com/f/6/f/f6f0e5566a32958540f7896563dc7761f10e4078.png)
The unit square's diagonal has a length of . Because
square is not parallel to the axis, the two points must be adjacent.
Now 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 of
within
such that
contains two adjacent lattice points is
.
By periodicity, this probability is the same for all and
. Therefore, the answer is
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12a/254
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.