Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AMC 12A Problems/Problem 23"

m (→‎Solution 1)
m (→‎Solution 1)
Line 8: Line 8:
  
 
=== Solution 1 ===
 
=== Solution 1 ===
−
This can be shown by considering the diagonal of <math>S</math>. The diagonal is <math>\sqrt{0.2^2 + 1.4^2} = \sqrt{2}</math>, which is the length of the diagonal of a unit square. Because <math>S</math> square is not parallel to the axis, the two points must be adjacent.
+
The unit square's diagonal has a length of <math>\sqrt{0.2^2 + 1.4^2} = \sqrt{2}</math>. Because <math>S</math> square is not parallel to the axis, the two points must be adjacent.
  
 
Because we have showed that the two lattice points contained in <math>T(v)</math> must be adjacent, let us consider the unit square <math>U</math> with vertices <math>(0,0), (1,0), (1,1)</math> and <math>(0,1)</math>. Let us first consider only two vertices, <math>(0,0)</math> and <math>(1,0)</math>. We want to find the area of the region within <math>U</math> that the point <math>v=(x,y)</math> will create the translation of <math>S</math>, <math>T(v)</math> such that it covers both <math>(0,0)</math> and <math>(1,0)</math>. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
 
Because we have showed that the two lattice points contained in <math>T(v)</math> must be adjacent, let us consider the unit square <math>U</math> with vertices <math>(0,0), (1,0), (1,1)</math> and <math>(0,1)</math>. Let us first consider only two vertices, <math>(0,0)</math> and <math>(1,0)</math>. We want to find the area of the region within <math>U</math> that the point <math>v=(x,y)</math> will create the translation of <math>S</math>, <math>T(v)</math> such that it covers both <math>(0,0)</math> and <math>(1,0)</math>. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.

Revision as of 20:51, 11 September 2013

Problem

Let $S$ be the square one of whose diagonals has endpoints $(0.1,0.7)$ and $(-0.1,-0.7)$. A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0\le y\le 2012$. Let $T(v)$ be a translated copy of $S$ centered at $v$. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coefficients in its interior?

$\textbf{(A)}\ 0.125\qquad\textbf{(B)}\ 0.14\qquad\textbf{(C)}\ 0.16\qquad\textbf{(D)}\ 0.25 \qquad\textbf{(E)}\ 0.32$

Solution

Solution 1

The unit square's diagonal has a length of $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$. Because $S$ square is not parallel to the axis, the two points must be adjacent.

Because we have showed that the two lattice points contained in $T(v)$ must be adjacent, let us consider the unit square $U$ with vertices $(0,0), (1,0), (1,1)$ and $(0,1)$. Let us first consider only two vertices, $(0,0)$ and $(1,0)$. We want to find the area of the region within $U$ that the point $v=(x,y)$ will create the translation of $S$, $T(v)$ such that it covers both $(0,0)$ and $(1,0)$. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.

For $T(v)$ to contain the point $(0,0)$, $v$ must be inside square $S$. Similarly, for $T(v)$ to contain the point $(1,0)$, $v$ must be inside a translated square $S$ with center at $(1,0)$, which we will call $S'$. Therefore, the area we seek is Area$(U \cap S \cap S')$.

To calculate the area, we notice that Area$(U \cap S \cap S') = \frac{1}{2} \cdot$ Area$(S \cap S')$ by symmetry. Let $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$. Let $M = (0.7, 0.4)$ be the midpoint of $S'_1S'_2$, and $N = (0.7, 0.7)$ along the line $S_1S'_1$. Let $I$ be the intersection of $S$ and $S'$ within $U$, and $J$ be the intersection of $S$ and $S'$ outside $U$. Therefore, the area we seek is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2]$. Because $S_2, M, N$ all have $x$ coordinate $0.7$, they are collinear. Noting that the side length of $S$ and $S'$ is $1$ (as shown above), we also see that $S_2M = MS'_1 = 0.5$, so $\triangle{S'_1NM} \cong \triangle{S_2IM}$. If follows that $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$ and $IS'2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$. Therefore, the area is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04$.

Because there are three other regions in the unit square $U$ that we need to count, the total area of $v$ within $U$ such that $T(v)$ contains two adjacent lattice points is $0.04 \cdot 4 = 0.16$.

By periodicity, this probability is the same if $v = (x,y)$, where $0 \le x \le 2012$ and $0 \le y \le 2012$. Therefore, the answer is $\boxed{\textbf{(C)}\ 0.16}$.

Note: the ranges of $x$ and $y$ in the problem are arbitrary as long as the maximum and minimum of the range are integers.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12A_Problems/Problem_23&oldid=57071"