2012 AMC 12A Problems/Problem 23

Problem

Let $S$ be the square one of whose diagonals has endpoints $(1/10,7/10)$ and $(-1/10,-7/10)$ . 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?

$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25}$

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$",Y,NE); [/asy]$

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.

Now 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 for all $0 \le x \le 2012$ and $0 \le y \le 2012$ . Therefore, the answer is $0.16 = \boxed{4/25 (C) }$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/254

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2012 AMC 12A Problems/Problem 23

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Problem

Solution

Video Solution by Richard Rusczyk

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