2012 AMC 12A Problems/Problem 23

Revision as of 20:25, 18 February 2012 by Aplus95 (talk | contribs) (Solution 1)


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 1

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 $S$. The diagonal is $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$, which is the length of the diagonal of a unit square. Because $S$ square is not parallel to the axis, square $T(v)$ cannot two points that are not 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 range of $x$ and $y$ in the problem is arbitrary as long as the maximum and minimum of the range are integers.

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions
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