AMC12 2012 A

AMC12 2012 A · Q23

AMC12 2012 A · Q23. It mainly tests Probability (basic), Coordinate geometry.

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?

设 $S$ 是一个正方形，其一条对角线的端点为 $(1/10,7/10)$ 和 $(-1/10,-7/10)$。点 $v=(x,y)$ 在所有满足 $0 \le x \le 2012$ 且 $0\le y\le 2012$ 的实数对 $(x,y)$ 中均匀随机选取。设 $T(v)$ 为以 $v$ 为中心的 $S$ 的平移拷贝。由 $T(v)$ 确定的正方形区域在其内部恰好包含两个整数坐标点的概率是多少？

(A) 0.125 0.125

(B) 0.14 0.14

(D) 0.25 0.25

(E) 0.32 0.32

Answer

Correct choice: (C)

正确答案：（C）

Solution

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 $0.16 = \boxed{\frac{4}{25} \textbf{(C)} }$

单位正方形的对角线长度为 $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$。由于正方形 $S$ 不与坐标轴平行，这两个点必须是相邻的。现在考虑单位正方形 $U$，其顶点为 $(0,0), (1,0), (1,1)$ 和 $(0,1)$。先只考虑两个顶点 $(0,0)$ 与 $(1,0)$。我们要找出 $U$ 内使得点 $v=(x,y)$ 所生成的平移正方形 $T(v)$ 同时覆盖 $(0,0)$ 与 $(1,0)$ 的区域面积。由对称性，覆盖其他三对相邻顶点的区域面积相同。要使 $T(v)$ 包含点 $(0,0)$，$v$ 必须在正方形 $S$ 内。同理，要使 $T(v)$ 包含点 $(1,0)$，$v$ 必须在以 $(1,0)$ 为中心的平移正方形内，记为 $S'$。因此所求面积为 Area$(U \cap S \cap S')$。为计算该面积，注意到由对称性 Area$(U \cap S \cap S') = \frac{1}{2} \cdot$ Area$(S \cap S')$。设 $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$。令 $M = (0.7, 0.4)$ 为 $S'_1S'_2$ 的中点，$N = (0.7, 0.7)$ 在直线 $S_1S'_1$ 上。令 $I$ 为 $S$ 与 $S'$ 在 $U$ 内的交点，$J$ 为 $S$ 与 $S'$ 在 $U$ 外的交点。则所求面积为 $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2]$。由于 $S_2, M, N$ 的 $x$ 坐标均为 $0.7$，它们共线。注意到 $S$ 与 $S'$ 的边长均为 $1$（如上所示），且 $S_2M = MS'_1 = 0.5$，所以 $\triangle{S'_1NM} \cong \triangle{S_2IM}$。因此 $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$，且 $IS'_2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$。于是面积为 $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04$。由于在单位正方形 $U$ 中还有另外三个需要计数的相同区域，使得 $T(v)$ 包含两相邻格点的 $v$ 的总面积为 $0.04 \cdot 4 = 0.16$。由周期性，这个概率对所有 $0 \le x \le 2012$ 且 $0 \le y \le 2012$ 都相同。因此答案为 $0.16 = \boxed{\frac{4}{25} \textbf{(C)} }$。

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