AMC12 2002 B

AMC12 2002 B · Q18

AMC12 2002 B · Q18. It mainly tests Coordinate geometry, Geometric probability (basic).

A point $P$ is randomly selected from the rectangular region with vertices $(0,0),(2,0),(2,1),(0,1)$. What is the probability that $P$ is closer to the origin than it is to the point $(3,1)$?

从顶点为$(0,0),(2,0),(2,1),(0,1)$的矩形区域中随机选取一点$P$。$P$到原点的距离比到点$(3,1)$的距离更近的概率是多少？

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{2}{3} \frac{2}{3}

(D) \frac{4}{5} \frac{4}{5}

(E) 1 1

Answer

Correct choice: (C)

正确答案：（C）

Solution

Assume that the point $P$ is randomly chosen within the rectangle with vertices $(0,0)$, $(3,0)$, $(3,1)$, $(0,1)$. In this case, the region for $P$ to be closer to the origin than to point $(3,1)$ occupies exactly $\frac{1}{2}$ of the area of the rectangle, or $1.5$ square units. If $P$ is chosen within the square with vertices $(2,0)$, $(3,0)$, $(3,1)$, $(2,1)$ which has area $1$ square unit, it is for sure closer to $(3,1)$. Now if $P$ can only be chosen within the rectangle with vertices $(0,0)$, $(2,0)$, $(2,1)$, $(0,1)$, then the square region is removed and the area for $P$ to be closer to $(3,1)$ is then decreased by $1$ square unit, left with only $0.5$ square unit. Thus the probability that $P$ is closer to $(3.1)$ is $\frac{0.5}{2}=\frac{1}{4}$ and that of $P$ is closer to the origin is $1-\frac{1}{4}=\frac{3}{4}$. $\mathrm{(C)}$ First, join points $(0,0)$ and $(3,1)$. This line $l_1$ has equation $y = \frac{1}{3}x$.

假设点$P$是在顶点为$(0,0)$、$(3,0)$、$(3,1)$、$(0,1)$的矩形内随机选取的。在这种情况下，使得$P$到原点比到点$(3,1)$更近的区域恰好占矩形面积的$\frac{1}{2}$，即$1.5$平方单位。如果$P$选在顶点为$(2,0)$、$(3,0)$、$(3,1)$、$(2,1)$的正方形内（面积为$1$平方单位），那么它一定更靠近$(3,1)$。现在若$P$只能在顶点为$(0,0)$、$(2,0)$、$(2,1)$、$(0,1)$的矩形内选取，则上述正方形区域被移除，使得$P$更靠近$(3,1)$的面积减少$1$平方单位，只剩下$0.5$平方单位。因此$P$更靠近$(3.1)$的概率为$\frac{0.5}{2}=\frac{1}{4}$，而$P$更靠近原点的概率为$1-\frac{1}{4}=\frac{3}{4}$。$\mathrm{(C)}$ 首先，连接点$(0,0)$与$(3,1)$。这条直线$l_1$的方程为$y=\frac{1}{3}x$。

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