2002 AMC 12B Problems/Problem 18

Problem

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 $(3,1)$ ?

$\mathrm{(A)}\ \frac 12 \qquad\mathrm{(B)}\ \frac 23 \qquad\mathrm{(C)}\ \frac 34 \qquad\mathrm{(D)}\ \frac 45 \qquad\mathrm{(E)}\ 1$

Solution

Solution 1

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 $1-\frac{1}{4}=\frac{3}{4}$ . $\mathrm{(C)}$

~ Nafer

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2002 AMC 12B Problems/Problem 18

Contents

Problem

Solution

Solution 1

See also