Difference between revisions of "2002 AMC 12B Problems/Problem 18"

Revision as of 15:42, 2 July 2019

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

The region containing the points closer to $(0,0)$ than to $(3,1)$ is bounded by the perpendicular bisector of the segment with endpoints $(0,0),(3,1)$ . The perpendicular bisector passes through midpoint of $(0,0),(3,1)$ , which is $\left(\frac 32, \frac 12\right)$ , the center of the unit square with coordinates $(1,0),(2,0),(2,1),(1,1)$ . Thus, it cuts the unit square into two equal halves of area $1/2$ . The total area of the rectangle is $2$ , so the area closer to the origin than to $(3,1)$ and in the rectangle is $2 - \frac 12 = \frac 32$ . The probability is $\frac{3/2}{2} = \frac 34 \Rightarrow \mathrm{(C)}$ .

Solution 2

<asy>

unitsize(36); draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); draw((0,2)--(0,3,5)); draw((6,0)--(7.5,0)); draw((4,0)--(4,2)); label("(0,0)",(-1,-1),W);

Assume that a point $P$ is randomly chosen inside the rectangle with vertices $(0,0)$ , $(3,0)$ , $(3,1)$ , $(0,1)$ .

In this case, the probability that $P$ is closer to the origin than to point $(3,1)$ is $\frac{1}{2}$ .

If $P$ is chosen within the square with vertices

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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