2002 AMC 12B Problems/Problem 18

Revision as of 16:42, 2 July 2019 by Nafer (talk | contribs) (Solution 2)

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

2002 12B AMC-18.png

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

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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