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

Revision as of 15:56, 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 $\frac{3/2}{2} = \frac 34 \Rightarrow \mathrm{(C)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 === Solution 1 ===
+<center>
 [[Image:2002_12B_AMC-18.png]]

Page

Toolbox

Search

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

Revision as of 15:56, 2 July 2019

Contents

Problem

Solution

Solution 1

See also