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

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\qquad\mathrm{(E)}\ 1</math>
 
\qquad\mathrm{(E)}\ 1</math>
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
=== Solution 1 ===
  
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[[Image:2002_12B_AMC-18.png]]
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Assume that the point <math>P</math> is randomly chosen within the rectangle with vertices <math>(0,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(0,1)</math>. In this case, the region for <math>P</math> to be closer to the origin than to point <math>(3,1)</math> occupies exactly <math>\frac{1}{2}</math> of the area of the rectangle, or <math>1.5</math> square units.
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If <math>P</math> is chosen within the square with vertices <math>(2,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(2,1)</math> which has area <math>1</math> square unit, it is for sure closer to <math>(3,1)</math>.
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Now if <math>P</math> can only be chosen within the rectangle with vertices <math>(0,0)</math>, <math>(2,0)</math>, <math>(2,1)</math>, <math>(0,1)</math>, then the square region is removed and the area for <math>P</math> to be closer to <math>(3,1)</math> is then decreased by <math>1</math> square unit, left with only <math>0.5</math> square unit.
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Thus the probability that <math>P</math> is closer to <math>(3.1)</math> is <math>\frac{0.5}{2}=\frac{1}{4}</math> and that of <math>P</math> is closer to the origin is <math>1-\frac{1}{4}=\frac{3}{4}</math>. <math>\mathrm{(C)}</math>
  
  
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The region containing the points closer to <math>(0,0)</math> than to <math>(3,1)</math> is bounded by the [[perpendicular bisector]] of the segment with endpoints <math>(0,0),(3,1)</math>. The perpendicular bisector passes through midpoint of <math>(0,0),(3,1)</math>, which is <math>\left(\frac 32, \frac 12\right)</math>, the center of the [[unit square]] with coordinates <math>(1,0),(2,0),(2,1),(1,1)</math>. Thus, it cuts the unit square into two equal halves of area <math>1/2</math>. The total area of the rectangle is <math>2</math>, so the area closer to the origin than to <math>(3,1)</math> and in the rectangle is <math>2 - \frac 12 = \frac 32</math>. The probability is <math>\frac{3/2}{2} = \frac 34 \Rightarrow \mathrm{(C)}</math>.
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~ Nafer
  
 
== See also ==
 
== See also ==

Latest revision as of 12:49, 8 June 2024

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

See also

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

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