Difference between revisions of "2002 AMC 12B Problems/Problem 18"
(template) |
Serengeti22 (talk | contribs) (→Solution) |
||
(51 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | A point <math>P</math> is randomly selected from the [[rectangle|rectangular]] region with vertices <math>(0,0),(2,0),(2,1),(0,1)</math>. What is the [[probability]] that <math>P</math> is closer to the origin than it is to the point <math>(3,1)</math>? | + | A point <math>P</math> is randomly selected from the [[rectangle|rectangular]] region with vertices <math>(0,0),(2,0),(2,1),(0,1)</math>. What is the [[probability]] that <math>P</math> is closer to the [[origin]] than it is to the point <math>(3,1)</math>? |
− | <math>\mathrm{(A)}\ | + | <math>\mathrm{(A)}\ \frac 12 |
− | \qquad\mathrm{(B)}\ | + | \qquad\mathrm{(B)}\ \frac 23 |
− | \qquad\mathrm{(C)}\ | + | \qquad\mathrm{(C)}\ \frac 34 |
− | \qquad\mathrm{(D)}\ | + | \qquad\mathrm{(D)}\ \frac 45 |
− | \qquad\mathrm{(E)}\ </math> | + | \qquad\mathrm{(E)}\ 1</math> |
== Solution == | == Solution == | ||
− | {{ | + | |
+ | |||
+ | |||
+ | === Solution 1 === | ||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == | ||
Line 14: | Line 31: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:49, 8 June 2024
Contents
Problem
A point is randomly selected from the rectangular region with vertices
. What is the probability that
is closer to the origin than it is to the point
?
Solution
Solution 1
Assume that the point is randomly chosen within the rectangle with vertices
,
,
,
. In this case, the region for
to be closer to the origin than to point
occupies exactly
of the area of the rectangle, or
square units.
If is chosen within the square with vertices
,
,
,
which has area
square unit, it is for sure closer to
.
Now if can only be chosen within the rectangle with vertices
,
,
,
, then the square region is removed and the area for
to be closer to
is then decreased by
square unit, left with only
square unit.
Thus the probability that is closer to
is
and that of
is closer to the origin is
.
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.