Difference between revisions of "2002 AMC 12B Problems/Problem 18"
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draw((4,0)--(4,2)); | draw((4,0)--(4,2)); | ||
label("(0,0)",(-1,-1),W); | label("(0,0)",(-1,-1),W); | ||
+ | <\asy><\center> | ||
Assume that a point <math>P</math> is randomly chosen inside the rectangle with vertices <math>(0,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(0,1)</math>. | Assume that a point <math>P</math> is randomly chosen inside the rectangle with vertices <math>(0,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(0,1)</math>. |
Revision as of 16:42, 2 July 2019
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
The region containing the points closer to than to
is bounded by the perpendicular bisector of the segment with endpoints
. The perpendicular bisector passes through midpoint of
, which is
, the center of the unit square with coordinates
. Thus, it cuts the unit square into two equal halves of area
. The total area of the rectangle is
, so the area closer to the origin than to
and in the rectangle is
. The probability is
.
Solution 2
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); <\asy><\center>
Assume that a point is randomly chosen inside the rectangle with vertices
,
,
,
.
In this case, the probability that is closer to the origin than to point
is
.
If is chosen within the square with vertices
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.