Difference between revisions of "2002 AMC 12B Problems/Problem 18"
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unitsize(36); | unitsize(36); | ||
− | draw((0,0)--(6,0)--(0, | + | draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); |
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | ||
label("$b$",(2.5,0),S); | label("$b$",(2.5,0),S); | ||
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label("$C$",(1,1),W); | label("$C$",(1,1),W); | ||
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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 15:44, 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
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.