Difference between revisions of "2002 AMC 12B Problems/Problem 18"
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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. | 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. |
Latest revision as of 12:49, 8 June 2024
Contents
[hide]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 |
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