Difference between revisions of "2002 AMC 12B Problems/Problem 18"
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=== Solution 2 === | === Solution 2 === | ||
− | Assume that | + | Assume that the 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>. |
− | In this case, the probability that <math>P</math> is closer to the origin than to point <math>(3,1)</math> is <math>\frac{1}{2}</math>. | + | In this case, the probability that <math>P</math> is closer to the origin than to point <math>(3,1)</math> is exactly <math>\frac{1}{2}</math>. |
− | If <math>P</math> is chosen within the square with vertices | + | 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>, it is closer to <math>(3,1)</math> because the longest distance to $( |
== See also == | == See also == |
Revision as of 17:03, 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 the point is randomly chosen inside the rectangle with vertices
,
,
,
.
In this case, the probability that is closer to the origin than to point
is exactly
.
If is chosen within the square with vertices
,
,
,
, it is closer to
because the longest distance to $(
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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