2002 AMC 12B Problems/Problem 18
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 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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