Difference between revisions of "2010 AIME II Problems/Problem 2"
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<math>\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}</math> | <math>\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}</math> | ||
− | Thus, the answer is <math>\boxed{281}</math> | + | Thus, the answer is <math>56 + 225 = \boxed{281}.</math> |
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=1|num-a=3|n=II}} | {{AIME box|year=2010|num-b=1|num-a=3|n=II}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:35, 9 June 2019
Problem 2
A point is chosen at random in the interior of a unit square
. Let
denote the distance from
to the closest side of
. The probability that
is equal to
, where
and
are relatively prime positive integers. Find
.
Solution
Any point outside the square with side length that have the same center as the unit square and inside the square side length
that have the same center as the unit square has
.
![[asy] unitsize(1mm); defaultpen(linewidth(.8pt)); draw((0,0)--(0,30)--(30,30)--(30,0)--cycle); draw((6,6)--(6,24)--(24,24)--(24,6)--cycle); draw((10,10)--(10,20)--(20,20)--(20,10)--cycle); fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray); fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white); [/asy]](http://latex.artofproblemsolving.com/2/f/8/2f82e5b678d789e6c2bfd670f46cff737577a21e.png)
Since the area of the unit square is , the probability of a point
with
is the area of the shaded region, which is the difference of the area of two squares
Thus, the answer is
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.