Difference between revisions of "2010 AIME II Problems/Problem 2"
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== Problem 2 == | == Problem 2 == | ||
A point <math>P</math> is chosen at random in the interior of a unit square <math>S</math>. Let <math>d(P)</math> denote the distance from <math>P</math> to the closest side of <math>S</math>. The probability that <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is equal to <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | A point <math>P</math> is chosen at random in the interior of a unit square <math>S</math>. Let <math>d(P)</math> denote the distance from <math>P</math> to the closest side of <math>S</math>. The probability that <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is equal to <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
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== Solution == | == Solution == | ||
− | Any point outside the square with side length <math>\frac{1}{3}</math> that | + | Any point outside the square with side length <math>\frac{1}{3}</math> that has the same center and orientation as the unit square and inside the square with side length <math>\frac{3}{5}</math> that has the same center and orientation as the unit square has <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math>. |
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− | Since the area of the unit square is <math>1</math>, the probability of a point <math>P</math> with <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is the area of the shaded region, which is the difference of the area of two squares | + | Since the area of the unit square is <math>1</math>, the probability of a point <math>P</math> with <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is the area of the shaded region, which is the difference of the area of two squares. |
<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> |
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+ | == Solution 2 == | ||
+ | First, let's figure out <math>d(P) \geq \frac{1}{3}</math> which is<cmath>\left(\frac{3}{5}\right)^2=\frac{9}{25}.</cmath>Then, <math>d(P) \geq \frac{1}{5}</math> is a square inside <math>d(P) \geq \frac{1}{3}</math>, so<cmath>\left(\frac{1}{3}\right)^2=\frac{1}{9}.</cmath>Therefore, the probability that <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is<cmath>\frac{9}{25}-\frac{1}{9}=\frac{56}{225}</cmath>So, the answer is <math>56+225=\boxed{281}</math> | ||
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+ | == Solution 3 == | ||
+ | First, lets assume that point <math>P</math> is closest to a side <math>S</math> of the square. If it is <math>\frac{1}{5}</math> far from <math>S</math>, then it should be at least <math>\frac{1}{5}</math> from both the adjacent sides of <math>S</math> in the square. This leaves a segment of <math>1 - 2 \cdot \frac{1}{5} = \frac{3}{5}</math>. If the distance from <math>P</math> to <math>S</math> is <math>\frac{1}{3}</math>, then notice the length of the side-ways segment for <math>P</math> is <math>1 - 2 \cdot \frac{1}{3} = \frac{1}{3}</math>. Notice that as the distance from <math>P</math> to <math>S</math> increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides <math>\frac{3}{5}</math> and <math>\frac{1}{3}</math> with height <math>\frac{1}{3} - \frac{1}{5} = \frac{2}{15}</math>. This trapezoid has area (or probability for one side) <math>\frac{1}{2} \cdot \left(\frac{1}{3}+\frac{3}{5}\right)\cdot \frac{2}{15} = \frac{14}{225}</math>. Since the square has <math>4</math> sides, we multiply by <math>4</math>. Hence, the probability is <math>\frac{56}{225}</math>. The answer is <math>\boxed{281}</math>. ~Saucepan_man02 | ||
== 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}} |
Latest revision as of 07:54, 7 February 2023
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 has the same center and orientation as the unit square and inside the square with side length that has the same center and orientation as the unit square has .
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
Solution 2
First, let's figure out which isThen, is a square inside , soTherefore, the probability that isSo, the answer is
Solution 3
First, lets assume that point is closest to a side of the square. If it is far from , then it should be at least from both the adjacent sides of in the square. This leaves a segment of . If the distance from to is , then notice the length of the side-ways segment for is . Notice that as the distance from to increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides and with height . This trapezoid has area (or probability for one side) . Since the square has sides, we multiply by . Hence, the probability is . The answer is . ~Saucepan_man02
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.