Difference between revisions of "2015 AMC 12A Problems/Problem 23"
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Each segment of half of the length of a side of the square is identical, so arbitrarily choose one. | Each segment of half of the length of a side of the square is identical, so arbitrarily choose one. | ||
− | The portion of the square within <math>0.5</math> units of a point on that segment is <math>0.5+d+\sqrt{.25-d^2}</math> where <math>d</math> is the distance from the corner. The integral from <math>0</math> to <math>0.5</math> of this formula resolves to <math> | + | The portion of the square within <math>0.5</math> units of a point on that segment is <math>0.5+d+\sqrt{.25-d^2}</math> where <math>d</math> is the distance from the corner. The integral from <math>0</math> to <math>0.5</math> of this formula resolves to <math>\frac{6+\pi}{8}</math>, so the probability of choosing a point within <math>0.5</math> of the first point is <math>\frac{6+\pi}{32}</math>. The inverse of this is <math>\frac{26-\pi}{32}</math>, so <math>a+b+c= \boxed{\textbf{(A)}\ 59}</math>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}} |
Revision as of 13:52, 5 February 2015
Problem
Let be a square of side length 1. Two points are chosen independently at random on the sides of . The probability that the straight-line distance between the points is at least is , where and are positive integers and . What is ?
$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}}\ 62 \qquad\textbf{(E)}\ 63$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
Solution
Each segment of half of the length of a side of the square is identical, so arbitrarily choose one.
The portion of the square within units of a point on that segment is where is the distance from the corner. The integral from to of this formula resolves to , so the probability of choosing a point within of the first point is . The inverse of this is , so .
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |