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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.
  
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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>6+\frac{\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>.
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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 $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$. The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$, where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$. What is $a+b+c$?

$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}}\ 62 \qquad\textbf{(E)}\ 63$ (Error compiling LaTeX. Unknown error_msg)

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 $0.5$ units of a point on that segment is $0.5+d+\sqrt{.25-d^2}$ where $d$ is the distance from the corner. The integral from $0$ to $0.5$ of this formula resolves to $\frac{6+\pi}{8}$, so the probability of choosing a point within $0.5$ of the first point is $\frac{6+\pi}{32}$. The inverse of this is $\frac{26-\pi}{32}$, so $a+b+c= \boxed{\textbf{(A)}\ 59}$.

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
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