Difference between revisions of "2015 AMC 12A Problems/Problem 23"

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==Problem==
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Let <math>S</math> be a square of side length 1. Two points are chosen independently at random on the sides of <math>S</math>. The probability that the straight-line distance between the points is at least <math>\frac12</math> is <math>\frac{a-b\pi}{c}</math>, where <math>a,b,</math> and <math>c</math> are positive integers and <math>\text{gcd}(a,b,c) = 1</math>. What is <math>a+b+c</math>?
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<math> \textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}}\ 62 \qquad\textbf{(E)}\ 63</math>
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==Solution==
 
==Solution==
 
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 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 6+π/8, so the probability of choosing a point within 0.5 of the first point is 6+π/32.  The inverse of this is 26-π/32, so a+b+c=(A) 59.
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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>6+\frac{\pi}{32}</math>.  The inverse of this is <math>26-\frac{\pi}{32}</math>, so <math>a+b+c= \boxed{\textbf{(A))}\ 59}</math>.
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== See Also ==
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{{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}}

Revision as of 01:58, 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. ! 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 $0.5$ 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 $6+\frac{\pi}{8}$, so the probability of choosing a point within $0.5$ of the first point is $6+\frac{\pi}{32}$. The inverse of this is $26-\frac{\pi}{32}$, so $a+b+c= \boxed{\textbf{(A))}\ 59}$.

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
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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