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

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==Solution==
 
==Solution==
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Each segment of half of the length of a side of the square is identical, so arbitrarily choose one.
+
Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
  
−
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>.
+
If the second point <math>B</math> is on the left-bottom segment, then if <math>A</math> is distance <math>x</math> away from the left-bottom vertex, then <math>B</math> must be at least <math>\dfrac{1}{2} - \sqrt{0.25 - x^2}</math> away from that same vertex. Thus, using an averaging argument we find that the probability in this case is
 +
<cmath>\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4(\frac{1}{4} - \frac{\pi}{16}) = 1 - \frac{\pi}{4}.</cmath>
 +
 
 +
(Alternatively, one can equate the problem to finding all valid <math>(x, y)</math> with <math>0 < x, y < \dfrac{1}{2}</math>  such that <math>x^2 + y^2 \ge \dfrac{1}{4}</math>, i.e. (x, y) is outside the unit circle with radius 0.5.)
 +
 
 +
Thus, averaging the probabilities gives
 +
<cmath>P = \frac{1}{8} (5 + \frac{1}{2} + 1 - \frac{\pi}{4}) = \frac{1}{32} (26 - \pi).</cmath>
 +
 
 +
Our answer is <math>\textbf{(A)}</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 12:53, 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

Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be at least $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from that same vertex. Thus, using an averaging argument we find that the probability in this case is \[\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4(\frac{1}{4} - \frac{\pi}{16}) = 1 - \frac{\pi}{4}.\]

(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$, i.e. (x, y) is outside the unit circle with radius 0.5.)

Thus, averaging the probabilities gives \[P = \frac{1}{8} (5 + \frac{1}{2} + 1 - \frac{\pi}{4}) = \frac{1}{32} (26 - \pi).\]

Our answer is $\textbf{(A)}$.

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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