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

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Our answer is <math>\textbf{(A)}</math>.
 
Our answer is <math>\textbf{(A)}</math>.
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==Case Solution==
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Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability <math>0.25, 0.5, 0.25</math>, respectively.
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 +
Opposite side: Probability is obviously <math>1</math>, no matter what.
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Same side: Pretend the points are on a line with coordinates <math>x</math> and <math>y</math>. If <math>|a-b|<=\frac{1}{2}</math>, drawing a graph will give probability <math>\frac{1}{4}</math>.
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Peripheral side: superimpose a coordinate system over the points; call them <math>(0, x)</math> and <math>(0, y)</math>. WLOG set <math>x, y >= 0</math> and <math>x, y <= 1</math>. We need <math>0.25x^2+0.25y^2>0.25</math>, and drawing the coordinate system with bounds <math>(0, 0), (1, 0), (0, 1), (1, 1)</math> gives probability <math>1-\frac{\pi}{16}</math>.
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Adding these up and finding the fraction gives us <math>\frac{1}{32} (26 - \pi)</math> for an answer of <math>\boxed{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}}
 
{{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}}

Revision as of 21:40, 2 February 2018

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 $a+b+c$?

$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63$

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 \[\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left(\frac{1}{4} - \frac{\pi}{16}\right) = 1 - \frac{\pi}{4}.\]

(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 \[P = \frac{1}{8} \left(5 + \frac{1}{2} + 1 - \frac{\pi}{4}\right) = \frac{1}{32} (26 - \pi).\]

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

Case Solution

Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability $0.25, 0.5, 0.25$, respectively.

Opposite side: Probability is obviously $1$, no matter what.

Same side: Pretend the points are on a line with coordinates $x$ and $y$. If $|a-b|<=\frac{1}{2}$, drawing a graph will give probability $\frac{1}{4}$.

Peripheral side: superimpose a coordinate system over the points; call them $(0, x)$ and $(0, y)$. WLOG set $x, y >= 0$ and $x, y <= 1$. We need $0.25x^2+0.25y^2>0.25$, and drawing the coordinate system with bounds $(0, 0), (1, 0), (0, 1), (1, 1)$ gives probability $1-\frac{\pi}{16}$.

Adding these up and finding the fraction gives us $\frac{1}{32} (26 - \pi)$ for an answer of $\boxed{A}$.

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
2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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 10 Problems and Solutions
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