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Difference between revisions of "2003 AMC 12B Problems/Problem 25"

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m (fixed typo, "all three pairwise distance" should be "all three pairwise distances")
 
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== Problem ==
 
== Problem ==
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Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distance between the points are less than the radius of the circle?
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Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?
  
 
<math>\mathrm{(A)}\ \dfrac{1}{36}
 
<math>\mathrm{(A)}\ \dfrac{1}{36}
Line 8: Line 8:
 
\qquad\mathrm{(E)}\ \dfrac{1}{9}</math>
 
\qquad\mathrm{(E)}\ \dfrac{1}{9}</math>
  
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==Solution==
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==Solution 1==
  
−
First: One can choose the first point anywhere on the circle.
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The first point is placed anywhere on the circle, because it doesn't matter where it is chosen.
 
   
 
   
−
Secondly: The Next point must lie within 60 degrees of arc on either side, a total of 120 degrees possible, 1/3 chance.
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The next point must lie within <math>60</math> degrees of arc on either side, a total of <math>120</math> degrees possible, giving a total <math>\frac{1}{3}</math> chance. The last point must lie within <math>60</math> degrees  of both points.  
  
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The last point must lie within 60 degrees  of both.  This ranges from 60 degrees arc to sit on (if the first two are 60 degrees apart) and a 1/6 probability, to 120 degrees (if they are negligibly apart) and a 1/3 chance. 
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The minimum area of freedom we have to place the third point is a <math>60</math> degrees arc(if the first two are <math>60</math> degrees apart), with a <math>\frac{1}{6}</math> probability.
−
As the second point moves from 60 degrees away to the first point, the probability changes linearly (every degree it moves, adds one degree to where the third could be), the probabilities at each end can be averaged to find 1/4.  
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The maximum amount of freedom we have to place the third point is a <math>120</math> degree arc(if the first two are the same point), with a <math>\frac{1}{3}</math> probability.
  
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Therefore the total probability is 1*1/3*1/4 = 1/12 or (D)
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As the second point moves farther away from the first point, up to a maximum of <math>60</math> degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be).
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(checked on amc.maa.org)
+
 
 +
Therefore, we can average probabilities at each end to find <math>\frac{1}{4}</math>, the average probability we can place the third point based on a varying second point.
 +
 
 +
Therefore the total probability is <math>1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}</math> or <math>\boxed{\text{(D)}}</math>
 +
 
 +
==Solution 2==
 +
 
 +
We will use geometric probability.
 +
 
 +
The first point can be anywhere. Each point must be <math>\frac{\pi}{3}</math> or less away from each other.
 +
 
 +
Define <math>x</math> be the amount of radians away the second point is from the first. We limit <math>x</math> to be in the interval <math>[-\pi, \pi]</math>. Define <math>y</math> be the amount of radians away the third point is from the first. We limit <math>y</math> to be in the interval <math>[-\pi, \pi]</math>. Now, we can deduct that: <cmath>|x| \le \frac{\pi}{3},</cmath> <cmath>|y| \le \frac{\pi}{3},</cmath> and <cmath>|x-y| \le \frac{\pi}{3}.</cmath> We now begin plotting these on the coordinate grid.
 +
 
 +
First note that the area of the points that <math>x</math> and <math>y</math> can be (ignoring the conditions) is <math>(2\pi)^2</math> (remember what we restricted <math>x</math> and <math>y</math> to).
 +
 
 +
Now, we can graph the equations we deduced on the coordinate grid. That should look like this:
 +
<asy>
 +
fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green);
 +
draw((-50, 0)--(50, 0));
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draw((0, -50)--(0, 50));
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draw((-50, -10)--(10, 50),red);
 +
draw((-10, -50)--(50, 10),red);
 +
draw((-40, 40)--(40, 40),red);
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draw((-40, -40)--(40, -40),red);
 +
draw((40, 40)--(40, -40),red);
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draw((-40, 40)--(-40, -40),red);
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label("$\frac{\pi}{3}$", (40, 0), SE);
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label("$\frac{\pi}{3}$", (-40, 0), SW);
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label("$\frac{\pi}{3}$", (0, 40), NE);
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label("$\frac{\pi}{3}$", (0, -40), NW);
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</asy>
 +
The area of the shaded region can be calculated in many ways. Eventually, you will find that the area is <math>\frac{\pi^2}{3}</math>.
 +
 
 +
Thus, the probability is <math>\frac{\frac{\pi^2}{3}}{(2\pi)^2} = \frac{1}{12}</math>, or <math>\boxed{\text{(D)}}</math>.
 +
 
 +
~superagh
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==Solution 3==
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 +
There are <math>\binom32 = 3</math> different ways to choose <math>2</math> points out of <math>3</math>. The probability that an angle between <math>2</math> points is smaller than <math>60^\circ</math> is <math>\frac{60^\circ}{360^\circ} = \frac16</math>. The probability that the third point is inside the <math>60^\circ</math> angle is <math>\frac{60^\circ}{360^\circ} = \frac16</math>
 +
 
 +
<math>3 \cdot \frac16 \cdot \frac16 = \boxed{\text{(D)} \frac{1}{12} }</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|ab=B|year=2003|num-b=24|after=Last Problem}}
 
{{AMC12 box|ab=B|year=2003|num-b=24|after=Last Problem}}
 +
{{MAA Notice}}

Latest revision as of 00:42, 14 October 2024

Problem

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

$\mathrm{(A)}\ \dfrac{1}{36} \qquad\mathrm{(B)}\ \dfrac{1}{24} \qquad\mathrm{(C)}\ \dfrac{1}{18} \qquad\mathrm{(D)}\ \dfrac{1}{12} \qquad\mathrm{(E)}\ \dfrac{1}{9}$

Solution 1

The first point is placed anywhere on the circle, because it doesn't matter where it is chosen.

The next point must lie within $60$ degrees of arc on either side, a total of $120$ degrees possible, giving a total $\frac{1}{3}$ chance. The last point must lie within $60$ degrees of both points.

The minimum area of freedom we have to place the third point is a $60$ degrees arc(if the first two are $60$ degrees apart), with a $\frac{1}{6}$ probability. The maximum amount of freedom we have to place the third point is a $120$ degree arc(if the first two are the same point), with a $\frac{1}{3}$ probability.

As the second point moves farther away from the first point, up to a maximum of $60$ degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be).

Therefore, we can average probabilities at each end to find $\frac{1}{4}$, the average probability we can place the third point based on a varying second point.

Therefore the total probability is $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$ or $\boxed{\text{(D)}}$

Solution 2

We will use geometric probability.

The first point can be anywhere. Each point must be $\frac{\pi}{3}$ or less away from each other.

Define $x$ be the amount of radians away the second point is from the first. We limit $x$ to be in the interval $[-\pi, \pi]$. Define $y$ be the amount of radians away the third point is from the first. We limit $y$ to be in the interval $[-\pi, \pi]$. Now, we can deduct that: \[|x| \le \frac{\pi}{3},\] \[|y| \le \frac{\pi}{3},\] and \[|x-y| \le \frac{\pi}{3}.\] We now begin plotting these on the coordinate grid.

First note that the area of the points that $x$ and $y$ can be (ignoring the conditions) is $(2\pi)^2$ (remember what we restricted $x$ and $y$ to).

Now, we can graph the equations we deduced on the coordinate grid. That should look like this: [asy] fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green); draw((-50, 0)--(50, 0)); draw((0, -50)--(0, 50)); draw((-50, -10)--(10, 50),red); draw((-10, -50)--(50, 10),red); draw((-40, 40)--(40, 40),red); draw((-40, -40)--(40, -40),red); draw((40, 40)--(40, -40),red); draw((-40, 40)--(-40, -40),red); label("$\frac{\pi}{3}$", (40, 0), SE); label("$\frac{\pi}{3}$", (-40, 0), SW); label("$\frac{\pi}{3}$", (0, 40), NE); label("$\frac{\pi}{3}$", (0, -40), NW); [/asy] The area of the shaded region can be calculated in many ways. Eventually, you will find that the area is $\frac{\pi^2}{3}$.

Thus, the probability is $\frac{\frac{\pi^2}{3}}{(2\pi)^2} = \frac{1}{12}$, or $\boxed{\text{(D)}}$.

~superagh

Solution 3

There are $\binom32 = 3$ different ways to choose $2$ points out of $3$. The probability that an angle between $2$ points is smaller than $60^\circ$ is $\frac{60^\circ}{360^\circ} = \frac16$. The probability that the third point is inside the $60^\circ$ angle is $\frac{60^\circ}{360^\circ} = \frac16$

$3 \cdot \frac16 \cdot \frac16 = \boxed{\text{(D)} \frac{1}{12} }$

~isabelchen

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 12 Problems and Solutions

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