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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>
  
−
==Solution==
+
==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 in an equilateral triangle formed by the center and the first point lying on either side of the radius formed
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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 this equilateral triangle
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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.
 +
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. 
  
−
Therefore the total probablity is 1*1/3*1/6 = 1/18 or (C)
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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).
 +
 
 +
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));
 +
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 <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==
 +
 
 +
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
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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