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

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(Solution)
 
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\qquad\mathrm{(E)}\ \dfrac{1}{9}</math>
 
\qquad\mathrm{(E)}\ \dfrac{1}{9}</math>
  
==Solution==
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==Solution 1==
  
 
The first point anywhere on the circle, because it doesn't matter where it is chosen.
 
The first point anywhere on the circle, because it doesn't matter where it is chosen.
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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>
 
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>
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==Solution 2==
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 +
We will use geometric probability.
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The first point can be anywhere. Each point must be <math>\frac{\pi}{3}</math> or less away from each other.
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 +
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.
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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).
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Now, we can graph the equations we deduced on the coordinate grid. That should look like this:
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<asy>
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fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green);
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draw((-50, 0)--(50, 0));
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draw((0, -50)--(0, 50));
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draw((-50, -10)--(10, 50),red);
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draw((-10, -50)--(50, 10),red);
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draw((-40, 40)--(40, 40),red);
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draw((-40, -40)--(40, -40),red);
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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>
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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>.
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Thus, the probability is <math>\frac{\frac{\pi^2}{3}}{(2\pi)^2} = \frac{1}{12}</math>, or <math>\boxed{\text{(D)}}</math>.
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 +
~superagh
  
 
==See Also==
 
==See Also==

Latest revision as of 20:05, 27 August 2020

Problem

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?

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

Solution 1

The first point 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.

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 $\frac{1}{4}$ to find 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

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
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All AMC 12 Problems and Solutions

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