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Difference between revisions of "2010 AMC 12B Problems/Problem 18"

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<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math>
 
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math>
  
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== Solution ==
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==Solution 1 (Complex Numbers)==
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===Solution 1 (Complex Numbers)===
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We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The starts on the origin. It is relatively easy to show that exactly one element in the set
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We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set
 
 
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath>
 
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath>
−
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.
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has magnitude less than or equal to <math> 1</math>. (Can you show how?) Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.
  
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===Solution 2 (Geometric)===
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==Solution 2 (Simple Calculus)==
 +
Yes, we are pulling out calculus...
 +
  
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The first frog hop doesn't matter because no matter where the frog hops, it lands on the border of the circle you want it to end in.  The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.
+
Represent every jump as a circle of radius 1. The first circle is a circle of radius 1 centered on the origin. WLOG, assume the first jump lands on <math>(1,0)</math>. Then, the third circle could be centered anywhere on the second circle, which is itself centered on <math>(1,0)</math>. Let us define <math>f(\theta)</math> as the value of the length of the first circle that lies within the area of the third circle in terms of the angle formed by the two points of intersection and either circle's center (symmetry, you chose!). The intersection of the two circles should form a geometrical lens shape. By sectors,
 +
<cmath>f(\theta)=\frac{\theta}{2\pi}\cdot2\pi=\theta</cmath>
 +
As the angle or angle of intersection continuously decreases from <math>\pi</math> (when the third circle is on top of the second circle) to <math>0</math> (when the third circle is only touching the first circle at one spot <math>(1,0)</math>, I just need to find the average value of this function to find the average arc length where the third jump could land to satisfy the problem. To do this, I can apply average function value with our old buddy calculus,
 +
<cmath>\frac{1}{\pi}\int_0^{\pi}\theta\;d\theta=\frac{\pi}{2}</cmath>
 +
The probability that the third jump will land on this arc length is just the arc length divided by the circumference, or <math>\frac{\frac{\pi}{2}}{2\pi}=\frac{1}{4} \implies \boxed{C}</math>
 +
 
 +
~BJHHar
 +
 
 +
==Solution 3 (Geometric)==
 +
 
 +
The first  hop doesn't matter because no matter where the hops, it lands on the border of the circle you want it to end in.  The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.
  
  
 
No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how <math>\frac{1}{2}x^2</math> and <math>x^2</math> are tangent. The area ratio of the two circles is <cmath>\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}</cmath>.
 
No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how <math>\frac{1}{2}x^2</math> and <math>x^2</math> are tangent. The area ratio of the two circles is <cmath>\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}</cmath>.
 +
 +
==Solution 4 (Easier Geometric Casework)==
 +
Firstly, we know that the first jump will land on a circle of radius one around the frog's starting position. Then, the frog has two possibilities for where he could jump next.
 +
 +
<math>\textbf{Case 1: }</math> The frog's next jump lands outside of the circle with radius 1. The frog's possible area he could land on is the area formed by a circle of radius 2, and the probability of him landing outside of the radius 1 circle into the radius 2 circle is <cmath>\frac{2^2 \cdot \pi - 1^2 \cdot \pi}{2^2 \cdot \pi} = \frac{3}{4}</cmath>.
 +
 +
To find the probability of the frog landing back in the circle of radius 1 on the third jump, we need to find the range that the circle could have traveled. After the second jump in case 1, the circle's position range from still on the circumference of the radius 1 circle, to being on the circumference of the radius 2 circle.
 +
 +
When on the radius 1 circle, the frog has, similarily, a <math>\frac{1}{4}</math> chance of jumping back in, while on the radius 2 circle, the frog has 0 chance of jumping back in the radius 1 circle, so the range is <math>[0, \frac{1}{4}]</math>. Taking the average, we get <math>\frac{1}{8}</math>, and that the frog have a total of <cmath>\frac{3}{4} \cdot \frac{1}{8} = \frac{3}{32}</cmath> percent chance to jump back in in this case.
 +
 +
<math>\textbf{Case 2: }</math> The frog's next jump lands outside of the circle with radius 1. The frog's possible area he could land on is the area formed by a circle of radius 2, and the probability of him landing inside of the radius 1 circle is simply <cmath>\frac{1^2 \cdot \pi}{2^2 \cdot \pi} = \frac{1}{4}</cmath>.
 +
 +
Similarily, to find the probability of the frog landing back in the circle of radius 1 on the third jump, we need to find the range that the circle could have traveled. After the second jump in case 2, the circle's position range from still on the circumference of the radius 1 circle, to being back to its starting point.
 +
 +
When on the radius 1 circle, the frog has, a <math>\frac{1}{4}</math> chance of jumping back in, while when the frog is on its starting point, the frog is certain to still be within the circle after one jump. Therefore, our range is going to be <math>[\frac{1}{4}, 1]</math>. Taking the average, we get <math>\frac{5}{8}</math>, and that the frog have a total of <cmath>\frac{1}{4} \cdot \frac{5}{8} = \frac{5}{32}</cmath> percent chance to jump back in in this case.
 +
 +
Adding the probability of the two cases, we get that the frog has a <cmath>\frac{5}{32} + \frac{3}{32} = \frac{8}{32} = \frac{1}{4}</cmath> chance of jumping back into no more than <math>1</math> meter from its starting position, therefore the answer is <math>\boxed{\frac{1}{4} \text{(C)}}</math>.
 +
 +
~erdaifuu
 +
 +
== Solution 4 ==
 +
Here's a funny solution: this problem reduces to picking three vectors on the unit circle and asking when the sum of these vectors also lies in the unit circle. But it's well known that the sum of these three vectors is the orthocenter of the triangle whose vertices are the heads of the three vectors. Also, it's well known that the orthocenter of a triangle lies inside its circumcircle iff the triangle is acute. Therefore, this problem is equivalent to asking when three randomly chosen points on a circle form an acute triangle. This is a classic problem, and it's well known that the answer is <math>\boxed{\frac{1}{4} \text{(C)}}</math>.
 +
 +
~ ihatemath123
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}
 
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:39, 9 January 2024

Problem

A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?

$\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}$

Solution 1 (Complex Numbers)

We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The starts on the origin. It is relatively easy to show that exactly one element in the set \[\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}\] has magnitude less than or equal to $1$. (Can you show how?) Hence, the probability is $\boxed{\text{(C)} \frac {1}{4}}$.

Solution 2 (Simple Calculus)

Yes, we are pulling out calculus...


Represent every jump as a circle of radius 1. The first circle is a circle of radius 1 centered on the origin. WLOG, assume the first jump lands on $(1,0)$. Then, the third circle could be centered anywhere on the second circle, which is itself centered on $(1,0)$. Let us define $f(\theta)$ as the value of the length of the first circle that lies within the area of the third circle in terms of the angle formed by the two points of intersection and either circle's center (symmetry, you chose!). The intersection of the two circles should form a geometrical lens shape. By sectors, \[f(\theta)=\frac{\theta}{2\pi}\cdot2\pi=\theta\] As the angle or angle of intersection continuously decreases from $\pi$ (when the third circle is on top of the second circle) to $0$ (when the third circle is only touching the first circle at one spot $(1,0)$, I just need to find the average value of this function to find the average arc length where the third jump could land to satisfy the problem. To do this, I can apply average function value with our old buddy calculus, \[\frac{1}{\pi}\int_0^{\pi}\theta\;d\theta=\frac{\pi}{2}\] The probability that the third jump will land on this arc length is just the arc length divided by the circumference, or $\frac{\frac{\pi}{2}}{2\pi}=\frac{1}{4} \implies \boxed{C}$

~BJHHar

Solution 3 (Geometric)

The first hop doesn't matter because no matter where the hops, it lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.


No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how $\frac{1}{2}x^2$ and $x^2$ are tangent. The area ratio of the two circles is \[\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}\].

Solution 4 (Easier Geometric Casework)

Firstly, we know that the first jump will land on a circle of radius one around the frog's starting position. Then, the frog has two possibilities for where he could jump next.

$\textbf{Case 1: }$ The frog's next jump lands outside of the circle with radius 1. The frog's possible area he could land on is the area formed by a circle of radius 2, and the probability of him landing outside of the radius 1 circle into the radius 2 circle is \[\frac{2^2 \cdot \pi - 1^2 \cdot \pi}{2^2 \cdot \pi} = \frac{3}{4}\].

To find the probability of the frog landing back in the circle of radius 1 on the third jump, we need to find the range that the circle could have traveled. After the second jump in case 1, the circle's position range from still on the circumference of the radius 1 circle, to being on the circumference of the radius 2 circle.

When on the radius 1 circle, the frog has, similarily, a $\frac{1}{4}$ chance of jumping back in, while on the radius 2 circle, the frog has 0 chance of jumping back in the radius 1 circle, so the range is $[0, \frac{1}{4}]$. Taking the average, we get $\frac{1}{8}$, and that the frog have a total of \[\frac{3}{4} \cdot \frac{1}{8} = \frac{3}{32}\] percent chance to jump back in in this case.

$\textbf{Case 2: }$ The frog's next jump lands outside of the circle with radius 1. The frog's possible area he could land on is the area formed by a circle of radius 2, and the probability of him landing inside of the radius 1 circle is simply \[\frac{1^2 \cdot \pi}{2^2 \cdot \pi} = \frac{1}{4}\].

Similarily, to find the probability of the frog landing back in the circle of radius 1 on the third jump, we need to find the range that the circle could have traveled. After the second jump in case 2, the circle's position range from still on the circumference of the radius 1 circle, to being back to its starting point.

When on the radius 1 circle, the frog has, a $\frac{1}{4}$ chance of jumping back in, while when the frog is on its starting point, the frog is certain to still be within the circle after one jump. Therefore, our range is going to be $[\frac{1}{4}, 1]$. Taking the average, we get $\frac{5}{8}$, and that the frog have a total of \[\frac{1}{4} \cdot \frac{5}{8} = \frac{5}{32}\] percent chance to jump back in in this case.

Adding the probability of the two cases, we get that the frog has a \[\frac{5}{32} + \frac{3}{32} = \frac{8}{32} = \frac{1}{4}\] chance of jumping back into no more than $1$ meter from its starting position, therefore the answer is $\boxed{\frac{1}{4} \text{(C)}}$.

~erdaifuu

Solution 4

Here's a funny solution: this problem reduces to picking three vectors on the unit circle and asking when the sum of these vectors also lies in the unit circle. But it's well known that the sum of these three vectors is the orthocenter of the triangle whose vertices are the heads of the three vectors. Also, it's well known that the orthocenter of a triangle lies inside its circumcircle iff the triangle is acute. Therefore, this problem is equivalent to asking when three randomly chosen points on a circle form an acute triangle. This is a classic problem, and it's well known that the answer is $\boxed{\frac{1}{4} \text{(C)}}$.

~ ihatemath123

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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