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

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~BJHHar
 
~BJHHar
  
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===Solution 3 (Geometric) (INCORRECT!!! probability is not evenly distributed)===
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===Solution 3 (Geometric) (INCORRECT!!! probability is not evenly distributed)(actually it is, except at the very edge and center, but since that has area 0, it is neglegible)===
  
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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.
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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 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.
  
  

Revision as of 21:28, 19 January 2021

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

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 be 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, \[f(\theta)=\frac{\theta}{2\pi}*2\pi=\theta\] 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) (INCORRECT!!! probability is not evenly distributed)(actually it is, except at the very edge and center, but since that has area 0, it is neglegible)

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 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 \[\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}\].

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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