Probability of Frog's Final Position <1m from Start: 2010 AMC 12

[naes213]

Homework Statement

This problem is from the 2010 AMC 12 High school math competition:

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?

Homework Equations

I drew circles representing all of the points which can be reached after two jumps. This is just two circles with one of the circle's centers located on the circumference of the other circle. I believe the probability is 1/4 after two jumps. S=theta*r ;theta=pi/2 for the portion of the second circle's circumference contained in the first circle.

The Attempt at a Solution

After the first two jumps the probability of being less than a meter from the starting point is 1/4. The problem is the third jump which can occur from any point on the second circle, each point with a different probability. If you could find the probability as a function of some angle theta around the second circle you could integrate it..yada yada...but this seems like the wrong way to go. Any suggestions? Thanks!

 

[awkward]

Let's say the frog starts at (0,0). By choosing coordinates appropriately, we can assume his position after the first hop is (1,0). If his second hop is at an angle theta, his position after the second hop is (1 + cos(theta), sin(theta)). If his third hop is then at an angle phi, his final position is at
(x,y) = (1+ cos(theta) + cos(phi), sin(theta) + sin(phi)).

He is within 1 meter of his starting point if x^2 + y^2 <= 1. Try substituting x = 1 + cos(theta) + cos(phi) and y = sin(theta) + sin(phi) in that inequality and see if you can find the region in the (theta, phi)-plane where it is satisfied. It may be simpler to start with finding the boundary of the region, where x^2 + y^2 = 1.

I think that's probably the way you are meant to go, but I must confess that my trigonometry is rusty and I took a different approach via complex variables. So I guess you can try that way if you're inclined-- it doesn't use any calculus, just the algebra of complex numbers, so I guess it's within the range of acceptable solutions. As before, the frog's initial position is at 0 in the complex plane. After the first hop, he is at 1. After the second hop he is at 1 + e^(i theta). After the third hop he is at 1 + e^(i theta) + e^(i phi). You can then work on
|1 + e^(i theta) + e^(i phi)| = 1
to find the boundary of the acceptable region.

 

[Mentallic]

Interesting problem! My working out wasn't quite what you could pull off in an exam room since I used a graphing calculator, and I'm honestly not even sure it's correct, but I'll place down my bet that the answer is 1/4 anyway.

 

[naes213]

Thanks for the responses. I'll have to try that, awkward, that's probably what they wanted. And Mentallic, yeah i tried to avoid using a calculator since i knew they couldn't use one, but it was tempting!