Comments on: Follow Up: Escaping the Beast http://mathfactor.uark.edu/2007/10/follow-up-escaping-the-beast/ The Math Factor Podcast Site Fri, 08 Aug 2014 12:52:06 +0000 hourly 1 https://wordpress.org/?v=4.9.25 By: jay ha http://mathfactor.uark.edu/2007/10/follow-up-escaping-the-beast/comment-page-1/#comment-1188 Mon, 24 Mar 2014 21:55:38 +0000 http://mathfactor.uark.edu/2007/10/23/follow-up-escaping-the-beast/#comment-1188 what about this variation?

    An escaped prisoner finds himself in the middle of a square swimming pool. The guard that is chasing him is at one of the corners of the pool. The guard can run faster than the prisoner can swim. The prisoner can run faster than the guard can run. The guard does not swim. Which direction should the prisoner swim in in order to maximize the likelihood that he will get away?

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By: stevestyle http://mathfactor.uark.edu/2007/10/follow-up-escaping-the-beast/comment-page-1/#comment-207 Wed, 09 Jan 2008 13:56:19 +0000 http://mathfactor.uark.edu/2007/10/23/follow-up-escaping-the-beast/#comment-207 I believe I can describe the solution.

The key points are:

1. The princess will never go inside the spiral/circle.

2. Wherever she ends up, she will have taken the shortest possible route.

Imagine tying one end of a piece of string to the centre of the lake (you can do that with this lake). Now hoop it around the circle and stretch it out to the edge of the lake, keeping it taught.

The path of the string is the shortest route the princess can take to reach that part of the shore.

The path is:

curve outwards for a bit;

continue in the same direction, but in a straight line, until reaching shore.

This is Chaim’s suggested path. He asks if we can improve by heading for shore earlier.

We can reach shore at the same point faster by cutting out the corner. Note that this is not the best route for the princess; it’s just a faster route for reaching the same point on shore.

We can give a range for where the optimal route lies.
If we travel a quarter circle and then head horizontally for shore then it’s easy to show this still beats the beast.
As the princess must travel at least 1, then the beast must travel at least four.
Calculating exactly where the best route lies is complex. I’ve have worked out some equations for it but not solved them. We don’t really need to know do we?

If the princess wants to land as far from the beast as possible, so she has the best safety margin, then she should swim out to _ radius then head to shore in the same direction.
The beast has to cover _ of a circle after the princess starts heading for shore, and an entire circle in total. When the princess lands the beast is still about 0.5 away.
She lands at angle theta, which has cosine _.
If we start the princess from any point on this path then this must still be the best path for her to take.
Note that this does not depend on exactly where the beast it, only on the direction from which he is coming.

Suppose we put the princess and the beast at random points, and the princess want to land as far from the beast as possible. Also assume it is best for her to head straight for shore, rather than retreat to the centre and start out from there.
She will lie on exactly one path of the type we have described, given the direction of the beast. This is her best path.
She should calculate how to land at an angle of theta while also heading away from the beast. This is her best path.

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By: strauss http://mathfactor.uark.edu/2007/10/follow-up-escaping-the-beast/comment-page-1/#comment-181 Mon, 31 Dec 2007 16:04:25 +0000 http://mathfactor.uark.edu/2007/10/23/follow-up-escaping-the-beast/#comment-181 Actually, this is only the optimal path of the form: curve out 1/4 of the way, then make a straight dash. Perhaps one can do better by starting to dash sooner…

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