bonanova

Bumping this thread. Does that include the argument the function takes? If not, f(x) = 0 + tan(tan-1(x)) where x is assigned the desired integer n. Or even simpler f(x) = 0 + x. No trig needed. I know that I'm missing something. Could you give an example as to form, one that is not the answer?.

OK, I'll bite: stuffing and gravy.

[spoiler=Grimbal has the path shape] Starting the lion at (0,0) and the tamer at (1,0) at same speed. The paths coincide after the tamer has gone 1/4 circle, or pi/2 radii, which, at unit speed, takes pi/2 seconds. The lion's semicircular path has the same length as the quarter circle, having 1/2 the diameter of the latter circle. The interesting thing about this chase is that the lion runs directly toward the tamer at only one moment in time: at the very start. At the point of capture they are running in precisely the same direction, the tamer being at the right side of the lion. A ray from the origin intersects the two paths where they are occupied at the same point in time. Another way of saying that in order to stay on a common radius, the lion must face increasingly away from the tamer toward the eventual point of capture. If the lion runs straight toward the tamer, the tamer escapes. Or does he? Suppose they are both on the perimeter, separated by a very small distance. Does the lion catch up? Perhaps. By running straight toward the tamer, the lion decreases his radius slightly, allowing him to reduce the angular separation. Hmm. Subject for a different puzzle perhaps. Edit: By running straight toward the tamer, the lion traces a path different from the semicircle (a path whose radius is greater at each point in time than the corresponding point on the semicircle.) But, lacking a linear speed advantage, the lion does not want the tamer directly in front of him. Nevertheless, so long as the lion's radius < 1, he has an angular speed advantage, and eventually he captures the tamer. So this does define a new calculation: how much extra time does that chase strategy on the part of the lion give to the tamer? Summary: The tamer becomes the lion's lunch, and he has pi/2 seconds to live once the chase is on. I'm marking the puzzle solved but invite Grimbal to describe an off-perimeter escape strategy for the tamer.

Yes, sort of. I don't have the equations and I expected the lion would approach the edge of the cage asymptotically, never actually reaching it. But now I think that's wrong. It is more like a parabola, and the lion's path is simply tangent to the circular edge, touching it at a single, well-defined point. The path has finite length and is retraceable.

Yes the lion can pounce, I.e. Feet can leave the ground. His top speed is still 1 radius/second. If the tamer's minutes, or seconds, are measured, how long does he have, with starting locations at center (lion) and perimeter (tamer)?

I agree. Nice! Is there a clever derivation?

Start the lion at the center of the cage and the tamer at the fence. along with a simple expression for the lion's path length (which is the tamer's (finite) expected survival time.) Which is correct? I'm interested to hear Grimbal's escape route.

Hi Grimbal. Care to share the tamer's escape path that precludes being caught in finite time?

So it seems the following statements are true? We can't have infinite pulls, but we can see trends as the number of pulls gets big. Introducing multiple players makes the game much like a favorable lottery with a group of participants: almost all the individuals will lose. But if enough players participate, (significantly more than N if the winning odds are 1/N,) there should be at least one winner; and the winnings turn out to be great enough that the group will win. That is, the winnings of the few will cover the $1 stakes of the many, with money left over. Introducing multiple players (each with his own fresh $1 stake and then looking at the aggregate payoffs to determine whether the game is won or lost) is indistinguishable from an individual person playing multiple times, periodically refreshing his $1 stake and aggregating his winnings. This is the essence of a Method 1 game whose only difference is the fact that it is played using a different set of payoffs. If the payoffs of multiple players are multiplied rather than aggregated, the probability with which the group loses is much closer to certainty than the loss probability of the individual players is. The group becomes in essence a single Method 2 player using a much larger number of pulls. Thus, the win probability of an individual player trends to zero with increased pulls. Since (2) mimics Method 1, and (4) mimics Method 2, it seems justifiable to say that the win probability for an individual Method-1 player trends to unity, while the win probability for an individual Method-2 player trends to zero, even though that probability multiplied by his likely payoff (if he wins) is greater than his original stake. (In the sense that it is justifiable to say that an individual $1 billion lottery ticket that has a one-in-a-million chance of winning is almost certainly a losing ticket, and that the statement becomes stronger if both the winnings and odds are escalated by a factor 1000.)

You are right about the lion's radial speed - it's faster than your approximation. Does the lion catch the tamer?.

Kudos!

This puzzle seemed to merit an analysis that went further than I had taken it. Especially since most of my analysis was dead wrong. geometric.pdf

There. Someone solve it for me. BTW, it was an Ogre last time, not a dog. Right. An ogre. It was a dog and rabbit.

You can do it in fewer.

Lion is not constrained. He wants to eat. Tamer does not want to be eaten. Starting positions could be opposite ends of diameter. After one unit of time lion could be at the center; tamer maintains maximum separation by not moving. We can start the chase from there.

The lion begins to pace; the tamer stands to his feet and looks nervously to his left.

Good start.

Wolfgang, there are five balls. Do you include E here?

No restrictions. The cases you list are all permitted. You may read the spoiler for clarification. It gives no clues.

Two envelopes, given one each to you and a friend, contain an amount of money between $5 and $160 inclusive, and one amount is twice the other amount. You look inside your envelope to see the amount you have received, but you do not disclose the amount. Your friend does the same. You see that you have received $x. You may suggest a swap (a one-time exchange of envelopes) but you don't converse directly with your friend. Instead, a facilitator comes to each of you, in turn, and asks what you want to do. If and only if you both ask for a swap the envelopes are exchanged. You and your friend make your decisions independently and in ignorance of the decision made by the other. For what values of x will you favor swapping envelopes? Assume your friend acts reasonably and in his/her own best interest.

A while back, we had fun rescuing a fair maiden in a boat on a circular lake from a dog. Here's another circular chase problem, but this time it's in a cage. A lion and its tamer are inside a circular cage of radius 1. They both can run effortlessly at a top speed of 1. If the lion gets hungry, does the tamer become his lunch?If so, what's the longest time the tamer has to live?Since this is a puzzle, they both can be represented as points.