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bonanova

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Everything posted by bonanova

  1. The puzzle is interesting, especially for those who are into number theory. Last week I looked at this puzzle and decided I would not be able to find the connection among the numbers. So I searched OEIS out of curiosity. For some reason I didn't find it. (Strange.) Next I Googled the numbers themselves, and that turned up the football reference. <moderator hat on> I agree with Mike's sentiments. The Forum is for solving puzzles; it's not an Internet scavenger hunt. If you can't solve it, say so and maybe start a (spoilered) group effort. Or ask the OP for a clue. The spoiler function permits multiple users to have a go at solving a puzzle. So solutions belong there. (Even funny ones about football.) <moderator hat off>
  2. I'll bet this is not the answer you're looking for, but it does qualify as a remarkable matchup:
  3. Good one. But I can't identify all of the movies, either.
  4. Assuming that they bought
  5. In this puzzle, limits can lead us astray.
  6. In the case of 80 mph the fly, Ferrari and truck all meet on the fly's first and only collision. The OP asks the fly to move to the truck, back to the Ferrari, back to the truck and finally back to the Ferrari. Then, "upon returning from the second trip," the three collide. The 80 mph case has them colliding upon reaching the truck for the first and only time.
  7. Yes. Sorry I didn't make that clear. In reply to your answer,
  8. I think those equations describe the conditions described in the OP. But I don't think there is a solution that satisfies the OP. Maybe that is just what you are saying. But I don't think it's the fault of the equations; it's the claim in the OP that the three all collide at the conclusion of the second round trip of the fly. That just can't happen if all three have constant speeds. Consider the situation at time1 + time2 + time3, when the fly has hit the truck for the second time. The fly completes his trip by returning to the Ferrari. But it doing so it must perform an impossible task. (1) It must fly back to the Ferrari at a speed that is greater than the speed of the truck (it's at least the speed of the Ferrari), but (2) it must arrive at the same time that the truck arrives. That is not possible. The fly will arrive at the Ferrari before the truck does.
  9. Here's a simulation without actually writing the program:
  10. I thought so at first. But simulation shows it doesn't get that high. So long as L is strictly less than sqrt(2) the probability is not close to unity.
  11. The problem is a paradox of sorts. At all times the fly is flying *away* from one of the vehicles. So the vehicles *can't* collide with each other during any part of the fly's journey! Yet of course the vehicles *do* collide in finite time. And they *do* collide, at the same time, with the fly. That's why the fly must make an infinite number of trips. After any finite number of fly-trips, the vehicles are a calculable distance apart.
  12. Inspired by BMAD's Pickup Sticks puzzle, I ask the following related question. Into a unit square you drop two needles (having negligible width) of length L. (0 < L < sqrt(2)). Consider the probability p( L ) that they will intersect. A moment's thought reveals that the probability approaches zero as L approaches zero. p( L ) is a strictly increasing function and approaches a limit as L approaches sqrt (2). What is the maximum value of p( L )? Clue
  13. After re-reading the problem, I'd agree with Baskaran. In fact,
  14. Getting this one started,
  15. I don't have an answer yet, but here is how I would attack the problem
  16. I don't recall that I posted it. The way I first heard it, tho, the question gives the speed of the fly and asks how far it flew. This form hasn't been posted that I can tell. What intrigues me about this version is that ...
  17. Working on it. I have a conjecture: If two pairs of circles are congruent the triangle is isosceles.
  18. It's an excellent puzzle. You can find some extended discussion of it here. To locate previous posts, enter key words in the search box -- upper right corner, just under your name. I found this link by entering the words "hats death row" I happened to remember that "death row" was in the title.
  19. A procedure that permits different interpretations does not help. If the counterfeit coin is lighter, then it's on the lighter side. If it's heavier then it's on the heavier side. All you can conclude is that the counterfeit coin is on the balance scale. But since you're weighing all the coins, you already knew it was on the scale. So you've learned nothing.
  20. I'm thinking the answer is different for rational and real numbers. Is your question purposely directed to rationals?
  21. I think your formula is correct. Do you have a proof?
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