CaptainEd

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CaptainEd last won the day on January 13 2016

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  1. My analysis was incomplete. If you skip the Center X, then a string of alternating C and F (with E between) is constrained by the number of E. With 12 E, at most 13 (C,F) can be interleaved with them. And, because there are only 6 F, that means the upper limit on a path would use 7 Cs and 6Fs. So as Rocdocmac says, if it's possible without touching X, it must start and end with C: CEFECEFECEFECEFECEFECEFEC. Either way, with or without the X cell, 25 appears to be the upper bound, and Thalia found one.
  2. (Sorry I don't know how to make spoiler on iPhone, but it seems Thalia is the winner, so I'm not surprising anyone) Here's why 25 is max: Lets label a corner C, Edge Center E, Face Center F, and the hidden Center X. There are 12 E, 8C, 6F, and 1X. Because of the required 90 degree turn at each cell, CE can only be followed by EF (or Exit). All possible moves are: CE : EF FE : EC or EF EC: CE EF : FE or FX FX : XF FE : EC Notice only F can reach X, so the Sequence of moves must include FXF, if the Center is to be reached at all. Also notice that once you have moved from C to E, you can only move from E to F. For each C, you can pass through one F. (In CEFE sequence). But while there are 8 C, there are only 6 F, so the total sequence can have only 6 C, and the maximum is only 25. Good news is, there are enough E to join all the C and F, so the sequence could look like ECEF ECEF ECEF X FECE FECE FECE A path that avoids X would have to have an E between two Fs, which would waste an E, leaving only 24.
  3. I hope this is clear: Oops! Plasmid beat me to it on the 8:13 version of this puzzle.
  4. Bonanova, cute puzzle! Pickett, great solve! Thank you both, that was fun to watch!
  5. Can't do spoilers on my phone.
  6. The more I tried to simulate Plasmid's argument, the more powerful it felt. While T is moving North, it doesn't matter how close L is to T, or how tiny the angle theta off vertical, with a sufficiently small stepsize, L will waste a portion of each step moving South, widening the angle theta, until eventually | T - L | > stepsize.
  7. You're stronger than I am; I edit the post, removing everything, and then I put in some quiet, gentle, lame stuff.
  8. @BMAD I can't prove this is the largest @bonanova: in my picture ( representing real folds on real paper ) a "tail" lapped over the Southeast corner, adding an extra two edges beyond the ones identified by your argument. I presume that they become part of the overall envelope of the figure, leading to my answer, greater than yours.
  9. My guess for Most sides:
  10. Another interesting puzzle! Thanks, BMAD! Here's a question from me: when you say "repeat this process", do you mean "pick any corner and bring that corner to the middle of the paper"? Or do we bring the corner to any place at all that we can choose (not just the absolute center of the paper)?
  11. I like Charlie's argument; I think Charlie earns the BGS,
  12. Rollie, your friend is "logical" and "competitive". But is your friend "greedy" as well? Does your friend prefer making more money than you? or more money, period? You suggested cooperating with your friend what if you...
  13. Oh, my! I got an A in Differential equations. That was 1965. I haven't used them since. What an interesting follow-on question!
  14. I'm embarrassed to admit that I don't know enough math to develop a closed form for the distance. So, simulating * the Tamer starting at ( 0,0 ), * the Lion starting at (0,1 ) * the Tamer moving along X-axis in increments of .01
  15. Suppose the point in the middle is called C . By "in the middle", do you mean "C is equidistant from the other two points"? or do you simply mean a line segment from one point to the other passes through C? I imagine different answers for the two cases. Assuming C is between A and B,