CaptainEd

I'm missing something.

I think that after we check 4, we know that fox is in an odd number right now, and will move to an even one that night.

Don't watch the video!!! it is a good video, it clearly states the problem, but then it gives the answer! (Clearly stated as well) if you watch the video, stop after the problem statement. I enjoy the problem, and I appreciated the ingenious solution, nice job!

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.

(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:

I hope this is clear: Oops! Plasmid beat me to it on the 8:13 version of this puzzle.

Bonanova, cute puzzle! Pickett, great solve! Thank you both, that was fun to watch!

Can't do spoilers on my phone.

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.

You're stronger than I am; I edit the post, removing everything, and then I put in some quiet, gentle, lame stuff.

@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.

My guess for Most sides:

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)?

I like Charlie's argument; I think Charlie earns the BGS,

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...

Oh, my! I got an A in Differential equations. That was 1965. I haven't used them since. What an interesting follow-on question!

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

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,

@Phaze: on my laptop, the two lines are the same color. I think you are seeing the same relationship between curves that I am seeing. Is the blue line the one which is on top of the picture at the right hand side?

I'll assume that you meant "parallel to the ground" rather than perpendicular, because you asked that we pull "back" rather than "up" or "down". I'm impressed at how this puzzle seems so tricky to someone who has spent nearly his entire life riding a bike. Here's my walk down the garden path. In all cases, I'm standing behind the bike, exerting force on the string towards me (ie, "pulling back")

oops, pumping involves raising the CG near the ground, and lowering it near the platform.

assuming zero initial velocity, If the platforms are the same height, then the acrobat leaving platform 1 will not be able to rise above the height of platform 2, whether still holding on to the trapeze or not, without additional input of energy (as plainglazed's motorized pinions could provide). However, as we all know from childhood, additional input CAN be added by "pumping"--standing up when near the platform height, and lowering the center of gravity when near the ground. I have no idea how to calculate how much additional energy (hence height, distance, etc.) can be added that way. For one thing, if you can get the trapeze to go higher than the platform, it will tend to fall closer to the platform...

@BMAD, I see I am being inconsistent. If the center is on the line that I gave, and x <= 1, the arc is at the top of the circle, and above the line from (0,3) to (2,0). If the center is exactly at (1,1.5), then the area is as I said above. if x > 1, then there is an arc of the bottom of the circle in the quadrant (x>0, y>0). Then the area under the arc is OUTSIDE the circle, so we'd have to know just what area you are interested in, in that case.

@BMAD, I've interpreted the goal as being to give the area meeting ALL of these constraints: ( a ) under the arc, ( b ) within the body of the circle, ( c ) above x-axis and to the right of the y-axis. Once the center of the circle moves to the right and up along the line I gave, this area shrinks until it is a thin lens connecting (0,3) and (2,0), with a limit of 0. Have I still got the formula wrong? Or is your question only about the domain of x?