# md65536

Senior Members

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3. ## Pirates

What do you do if there are 2 pirates who want it, and then neither wants it after the adjustment, or vice versa? What if there's 3 pirates on one side and 1 on the other, and you move some treasure and two (or three) pirates move to the other side? How many adjustments will it take to guarantee there are 2 on each side?

5. ## Pirates

Yes, that's right. I'm assuming that if someone wants a share, and someone else takes it, and the former doesn't also get a share that they're satisfied with from that same division of shares, then they're dissatisfied. Like a child saying they wanted the purple-flavour yogurt that their sibling took and the parent opening a new purple-flavour yogurt and trying to argue that it's the same amount... it's hopeless! Only their own judgment can determine if they're satisfied.

7. ## Pirates

What I wrote is ambiguous... The solution I'm working on seems to work, but I'm having trouble proving it:
8. ## Pirates

Nope, I'm wrong again! This is bad reasoning. Lol, this puzzle has broken my brain. Every time you say I'm right I disagree. Every time I realize something I just said was wrong, I forget some other detail and say something else wrong, all the time flipping back and forth between 2 incorrect positions. What I forgot is that is that it doesn't matter if someone is not satisfied with letting some other share be taken, IF they're also taking a share that they're satisfied with at the same time. So it might (should? but I don't want to give any more answers unless I figure it out) be possible to have someone split up the treasure, and have either everyone satisfied with the shares everyone else is taking, OR they're taking a share they're satisfied with. (However, my previous solution still doesn't work, sometimes not fulfilling either condition.)
9. ## Pirates

My answer before isn't good enough. Consider a worst-case scenario: One pirate divides the loot, and the others evaluate. Say each of the other 4 think there is only one share that is less than equal, and that is the only share that they'd be satisfied with someone taking, but also each of the 4 think that a different share is the smallest. Then there is no share that can be kept by anyone, where they all agree on. Therefore I think that no solution that involves the piles being split up, and then allowing one of them to be kept, can work. I think any solution that involves one person splitting up the loot, would require that the shares be adjusted before guaranteeing that they all agree.
10. ## Pirates

Then my previous answer is no good. I guess any pirate is satisfied if they get to divide the treasure, and any pirate is satisfied if they can pick, but obviously the others wouldn't be happy letting someone else pick.
11. ## Pirates

There are a lot of extra assumptions here. 1. the treasure can be divided to any precision. This wouldn't work if the treasure was 3 large diamonds. 2. The one dividing believes they can do so exactly. 3. There is no conflict in choosing who divides and/or who picks. (I guess pirates are more peaceful than children.)

13. ## Average speed

Sure, apologies for veering off-topic but the ideas still relate! If you look at the Doppler factor formula, you'll see why a sign change of v gives a reciprocal factor. Before or while looking at examples, here's a challenge. Using v=+/-0.6c, the Doppler factors are 2 and 0.5 (perhaps opposite of intuition?) and Lorentz factor is 1.25. Given a statement like, "B spends 1 year traveling away while seeing A age 0.5 years. B turns around, and spends 1 year returning during which it sees A age 2 years, for total of A aging 2.5 years to B's 2," can you similarly (with no more math than that) describe what inertial twin A sees? There's no need to calculate distance, delay of light etc. https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html https://hepweb.ucsd.edu/ph110b/110b_notes/node60.html
14. ## Average speed

The relativistic Doppler effect includes time dilation, and it gives a complete solution to the basic twin paradox (hard to hide aging when you can see each other age the entire time). https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
15. ## Average speed

The same thing happens with the relativistic Doppler effect in the twin paradox, with an outbound and return trip at the same speed. If the clocks appear to tick at 0.5x the rate of a local clock on the outbound trip, they'll appear to tick 2x on the inbound trip. Someone once used incorrect intuition to argue on these forums that this shows that the clocks would age the same amount during the trip, thus disproving special relativity.

I thought this might be incorrect reasoning, because you could change the east/west direction of the shadow simply by changing the rotation rate of the Earth, without affecting how the moon moves relative to the sun. But I think your reasoning must be right. It seems then that if the Earth were spinning much faster, then even though day-to-day the moon appears to have lagged behind the sun, during a single day the moon would appear to be overtaking the sun at lower latitudes. This would be due to parallax. The animation at https://en.wikipedia.org/wiki/Solar_eclipse_of_August_23,_2044 shows how the shadow can go "backwards" at high latitudes. It seems that in this case, the sun and moon are "to the North", eg. during "night time" where there is midnight sun. The sun still appears to be overtaking the moon, but they're both moving in a west to east direction that late in the evening! This should happen anywhere at high latitudes when there is an eclipse in the evening after the sun has passed the westernmost point in the sky and begins moving eastward again before setting, or at dawn before the sun reaches the easternmost point.

19. ## Another missing area

I can't imagine ever noticing that, but now I see why it works for other triangles with the same areas. If you skew the triangle eg. to make a right-angle triangle, you don't change the areas. If you uniformly scale the triangle vertically by r (preserving the ratios between the areas), and then scale horizontally by 1/r, each region's area is scaled by the same factor.

22. ## Find the missing area

That makes it a lot easier! Is it
23. ## Find the perimeter

I see there's other ways to figure this out, but I noticed that there are lots of ways the DE line can be chosen... Anyway the answer I get is
24. ## A grid and a shape

Intuition is that rotation will complicate things because it allows for more ways in which a shape can be moved to dodge the grid, but it seems not to matter because it can still be placed off the grid without rotation.
25. ## A grid and a shape

No fair asking for something that's impossible in the original problem ðŸ˜  "Proof": I think there might be a problem here talking about the area of points, like with the Banach-Tarski paradox.
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