19

There are at least two possible ways to do this, depending on your definition of a polygon. 1: 2:


15

Lot of interesting answers here. My attempt was this: Admittedly, there are several pretty good definitions of corner which would not deem this as a solution.


14

25 - antidisestablishmentarian


11

Here are 22 names with 79 changes: [EDITED to add: It's been pointed out in comments that I carelessly included one user who is slightly below the 3k mark. Here's how to fix that up. The number of users goes down by 1; the number of changes is unaltered.] These are more or less all the short names of 30k+ users that seems like they have much chance of ...


11

First of all, note that So Here are two versions of this. First, But


9

To a real-life problem I had to give a real-life answer: But you asked for an actual tiling, without gaps, so here it is. PS: there is a simpler pattern where pairs disassemble with a single translation:


9

Following in the footsteps of @agnishom-chattopadhyay, here are a couple more ideas that rely on stretching the idea of what counts as a corner: The polygon on the left is so the midpoint counts as zero, one, two or four corners, entirely depending on your definition. (I don't think there's any reasonable way to count it as three, though.) The polygon on ...


8

Just to give a baseline answer: It's certainly possible that


8

I believe I can get using the following...


8

I believe this is the highest score possible with 100 or fewer operations: 99 operations; Score > $10^{10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}}$ As the rather absurd score above suggests, arbitrarily large scores are possible. In particular, we can construct a sequence of approximations $A_k$ that converge extremely rapidly to $\pi$, giving scores ...


8

Fact 1: Fact 2: Proof: Exhaustion... Maximizing possibilities: Notation: (x,y,z) : highest possible value Possibility 1: (1,2,?) Possibility 2: (1,3,?) Possibility 3: (1,4,?) Overall:


8

5 names in 15 moves:


8

Current attempt: 36 names (128 moves) Past attempts: 27 names (78 moves) 27 names (92 moves)


7

I have a fun solution:


7

Anders Kaseorg's answer on the 8/8 list has the most optimal N/N answers up to N = 7 in a spoiler. And he has... With a little enumeration, it's easy to see that you can use that set to get up to... EDIT It's actually pretty simple to reason this out without brute forcing it (too much).


7

Not sure if this counts: 27 letters - Honorificabilitudinitatibus NOTE: The above is rejected I believe mainly because of kbar - short for kilobar. nth is in wikitionary while zth is not. Note sure if this works either 27 letters - chemolithoheterotrophically


7

Here's an 18: And a 20: And a 23: (Oops, no, there was an error in this one, pointed out by Rubio in TSL. Since I now have a longer one I shan't bother trying to fix it. And a 24: The dictionary file on my computer has no words longer than that other than "antidisestablishmentarianism", which I think can't be done -- but of course it isn't exhaustive...


7



7

I seem to recall that the solution to this problem for $n$ lines is The idea is that For example, here is the solution for 6 lines:


6

This is a The message is How I found the key


6

Here's my contribution 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 34) 35) 36) 37) 38) This is currently 74 points


6

This is just for the teaser question (do not know the answer for the real one). I think you can pull the puzzle apart by


5

Another answer, which is probably closer to what the OP intended: The problem with this is that To make a


5

Slightly simplifying Dark Malthorp's remarkable solution, we get: $$\pi \approx A_k = \left[1 - \left( 2 \cdot \log\lceil \sqrt{3!_k}\rceil + \sqrt{4}\right)\div(\lfloor\sqrt{\lceil\sqrt 5\rceil!_k}\rfloor!)\right]^{\lceil\sqrt{6!_{k-1}}\rceil!} \div \lfloor\sqrt 7\rfloor \cdot \lceil\sqrt{\lceil\sqrt 8\rceil!_k} \rceil!\cdot \lfloor\sqrt{\sqrt{9}!_k}\rfloor!...


5

This might be not what an engineer likes to hear, but Then, If the counterfeit is lighter than the rest else


5

My best solution for two equal-sized armies gives each army an area of exactly $\frac{7}{48} \approx 0.1458$. 3 I am not aware of any proof that this is the best solution for two armies. However, you can approximate our point-sized soldiers with regular chess queens on an NxN board, and it turns out that many provably best solutions to that discrete ...


5

Some grids for question 1: 44 circles 45 circles 45 circles


5

Although I was too lazy to get the angles exact, the idea should hold in principle if the picture isn't quite right. They interlock.


5

Here are a few to begin with. I'm not sure how you define 'different' mates - is that the key move only? Mate in 1, White to move: Mate in 1, Black to move: Mate in 2, White to move: Mate in 2, Black to move:


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