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# Tag Info

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.

11

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

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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 ...

4

If I understand the rules correctly, you can get arbitrarily large scores: We have a sequence $A_k$ where for each $k$, the number of operations is $42 + 5k$, and the limit of the score of $A_k$ is infinity as $k$ goes to infinity: Claim: There exists $r\in \mathbb{R}$ such that $$\frac{-\log|1 - A_k/\pi|}{42 + 5k} > \frac{r}{42 + 5k} + 3!_{k-3}$$ for ...

4

One possibility is to simply have a In the attached image one could argue that the building, if so designed, fits that criteria even though the outside of the building has only four corners. This is a technicality, only, but still..

3

If the trickery is in the word "polygon" one other option is to make it for example, a polygon but only have right angles

3

Another solution is the answer to this age-old problem Where part of the the answer is This makes a For completeness, the other half of the answer to that riddle is A video explanation of both halves can be found here

1