# Tag Info

### What is the name of the 65537-gon?

"$65537$-gon" is the name. Likewise "$257$-gon": writing (let alone saying) something like "diacosipentacontaheptagon" serves less to communicate $-$ if indeed it succeeds in communicating at all $-$...
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### Can one "hear" the shape of a polygon via external reflections?

For question 3, the answer is yes: take a solid disc and excavate half of the Penrose unilluminable room from it. Then, there are boundary arcs which can never be touched, and you can perturb them ...
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### Acute triangles in "obtuse" polygons?

Take a very obtuse isosceles triangle and chop its acute angles.
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As Sam Hopkins commented, 8 vertices are enough. Let $Q$ be the pentagon from the picture and let $\pi$ be the plane containing it. Now we can define the triangle $P$ as a triangle of less diameter ...
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### How to characterize the regularity of a polygon?

Internal angles are not enough to determine the regularity of a polygon. E.g., angles of $2\pi/3$ between sides of length $1,1,4,1,1,4$ make an irregular hexagon. For a metric of regularity, I suggest ...
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### Limiting shape for Brillouin zones

Yes, it is. Take a point P, and let us check if it belongs to Nth Brillouin zone. The Bragg planes (rather, lines, as we are in $R^2$) that we have to cross while going from the origin $O$ to $P$, ...
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### Necessary and sufficient condition for quadrilateral to be cyclic

Regarding the "only if" part, a stronger result actually holds: if the quadrilateral $ABCD$ is cyclic, then $PQRS$ is similar to $ABCD$, and so it is cyclic, too. This is stated, without ...
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### Collinearity in tangential pentagon

This follows from Brianchon's theorem. Note that in order to use that you need to consider the degenerate hexagon, $ABCFDE$. The theorem implies your conclusion.
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### Geometry, Number Theory and Graph Theory of n-gon, permutation and graph labeling?

For $t=n-1$ you are asking if the cyclic group of order $n$ is sequenceable, where a finite group $G$ is sequenceable if you can write down the non-identity elements $g_1,\ldots,g_{n-1}$ in such a way ...
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