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40 votes

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 $-$...
Noam D. Elkies's user avatar
34 votes

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 ...
Adam P. Goucher's user avatar
24 votes
Accepted

Acute triangles in "obtuse" polygons?

Take a very obtuse isosceles triangle and chop its acute angles.
Anton Petrunin's user avatar
14 votes
Accepted

Unlinked interlocking planar polygons

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 ...
Saúl RM's user avatar
  • 10.4k
10 votes

Maximum area of the intersection of a parallelogram and a triangle

I don't know if this is the optimal, but an isosceles triangle with base and height $\sqrt{2}$ overlaps $2 \left(\sqrt{2}-1\right) \approx 0.828427$ when placed as below, and so improves over $\frac{...
Joseph O'Rourke's user avatar
10 votes

An inequality related to area and sidelengths of a polygon $Area(A_1A_2....A_n) \le \frac{1}{n}cotg{\frac{\pi}{n}} \sum_{i=1}^nA_iA_{i+1}^2$

Cauchy–Schwarz tells you that $$\sum_{i=1}^n \lvert A_iA_{i+1}\rvert^2\geq \frac{P^2}{n}$$ where $P$ is the perimeter of the polygon. Then we need the inequality $$P^2\geq 4n\tan(\pi/n)A$$ which is ...
Gjergji Zaimi's user avatar
10 votes
Accepted

Strange formula for area of a convex polygon

Denote $A_i=(0,b_i)$. It is a point on $\ell_i=\{(x,y):y=k_ix+b_i\}$, and let $$P_i=\ell_i\cap \ell_{i+1}=\left(\frac{b_{i+1}-b_i}{k_i-k_{i+1}},\frac{k_ib_{i+1}-k_{i+1}b_i}{k_i-k_{i+1}}\right)$$ be ...
Fedor Petrov's user avatar
9 votes

Need help with finding all angles of 11 sided 3D object

There is a problem in the design, if I understand the phrases "the height I want the spire to be" and "how tall I choose to make the spire": these suggest there is freedom to ...
Joseph O'Rourke's user avatar
8 votes
Accepted

Necessary and sufficient condition for tangential polygon to be cyclic

$H_i$ lies on the ray $IA_i$ and $IH_i\cdot IA_i=2r^2$ (where $r=IB_i$), since the midpoint of $IH_i$ is the midpoint of $B_iB_{i-1}$. Hence $H_i$ is the image of $A_i$ under the inversion with ...
Fedor Petrov's user avatar
7 votes

First Dirichlet eigenvalue on regular polygons

Ignoring the normalization, your inequality writes $$\lambda(P)\|u\|_{L^2(P)}^2\ge\|\partial_\nu u\|_{L^\infty(\partial P)}^2|P|\qquad?$$ Good news: this is scaling invariant. Bad news: this is false ...
Denis Serre's user avatar
  • 51.9k
7 votes

How to characterize the regularity of a polygon?

All indices are in $\mathbb Z\bmod6$. Let $z_k=\frac{\ell_{k,k+1}}{\ell_{k-1,k}}e^{i(\pi-\theta_k)}=\frac{v_{k+1}-v_k}{v_k-v_{k-1}}$, where $\ell$ is edge length, $\pi-\theta$ is vertex exterior angle,...
mr_e_man's user avatar
  • 191
6 votes

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 ...
Francesco Polizzi's user avatar
6 votes

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.
user127776's user avatar
  • 5,861
6 votes

Unlinked interlocking planar polygons

It is not possible with 7 (i.e., with a triangle $T$ and a quadrilateral $Q$). I write a rough proof. First, any quadrilateral $Q$ lying in a plane $\pi$ can be partitioned in two triangles $Q_1$ and $...
Del's user avatar
  • 380
6 votes

First Dirichlet eigenvalue on regular polygons

There is a more general reason why any such statement will fail: If you consider a $P$ consisting of $N$ separate copies of the same basic region $P_0$, then $|P|=N|P_0|$, while everything else in ...
Christian Remling's user avatar
5 votes
Accepted

Point generation in polygon

To generate points with the same distribution as the Halton sequence in a polygon you can just take a rectangle enclosing the polygon - for example take the convex hull of the polygon (assuming it may ...
Ivan Meir's user avatar
  • 4,822
5 votes

Strange formula for area of a convex polygon

Fedor's derivation is very slick and elegant - surely the best way to guess the formula if you didn't know it already. It does however require additional justification to show that the oriented ...
Ivan Meir's user avatar
  • 4,822
5 votes
Accepted

A closed chain of $2n+1$-gon around $2n+1$-points

You may easily calculate everything in complex numbers. Denote $m=2n+1$, $w=e^{2\pi i/n}$, $Q_i=A_{i,3}=A_{i+1,2}$ for $i=i,2,\ldots$. We may suppose that $P_{m+i}=P_i$ for $i=1,\ldots,m$ and we have ...
Fedor Petrov's user avatar
5 votes
Accepted

An inequality related to area and sidelengths of a polygon $Area(A_1A_2....A_n) \le \frac{1}{n}cotg{\frac{\pi}{n}} \sum_{i=1}^nA_iA_{i+1}^2$

Presumably the indices $i$ in $A_i$ are taken mod $n$, so "$A_{n+1}$" is to be identified with $A_1$. This must be a known isoperimetric inequality, but it's easier to prove than to find in the ...
Noam D. Elkies's user avatar
5 votes
Accepted

What is the name of the 65537-gon?

Following the portuguese nomenclature (I am from Brazil) and translating to english its results: hexacontakaipentachiliakaipentahectakaitriacontakaiheptagon. Best regards!!
Daniel Corrêa's user avatar
5 votes

Necessary and sufficient condition for quadrilateral to be cyclic

This is not an answer but a comment. However, it will be too long and it would be awkward to break it up. It is of course natural to ask what is special about the nine point centre here. One can put ...
bathalf15320's user avatar
5 votes
Accepted

Billiard circuits in pentagons

We do not need the pentagon to be cyclic. But we do have a requirement for the angles. Begin with two cases where a cyclic path can be defined. In the first case it does not exactly meet the ...
Oscar Lanzi's user avatar
  • 1,865
4 votes

Reordering vertices of a polygon

I'm not certain, but it may be that the expansive motions in the Connelly-Demaine-Rote paper cited below can provide a route to your $Q''$. An expansive motion is one in which the distance between ...
Joseph O'Rourke's user avatar
4 votes

Fitting one Polygon in another

Here are two references. The first, a 1981 paper, provides an algorithm to solve the problem allowing translations and rotations, but not scaling: Chazelle, Bernard. "The polygon containment ...
Joseph O'Rourke's user avatar
4 votes

Algo for covering maximum surface of a polygon with rectangles

If the rectangles must be aligned (Gerry Myerson's question), then perhaps this approximation might suffice, depending on your needs. Let the width $w$ of your rectangles be $1$. Orient your polygon $...
Joseph O'Rourke's user avatar
4 votes
Accepted

Problem on distances in a polygon

Let me prove a bit more general statement. Let $P=[v_1\dots v_n]$ and $P'=[v_1'\dots v_n']$ be two solid polygons such that if $[vw]$ is a side of $P$ or a diagonal which lies in $P$ completely ...
Anton Petrunin's user avatar
4 votes
Accepted

Construct closed chain of $k$-gon around $n$ points-$n, k$ are odd primes number

We can consider all given points as complex numbers, points of the complex plane. Then, as far as I understood, for all integers $m\ge 1$, $1\le j\le k$ we we have $$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2},$$ ...
Alex Ravsky's user avatar
  • 4,282
4 votes
Accepted

A generalization of Harcourt's theorem

So we need to show $r\cdot \sum d_i=\sum n_id_i$, where $r$ denotes the inradius. Rewrite this as $0=\sum (n_i-r)d_i$, denote $n_i-r=m_i$. Then $m_i=:f(A_i)$ is the signed distance from $A_i$ to the ...
Fedor Petrov's user avatar

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