35
votes
Accepted
Can a row of five equilateral triangles tile a big equilateral triangle?
I suppose I should post: I solved this on MathOverflow. The answer is YES: a size-45 triangle can be tiled.
I thank two insights from Josh B here: first that a rhombus with side length 15 can be ...
35
votes
Accepted
Convex polygons that do not tile the plane individually, but together they do
There is a tiling of the plane made from regular heptagons and irregular pentagons.
We know that regular heptagons cannot tile the plane.
The irregular pentagon has four equal sides and one shorter ...
31
votes
How few disks are needed to cover a square efficiently?
If different circle sizes are allowed, we may reach an efficiency $>76.3\%$ with just five disks:
With $13$ disks the maximum efficiency is already $>80.4\%$. It is enough to replace each "...
28
votes
Accepted
How few disks are needed to cover a square efficiently?
With a bit of Googling I found this paper: Covering a Square with up to 30 Equal Circles, by Kari J. Nurmela and Patric R. J. Östergård.
They used a computer search to find coverings of a square by $...
18
votes
Can a row of five equilateral triangles tile a big equilateral triangle?
Here's the minimal solution, a side-30 triangle. Also posted to MathOverflow here https://mathoverflow.net/questions/267095/can-a-row-of-five-equilateral-triangles-tile-a-big-equilateral-triangle.
18
votes
Accepted
What is the exact value of the radius in the Six Disks Problem?
The minimal polynomial for $r(6)$ is
$$\color{red}{7841367r^{18}-33449976r^{16}+62607492r^{14}-63156942r^{12}+41451480r^{10}-19376280r^8+5156603r^6-746832r^4+54016r^2+3072}$$
I obtained this result in ...
16
votes
Convex polygons that do not tile the plane individually, but together they do
HINT:
Consider a convex hexagon that can tile the plane. There are three types of tiling hexagons, we take one of type 1, which has two opposite sides parallel and equal
Cut it into two pentagons. ...
16
votes
Accepted
Number of ways to stack LEGO bricks
Your result $4^{n-1}$ is correct. Bóna's proof goes through with a single modification. In the last step that counts the half-pyramids, there is one more option: A bottom brick with a half-pyramid on ...
14
votes
Tiling the plane with consecutive squares
Here is a solution with $n=7$. Edit. I added some comments expressing my belief that $n=7$ may well be the largest solution. (Well, I keep editing my answer, but at the end I came up with a solution ...
14
votes
Accepted
What is the minimum area of a rectangle containing all circles of radius $1/n$?
One can tackle this question by forming an offset grid of squares of radii $1/k$ in a roughly triangular fashion that fits inside the top left sector of the rectangle, whereby the squares can be ...
12
votes
What is the minimum area of a rectangle containing all circles of radius $1/n$?
For suitable $a$ and $N$, we can place disjoint disks of radii $\frac 1n$, $n\ge N$ into an $a\times a$ square, column by column, where column $k$ ($k=0,1,2,\ldots$), consists only of disks of radius $...
11
votes
Greatest number of parts in which n planes can divide the space
On 2d
Question: What is the maximum number of regions that can be formed with n lines?
The main idea: A line can cut another line in at most 1 point.
As we want to form maximum number of regions, ...
11
votes
Greatest number of parts in which n planes can divide the space
Your question is equivalent to the following:
What is the greatest number of parts a cake (cylinder/cube/any other convex shape) can be divided into with n straight cuts?
Appropriately these are ...
11
votes
Can we partition the plane like this?
Yes it can, even for the rectangles with vertical and horizontal sides. On $k$-th step the union of $I_1,\dots,I_{n_k}$ is a rectangle $[a_k,b_k]\times [c_k,d_k]$ where $a_k<-k<k<b_k,c_k<-...
10
votes
Projection of a point onto a convex polyhedra
As mentioned in other answers and comments, the problem you need to solve is an inequality constrained, strongly convex QP:
$$
\begin{aligned}
\mathrm{minimize}\ &\tfrac{1}{2}\|x\|^2_2 \\
\...
10
votes
Tiling the plane with consecutive squares
$n=3$, $n=4$, $n=5$ all tile the plane:
Each of these 'symmetrically bitten rectangle' shapes tiles the plane by translation (e.g., attach them along opposite long sides to form diagonal bands, then ...
9
votes
Can we partition the plane like this?
A sequence of rectangles $R_i$ of size $1\times a_i$ for $i=1,2,3\ldots$ with $a_i\ge 1$ can be filled with rectangles $1\times \frac{1}{n}$ generated by the harmonic series. Then the rectangles $R_i$ ...
9
votes
Accepted
Why do Pentagon tilings not solve the Einstein Problem?
They key is that such a tile can only tile the plane aperiodically. The other tiles you cite do not satisfy this criterion because while they can tile the plane in an aperiodic fashion, there exist ...
8
votes
Polyhedra vs Polytope
In his classic text on Convex Polytopes, Grünbaum gave three incompatible definitions of a polyhedron, each used in a different branch of mathematics. There are plenty more. He later wrote somewhat ...
8
votes
Accepted
Puzzle: Cut regular tetrahedron into distinct-sized regular tetrahedra?
It is not possible to dissect a regular tetrahedron into any finite number${}>1$ of regular tetrahedra at all. This is simply because the angle between the faces of a regular tetrahedron, which is ...
8
votes
Accepted
Reference request: Good introduction to Sphere Packing
The literature hints in the comments are already very good. When it comes to Sphere Packings, you can hardly get around Conway's & Sloane's book "Sphere packings, lattices and groups".
...
8
votes
Polyhedra has more corners than facets
In $\Bbb R^2$ the claim is trivial, since every polygon always has the same number of sides and vertices.
In $\Bbb R^3$, it's definitely false. For every polyhedron that has $F$ facets and $C$ ...
7
votes
Accepted
What's the upper bound for sofa problem?
It does not seem useful to me to just repeat the solution in e.g. Ian Stewart's book Another Fine Math You've Got Me Into... as linked from the Wikipedia Moving sofa problem page, so I'll just ...
7
votes
Accepted
Tetrahedron circumradius in high dimensions
The standard $n$ simplex is the convex hull in $\mathbb{R}^{n+1}$ of the points $$(1,0,0,\dots,0),\\
(0,1,0,\dots,0)\\
\vdots\\
(0,0,\dots,0,1)$$
That is, the set of all points in $\mathbb{R}^{n+1}$ ...
7
votes
Accepted
Coloring every point of the plane with 4 colors
Let $P$ be an arbitrary regular hexagon with side length $1$. Note that there are at least two vertices of $P$ given the same colour. We shall consider some cases:
Case 1: The vertices of $P$ are ...
7
votes
Frame challenge: Find the maximum $n$ such that circles of radius $1, \frac12, \frac13, ..., \frac1n$ can be held immobile by a convex frame.
You can start by using the Descartes Circle Theorem. For four mutually touching circles $2 (a^2+b^2+c^2+d^2) = (a+b+c+d)^2$, where the values are 1/radius (the bend). For example, circles with radii {...
6
votes
Accepted
For which dimensions does there exist a regular $n$-polytope such that the distance of its vertex to its center equals the length of its side?
We can work with standard embeddings of the $n$-cube (as $[-1,1]^n$) and cross-polytope (vertices $(\pm1,0,0,\ldots) $, $(0,\pm1,0,\ldots) $, etc.) and the embedding of the $n$-simplex in $\mathbb{R}^...
6
votes
For which dimensions does there exist a regular $n$-polytope such that the distance of its vertex to its center equals the length of its side?
You might be interested to know, that, when loosening the condition of being a regular polytope, then not only sporadically some oddballs do fit your property of having circumradius = edge length, but ...
6
votes
How close are the closests cells of the same color in a periodically colored grid?
There's a continuous variant of this problem that tells you pretty well why the patterns you see exist:
First, let $|\cdot |:\mathbb R \rightarrow [0,1/2]$ be the function that takes a real number ...
5
votes
Accepted
Do the last three remaining cards in a game of Set always form a set?
I'll just note that it is possible for a game to end with more than 3 cards and no sets on the table, but if there are three cards left at the end then indeed they must form a set.
Under the ...
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