graphs and networks – Generate Aztec triangle of size n automatically?

In the paper titled “Perfect Matchings of Cellular Graphs” by Mihai Ciucu, the Aztec triangle of size n (n= 1, 2, 3, 4, 5, ...) is equivalent to a triangular grid of n1 squares (n1 = 1, 4, 9, 16, 25, ...) .

See the following example:
image

Is there a automatical way to generate such patterns?

Also are there some general method to generate Aztec diamond of order n (not just triangle)?

Thank you very much!

mathematics – Ray-Triangle Intersection: does the direction of the triangle normal matter?

Remember that A, B, and C come from the normal too. So negating the normal flips the sign of all four terms, making:

$$begin{align}
(-A) x + (- B) y + (- C) z + (- D) &= 0\
(-1)(Ax + By + Cz + D) &= 0\
(-1) times (-1)(Ax + By + Cz + D) &= (-1) times 0\
Ax + By + Cz + D &= 0
end{align}$$

So you can see, negating the normal still gives us an equivalent equation, with all the same solutions as the original. Whether we start with $vec n = (A, B, C)$ or $vec n = (-A, -B, -C)$ does not alter the plane equation in any way that matters.

The direction of the normal can matter for lighting, or whether you consider the triangle “front facing” for your purposes, but does not change anything about the math for detecting the intersection.

The resource you’re using contains an error, though. if $ vec n = (A, B, C)$ is the plane normal, then $D = – vec n cdot vec p$ for a point $vec p$ in the plane. The article you linked neglects that minus sign.

Another way of thinking of this: if your normal is a unit vector, then $D$ is a signed distance from the plane to the origin, along the direction of the normal. If the normal points toward the origin, this signed distance is measured in the same direction as the normal, and gives a positive value. If the normal points away from the origin, then this distance is measured in the direction opposite the normal, and gives a negative value.

Meanwhile, $Ax + By + Cz$ measures a signed distance from the origin to the plane perpendicular to $(A, B, C)$ containing the point $(x, y, z)$.

Since the “from” and “to” are flipped between these two parts, the two values will have the same magnitude but opposite sign for any $(x, y, z)$ on the plane, cancelling out to the zero on the right hand side.

(If the normal is not a unit vector, the same argument applies, except now we’re dealing with a scaled distance – like we measured it with a different choice of units)

Meaning of inverted vs non-inverted ISA triangle in EER diagram

I found two different ISA triangles in various books. Are they give the same meanings? I am currently studying about specialization and generalization in DBMS.

calculus – Finding maximum area when the triangle is isosceles?

The two equal sides are of length 3. I tried to solve using the Pythagorean theorem, finding the derivative to the area function and equaling to zero. When I equal to zero I get a negative value and I’m unsure of how to proceed.

begin{equation}
A=frac{bh}{2}
end{equation}

begin{equation}
h=sqrt{frac{b^2}{4}+9}
end{equation}

Completing the area function and finding the derivative
begin{equation}
A=frac{b}{2}sqrt{frac{b^2}{4}+9}
end{equation}

begin{equation}
A’=frac{b^2+18}{2sqrt{{b^2}+36}}
end{equation}

Equal to zero to find maximum value
begin{equation}
A’=0
end{equation}

begin{equation}
b=sqrt{-9}=3i
end{equation}

How should I interpret this? I was expecting a positive value, not a negative value.

geometry – Finding an angle in a triangle

enter image description here

I want to find angle x in this picture.

enter image description here

And this is what I’ve done so far. Without loss of generality, assume $overline{rm BC}=1$

then, $$overline{rm BD}= 2sin{frac{x}{2}}$$, $$overline{rm BH}= 4sin^2{frac{x}{2}}= 2(1-cos{x}), quad overline{rm CH} = 2cos{x}-1$$
$$overline{rm CE}=frac{2cos{x}-1}{sqrt{2-2cos{x}}}$$
Let $overline{rm DE}=y$,

since $bigtriangleup DCE = bigtriangleup HCE$,
$$frac{1}{2}ysin{50^{circ}}=frac{1}{2}sin{x}frac{(2cos{x}-1)^2}{2-2cos{x}}$$
Then by applying law of cosines to $bigtriangleup DEC$,
$$y^2+1-2ycos{50^{circ}}=frac{(2cos{x}-1)^2}{2-2cos{x}}$$
So we have a system of equations
$$begin{cases}ysin{50^{circ}}=sin{x}frac{(2cos{x}-1)^2}{2-2cos{x}}\y^2+1-2ycos{50^{circ}}=frac{(2cos{x}-1)^2}{2-2cos{x}} end{cases}$$
But it’s too messy to solve since 50 is not special angle.
How can I solve this problem?

Describe a Monte Carlo algorithm for the Triangle Packing problem

Book: Parameterized Algorithms by Marek Cygan (free to download legally)

Chapter about Multivariate polynomials on Page 353 (In the book not the pdf) Question 10.19:

Describe a Monte Carlo $2^{3k}n^{O(1)}$ – time polynomial-space algorithm for the Triangle Packing problem: given a graph $ G $ and a number $ k in N $, decide whether $ G $ contains
$ k $ disjoint triangles (subgraphs isomorphic to K3).

Hint 10.19 in Page 354: Use a similar approach as in the $2^{k}n^{O(1)}$ -time algorithm for $k$-path from this
chapter (Page 333).

I tried to solve the problem using the hint, but without success unfortunately, there are solutions to the book? I would be happy to get help if it can be solved even without official solutions of the book but it is important to solve it using the hint they brought in order to use the tools learned in the same chapter.

Thanks in advance.

mg.metric geometry – A generic question on triples of circles associated with a triangle

This question is inspired by two posed by P.Terzić (both given elegant synthetic proofs by F. Petrov). The starting point is a triangle $ABC$ and a triangle centre $G_1$. There are two classical ways to use this to generate a new triangle $A_1B_1C_1$ where the new vertices are

  1. the reflections of the original ones in $G_1$;

or

  1. the feet of the Cevians through this point.

We now add a second centre $G_2$ to the ingredients and consider the three circles through $AG_2 A_1$, $BG_2B_1$ and $CG_2C_1$.

These three circles have the one common point $G_2$, of course, and our question is to determine under which conditions they have a second one. More specifically, our question is whether the following dichotomy holds:

For a given (distinct) pair of triangle centres, then either the above result holds or the triangle is isosceles.

Remarks. 1. The results of Petrov display two situations where the first situation is true.

  1. we are using the term “circle” in the generalised sense which includes lines. These occur in the specal case of an isosceles triangle (for an equilateral one we have three lines through the centre–the second point of intersection is at infinity).

  2. we refer to the online Encyclopedia of Triangle Centers for the concepts and notations we are using.

parity – The Sierpiński triangle and the number of $(0,1)$-polynomials $p(x)$ where $p(x)^2$ has largest coefficient $k$.

My Twitter bot @oeisTriangles randomly selects an OEIS “table”-style sequence and draws an image, where even terms are light-colored and odd terms are dark-colored. Today it tweeted an image for OEIS sequence A169950:

(T)riangle read by rows, in which $T(n,k)$ (…) is the number of (monic degree $n$ (0,1)-)polynomials of thickness $k$.

where

The thickness of a polynomial $f(x)$ is the magnitude of the largest coefficient in the expansion of $f(x)^2$.

The table begins:

$$
1 \
1~~~~1 \
1~~~~2~~~~1 \
1~~~~5~~~~1~~~~1 \
1~~~~8~~~~4~~~~2~~~~1 \
1~~~~13~~~~8~~~~8~~~~1~~~~1 \
1~~~20~~~15~~~18~~~~7~~~~2~~~~1
$$


Conjecture

This “parity triangle” looks like a Sierpiński triangle, which makes me conjecture that for all $n, k in mathbb N_{geq 0},$ $$T(n,k) equiv binom{n}{k} bmod 2.$$

Parity triangle of A169950, which looks like Pascal's triangle mod 2.

Is this conjecture true? If so, is there a recurrence that explains this?

geometry – What Height can Divide the Area of a Triangle by 2


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Help with a geometry problem with a triangle and its orthocentre

Let ABC a traingle inscribed in a circle with radius 1 and center O. Let the angle AOM=150 where M is the middle BC. Let H the orthocentre of the triangle. If A,B,C are selected such that Oh has the minimum lenght, than the lenght of BC is
A: $sqrt{15}$
B: $sqrt{13}/2$
C: $sqrt{3}/2$
D:$sqrt{13}/4$
I made a sketch and tried to apply the Sylvester’s theoreme and to solve the problem with vectors but did not succed. Could you please help me?