About the A099938 sequence

This sequence contains the inverses of the radii of the circles in this chain:

See details at About the A099938 sequence

Simplificaciones curiosas

Hay algunas fracciones que podemos simplificar eliminando el mismo dígito del numerador y del denominador, lo cual es cierto, por simple casualidad, pero no porque estemos efectuando una operación con los dígitos en cuestión. Por ejemplo, \[\frac{136}{340}=\frac{16}{40},\] donde han desaparecido un 3 de arriba y un 3 de abajo.

En el este documento en formato pdf incluyo el código de Mathematica que obtiene estas fracciones.

Simplificaciones curiosas

Análisis de un problema geométrico

Analizamos un problema geométrico, encontrando una generalización y una sencilla solución del mismo.

Análisis de un problema

An elimination problem

Given a triangle $ABC$ and two isogonal points $P$ and $P*$, call $A'B'C'$ and $A''B''C''$ the antipedal triangles of $P, P*$, respectively.

The triangle bounded by $A'A''$, $B'B''$, $C'C''$ is always perspective with ABC (see Hyacinthos message #21782).

The perspector is complicated, although we can see that it is the isotomic conjugate of a simpler point.
We want to calculate the locus of this isotomic conjugate when the point $P$ moves along the Euler line. 


An elimination problem 

Area of a conic section

This problem was proposed by Shafiqur Rahman on Facebook: Find the area and length of the semi axes of the section of the paraboloid $2x^2+y^2=z$ by the plane $x + 2 y + z = 4$.

We find a rigid motion that maps the given plane to the plane $z = 0$. We apply the same transformation to the given paraboloid and make $z=0$ then we get a equation of the conic on the $z=0$ plane. Next we get the reduced equation of the conic and its semiaxes.

Read the details here: Area of a conic section

Are they equal?

This question was proposed by Tony García on Facebook. Here is a solution that includes Mathematica calculations. The Descartes formula for circles touching other three mutually tangent circles is also used.

Solution

These are the values of R₁ (radius of upper red circle) and R₂ (radius of lower red circle) in terms of the side a of the square.

Locus of centroid of equilateral triangles in a parabola

Problem. Find the locus of centroid of equilateral triangles inscribed in the parabola $x^2=4ay$.

To solve this problem we use complex numbers and Mathematica's Eliminate. We find that the locus is another parabola.

 Locus of centroid of equilateral triangles inscribed in a parabola

La integral definida

La integral definida se define como un límite de sumas superiores e inferiores, como indica la siguiente figura.

Aquí veremos cómo realizar una figura como ésta con Mathematica. En primer lugar vamos a definir las funciones: Maximo[f_, a_, b_] := Apply[Max, Map[f, Table[a + i/10 (b - a), {i, 0, 10}]]] Minimo[f_, a_, b_] := Apply[Min, Map[f, Table[a + i/10 (b - a), {i, 0, 10}]]] Para cierta función $f$ definida sobre el intervalo $[a,b]$, estas funciones calculan los máximos $M_1, M_2, \dots, M_n$ y mínimos $m_1, m_2, \cdots, m_n$ en los intervalos $[x_0, x_1], [x_1, x_2], \dots, [x_{n-1},x_n]$ resultantes de dividir el intervalo $[a,b]$ en $n$ partes iguales. A continuación definimos la función: CrearGrafico[f_, a_, b_, n_] := Module[{plot, a1}, plot = Plot[f[x], {x, a, b}, PlotStyle -> {Thick, Red}]; xs = Table[a + i/n (b - a), {i, 0, n}]; Ms = Table[Maximo[f, xs[[i]], xs[[i + 1]]], {i, 1, n}]; ms = Table[Minimo[f, xs[[i]], xs[[i + 1]]], {i, 1, n}]; M = Apply[Max, Ms]; m = Apply[Min, ms]; instr = Join[ {RGBColor[1, 1, 0.6]}, Table[Rectangle[{xs[[i]],0}, {xs[[i+1]], Ms[[i]]}], {i,1,n}], {Yellow}, Table[Rectangle[{xs[[i]],0}, {xs[[i+1]], ms[[i]]}], {i,1,n}], {Black}, Table[ Line[{{xs[[i]], ms[[i]]}, {xs[[i+1]], ms[[i]]}}], {i,1,n}], Table[ Line[{{xs[[i]], Ms[[i]]}, {xs[[i+1]], Ms[[i]]}}], {i,1,n}], Table[Line[{{xs[[i]], Min[0, ms[[i]], ms[[i - 1]]]}, {xs[[i]], Max[Ms[[i]], Ms[[i - 1]]]}}], {i, 2, n}], {Line[{{b, 0}, {b, Ms[[n]]}}]} ]; Show[{Graphics[instr], plot}, AspectRatio -> Automatic, Axes -> True, PlotRange -> {{a, b}, {m, M}}] ] Esta función crea la figura para una función $f$ en un intervalo $[a,b]$ en el que se han hecho $n$ divisiones. Podemos hacer interactivo el gráfico (permitiendo al usuario manipular el valor de $n$) definiendo CrearGrafico[f_, a_, b_] := Manipulate[ CrearGrafico[f, a, b, n], {n, 5, 40, 1}] y ahora, para representar $f(x)=1-x^2$ en $[0,1]$ podemos introducir CrearGrafico[1 - #^2 &, 0, 1]

Concurrent Brocard Axes

Let $ABC$ be a triangle, $P$ a point and $A' = AP \cap BC$. If the Euler lines of three triangles from the set $\{A'AB, A'BP, A'PC, A'CA\}$ are concurrent, then the Euler lines of all four triangles are concurrent (see Forum Geometricorum 2001).

Antreas Hatzipolakis asks for the same for Brocard axes (see Hyacinthos message 20664). Let's investigate it using barycentric coordinates and Mathematica.

Concurrent Brocard Axes