mg.metric geometry – The product of the lengths of two line segments that belong to Newton line

I am looking for the proof of the following claim:

Consider a family of bicentric quadrilaterals with the same inradius length. Denote by $P$ and $Q$ the midpoints of the diagonals, and by $I$ the incenter. Then, $|PI| cdot |QI|$ has the same value for all quadrilaterals in the family.

enter image description here

The GeoGebra applet that demonstrates this claim can be found here.

mg.metric geometry – A circle with radius R, find the total area of all the circles added together

A circle with radius R is shown below in figure a. In figure b, two circles with a radius of 1/2 R are placed on top of the original circle from figure a. In figure c, four circles with a radius of 1/4 R are placed on top of the circles from figure b.

The image

Assuming this pattern continues indefinitely, find the total area of all the circles added together.

(With steps would be wonderful, I really don’t understand)

mg.metric geometry – What is the correct name of points with this property?

Let $(X, d)$ be a metric space and $x in X$.
Suppose for all $x_1, x_2 in X$ the following inequality holds:
d(x_1, x_2) le max bigl{ d(x, x_1), d(x, x_2) bigr}.

For example, singleton in the Gromov-Hausdorff space satisfies this condition.
And every point in an ultrametric space too.
What is the correct name of points $x$ with this property?

mg.metric geometry – Does codimension-1 collapsing with bounded curvature have boundary?

Let $(M^n,g_i)$ be a sequence of smooth complete Riemannian manifold with $|sec_{g_i}| le 1$. Suppose $(M_i^n,g_i)$ converges to a limit space $(X^{n-1},d)$ in the Gromov-Hausdorff sense, where the Hausdoff dimension of $X$ is $n-1$.

Can we show that $X$ contains no boundary point? Here, a point is a boundary point of $X$ if its tangent cone is isometric to $mathbb R^{n-2} times mathbb R_+$.

mg.metric geometry – Banach fixed point theorem / convergence squeeze

I am currently investigating an iterative learning algorithm and its convergence time. If we let $x_1 = g(x_0)$ and let $epsilon := |x_t – x^*|$ be our desired error bound from the fixed state, then we have
$$t geq lnleft( frac{epsilon(1-L)}{|g(x_0) – x_0|} right) / ln(L)$$
where L is the Lipschitz constant. My question is this: our function is of the form $g(x) = frac{1 – s(x)}{Ccdot s(x)}$ where $C$ is some constant and $s(x)$ is a function is not always known. If I can verify that the unknown $s(x)$ is sandwiched between two polynomials, does this guarantee that $g(x)$‘s convergence time can thus be bounded as well? For example if I prove
$$C_1x^{k_1} leq s(x) leq C_2x^{k_2}$$
then can I say
$$text{Conv. time of } C_1x^{k_1} leq text{Conv. time of } s(x) leq text{Conv. time of } C_2x^{k_2}$$

mg.metric geometry – Is the Jaccard distance between probability vectors a metric?

Let X and Y be probability vectors, meaning that X = $(x_1, x_2, …, x_n)^T$, where $x_ileq 1$ and $sum_{i=1}^{n}x_i=1$ (Y is defined similarly).

Define the Jaccard distance as

J_d = 1 – frac{textbf{X}cdottextbf{Y}}{textbf{X}cdot textbf{X}+ textbf{Y}cdottextbf{Y} – textbf{X}cdottextbf{Y}}

Is $J_d$ a proper distance (i.e., metric)?

mg.metric geometry – Equal products of triangle areas

Claim. Given hexagon circumscribed about an ellipse. Let $A_1,A_2,A_3,A_4,A_5,A_6$ be the vertices of the hexagon and let $B$ be the intersection point of its principal diagonals. Denote area of triangle $triangle A_1A_2B$ by $K_1$, area of triangle $triangle A_2A_3B$ by $K_2$,area of triangle $triangle A_3A_4B$ by $K_3$,area of triangle $triangle A_4A_5B$ by $K_4$,area of triangle $triangle A_5A_6B$ by $K_5$ and area of triangle $triangle A_1A_6B$ by $K_6$ .Then, $$K_1 cdot K_3 cdot K_5=K_2 cdot K_4 cdot K_6$$

mg.metric geometry – Stability of isoperimetric inequality

Let $S$ be subset of $mathbb{R}^n$ with perimeter 1.

Isoperimetric inequality states that then the volume of $S$ is not greater than $V_n$,

where $V_n$ is the volume of a ball in $mathbb{R}^n$ with perimeter 1.

Assume that $C cdot text{(Volume of }S) ge V_n$, where $C$ is some constant.

Is it true that 99% of $S$ can be covered by union of constant number of balls with constant
radius? (The constants can be defense on $n$ and $C$.)

P.S. The question is motivated by the similar question about boolean cube.

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

This question is inspired by two posed by P.Terzińá (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 $G_1$;


  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.

  3. since there are now over $40,000$ registered centres there are potentially that number squared such results. Hence a synthetic proof can’t be expected and one is forced to apply Tate’s maxim–think geometrically, prove algebraically.

mg.metric geometry – Acute triangles in “obtuse” polygons?

Let $P$ be a convex polygon. Suppose every interior angle of $P$ is obtuse. Is it always the case that there exists three vertices $p, q, r$ of $P$ such that $triangle pqr$ is acute?

I conjecture that the answer is yes. I have tried different types of triangulations of $P$ (e.g. fan triangulation, triangulation by repeatedly connecting every other vertex, etc.) However, I haven’t come up with a proof yet.