Inclusion Properties of the Triangular Ratio Metric Balls

Inclusion properties are studied for balls of the triangular ratio metric, the hyperbolic metric, the j∗\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$j^*$$\end{document}-metric, and the distance ratio metric defined in the unit ball domain. Several sharp results are proven and a conjecture about the relation between triangular ratio metric balls and hyperbolic balls is given. An algorithm is also built for drawing triangular ratio circles or three-dimensional spheres.


Introduction
In the field of geometric function theory, the study focuses on different geometrical properties of several types of functions, including for instance conformal, quasiconformal, and quasiregular mappings.It is often useful to construct a metric space to investigate behaviour of metrics under these mappings to understand better how they distort distances.Especially, the hyperbolic metric is an important subject of study because of its conformal invariance, sensitivity to boundary variation, monotonicity with respect to domain, and intricate features of hyperbolic geometry [3, p. 191-192].Numerous generalizations of the hyperbolic metric called hyperbolic type metrics have also been defined to create geometrical systems sharing some of these properties, including the way the hyperbolic metric measures distances between two points x, y in a domain G by taking into account their position with respect to the boundary ∂G of the domain.
For a domain G R n , one of the hyperbolic type metrics is the triangular ratio metric s G : G × G → [0, 1], [1, (1.1), p. 683] which was originally introduced by P. Hästö in 2002 [6].This metric has been studied much [1,2,4,10,11,12,13,14] because, despite its relatively simple definition, computing the value of the infimum in the denominator is a non-trivial task in the case where the domain G is, for instance, the unit ball B n .In fact, if G = B 3 and the boundary of this domain is a spherical mirror, the correct point z coincides with the point that the light ray sent from the point x must hit to reflect into the point y and, because of this relation to optics, the problem has a very long history explained in [2].
Given the point x in a domain G R n and some radius 0 < t < 1, the x-centered triangular ratio metric ball with this radius t is defined as B s (x, t) = {y ∈ G | s G (x, y) < t}.To study the different mappings, it is often useful to know what is the largest ball B d (x, r 0 ) defined with some other metric d that is included in the triangular ratio metric ball B s (x, t) or the smallest ball B d (x, r 1 ) containing the ball B s (x, t).This type of an inclusion question has been recently studied for the triangular metric by S. Hokuni, R. Klén, Y. Li, and M. Vuorinen [7] in such cases where the domain G is the upper half-space or the punctured real space, and Klén and Vuorinen have also researched the inclusion properties of several hyperbolic metrics other than the triangular ratio metric in the unit ball [9] and other domains [8].
In this article we continue the earlier research by studying the inclusion properties of the balls of the triangular ratio metric defined in the unit ball.In Section 3, we focus on triangular ratio metric balls in terms of the Euclidean geometry, give the sharp inclusion result between triangular ratio metric and Euclidean balls, and also offer an algorithm for drawing triangular ratio metric circles or three-dimensional spheres.In Section 4, we study the ball inclusion for the triangular ratio metric and two other hyperbolic type metrics called the distance ratio metric and the j * -metric, find an explicit formula for j * -metric balls and also prove one earlier conjecture from [7].Finally, in Section 5, we research the connection between triangular ratio metric balls and hyperbolic balls.

Preliminaries
An n-dimensional Euclidean ball with center x ∈ R n and radius r > 0 is denoted by B n (x, r), its closure by B n (x, r) and its sphere by S n−1 (x, r).For the unit ball and sphere, the simplified notations B n and S n−1 are used.The notation L(x, y) means a Euclidean line through points x, y ∈ R n .For any point x in a domain G R n , d G (x) denotes the Euclidean distance from x to the boundary ∂G by x be the complex conjugate of x.For three points x, y, z ∈ R n , let XY Z ∈ [0, π] be the magnitude of the angle between the vector from y to x and the vector from y to z and denote the origin with the letter o in these angle notations.
The hyperbolic metric can be computed in the unit ball with the following formula [3, (4.16), p. 55] In (2.1), sh denotes the hyperbolic sine.Similarly, the hyperbolic cosine and tangent are denoted here by ch and th, and the inverse hyperbolic functions are arsh, arch, and arch, while acos is the arccosine function.For any hyperbolic type metric d defined in a domain G R n , x ∈ G, and r > 0, the n-dimensional ball is The inclusion result between the hyperbolic and the distance ratio metric is known in the case of the unit ball.
and, for The following inequality holds between the triangular ratio metric and the j * -metric.
There are a few results that help to find the triangular ratio distance in the unit ball.

Triangular ratio metric balls in Euclidean geometry
If triangular metric balls defined in G = B n are origin-centered, they are equivalent to Euclidean balls, as stated by the following lemma.However, if x = 0, expressing the triangular ratio metric ball B s (x, t) for 0 < t < 1 in terms of Euclidean geometry becomes considerably more difficult, but with the help Lemma 2.8 and Theorem 2.9, we can prove Lemmas 3.2 and 3.3 that reveal certain points always included in the sphere S s (x, t).Lemma 3.1.For G = B n and 0 < t < 1, the origin-centered triangular ratio metric ball B s (0, s) equals the Euclidean ball B n (0, 2t/(1 + t)).
Next, we will find the intersection points of the triangular ratio metric sphere S s (x, t) and the line L(0, x) that can be also seen in Figure 1.Lemma 3.2.For G = B n , x ∈ B n \ {0} and 0 < t < 1, the triangular ratio metric sphere S s (x, t) intersects with the line L(0, x) at the points Proof.Any point y ∈ L(0, x) can be written as y = x + rx/|x| with r ∈ R. Since the points x, y, 0 are collinear, by Lemma 2.8, For s B n (x, y) = t, we can solve Clearly, one intersection point is found with r < 0 and the other one with r > 0 and, since (1 − |x|)/(1 − t) ≤ (1 + |x|)/(1 + t) is equivalent to t ≤ |x|, r can be written as a minimum expression so that substituting these values of r in y = x + rx/|x| proves the lemma.
By the quadratic formula, we can solve that, Above, the sign ± needs to be a plus sign.This is because where the difference is |x| 2 − t 2 non-negative for t ≤ |x|.Thus, the root with a minus sign fulfills the condition |x| < cos(u) < 1 if and only if t = |x| and, in this special case, it is equal to the root with a plus sign.In turn, which holds for all x ∈ B 2 \ {0} and 0 < t ≤ |x|.Consequently, we have and the points y = |x|e (k±2u)i can be written as x(e ui ) ±1 , where Let us yet prove the rest of the lemma.For all such points x, y ∈ B 2 that |y| = |x|, the inequality s B 2 (x, y) ≤ |x| holds because, if x, y are collinear with the origin, then by Lemma 2.8 and otherwise this inequality can be directly seen from Theorem 2.9.Therefore, if t > |x|, s B n (x, y) < t for all x ∈ B 2 \ {0} and y ∈ S 1 (0, |x|).
Lemmas 3.2 and 3.3 can be used to prove that certain inclusion results are sharp, as done also in the following theorem.
Proof.For all y ∈ B n (x, r 0 ) with r 0 as above, it follows from the fact that Consequently, for all y ∈ B n (x, r 0 ), the triangular ratio distance between x and y is less than t so B n (x, r 0 ) ⊆ B s (x, t).Note that the upper limit of r 0 is also sharp.Namely, if the radius r 0 could be any greater than 2t(1 − |x|)/(1 + t), the ball B n (x, r 0 ) would contain the point y 0 ∈ S s (x, t) of Lemma 3.2 and this point is trivially outside of the ball B s (x, t).For all y ∈ B n \ B n (x, r 1 ) where r 1 is as in the theorem, it follows from the fact that  The following theorem is used to create an algorithm for drawing triangular ratio circles.
there is no such point y ∈ B 2 for which the equality would hold and L(0, z) would bisect the angle between L(x, z) and L(y, z).Otherwise, a point y ∈ B 2 fulfills both of these conditions if and only if Proof.Fix t, x, u, and z as in the lemma.Note that u = XOZ.Now, For y ∈ B 2 , L(0, z) bisects the angle between L(x, z) and L(y, z), if and only if OZY = v but y / ∈ L(x, z) unless v = 0. See Figure 2. Consequently, we can write Suppose then that t = |x − y|/(|x − z| + |z − y|).By using trigonometric identities cos(2v) = 2 cos 2 (v) − 1 and cos 2 (v) + sin 2 (v) = 1, we will have The roots above are defined if and only if If t ≥ |x|, the inequality above holds for all cos(u).If t < |x|, we can solve that and we need to choose |x| for h to be well-defined.The proof is now finished.
Note that even if x, y ∈ B 2 and z ∈ S 1 so that L(0, z) bisects the angle between the lines L(x, z) and L(y, z), this does not necessarily mean that However, if this equality holds for some x, y ∈ B 2 and z ∈ S 1 , then L(0, z) must bisect the angle between the lines L(x, z) and L(y, z) by Theorem 2.7.In other words, when given x ∈ B 2 and u ∈ [0, π], Theorem 3.6 can be only used to find potential candidates for such points y ∈ B 2 that the equality (3.7) holds with the point z = e ui .This is also the idea behind the following algorithm, which finds efficiently such points y that belong to the circle S s (x, t).This algorithm has been used to create most of the images of this article, including Figure 3 that shows how the radius t affects the shape of the triangular ratio metric circles.

Algorithm for drawing triangular ratio circles in the unit disk
The following instructions can be used to write an algorithm that plots the triangular ratio circle S s (x, t) in the domain G = B 2 when the user gives the center x ∈ B 2 and the radius 0 < t < 1 of as an input.
1.If x = 0, plot the Euclidean circle S 1 (x, 2t/(1 + t)) and stop after that.2. Otherwise, initialize a vector Y , fix for instance N = 10, 000 and = 10 −5 , and define the functions , and 3. If t ≥ |x|, create a vector U containing N values from the interval [0, π] and, if t < |x| instead, choose the N values of the vector U from the set 0, acos 5. Then for each value y ∈ Y , compute the triangular ratio distance s B 2 (x, y) by using for instance the algorithm [1, Algorithm 2.5, p. 686] and remove this value y from Y if |s B 2 (x, y) − t| > .
6.For each remaining element y ∈ Y to the vector Y , add the value xy/x to the vector Y because this is a reflection of y over the line L(0, x) by [3, B.11, p. 460] and the triangular ratio metric is invariant under reflection.
7. Organize all the values y ∈ Y so that the complex argument of the difference y − x increases monotonically on [0, 2π) as the one lists the elements of the vector Y and plot the triangular ratio circle S s (x, t) by connecting the organized points of Y .
For x ∈ G = B 3 and 0 < t < 1, the three-dimensional triangular ratio sphere S s (x, t) can be drawn by using the algorithm above to plot the two-dimensional circle S s (x, t) in any two-dimensional plane containing both x and the origin, and then plot the threedimensional sphere by rotating this circle around either the line L(0, x) or any line that contains the origin if x = 0.

Triangular ratio metric and j * -metric balls
Unlike for triangular ratio metric balls, an explicit expression can be found for any j * -metric sphere S j * (x, k) in terms of Euclidean geometry in the domain G = B n , and the results of Theorem 4.1 and Corollary 4.4 can be used to create an algorithm drawing j * -metric balls in the unit disk or ball.Theorem 4.1.For x ∈ B n \ {0} and 0 < k < 1, the sphere S Υ (x, k) defined with the function , Proof.For y ∈ B n \ {0}, let u = XOY .By the cosine formula, Consequently, The root above is defined if and only if The discriminant in the inequality above is Suppose that the condition (4.3) holds and the root (4.2) is defined.Let us first consider the case where the sign ± of this root is a plus.Clearly, y ∈ B n \ {0} if and only if ).If this inequality holds and k < 1/3, By combining the inequalities k < 1/3, |x| > 2k/(1 − k), and cos(u to the condition (4.3), it follows that we can choose a plus for the sign ± of the root (4.2) only if Next, we will study the root (4.2) with minus.Suppose again that the condition (4.3) hold.It can be verified that By combining these observations with the condition (4.3), we have , and cos(u The theorem follows now.
The result of the following corollary can be better understood by looking at Figure 4.
where S Υ (x, k) is as in Theorem 4.1.
Proof.The first part of the corollary follows directly from Lemma 3.1 because j * B n (y, 0) = s B n (y, 0) for all y ∈ B n .
Suppose below that x = 0 and denote r = |x − y|.Consider first the part of the j * -metric sphere S j * (x, k) in the closure of the Euclidean metric ball B n (0, |x|).For all y ∈ B n (0, |x|), |y| ≤ |x| and j * If |y| > |x|, then the distance j * (x, y) is equivalent to the value Υ B n (x, y) of the function introduced in Theorem 4.1, so and the corollary follows.
Remark 4.5.Note that Corollary 4.4 can be used to find the balls in the distance ratio metric, too, given Proof.The j * -metric ball B j * (x, k 0 ) is included in the Euclidean ball B n (x, r) if and only if k 0 ≤ j * B n (x, y) for all y ∈ S n−1 (x, r), and B j * (x, k 1 ) contains B n (x, r) if and only if j * B n (x, y) ≤ k 1 for y ∈ S n−1 (x, r).For all y ∈ S n−1 (x, r), where u = OXY or, if x = 0, u = ZOY with z = 1.Clearly, the minimum value of the distance j * B n (x, y) with respect to y and the maximum value r/(2 − 2|x| − r), from which the lemma follows.
The following inequality helps to formulate a result about the ball inclusion between the triangular ratio metric and the j * -metric.
. Lemma 4.9.For any domain G R n , x ∈ G, and 0 < t < 1, where the latter inclusion above is sharp for all possible choices of a domain G, and the former inclusion can be replaced with Proof.The inequality of Corollary 4.8 is equivalent to j * G (x, y) ≤ s G (x, y)/(1 + s G (x, y)) and, by Theorem 2.6, the inequality j * G (x, y) ≤ s G (x, y) holds for all x, y ∈ G.It now follows that To show the inclusion B s (x, t) ⊆ B j * (x, t) is sharp, fix x ∈ G, z ∈ S n−1 (x, d G (x)) ∩ ∂G, and y = x + 2t(z − x)/(1 + t) and note that The last part of the lemma follows from Theorem 2.6.
Given j * G (x, y) = th(j G (x, y)/2), the j * -metric in the former result can be easily replaced with the distance ratio metric.Corollary 4.10.For any domain G R n , x ∈ G, and 0 < t < 1, where latter inclusion above is sharp for all possible choices of a domain G and the former inclusion can be replaced with Proof.Follows from Lemma 4.9 and the formula (2.3).
Remark 4.11.Corollary 4.10 proves an earlier conjecture in [7,Conj. 7.7,p. 122], and also improves the constant in the first of this conjecture.
As noted already in Lemma 3.5, there is not always a Euclidean ball that is included in the unit ball, contains another ball defined with some hyperbolic type metric, and has the same center as the other ball, the following lemma offers a non-x-centered Euclidean ball that contains the balls B s (x, t) and B j * (x, t) and is included in the unit ball if t ≥ |x|.Lemma 4.12.For G = B n , x ∈ B n \ {0}, and |x| ≤ t < 1, B s (x, t) ⊆ B j * (x, t) ⊆ B n (q, r) B n if q = x − 2tx 1 + t and r = 2t 1 + t and B n (q, r) here is the smallest Euclidean ball containing the ball B s (x, t).
Proof.By Lemma 4.9, B s (x, t) ⊆ B j * (x, t).Let y 0 and y 1 be the intersection points in S s (x, t) ∩ L(0, x) found in Lemma 3.2.Since t ≥ |x|, and it can be easily verified that q = (y 0 − y 1 )/2 and r = |q − y 0 |/2.Consequently, B n (q, r) B n because |q| + r = |y 0 | < 1.Let us now prove that B j * (x, t) ⊆ B n (q, r) by showing that, for all y ∈ S n−1 (q, r), If u = XQY , OQY = π − u since q is between the origin and the point x.By cosine formula, Clearly, the distances |x|, |y|, and |x−y| are all increasing with respect to |x| and therefore Thus, the inclusion of the lemma holds and, since y 0 , y 1 ∈ S s (x, t) and |y 0 − y 1 | = 2r, there is no such Euclidean ball with smaller radius than r that contains B s (x, t).

Triangular ratio metric balls in hyperbolic geometry
Since the hyperbolic metric is conformally invariant unlike the j * -metric or the triangular ratio metric, it is useful to have some ball inclusion results between the hyperbolic metric and these two metrics. and , Proof.Follows from Theorem 2.5 and the formula (2.3).
Proof.By formula (2.1) and Lemma 3.1, Recall that, if G = B n , we have explicit formulas for two points y 0 , y 1 ∈ S s (x, t) with any x ∈ B n \ {0} and 0 < t < 1 by Lemma 3.2, from which follows that we can create upper and lower bounds for the radii in the inclusion of the triangular ratio metric balls and hyperbolic balls. , , where Proof.Let us first consider the latter part of the result.Suppose first that x ∈ B n \ {0}.If y 0 , y 1 ∈ S s (x, t) are as in Lemma 3.2, , which proves the lower bound of t 1 .For x ∈ B n \ {0}, suppose that R = ρ B n (x, y 1 ) with y 1 as in Lemma 3.2 for 0 < t < 1.Now, the circles S ρ (x, R) and S s (x, t) intersect at the point y 1 .Since y 1 = S ρ (x, R) ∩ L(0, x) ∩ B(0, |x|), it follows from Lemma 2.4 that Thus, the upper bound of t 0 follows, too.
The next result follows directly from Corollary 5.1 and, while all of its bounds might not be sharp, we know by Lemma 5.3 and its proof that the upper bound of t 0 and the lower bound of R 1 below are the best ones possible in the case t ≥ |x|.Proof.Follows from Lemma 4.9 and Corollary 5.1.
Computer test suggest that the following conjecture holds and, by comparing it to Lemma 5.3 and Corollary 5.4, we see that, to prove this result, it is enough to show that the inclusion B ρ (x, R 0 ) ⊆ B s (x, t) holds for 0 < t < 1 and the inclusion B s (x, t) ⊆ B ρ (x, R 1 ) for 0 < t < |x|. .The triangular ratio metric circle S s (x, t) for G = B 2 , x = 0.3 + 0.45i and t = 0.5 as solid line and the hyperbolic circles S ρ (x, R 0 ) and S ρ (x, R 1 ) as dotted line, when R 0 is as large as possible and R 1 as small as possible within the bounds of Conjecture 5.5.
Figure 5 depicts the sharp inclusion of a triangular ratio metric balls and two hyperbolic metric balls found with Conjecture 5.5.

1 0Figure 1 .
Figure1.The intersection points y 0 and y 1 of Lemma 3.2 for the triangular ratio metric circle S s (x, t) and the line L(0, x) in the domain G = B 2 , when x = 0.3 + 0.7i and t = 0.5.

Figure 3 .
Figure3.The triangular ratio metric circles S s (x, t) for the radii t = 0.1, 0.3, 0.5, 0.7, when the center x is 0.6 and the domain G is the unit disk B 2 .

Conjecture 5 . 5 .Figure 5
Figure 5.The triangular ratio metric circle S s (x, t) for G = B 2 , x = 0.3 + 0.45i and t = 0.5 as solid line and the hyperbolic circles S ρ (x, R 0 ) and S ρ (x, R 1 ) as dotted line, when R 0 is as large as possible and R 1 as small as possible within the bounds of Conjecture 5.5.