On the Erdos-Mordell Inequality for Triangles in Taxicab Geometry

TL;DR

This paper investigates the Erdos-Mordell inequality within taxicab geometry, establishing conditions under which the inequality holds with a specific weight factor, expanding geometric inequalities into non-Euclidean contexts.

Contribution

It extends the Erdos-Mordell inequality to taxicab geometry, identifying the conditions and a specific weight factor where the inequality remains valid.

Findings

The inequality holds for certain triangle configurations in taxicab geometry.

The weight factor w = 3/2 ensures the inequality's validity.

Conditions on points A, B, C are specified for the inequality to hold.

Abstract

In this work the Erdos-Mordell's inequality is examined for the case of a triangle $ABC$ in the taxicab plane geometry. It is shown that the Erdos-Mordell's inequality $R_A + R_B + R_C \, \geq \, w \, (r_a + r_b + r_c)$ holds for triangles with appropriate positions for its points $A$, $B$ and $C$, if $w = 3/2$.

Equations

$$R_A+R_B+R_C\ge 2\left(r_a+r_b+r_c\right)\text{vskip}3.0ptplus1.0ptminus1.0pt$$

$$d_1\left(A,B\right)=|x_A-x_B|+|y_A-y_B|.$$

$$d_k\left(A,B\right)={\left(|x_A-x_B{|}^k+|y_A-y_B{|}^k\right)}^{\frac{1}{k}}\text{vskip}3.0ptplus1.0ptminus1.0pt$$

$$\begin{array}{r}{R}_A=d_1\left(M,A\right)=|x|+|y-r|,\\ R_B=d_1\left(M,B\right)=|x-p|+|y|,\\ R_C=d_1\left(M,C\right)=|x-q|+|y|.\end{array}$$

$$d_1\left(M,\ell \right)=\frac{|a{x}_M+b{y}_M+c|}{max\{|a|,|b|\}}.$$

$$r_{a}=d_{1}(M,\ell_{\!\mbox{\tiny$BC$}}),\;r_{b}=d_{1}(M,\ell_{\!\mbox{\tiny$AC$}}),\;r_{c}=d_{1}(M,\ell_{\!\mbox{\tiny$AB$}}).$$

$$|x|+|y-r|+|x-p|+|x-q|+2|y|\ge 2\left(|y|+\frac{|qr-rx-qy|}{max\{|r|,|q|\}}+\frac{|pr-rx-py|}{max\{|r|,|p|\}}\right).$$

$$\begin{split}R_{A}&=d_{1}\left(M,A\right)=28,\,\,\,\,\,\,\,\,\,R_{B}=d_{1}\left(M,B\right)=22,\,\,\,\,\,\,R_{C}=d_{1}\left(M,C\right)=42,\\[-1.93748pt] r_{a}&=d_{1}\left(M,\ell_{\!\mbox{\tiny$BC$}}\right)=2,\,\,\,\,\,\,\,\,\,\,\,\,\,r_{b}=d_{1}\left(M,\ell_{\!\mbox{\tiny$AC$}}\right)=28,\,\,\,\,\,\,\,\,\,r_{c}=d_{1}\left(M,\ell_{\!\mbox{\tiny$AB$}}\right)=\mbox{\small$\displaystyle\frac{56}{3}$}.\end{split}$$

$$R_A+R_B+R_C\ge w\left(r_a+r_b+r_c\right),\text{vskip}3.0ptplus1.0ptminus1.0pt$$

$$|x|+|y-r|+|x-p|+|x-q|+2|y|\ge w\left(|y|+\frac{|qr-rx-qy|}{max\{|r|,|q|\}}+\frac{|pr-rx-py|}{max\{|r|,|p|\}}\right).\text{vskip}3.0ptplus1.0ptminus1.0pt$$

$$\begin{split}|x|=x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|x-p|=\left\{\begin{array}[]{ccc}p-x&:&x<p\\[4.30554pt] x-p&:&x\geq p\end{array},\right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|x-q|=q-x,\\ |y|=y,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|y-r|=r-y,\\ |qr-rx-qy|=qr-rx-qy,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|pr-rx-py|=-pr+rx+py.\vskip 3.0pt plus 1.0pt minus 1.0pt\end{split}$$

$$\left\{\begin{array}[]{ccc}q+r+y+p-x\geq w\left(y+\displaystyle\frac{qr-rx-qy}{max\{r,q\}}+\displaystyle\frac{-pr+rx+py}{max\{r,p\}}\right)&:&x<p\\[10.76385pt] q+r+y+x-p\geq w\left(y+\displaystyle\frac{qr-rx-qy}{max\{r,q\}}+\displaystyle\frac{-pr+rx+py}{max\{r,p\}}\right)&:&x\geq p\end{array}\right.\vskip 3.0pt plus 1.0pt minus 1.0pt$$

$$\begin{split}|x|=\left\{\begin{array}[]{ccc}-x&:&x<0\\[4.30554pt] x&:&x\geq 0\end{array},\right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|x-p|=x-p,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|x-q|=q-x,\\ |y|=y,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|y-r|=r-y,\\ |qr-rx-qy|=qr-rx-qy,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|pr-rx-py|=-pr+rx+py.\vskip 3.0pt plus 1.0pt minus 1.0pt\end{split}$$

$$\left\{\begin{array}[]{ccc}-p+q+r+y-x\geq w\left(y+\displaystyle\frac{qr-rx-qy}{max\{r,q\}}+\displaystyle\frac{-pr+rx+py}{max\{r,-p\}}\right)&:&x<0\\[10.76385pt] -p+q+r+y+x\geq w\left(y+\displaystyle\frac{qr-rx-qy}{max\{r,q\}}+\displaystyle\frac{-pr+rx+py}{max\{r,-p\}}\right)&:&x\geq 0\end{array}\right.\vskip 3.0pt plus 1.0pt minus 1.0pt$$

$$\!\!\!\!\begin{array}[]{clll}1^{\circ}&\langle\textbf{{a}}\rangle\;0<r\leq p<q,&\langle\textbf{{b}}\rangle\;0\leq p<r\leq q,&\langle\textbf{{c}}\rangle\;0\leq p<q<r;\\[7.74997pt] \begin{array}[]{l}\textrm{\footnotesize For this subcase, see Fig. $4/1^{\circ}$}\\[-1.93748pt] \textrm{\footnotesize with representations:}\end{array}&\langle\textbf{{a}}\rangle\;\mbox{$\begin{array}[]{l}\textrm{\footnotesize long and double-short}\\[-1.93748pt] \textrm{\footnotesize dashed line,}\end{array}$}&\langle\textbf{{b}}\rangle\;\textrm{\footnotesize dashed line,}&\langle\textbf{{c}}\rangle\;\textrm{\footnotesize continuous line};\\[15.49994pt] 2^{\circ}&\langle\textbf{{a}}\rangle\;0<r\leq-p\leq q,&\langle\textbf{{b}}\rangle\;0<-p\leq r\leq q,&\langle\textbf{{c}}\rangle\;0<-p\leq q<r;\\[7.74997pt] \begin{array}[]{l}\textrm{\footnotesize For this subcase, see Fig. $4/2^{\circ}$}\\[-1.93748pt] \textrm{\footnotesize with representations:}\end{array}&\langle\textbf{{a}}\rangle\;\mbox{$\begin{array}[]{l}\textrm{\footnotesize long and double-short}\\[-1.93748pt] \textrm{\footnotesize dashed line,}\end{array}$}&\langle\textbf{{b}}\rangle\;\textrm{\footnotesize dashed line,}&\langle\textbf{{c}}\rangle\;\textrm{\footnotesize continuous line}.\end{array}$$

$$\!R_{A}+R_{B}+R_{C}\!\geq\!w\left(r_{a}+r_{b}+r_{c}\right)\,\Longleftrightarrow\left\{\begin{array}[]{cccc}\!\!\alpha_{i1}x+\beta_{i1}y+\gamma_{i1}\geq 0&:&x<x_{i}&\left[\Pi_{i1}\right]\\[4.30554pt] \!\!\alpha_{i2}x+\beta_{i2}y+\gamma_{i2}\geq 0&:&x\geq x_{i}&\left[\Pi_{i2}\right]\end{array}\right.\vskip 3.0pt plus 1.0pt minus 1.0pt$$

$$f_{ij}(x,y)={\alpha }_{ij}x+{\beta }_{ij}y+{\gamma }_{ij}\ge 0,$$

$$⟨\text{a}⟩0<r\le p<q\vee ⟨\text{b}⟩0\le p<r\le q\vee ⟨\text{c}⟩0\le p<q<r⟹w\le 2+\frac{p+q}{r}.$$

$$\begin{array}[]{ccl}f_{11}(0,r)\geq 0\!&\!\;\Longleftrightarrow\;\!&\!((p-q)wr-pq)\cdot 0-pq(w-1)\cdot r+pq(p+q+r)\geq 0\\[10.76385pt] \!&\!\;\Longleftrightarrow\;\!&\!-pq(w-1)\cdot r+pq(p+q+r)\geq 0\\[10.76385pt] \!&\!\;\mathop{\Longrightarrow}\limits_{\mbox{\tiny$pq>0$}}\;\!&\!-wr+p+q+2r\geq 0\\[6.45831pt] \!&\!\;\mathop{\Longrightarrow}\limits_{\mbox{\tiny$r>0$}}\;\!&\!w\leq 2+\displaystyle\frac{p+q}{r};\end{array}$$

$$⟨\text{a}⟩0<r\le -p\le q\vee ⟨\text{b}⟩0<-p\le r\le q\vee ⟨\text{c}⟩0<-p\le q<r⟹w\le 2+\frac{q-p}{r}.$$

$$⟨\text{a}⟩0<r\le p<q\vee ⟨\text{b}⟩0\le p<r\le q⟹w\le 1+\frac{q^2+pr}{r(q-p)};$$

$$⟨\text{c}⟩0\le p<q<r⟹w\le 1+\frac{p+r}{q-p}.$$

$$⟨\text{a}⟩0<r\le -p\le q\vee ⟨\text{b}⟩0<-p\le r\le q⟹w\le \frac{q}{r}\left(1+\frac{r-p}{q-p}\right);$$

$$⟨\text{c}⟩0<-p\le q<r⟹w\le 1+\frac{r-p}{q-p}.$$

$$⟨\text{a}⟩0<r\le p<q⟹w\le \frac{p}{r}\left(1+\frac{q+r}{q-p}\right);$$

$$⟨\text{b}⟩0\le p<r\le q\vee ⟨\text{c}⟩0\le p<q<r⟹w\le 1+\frac{q+r}{q-p}.$$

$$⟨\text{a}⟩0<r\le -p\le q⟹w\le \frac{-p}{r}\left(1+\frac{q+r}{q-p}\right);$$

$$⟨\text{b}⟩0<-p\le r\le q\vee ⟨\text{c}⟩0<-p\le q<r⟹w\le 1+\frac{q+r}{q-p}.$$

$$⟨\text{a}⟩0<r\le p<q⟹w\le 1+\frac{q^2+pr}{2r(q-p)};$$

$$⟨\text{b}⟩0\le p<r\le q\vee ⟨\text{c}⟩0\le p<q<r⟹w\le 1+\frac{q-p}{r+p}+\frac{q}{q-p}.$$

Full text

1 Introduction

Let the distance between two points, as well as the distance between a line and a point be defined in the Euclidean plane. Then, for a triangle $ABC$ in such a plane the Erdös-Mordell’s inequality holds [4], [18]:

[TABLE]

where $R_{A}$, $R_{B}$ and $R_{C}$ are distances from the interior point $M$ of $\triangle ABC$ to vertices $A$, $B$ and $C$ respectively and $r_{a}$, $r_{b}$ and $r_{c}$ are distances from the point $M$ of the triangle to the corresponding edges which contain the vertices of $\triangle\,ABC$ (Fig. 1).

Figure 1: A geometric illustration of the Erdös-Mordell inequality in $\triangle ABC$

Let there be two points, $A\left(x_{A},y_{A}\right)$ and $B\left(x_{B},y_{B}\right)$, then the distance between them in taxicab geometry is defined as:

[TABLE]

This distance is also called the Manhattan or city block distance. This metric is a special case of the Minkowski metric of order $k$ (where $k\geq 1$) which is defined by the following formula:

[TABLE]

The Minkowski metric contains in itself the taxicab metric for the value $k=1$ and the Euclidean metric for $k=2$ [9]. The term ”taxicab” was first introduced by K. Menger [16]. A graphical representation of distances between points $A$ and $B$ is given in Fig. 2, in taxicab metric with $d_{1}$ (dashed/long dashed lines) and in Euclidean metric with $d_{2}$ (continuous line).

*Figure 2: A geometric illustration of the Minkowski and the Euclidean distances

between two points*

In the rest of this work, only taxicab distances are considered.

Let the $\triangle ABC$ be a triangle with vertices $A\left(0,r\right),\,B\left(p,0\right),\,C\left(q,0\right),\,p\neq q,\,r\neq 0$. Without diminishing generality, let $p<q$. We denote by $M\left(x,y\right)$ an arbitrary point in the plane of the triangle $\triangle ABC$ (Fig. 1). The Taxicab distance from the point $M$ to the points $A$, $B$ and $C$, are given by functions:

[TABLE]

Recently, general formulae for distance in taxicab geometry were analyzed in the paper [2]. Authors Kaya et al. [7] define the distance of a point to a line in taxicab plane geometry with the following statement:

Lemma 1**.**

Distance of point $M\left(x_{M},y_{M}\right)$ to the line $\ell\!:ax+by+c=0$ in the Taxicab plane is$:$

[TABLE]

Let us notice that

[TABLE]

Based on (4) and (6), the Erdös-Mordell’s inequality (1) for $\triangle\,ABC$ in taxicab metric is defined by the following relation:

[TABLE]

Inequalities in the taxicab geometry are the topic of recent research, see e.g [8]. Let us emphasize that the topic of the Erdös-Mordell inequality is current, as it has been shown in the papers [3], [5], [10] – [14], [22] and books [1] and [17]. V. Pambuccian proved that, in the plane of absolute geometry, the Erdös-Mordell inequality is an equivalent to the non-positive curvature [20]. In the paper [15] is given an extension of the Erdös-Mordell inequality on the interior of the Erdös-Mordell curve. In relation to the Erdös-Mordell inequality N. Dergiades in the paper [3] proved one extension of the Erdös-Mordell type inequality. Most notably, the Erdös-Mordell inequality has been considered in the taxicab plane geometry by N. Sönmez who has shown that (1 is a strict inequality: $R_{A}+R_{B}+R_{C}>2\left(r_{a}+r_{b}+r_{c}\right)$, [21]. In this work we prove that the conclusion reached by N. Sönmez is incorrect. That shall be shown through the following example.

Example 1**.**

(counterexample) Let the vertices of $\triangle ABC$ be given with $p=-20,\,q=40,\,r=30$ and let point $M(0,m)$ be defined with $m=2$ (Fig. 3). The taxicab distance from the point $M$ to the vertices of $\triangle ABC$ is given by (4) and the distance from point $M$ to the lines \ell_{\!\mbox{\tinyBC}}\!:y=0,\,\ell_{\!\mbox{\tinyAC}}\!:-rx-qy+qr=0 and \ell_{\!\mbox{\tinyAB}}\!:-rx-py+pr=0 is given by (5):

[TABLE]

Figure 3: A geometric illustration of the counterexample

From (8) we obtain $L=R_{A}+R_{B}+R_{C}=92$ and $R=r_{a}+r_{b}+r_{c}=\frac{146}{3}$. In the case of the Erdös-Mordell inequality, it holds that $L\geq 2R$ i.e $92\geq 97.\overline{3}\,$. From this follows that the Erdös-Mordell inequality does not hold for all interior points of $\triangle\,ABC$. $\Box$

In the rest of this paper, the Erdös-Mordell inequality in taxicab geometry is considered in the form:

[TABLE]

where the positive real number w is defined as such that the previous inequality holds for all interior points of $\triangle ABC$. The main goal of the work is to, for all positive values of the weight coefficient $w$, determine a upper bound $\mathfrak{M}$ such that the Erdös-Mordell inequality holds for $0<w\leq\mathfrak{M}$.

2 The Main Results

The Erdös-Mordell inequality in taxicab plane geometry has the following form:

[TABLE]

It should be noted that the Erdös-Mordell inequality in the taxicab plane geometry defined by (10) refers to triangles ABC with the appropriate positions of points $A(0,r)$, $B(p,0)$ and $C(q,0)$ in two cases. The first case is when coordinates $p$, $q$ and $r$ are positive and the second case is when the $p$ coordinate is negative, with positive $q$ and $r$ coordinates. Furthermore, we do not consider the general position of the triangle in the taxicab plane nor the rotation of such a triangle to $\triangle ABC$.

$1^{\circ}$ We analyze $\triangle\,ABC$ with $p,q,r>0$ (see Fig. 4), then, for all interior points of the triangle holds:

[TABLE]

Then, the form of the Erdös-Mordell inequality (10) becomes:

[TABLE]

Symmetric positions of $\triangle ABC$ relative to the coordinate axes can be analogously considered.

$2^{\circ}$ We analyze $\triangle\,ABC$ with $p<0$   and   $q,r>0$ (see Fig. 4), then, for all interior points of the triangle holds:

[TABLE]

Then, the form of the Erdös-Mordell inequality (10) becomes:

[TABLE]

As in case $1^{\circ}\!\!$, symmetric positions of $\triangle ABC$ relative to the coordinate axes can be analogously considered.

Let us notice that for point $A(0,r)$, there exist the following subcases:

[TABLE]

Figure 4: The two types of triangles $ABC$ with subcasses

In formula (11), for the first triangle type $(i=1)$, branching is achieved for $x=p$, where $p$ will then be denoted with $x_{1}.\,$In formula (13), for the second triangle type $(i=2)$, branching is achieved for $x=0$, where [math] will then be denoted with $x_{2}$. Then, the Erdös-Mordell inequality (10), with weight coefficient $w>0$, is considered with the following theorem:

Theorem 1**.**

It holds:

[TABLE]

where coefficients $\alpha_{ij},\beta_{ij},\gamma_{ij}$ $(j\!=\!1,2)$, are given by Tab. 1 for $i\!=\!1$ and Tab. 2 for $i\!=\!2$.

[TABLE]

Table 1: The Erdös-Mordell inequality in the taxicab plane geometry for case $1^{\circ}$

[TABLE]

Table 2: The Erdös-Mordell inequality in the taxicab plane geometry for case $2^{\circ}$

Let us notice that the Erdös-Mordell inequality reduces to a problem of the positivity of the linear function

[TABLE]

for some choice of interior points $(x,y)$ of a triangle, for concretely defined values of parameters $\alpha_{ij},\beta_{ij}$ and $\gamma_{ij}$ given by the above tables. The problem of determining the minimum and maximum of linear functions $f_{ij}(x,y)$ reduces down to the determining of the minimum and maximum in the vertices of the considered triangles, according to [6]. Given that, it is enough to consider the cases of the minima and maxima of linear functions $f_{ij}(x,y)$ in vertices of $\triangle\,ABD$ and $\triangle\,BCD$ for $A(0,r)$, $B(p,0)$, $C(q,0)$ and $D(p,\frac{r}{q}(q-p))$ when $i=1$ and in vertices of $\triangle\,ABO$ and $\triangle\,ACO$ for $A(0,r)$, $B(p,0)$, $C(q,0)$ and $O(0,0)$ when $i=2$.

The following statements hold:

Statement 1**.**

Let $A(0,r)\in\left[\Pi_{11}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

• Proof.

From Table 1:

$\langle\textbf{{a}}\rangle$   By substituting coordinates $x=0$ and $y=r$ into $f_{11}(x,y)=\alpha_{11}x+\beta_{11}y+\gamma_{11}$ the following is obtained:

[TABLE]

$\langle\textbf{{b}}\rangle$  $q(r-pw)\cdot r+qr(w(p-r)+p+q+r)\geq 0$, from which follows $w\leq 2+\displaystyle\frac{p+q}{r}$;

$\langle\textbf{{c}}\rangle$  $w(q\!-\!p\!-\!r)\cdot r+r(w(p\!-\!q)\!+\!p\!+\!q\!+\!r)\!+\!r\geq 0$, from which follows $w\leq 2+\displaystyle\frac{p\!+\!q}{r}$.

$\Box$ ∎

Statement 2**.**

Let $A(0,r)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

• Proof.

By Table 2:

$\langle\textbf{{a}}\rangle$  $-pq(w+1)r+pq(2rw+p-q-r)\geq 0$, from which follows $w\leq 2+\displaystyle\frac{q-p}{r}$;

$\langle\textbf{{b}}\rangle$  $q(r-pw)r+qr(w(p-r)-p+q+r)\geq 0$, from which follows $w\leq 2+\displaystyle\frac{q-p}{r}$;

$\langle\textbf{{c}}\rangle$  $(w(q\!-\!p\!-\!r)+r)r+r(w(p\!-\!q)\!-\!p\!+\!q\!+\!r)\geq 0$, from which follows $w\leq 2+\displaystyle\frac{q-p}{r}$.

$\Box$ ∎

Statement 3**.**

Let $B(p,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusions hold for the weight coefficient $w\!:$

[TABLE]

[TABLE]

• Proof.

By Table 1:

$\langle\textbf{{a}}\rangle$  $((p-q)wr+pq)p+pq(-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{q^{2}+pr}{r(q-p)}$;

$\langle\textbf{{b}}\rangle$  $r(w(r\!-\!q)\!+\!q)p\!+\!qr(w(p\!-\!r)\!-\!p\!+\!q\!+\!r)\!\geq\!0$, from which follows $w\!\leq\!1\!+\!\displaystyle\frac{q^{2}\!+\!pr}{r(q\!-\!p)}$;

$\langle\textbf{{c}}\rangle$  $rp+r(w(p-q)-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{p+r}{q-p}$. $\Box$ ∎

Statement 4**.**

Let $B(p,0)\in\left[\Pi_{21}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusions hold for the weight coefficient $w\!:$

[TABLE]

[TABLE]

• Proof.

By Table 2:

$\langle\textbf{{a}}\rangle$  $(pq-(p+q)wr)p+pq(2rw+p-q-r)\geq 0$, from which follows $w\leq\displaystyle\frac{q}{r}\left(\!1+\frac{r-p}{q-p}\!\right)$;

$\langle\textbf{{b}}\rangle$  $r(w(r-q)-q)p+qr(w(p-r)-p+q+r)\geq 0$, from which follows $w\leq\displaystyle\frac{q}{r}\left(\!1+\frac{r-p}{q-p}\!\right)$;

$\langle\textbf{{c}}\rangle$  $-rp+r(w(p-q)-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{r-p}{q-p}$. $\Box$ ∎

Statement 5**.**

Let $C(q,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusions hold for the weight coefficient $w\!:$

[TABLE]

[TABLE]

• Proof.

By Table 1:

$\langle\textbf{{a}}\rangle$  $((p-q)wr+pq)q+pq(-p+q+r)\geq 0$, from which follows $w\leq\displaystyle\frac{p}{r}\left(1+\displaystyle\frac{q+r}{q-p}\right)$;

$\langle\textbf{{b}}\rangle$  $r(w(r\!-\!q)\!+\!q)q\!+\!qr(w(p\!-\!r)\!-\!p\!+\!q\!+\!r)\!\geq\!0$, from which follows $w\!\leq\!1\!+\!\displaystyle\frac{q\!+\!r}{q\!-\!p}$;

$\langle\textbf{{c}}\rangle$  $rq+r(w(p-q)-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{q+r}{q-p}$. $\Box$ ∎

Statement 6**.**

Let $C(q,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusions hold for the weight coefficient $w\!:$

[TABLE]

[TABLE]

• Proof.

By Table 2:

$\langle\textbf{{a}}\rangle$  $(-pq\!-\!(p\!+\!q)wr)q\!+\!pq(2rw\!+\!p\!-\!q\!-\!r)\!\geq\!0$, from which follows $w\!\leq\!\displaystyle\frac{-p}{r}\left(\!1\!+\!\displaystyle\frac{q\!+\!r}{q\!-\!p}\!\right)$;

$\langle\textbf{{b}}\rangle$  $r(w(r\!-\!q)\!+\!q)q\!+\!qr(w(p\!-\!r)\!-\!p\!+\!q\!+\!r)\!\geq\!0$, from which follows $w\!\leq\!1\!+\!\displaystyle\frac{q\!+\!r}{q\!-\!p}$;

$\langle\textbf{{c}}\rangle$  $rq+r(w(p-q)-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{q+r}{q-p}$. $\Box$ ∎

Statement 7**.**

Let $D(p,\frac{r}{q}(q-p))\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $D(p,\frac{r}{q}(q-p))$, then the following conclusions hold for the weight coefficient $w\!:$

[TABLE]

[TABLE]

• Proof.

By Table 1:

$\langle\textbf{{a}}\rangle$  $((p-q)wr+pq)p-pq(w-1)\displaystyle\frac{r}{q}(q-p)+pq(-p+q+r)\geq 0$,

from which follows $w\leq 1+\displaystyle\frac{q^{2}+pr}{2r(q-p)}$;

$\langle\textbf{{b}}\rangle$  $r(w(r-q)+q)p+q(r-pw)\displaystyle\frac{r}{q}(q-p)+qr(w(p-r)-p+q+r)\geq 0$,

from which follows $w\leq 1+\displaystyle\frac{q-p}{r+p}+\displaystyle\frac{q}{q-p}$;

$\langle\textbf{{c}}\rangle$  $rp+(w(q-p-r)+r)\displaystyle\frac{r}{q}(q-p)+r(w(p-q)-p+q+r)\geq 0$,

from which follows $w\leq 1+\displaystyle\frac{q-p}{r+p}+\displaystyle\frac{q}{q-p}$. $\Box$ ∎

Statement 8**.**

Let $O(0,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $O(0,0)$, then the following conclusions hold for the weight coefficient $w\!:$

[TABLE]

[TABLE]

[TABLE]

• Proof.

By Table 2:

$\langle\textbf{{a}}\rangle$ $pq(2rw+p-q-r)\geq 0$, from which follows $w\leq\displaystyle\frac{1}{2}+\displaystyle\frac{q-p}{2r}$;

$\langle\textbf{{b}}\rangle$  $qr(w(p-r)-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{q}{r-p}$;

$\langle\textbf{{c}}\rangle$  $r(w(p-q)-p+q+r)\geq 0$, from which follows $w\leq 1+\displaystyle\frac{r}{q-p}$. $\Box$ ∎

Let the positions of points $B$ and $C$ be given. Then, let us consider the positions of point $A(0,r)$ in the concrete cases $\langle\textbf{{a}}\rangle$, $\langle\textbf{{b}}\rangle$, $\langle\textbf{{c}}\rangle$ which were considered in Statements 1–8. Through the aforementioned Statements the functions of upper bounds $\omega$ for the weight coefficient $w$ were obtained:

[TABLE]

Our goal is to, for the functions $\omega(p,q,r)$, dependent on concrete subcases \langle\mbox{\boldmath\theta}\rangle, where \mbox{\boldmath\theta}\in\{\textbf{{a}},\textbf{{b}},\textbf{{c}}\}, find the values:

[TABLE]

In this way, the Erdös-Mordell inequality (9) holds for $w=\mathfrak{M}$ for all interior points of $\triangle\,ABC$. If $\mathfrak{M}$ is a minimum in this area, then an equality is also possible in (9).

2.1  Determining value of $\mathfrak{M}$ by areas

In this section of the work, the values of $\mathfrak{M}$ by areas of $\triangle ABC$ are determined in dependence on cases \langle\mbox{\boldmath\theta}\rangle, where \mbox{\boldmath\theta}\in\{\textbf{{a}},\textbf{{b}},\textbf{{c}}\}.

The following three propositions are obtained on the basis of Statement 1.

Proposition 1**.**

Let $A(0,r)\in\left[\Pi_{11}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!p\!<\!q$. Then, we notice the following expression holds:

[TABLE]

The above conclusion is correct because the real number $\displaystyle\frac{q}{p}$ fulfills $\displaystyle\frac{q}{p}>1$ and it is possible to choose a number $\displaystyle\frac{q}{p}$ such that it is arbitrarily close to $1$. $\Box$ ∎

Proposition 2**.**

Let $A(0,r)\in\left[\Pi_{11}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!\leq\!p\!<\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 3**.**

Let $A(0,r)\in\left[\Pi_{11}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!\leq\!p\!<\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

The following three propositions are obtained on the basis of Statement 2.

Proposition 4**.**

Let $A(0,r)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!-p\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 5**.**

Let $A(0,r)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!<\!-p\!\leq\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 6**.**

Let $A(0,r)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $A(0,r)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!<\!-p\!\leq\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Similar to previous propositions, the following three propositions are obtained from Statement 3.

Proposition 7**.**

Let $B(p,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!p\!<\!q$. Then, we notice the following expression holds:

[TABLE]

because $t=\displaystyle\frac{p}{q-p}>0$ holds 2\,t+\mbox{\small\displaystyle\frac{1}{t}}\geq 2\sqrt{2}. $\Box$ ∎

Proposition 8**.**

Let $B(p,0)\in\left[\Pi_{12}\right]$. If inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!\leq\!p\!<\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 9**.**

Let $B(p,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!\leq\!p\!<\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

The following three propositions are obtained on the basis of Statement 4.

Proposition 10**.**

Let $B(p,0)\in\left[\Pi_{21}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!-p\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 11**.**

Let $B(p,0)\in\left[\Pi_{21}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!<\!-p\leq r\leq q$. Then, we notice the following expression holds:

[TABLE]

because $q(r+(-p))\geq r(q+(-p))\;\Longleftrightarrow\;q\geq r$. $\Box$ ∎

Proposition 12**.**

Let $B(p,0)\in\left[\Pi_{21}\right]$. If the inequality $(\ref{E8})$ holds for $B(p,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!<\!-p\leq q<r$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

The following three propositions are obtained on the basis of Statement 5.

Proposition 13**.**

Let $C(q,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!p\!<\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 14**.**

Let $C(q,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!\leq\!p\!<\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 15**.**

Let $C(q,0)\in\left[\Pi_{12}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!\leq\!p\!<\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Similar to previous propositions, the following three propositions are obtained from Statement 6.

Proposition 16**.**

Let $C(q,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!-p\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

because $-pq+(-p)r\geq rq+(-p)r\;\Longleftrightarrow\;-p\geq r$. $\Box$ ∎

Proposition 17**.**

Let $C(q,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!<\!-p\!\leq\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 18**.**

Let $C(q,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $C(q,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!<\!-p\!\leq\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

because $2q\geq q-p\;\Longleftrightarrow\;q\geq-p$. $\Box$ ∎

The following three propositions are obtained on the basis of Statement 7.

Proposition 19**.**

Let D{\big{(}}p,\frac{r}{q}(q-p){\big{)}}\in\left[\Pi_{12}\right]. If the inequality $(\ref{E8})$ holds for D{\big{(}}p,\frac{r}{q}(q-p){\big{)}}\in\left[\Pi_{12}\right], then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!p\!<\!q$. Then, we notice the following expression holds:

[TABLE]

because $t=\displaystyle\frac{p}{q-p}>0$ for \mbox{\small\displaystyle\frac{1}{2t}}+t\geq\sqrt{2}. $\Box$ ∎

Proposition 20**.**

Let D{\big{(}}p,\frac{r}{q}(q-p){\big{)}}\in\left[\Pi_{12}\right]. If the inequality $(\ref{E8})$ holds for D{\big{(}}p,\frac{r}{q}(q-p){\big{)}}\in\left[\Pi_{12}\right], then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!\leq\!p\!<\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

because $t=\displaystyle\frac{q-p}{q+p}>0$ holds 2\,t+\mbox{\small\displaystyle\frac{1}{t}}\geq 2\sqrt{2}. $\Box$ ∎

Proposition 21**.**

Let D{\big{(}}p,\frac{r}{q}(q-p){\big{)}}\in\left[\Pi_{12}\right]. If the inequality $(\ref{E8})$ holds for D{\big{(}}p,\frac{r}{q}(q-p){\big{)}}\in\left[\Pi_{12}\right], then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!\leq\!p\!<\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

because \mbox{\small\displaystyle\frac{q}{q-p}}\geq 1. $\Box$ ∎

Similar to previous propositions, the following three propositions are obtained from Statement 8.

Proposition 22**.**

Let $O(0,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $O(0,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{a}}\rangle\;0\!<\!r\!\leq\!-p\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 23**.**

Let $O(0,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $O(0,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{b}}\rangle\;0\!<\!-p\!\leq\!r\!\leq\!q$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Proposition 24**.**

Let $O(0,0)\in\left[\Pi_{22}\right]$. If the inequality $(\ref{E8})$ holds for $O(0,0)$, then the following conclusion holds for the weight coefficient $w\!:$

[TABLE]

and in that case

[TABLE]

• Proof.

Let us consider $\langle\textbf{{c}}\rangle\;0\!<\!-p\!\leq\!q\!<\!r$. Then, we notice the following expression holds:

[TABLE]

$\Box$ ∎

Let us emphasize that the results of the previous three Propositions provide an improvement over some results from paper [5].

3 Summa summarum

Based on the propositions above, a theorem follows:

Theorem 2**.**

In taxicab geometry for an interior point of $\triangle ABC$ in an appropriate position, the Erdös-Mordell’s inequality holds

[TABLE]

It is well known that taxicab distance depends on the rotation of the coordinate system, but does not depend on its translation or its reflection over a coordinate axis [19]. For an arbitrary triangle $ABC$ we set the following open problem (illustrated by Fig. 5).

Conjecture 1**.**

In taxicab geometry for an interior point of any triangle $ABC$ the Erdös-Mordell’s inequality holds

[TABLE]

Figure 5. A geometric illustration of conjecture 1

Acknowledgment. Research of the second author was supported in part by the Serbian Ministry of Education, Science and Technological Development, under Projects ON 174032 and III 44006.

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