When the cyclic kite configuration is formed by the combination of one equilateral triangle and one isosceles triangle with the common base $P_{2}N{P}_4$ (First Kind), as shown in Figure 1 .

Statement: The necessary and sufficient conditions for the existence of a cyclic kite configuration with four positive point masses $m_k\left(k=1,2,3,4\right)$ located at the respective vertices $P_k$ of the kite are the existence of a non-dimensional mass parameter $\mu \in \left]0,1/3\right[$ such that $m_2=m_4=M\mu$, $m_1=M/2\left(1-\mu \right)$, $m_3=M/2\left(1-3\mu \right)$ where $m_1+m_2+m_3+m_4=M$ and $P_1{P}_2=P_1{P}_4=P_2{P}_4=\sqrt{3}R$, $P_2{P}_3=P_3{P}_4=R$, $P_1{P}_3=2R$, where R is the radius of the common circular orbit.

Proof: Let $P_1{P}_2{P}_3{P}_4$ be a cyclic kite formed by the combination of one equilateral triangle $P_1{P}_2{P}_4$ and one isosceles triangle $P_2{P}_3{P}_4$ with the common base $P_{2}N{P}_4$ and axis of symmetry $P_{1}O{P}_3=2R$. Let $P_k$ be the positions of the four positive point masses $m_k$ moving on a common circular orbit with centre at O, radius R and diameter $P_{1}O{P}_3=2R$ as the axis of symmetry. Considering the centre of the circle as the origin, the axis of symmetry as the X-axis and a line $YO{Y}^{\prime }$ parallel to $P_{2}N{P}_4$ and perpendicular to $P_{1}O{P}_3$ as the Y-axis. Using the properties of a cyclic quadrilateral $P_1{P}_2{P}_3{P}_4$ in the form of a cyclic kite configuration and the cyclic equilateral triangle $P_1{P}_2{P}_4$, the coordinates of the four point masses can be written as $P_k\left(R\cos {\theta }_k,R\sin {\theta }_k\right)$, where ${\theta }_1=0$, ${\theta }_2=2\pi /3$, ${\theta }_3=\pi$, ${\theta }_4=4\pi /3$.

Thus, the position vectors of the four-point masses are given by ${\stackrel{\to }{r}}_k=R\cos {\theta }_k\hat{i}+R\sin {\theta }_k\hat{j}=\stackrel{\to }{O{P}_k}$

$\Rightarrow {\stackrel{\to }{r}}_1=R\hat{i},{\stackrel{\to }{r}}_2=-\frac{R}{2}\hat{i}+\frac{\sqrt{3}}{2}R\hat{j},{\stackrel{\to }{r}}_3=-R\hat{i},{\stackrel{\to }{r}}_4=-\frac{R}{2}\hat{i}-\frac{\sqrt{3}}{2}R\hat{j}.$(1)

The position vector of the centre of mass of the system of four-point masses is given by

$\stackrel{\to }{c}=M^{-1}{\displaystyle \underset{k=1}{\overset{4}{\sum }}{m}_k{\stackrel{\to }{r}}_k}.$(2)

As our kite is a central configuration, we consider the centre of mass as the origin, then $\stackrel{\to }{c}=\hat{0}$

$\Rightarrow m_1{\stackrel{\to }{r}}_1+m_2{\stackrel{\to }{r}}_2+m_3{\stackrel{\to }{r}}_3+m_4{\stackrel{\to }{r}}_4=\hat{0},$

$\Rightarrow m_{1}R\hat{i}+m_2\left(-\frac{R}{2}\hat{i}+\frac{\sqrt{3}}{2}R\hat{j}\right)+m_3\left(-R\hat{i}\right)+m_4\left(-\frac{R}{2}\hat{i}-\frac{\sqrt{3}}{2}R\hat{j}\right)=0,$

$\Rightarrow R\left\{2\left(m_1-m_3\right)-\left(m_2+m_4\right)\right\}\hat{i}+\sqrt{3}R\left\{m_2-m_4\right\}\hat{j}=0.$

Taking the scalar product of $\hat{i}$ and $\hat{j}$ with the above equation, we get

$\begin{array}{l}2\left(m_1-m_3\right)-\left(m_2+m_4\right)=0,\\ m_2=m_4.\end{array}\}$(3)

Also, we know that

$m_1+m_2+m_3+m_4=M.$(4)

Equations (3) and (4) yield

$m_1-m_2-m_3=0,$ (5)

$m_1+2m_2+m_3=M.$ (6)

Let us introduce a non-dimensional mass parameter $\mu \in \left]0,1/3\right[$ such that $m_2=m_4=M\mu$ then from Equations (5) and (6), we get

$m_1=\frac{M\left(1-\mu \right)}{2},m_3=\frac{M\left(1-3\mu \right)}{2}.$

Also,

$\begin{array}{l}\left|\stackrel{\to }{P_1{P}_2}\right|=\left|\stackrel{\to }{O{P}_2}-\stackrel{\to }{O{P}_1}\right|=\left|-\frac{R}{2}\hat{i}+\frac{\sqrt{3}}{2}R\hat{j}-R\hat{i}\right|=\sqrt{3}R=\left|\stackrel{\to }{P_1{P}_4}\right|=\left|\stackrel{\to }{P_2{P}_4}\right|,\\ \left|\stackrel{\to }{P_2{P}_3}\right|=\left|\stackrel{\to }{P_3{P}_4}\right|=R,\left|\stackrel{\to }{P_1{P}_3}\right|=2R.\end{array}$

Thus, the necessary conditions for four positive point masses $m_k$ situated at the vertices of a cyclic kite configuration formed by an equilateral triangle $P_1{P}_2{P}_4$ and an isosceles triangle $P_2{P}_3{P}_4$ are

$\begin{array}{l}\text{(a)}{m}_2=m_4=\mu ,m_1=M-\mu /2,m_3=M-3\mu /2,\\ \text{(b)}{m}_1=m_3+2\mu ,\\ \text{(c)}{P}_1{P}_2=P_2{P}_4=P_1{P}_4=\sqrt{3}R\text{i}\text{.e}\text{.,}{r}_{12}=r_{24}=r_{14}=\sqrt{3}R,\\ \text{(d)}{P}_2{P}_3=P_3{P}_4=R\text{i}\text{.e}\text{.},r_{23}=r_{34}=R,\\ \text{(e)}{P}_1{P}_3=2R\text{i}\text{.e}\text{.,}{r}_{13}=2R.\end{array}\}$ (7)

Conversely, the results of Equation (7a), $m_2=m_4=M\mu$ represents that the diagonal $P_{1}O{P}_3$ is the line of symmetry. Equation (7b) shows that $m_1>m_3$ and from the results of Equations (7c), (7d) and (7e) $P_1{P}_2^2+N{P}_2^2=P_1{N}^2$ i.e., the diagonals are perpendicular to each other. Thus, the results of Equation (7) represent the sufficient conditions for the existence of a cyclic kite configuration.

When the kite is formed by two congruent isosceles right-angled triangles (Second Kind), as shown in Figure 2 .

Statement: The necessary and sufficient conditions for the existence of a cyclic kite configuration with four positive point masses $m_k\left(k=1,2,3,4\right)$ at the respective vertices $P_k$ of the kite are that there exists a non-dimensional mass parameter $\mu \in \left]0,1/2\right[$such that $m_2=m_4=M\mu$, $m_1=M\left(1/2-\mu \right)=m_3$, where $m_1+m_2+m_3+m_4=M$ and $P_1{P}_2=P_2{P}_3=P_3{P}_4=P_1{P}_4=\sqrt{2}R$, $P_1{P}_3=P_2{P}_4=2R$, where R is the radius of the common circular orbit.

Proof: Let $P_1{P}_2{P}_3{P}_4$ be the cyclic kite formed by the combination of two congruent isosceles right-angled triangles $P_1{P}_2{P}_4$ and $P_2{P}_3{P}_4$ with the common base $P_{2}O{P}_4=2R$ and axis of symmetry $P_{1}O{P}_3=2R$. Let $P_k$ be the positions of the four positive point masses $m_k$ on the cyclic kite. Taking the centre O of the circle as the origin and two perpendicular diameters $P_{1}O{P}_3$ and $P_{2}O{P}_4$ as the X-axis and Y-axis, respectively, then the position vectors of the four point masses are $\stackrel{\to }{O{P}_1}=R\hat{i}={\stackrel{\to }{r}}_1$, $\stackrel{\to }{O{P}_2}=R\hat{j}={\stackrel{\to }{r}}_2$, $\stackrel{\to }{O{P}_3}=-R\hat{i}={\stackrel{\to }{r}}_3$, $\stackrel{\to }{O{P}_4}=-R\hat{j}={\stackrel{\to }{r}}_4$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along the X-axis and Y–axis respectively. If we take the centre of mass as the origin, then by using Theorem 1, we have, $m_{1}R\hat{i}+m_{2}R\hat{j}-m_{3}R\hat{i}-m_{4}R\hat{j}=\hat{0}$, Here either of $P_{1}O{P}_3\text{or}{P}_{2}O{P}_4$ may be taken as the axis of symmetry. Taking the scalar product of $\hat{i}$ and $\hat{j}$, we get

$m_1=m_3\text{and}{m}_2=m_4.$(8)

Introduction of Equation (8) in Equation (4) gives

$m_1+m_2=M/2$(9)

Let us introduce a non-dimensional mass parameter $\mu \in \left]0,1/2\right[$ such that $m_2=m_4=M\mu$, then from Equations (2) and (7) $m_1=m_3=M\left(1/2-\mu \right)$. Both the masses of each pair are equal, so both the diagonals are the axes of symmetry of the cyclic kite configuration. Thus, the necessary conditions for the existence of a cyclic kite configuration are

$\begin{array}{l}\text{(a)}{m}_2=m_4=\mu ,m_1=m_3=M\left(1/2-\mu \right),\\ \text{(b)}{m}_1=m_3\ge m_2=m_4,\\ \text{(c)}{P}_1{P}_2=P_2{P}_3=P_3{P}_4=P_1{P}_4=\sqrt{2}R\text{i}\text{.e}\text{.,}{r}_{12}=r_{23}=r_{34}=r_{14}=\sqrt{2}R,\\ \text{(d)}{P}_1{P}_3=P_2{P}_4=2R\text{i}\text{.e}\text{.,}{r}_{13}=r_{24}=2R.\end{array}\}$ (10)

Conversely, the results of Equation (10a) (similar to Long et al. [4] ) and Equation (10b) justify that both the diagonals are the axes of symmetry, and Equations (10c), and (10d) are the sufficient conditions for the perpendicularity of the diagonals.

When the kite is formed by the combination of two non-congruent arbitrary isosceles triangles (Generalized Kind), as shown in Figure 3 .

Statement: The necessary and sufficient conditions for the existence of a cyclic kite configuration are that there exists a parameter $\mu \in \left[0,1/2\left(1-\cos \theta \right)\right]$ such that $m_2=m_4=M\mu$, $m_1=M\left[1/2-\mu \left(1+\cos \theta \right)\right]$, $m_3=M\left[1/2-\mu \left(1-\cos \theta \right)\right]$ where $\theta \in \left(0,\pi \right)$, $m_1+m_2+m_3+m_4=M$ and $P_1{P}_2=P_1{P}_4=2R\sin \theta /2$, $P_2{P}_4=2R\sin \theta$, $P_2{P}_3=P_3{P}_4=2R\cos \theta /2$, where $P_1{P}_3=2R$.

Proof: Let $P_1{P}_2{P}_3{P}_4$ be an arbitrary cyclic kite configuration formed by the combination of two non-congruent arbitrary isosceles triangles $P_1{P}_2{P}_4$ and $P_2{P}_3{P}_4$ with the common base $P_{2}N{P}_4$. Let the four point masses $m_1,m_2,m_3,m_4$ be moving on a common circular orbit of radius R and centre O. Considering $P_1{P}_2=P_1{P}_4$ and $P_2{P}_3=P_3{P}_4$ as the pair of adjacent equal sides. Let $\angle P_{1}O{P}_2=\angle P_{1}O{P}_4=\theta$ then $\angle P_{2}O{P}_3=\angle P_{4}O{P}_3=\pi -\theta$. Taking the axis of symmetry $P_{1}O{P}_3$ as the X-axis, O as the origin (the barycentre) and the line YOY’ as the Y-axis, then the position vectors of the four-point masses are given by

$\begin{array}{l}\stackrel{\to }{O{P}_1}=R\hat{i}={\stackrel{\to }{r}}_1,\stackrel{\to }{O{P}_2}=R\cos \theta \hat{i}+R\sin \theta \hat{j}={\stackrel{\to }{r}}_2,\\ \stackrel{\to }{O{P}_3}=-R\hat{i}={\stackrel{\to }{r}}_3,\stackrel{\to }{O{P}_4}=R\cos \theta \hat{i}-R\sin \theta \hat{j}={\stackrel{\to }{r}}_4.\end{array}$

As in previous theorems, $m_1{\stackrel{\to }{r}}_1+m_2{\stackrel{\to }{r}}_2+m_3{\stackrel{\to }{r}}_3+m_4{\stackrel{\to }{r}}_4=\hat{0}$

$\begin{array}{l}\Rightarrow m_{1}R\hat{i}+m_2\left(R\cos \theta \hat{i}+R\sin \theta \hat{j}\right)+m_3\left(-R\hat{i}\right)+m_4\left(R\cos \theta \hat{i}-R\sin \theta \hat{j}\right)=\hat{0},\\ \Rightarrow R\left(m_1+m_2\cos \theta -m_3+m_4\cos \theta \right)\hat{i}+R\left(m_2-m_4\right)\sin \theta \hat{j}=\hat{0}.\end{array}$

By taking the scalar products of $\hat{i}$ and $\hat{j}$, we get

$m_1+\left(m_2+m_4\right)\cos \theta -m_3=0.$(11)

$m_2=m_4.$(12)

The collaboration of Equations (11) and (12) gives

$m_1+2m_2\cos \theta -m_3=0.$(13)

Introduction of Equation (12) in Equation (4) gives

$m_1+2m_2+m_3=M.$(14)

The addition of Equations (13) and (14) gives

$m_1+m_2\left(1+\cos \theta \right)=M/2.$ (15)

Let us introduce a parameter $\mu \in \left[0,1/2\left(1-\cos \theta \right)\right]$ such that $m_2=m_4=M\mu$ where $\theta \in \left(0,\pi \right)$ then from Equations (14) and (15), we get

$m_1=M\left[1/2-\mu \left(1+\cos \theta \right)\right]\text{and}{m}_3=M\left[1/2-\mu \left(1-\cos \theta \right)\right].$

Thus, the necessary conditions for the existence of a cyclic kite configuration are

$\begin{array}{l}\text{(a)}{m}_2=m_4=\mu ,\\ \text{(b)}{m}_1\ge m_2=m_4\ge m_3,\\ \text{(c)}{m}_3=m_1+2\mu \cos \theta ,\\ \text{(d)}{m}_1=\left(M/2\right)-\mu \left(1+\cos \theta \right)\text{,}{m}_3=\left(M/2\right)-\mu \left(1-\cos \theta \right)\text{,}\\ (\text{e})P_1{P}_2=P_1{P}_4=2R\sin \theta /2,P_2{P}_3=P_3{P}_4=2R\cos \theta /2,\\ P_2{P}_4=2R\sin \theta ,P_1{P}_3=2R.\end{array}\}$(16)

Conversely, the results of Equations (16a) and (16b) will justify that the line $P_{1}O{P}_3$ is a line of symmetry, and the results of Equation (16e) will justify the perpendicularity of the diagonals of the kite.

Suppose the equal point masses lying on opposite sides of the axis of symmetry of the kite configuration are very close to either of the point masses of the axis of symmetry. In that case, a collision may happen, and hence the kite configuration may collapse. So, there should be a restricted domain of $\theta$ for the existence of a kite configuration. The necessary condition may be taken as $2P_2{P}_3\ge R$

$\text{i}\text{.e}\text{.,}4R\cos \theta /2\ge R\Rightarrow \cos \theta /2\ge 1/4\Rightarrow \cos \theta /2\ge 0.25\Rightarrow {14.5}^{\circ }\le \theta /2\le {75.5}^{\circ }$

$\Rightarrow {29}^{\circ }\le \theta \le {151}^{\circ }$.

Thus, the kite configuration will exist for $\theta \in \left[\pi /6,5\pi /6\right]$ and will collapse if $\theta \le {29}^{\circ }$ and $\theta \ge {151}^{\circ }$ approximately.

The foregoing results of the three theorems represent the sufficient conditions for the existence of at least one line of symmetry and for the perpendicularity of the diagonals of the kite. In particular, we have verified and extended certain results established by earlier researchers, including Long et al. [4] , Chavela et al. [5] , Albouy et al. [6] , Pina et al. [7] , Pina [8] , Cors et al. [9] , and Corbera et al. [12] . Specifically, the results expressed in Equations (7a), (7b) and (7c) corroborate the theories of Chavela et al. [5] , Albouy et al. [6] , Pina et al. [7] , and Cors et al. [9] . Furthermore, the results of Equations (7c), (7d) and (7e) are consistent with the conditions provided by Long et al. [4] as

$\left(r_{24}^3-r_{12}^3\right)\left(r_{23}^3-r_{34}^3\right)\left(r_{13}^3-r_{14}^3\right)=\left(r_{12}^3-r_{14}^3\right)\left(r_{13}^3-r_{34}^3\right)\left(r_{24}^3-r_{23}^3\right).$(17)

This is the necessary and sufficient condition of a four-body planar central configuration that the six mutual distances $r_{ij}$ determine a geometrically realizable planar configuration. The results of Equation (7) vanish the Caley-Menger determinant.

$\text{i}\text{.e}\text{.,}V\left(r\right)=\left|\begin{array}{ccccc}0& 1& 1& 1& 1\\ 1& 0& r_{12}^2& r_{13}^2& r_{14}^2\\ 1& r_{12}^2& 0& r_{23}^2& r_{24}^2\\ 1& r_{13}^2& r_{23}^2& 0& r_{34}^2\\ 1& r_{14}^2& r_{24}^2& r_{34}^2& 0\end{array}\right|=3R^6\left|\begin{array}{cccc}1& 1& 1& 1\\ -1& 1& 1& 0\\ -1& -1& 1& 2\\ 1& -2& -1& 1\end{array}\right|=-3R^6\left|\begin{array}{ccc}-2& 0& 1\\ 0& -2& -1\\ 3& -1& -2\end{array}\right|=0.$(18)

$V\left(r\right)=0$ can also be satisfied by the results of Equations (7) and (10). Thus, the square of six mutual distances $r_{ij}$ satisfy $V\left(r\right)=0$ i.e., the volume of the tetrahedron formed by four-point masses is zero, i.e., a cyclic kite configuration is planar.

Again,

$P\left(r\right)=r_{12}{r}_{34}+r_{14}{r}_{23}-r_{24}{r}_{13}=0.$

Thus, the four-point masses forming a cyclic kite configuration and numbered sequentially lie on a common circle with diagonals, $r_{13}$ and $r_{24}$. where $r_{24}\le r_{13}$. The twice of the directed areas of the four parts of the kite configuration $P_1{P}_2{P}_3{P}_4$ are given by

$\begin{array}{l}{\stackrel{\to }{S}}_1={\stackrel{\to }{r}}_2^{\stackrel{\to }{r}}_3+{\stackrel{\to }{r}}_3^{\stackrel{\to }{r}}_4+{\stackrel{\to }{r}}_4^{\stackrel{\to }{r}}_2,{\stackrel{\to }{S}}_2={\stackrel{\to }{r}}_1^{\stackrel{\to }{r}}_4+{\stackrel{\to }{r}}_4^{\stackrel{\to }{r}}_3+{\stackrel{\to }{r}}_3^{\stackrel{\to }{r}}_1,\\ {\stackrel{\to }{S}}_3={\stackrel{\to }{r}}_1^{\stackrel{\to }{r}}_2+{\stackrel{\to }{r}}_2^{\stackrel{\to }{r}}_4+{\stackrel{\to }{r}}_4^{\stackrel{\to }{r}}_1,{\stackrel{\to }{S}}_4={\stackrel{\to }{r}}_1^{\stackrel{\to }{r}}_3+{\stackrel{\to }{r}}_3^{\stackrel{\to }{r}}_2+{\stackrel{\to }{r}}_2^{\stackrel{\to }{r}}_1,\end{array}$

$\begin{array}{l}\Rightarrow {\stackrel{\to }{S}}_1=\left(-\frac{R}{2}\hat{i}+\frac{\sqrt{3}}{2}R\hat{j}\right)^\left(-R\hat{i}\right)+\left(-R\hat{i}\right)^\left(-\frac{R}{2}\hat{i}-\frac{\sqrt{3}}{2}R\hat{j}\right)\\ +\left(-\frac{R}{2}\hat{i}-\frac{\sqrt{3}}{2}R\hat{j}\right)^\left(-\frac{R}{2}\hat{i}+\frac{\sqrt{3}}{2}R\hat{j}\right)=\frac{\sqrt{3}}{2}{R}^2,\end{array}$

$\Rightarrow {\stackrel{\to }{S}}_1=\frac{\sqrt{3}}{2}{R}^2\hat{k},{\stackrel{\to }{S}}_2=-\sqrt{3}{R}^2\hat{k},{\stackrel{\to }{S}}_3=3\frac{\sqrt{3}}{2}{R}^2\hat{k},{\stackrel{\to }{S}}_4=-\sqrt{3}{R}^2\hat{k}.$(19)

Thus, the directed areas ${\stackrel{\to }{S}}_1$ and ${\stackrel{\to }{S}}_3$ are normals to the plane of motion of the point masses, whereas ${\stackrel{\to }{S}}_2$ and ${\stackrel{\to }{S}}_4$ are the anti-normals to the plane of motion that is all directed areas parallel to each other.

Also,

${\displaystyle \underset{k=1}{\overset{4}{\sum }}{\stackrel{\to }{S}}_k}={\stackrel{\to }{S}}_1+{\stackrel{\to }{S}}_2+{\stackrel{\to }{S}}_3+{\stackrel{\to }{S}}_4=0.$(20)

as $P_k\left(x_k,y_k\right)\Rightarrow P_1\equiv \left(R,0\right),P_2\equiv \left(-\frac{R}{2},\frac{\sqrt{3}}{2}R\right),P_3\equiv \left(-R,0\right),P_4\equiv \left(-\frac{R}{2},-\frac{\sqrt{3}}{2}R\right)$,

then ${\displaystyle \underset{k=1}{\overset{4}{\sum }}{S}_k}{x}_k={\displaystyle \underset{k=1}{\overset{4}{\sum }}{S}_k}{y}_k=0$,

and consequently

${\displaystyle \underset{k=1}{\overset{4}{\sum }}{S}_k}{\stackrel{\to }{r}}_k=0,$

$\begin{array}{c}{\displaystyle \underset{m=1}{\overset{4}{\sum }}{\displaystyle \underset{n=1}{\overset{4}{\sum }}{r}_{mn}^2}}{S}_m{S}_n={\displaystyle \underset{m=1}{\overset{4}{\sum }}{\displaystyle \underset{n=1}{\overset{4}{\sum }}|{\stackrel{\to }{r}}_m}}-{{\stackrel{\to }{r}}_n|}^2{S}_m{S}_n={\displaystyle \underset{m=1}{\overset{4}{\sum }}{\displaystyle \underset{n=1}{\overset{4}{\sum }}\left({\stackrel{\to }{r}}_m^2+{\stackrel{\to }{r}}_n^2-2{\stackrel{\to }{r}}_m\cdot {\stackrel{\to }{r}}_n\right)S_m{S}_n}},\\ ={\displaystyle \underset{m=1}{\overset{4}{\sum }}{S}_m}{\stackrel{\to }{r}}_m^2{\displaystyle \underset{n=1}{\overset{4}{\sum }}{S}_n}+{\displaystyle \underset{m=1}{\overset{4}{\sum }}{S}_m}{\displaystyle \underset{n=1}{\overset{4}{\sum }}{S}_n}{\stackrel{\to }{r}}_n^2-2{\displaystyle \underset{m=1}{\overset{4}{\sum }}{S}_m{\stackrel{\to }{r}}_m}\cdot {\displaystyle \underset{n=1}{\overset{4}{\sum }}{S}_n}{\stackrel{\to }{r}}_n,\end{array}$

${\displaystyle \underset{m=1}{\overset{4}{\sum }}{\displaystyle \underset{n=1}{\overset{4}{\sum }}{r}_{mn}^2}}{S}_m{S}_n=\left({\displaystyle \underset{m=1}{\overset{4}{\sum }}{S}_m}{\stackrel{\to }{r}}_m{}^2\right)0+0\left({\displaystyle \underset{n=1}{\overset{4}{\sum }}{S}_n}{\stackrel{\to }{r}}_n{}^2\right)-2\left({\displaystyle \underset{m=1}{\overset{4}{\sum }}{S}_m{\stackrel{\to }{r}}_m}\right)\cdot \left({\displaystyle \underset{n=1}{\overset{4}{\sum }}{S}_n}{\stackrel{\to }{r}}_n\right)=0$ (21)

where ${\stackrel{\to }{r}}_m=x_m\hat{i}+y_m\hat{j}$.

Using Equations (7c), (7d), (10c) and (10d), we get

$\begin{array}{l}\left|{\stackrel{\to }{S}}_1\right|=\frac{r_{23}{r}_{34}{r}_{24}}{4R}=\frac{3\sqrt{3}}{2}{R}^2={\Delta }_1,\left|{\stackrel{\to }{S}}_2\right|=\frac{r_{13}{r}_{34}{r}_{14}}{4R}=\sqrt{3}{R}^2={\Delta }_2,\\ \left|{\stackrel{\to }{S}}_3\right|=\frac{r_{12}{r}_{24}{r}_{14}}{4R}=\frac{\sqrt{3}}{2}{R}^2={\Delta }_3,\left|{\stackrel{\to }{S}}_4\right|=\frac{r_{12}{r}_{23}{r}_{13}}{4R}=\sqrt{3}{R}^2={\Delta }_4.\end{array}\}$(22)

$\Rightarrow {\Delta }_1\ge {\Delta }_2={\Delta }_4\ge {\Delta }_3.$(23)

Now using the results of Equations (7c), (7d), (7e) and a matrix A, we have

$\begin{array}{c}\left[r_{23}^2{r}_{34}^2{r}_{24}^2\right]A{\left[r_{23}^2{r}_{34}^2{r}_{24}^2\right]}^{\text{T}}=\left[4{\Delta }_1^2\right]\\ \left[r_{13}^2{r}_{34}^2{r}_{14}^2\right]A{\left[r_{13}^2{r}_{34}^2{r}_{14}^2\right]}^{\text{T}}=\left[4{\Delta }_2^2\right]\\ \left[r_{12}^2{r}_{24}^2{r}_{14}^2\right]A{\left[r_{12}^2{r}_{24}^2{r}_{14}^2\right]}^{\text{T}}=\left[4{\Delta }_3^2\right]\\ \left[r_{12}^2{r}_{23}^2{r}_{13}^2\right]A{\left[r_{12}^2{r}_{23}^2{r}_{13}^2\right]}^{\text{T}}=\left[4{\Delta }_4^2\right]\end{array}\}\text{where}A=\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$ (24)

Thus, the results of Equations (7c), (7d) and (7e) justified the results of previous authors. Similarly, all the results from Equations (17) to (24) can be verified by the results of Equations (10) and (16) also. The partial derivatives of the Caley-Menger determinant with respect to the variables $r_{ij}^2\left(i\ne j=1,2,3,4\right)$ are given by

$\frac{\partial V\left(r\right)}{\partial r_{ij}^2}={\left(-1\right)}^{i+j-1}32{\Delta }_i{\Delta }_j.$ (25)