When Four Cyclic Antipodal Points Are Ordered Counterclockwise

1. Four Cyclic Antipodal Points

A cyclic antipodal points of a circle in a Euclidean plane ${\mathbb{R}}^2$ is a pair $(A,A^{\prime })$ of points $A,A^{\prime }\in$${\mathbb{R}}^2$ that are the intersections of the circle with a diameter of the circle. A circle with four cyclic antipodal points is shown in Figure 1. The distance $\left|AB\right|$ between two points $A,B\in$${\mathbb{R}}^2$ is given by

$$\left|AB\right|=|-A+B|.$$

(1)

Theorem 1.

(A Four Cyclic Antipodal Points Theorem)..Let $\Sigma (O,r)$ be a circle in the Euclidean plane ${\mathbb{R}}^2$ with radius r, centered at $O\in$${\mathbb{R}}^2$, with four cyclic antipodal points $(A,A^{\prime })$, $(B,B^{\prime })$, $(C,C^{\prime })$ and $(D,D^{\prime })$, such that the eight points $\{A,B,C,D,A^{\prime },B^{\prime },C^{\prime },D^{\prime }\}\subset$${\mathbb{R}}^2$ are arbitrarily ordered counterclockwise (or clockwise), as shown in Figure 1.

Then, the four cyclic antipodal points satisfy the identity

$$\begin{array}{cc}\hfill |A{B}^{\prime }\left|\right|B{C}^{\prime }\left|\right|C{D}^{\prime }|& -|A{B}^{\prime }\left|\right|BC\left|\right|CD|-\left|AB\right||B{C}^{\prime }\left|\right|CD|\hfill \\ & -\left|AB\right|\left|BC\right||C{D}^{\prime }|=4r^2\left|A^{\prime }D\right|.\hfill \end{array}$$

(2)

Proof. 

The proof is based on the elegant trigonometric identity

$$\begin{array}{cc}\hfill sin\frac{\alpha +\pi }{2}sin\frac{\beta +\pi }{2}sin\frac{\gamma +\pi }{2}& -sin\frac{\alpha +\pi }{2}sin\frac{\beta }{2}sin\frac{\gamma }{2}\hfill \\ \hfill & -sin\frac{\alpha }{2}sin\frac{\beta +\pi }{2}sin\frac{\gamma }{2}\hfill \\ \hfill & -sin\frac{\alpha }{2}sin\frac{\beta }{2}sin\frac{\gamma +\pi }{2}=sin\frac{\delta }{2}\hfill \end{array}$$

(3)

which holds for all $\alpha ,\beta ,\gamma ,\delta \in$${\mathbb{R}}^2$ that satisfy the condition

$$\alpha +\beta +\gamma +\delta =\pi .$$

(4)

The trigonometric identity in (3) – (4) can readily be verified by computer assisted computation.

In order to realize (3) – (4) geometrically by a circle with four cyclic antipodal points, shown in Figure 1, we define the four O-vertex angles in Figure 1 as follows.

$$\begin{array}{cc}\hfill \alpha & =\angle AOB\hfill \\ \hfill \beta & =\angle BOC\hfill \\ \hfill \gamma & =\angle COD\hfill \\ \hfill \delta & =\angle DO{A}^{\prime }.\hfill \end{array}$$

(5)

We note that $\alpha +\beta +\gamma +\delta =\pi$, as required by Condition (4), since the points $A,B,C,D,A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ are ordered counterclockwise.

Then, consequently, the remaining three O-vertex angles in Figure 1 are

$$\begin{array}{cc}\hfill \angle AO{B}^{\prime }& =\alpha +\pi \hfill \\ \hfill \angle BO{C}^{\prime }& =\beta +\pi \hfill \\ \hfill \angle CO{D}^{\prime }& =\gamma +\pi .\hfill \end{array}$$

(6)

Applying the law of cosines to triangle AOB yields

$$\begin{array}{cc}\hfill {\left|AB\right|}^2& =2r^2-2r^{2}cos\alpha \hfill \\ \hfill & =2r^2(1-cos\alpha )\hfill \\ \hfill & =4r^2{sin}^2\frac{\alpha }{2}\hfill \end{array}$$

(7)

so that

$$\left|AB\right|=2rsin\frac{\alpha }{2}.$$

(8)

Similarly to (8), by means of (5) – (6), we obtain the following seven results:

$$\begin{array}{cc}\hfill sin\frac{\alpha }{2}& =\frac{1}{2r}\left|AB\right|\hfill \\ \hfill sin\frac{\beta }{2}& =\frac{1}{2r}\left|BC\right|\hfill \\ \hfill sin\frac{\gamma }{2}& =\frac{1}{2r}\left|CD\right|\hfill \\ \hfill sin\frac{\delta }{2}& =\frac{1}{2r}\left|A^{\prime }D\right|\hfill \\ \hfill sin\frac{\alpha +\pi }{2}& =\frac{1}{2r}\left|A{B}^{\prime }\right|\hfill \\ \hfill sin\frac{\beta +\pi }{2}& =\frac{1}{2r}\left|B{C}^{\prime }\right|\hfill \\ \hfill sin\frac{\gamma +\pi }{2}& =\frac{1}{2r}\left|C{D}^{\prime }\right|.\hfill \end{array}$$

(9)

Substituting the sines from (9) into the trigonometric identity (3) yields (2), as desired. □

The proof of Theorem 1 is motivated by the proof of Ptolemy’s Theorem in [1]. In [1,2] Ptolemy’s Theorem is extended to hyperbolic geometry and, similarly, Theorem 1 can be extended to hyperbolic geometry as well.

2. Special Cases

In the special case when $A=B$ and, hence, $A^{\prime }=B^{\prime }$, the result (2) of Theorem 1 descends to

$$\begin{array}{cc}\hfill |B{B}^{\prime }|B{C}^{\prime }\left|C{D}^{\prime }\right|& -|B{B}^{\prime }\left|\right|BC\left|\right|CD|-\left|BB\right||B{C}^{\prime }\left|\right|CD|\hfill \\ & -\left|BB\right|\left|BC\right||C{D}^{\prime }|=4r^2\left|B^{\prime }D\right|.\hfill \end{array}$$

(10)

Noting that $|B{B}^{\prime }|=2r$ and $\left|BB\right|=0$, (10) yields

$$|B{C}^{\prime }\left|\right|C{D}^{\prime }|-|BC\left|\right|CD|=2r|B^{\prime }D|.$$

(11)

Formalizing the result in (11) we obtain the result (13) of Corollary 1.

In a second special case, when $D=A^{\prime }$ and, hence, $D^{\prime }=A$, the result (2) of Theorem 1 descends to

$$\begin{array}{cc}\hfill |A{B}^{\prime }\left|\right|B{C}^{\prime }\left|\right|AC|& -|A{B}^{\prime }\left|\right|BC\left|\right|A^{\prime }C|-|AB\left|\right|B{C}^{\prime }\left|\right|A^{\prime }C|\hfill \\ & -\left|AB\right|\left|BC\right|\left|AC\right|=4r^2\left|A^{\prime }{A}^{\prime }\right|=0.\hfill \end{array}$$

(12)

Formalizing the result in (12) we obtain the result (14) of Corollary 1.

Corollary 1.

(A Three Cyclic Antipodal Points Theorem). Let $\Sigma (O,r)$ be a circle in the Euclidean plane ${\mathbb{R}}^2$ with radius r, centered at $O\in$${\mathbb{R}}^2$, with three cyclic antipodal points $(A,A^{\prime })$, $(B,B^{\prime })$ and $(C,C^{\prime })$. The six points $\{A,B,C,A^{\prime },B^{\prime },C^{\prime }\}\subset$${\mathbb{R}}^2$ are arbitrarily ordered counterclockwise (or clockwise), as shown in Figure 2.

Then,

$$|A{B}^{\prime }\left|\right|B{C}^{\prime }|-|AB\left|\right|BC|=2r|A^{\prime }C|$$

(13)

and

$$\begin{array}{cc}\hfill |A{B}^{\prime }\left|\right|B{C}^{\prime }\left|\right|AC|& -|A{B}^{\prime }\left|\right|BC\left|\right|A^{\prime }C|-|AB\left|\right|B{C}^{\prime }\left|\right|A^{\prime }C|\hfill \\ & -\left|AB\right|\left|BC\right|\left|AC\right|=0.\hfill \end{array}$$

(14)