# TRIGONOMETRIC IDENTITIES AND URQUHART'S THEOREM

Antoine Mhanna

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Antoine Mhanna. TRIGONOMETRIC IDENTITIES AND URQUHART'S THEOREM. International Journal of Computer Discovered Mathematics, 2023, 8, pp.9-14. http://03291910v2

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Distributed under a Creative Commons Attribution 4.0 International License
# TRIGONOMETRIC IDENTITIES AND URQUHART'S THEOREM

ANTOINE MHANNA¹*

ABSTRACT. The purpose of this note is to represent a general analytic geometry problem implying Urquhart's Theorem. The proofs are elementary and rely on some calculations and trigonometric identities.

## 1. INTRODUCTION

We start by stating this simple problem: In the Euclidean plane ($O, \vec{i}, \vec{j}$), take two concentric circles ($A$) respectively ($B$) of radius $a$ respectively $b$ centered at $O$. Let $A \in (A)$ with $A(a \cos(\alpha); a \sin(\alpha))$ and $B \in (B)$ with $B(b \cos(\beta); b \sin(\beta))$ for some reals $\alpha$ and $\beta$ in $[-\pi, \pi]$. For $C(r, 0)$, $r > 0$ the perpendicular bisector of $[AC]$ resp. $[BC]$ intersect $(OA)$ at $A'$ resp. $(OB)$ at $B'$. Assuming that no opposite sides of the quadrilateral $OA'CB'$ are parallel, let $I = (A'C) \cap (OB')$ and $J = (B'C) \cap (OA')$. We fix all parameters except the radius $b$ considered as a variable $b \ge 0$ and want to find $b$ for which

$$ (1.1) \qquad |IO \pm IC| = |JO \pm JC|. $$

In this context we state the known Urquhart's theorem (see also [4]) as follows:

**Theorem 1** ([1, 2]). For $\alpha\beta \le 0$, $a \ge r$ and $b = a$ we have $IO + IC = JO + JC$ (see Figure 1).

<footnote>Date: 2022</footnote>

<footnote>* Corresponding author.</footnote>

<footnote n="2010">2010 Mathematics Subject Classification. 51N20; 33B10; 26B99.</footnote>

<footnote>Key words and phrases. Urquhart's Theorem; Trigonometric Identities.</footnote>
<figure><img src="image_2.png" /><figcaption>FIGURE 1. Urquhart's theorem</figcaption></figure>

## 2. MAIN RESULTS

Considering the line (OA) resp. (OB) of equation $y = \tan(\alpha)x$ resp. $y = \tan(\beta)x$; by straight calculations we have:

$A' \left( \frac{\cos(\alpha)(a^2 - r^2)}{2a - 2r \cos(\alpha)}, \frac{\sin(\alpha)(a^2 - r^2)}{2a - 2r \cos(\alpha)} \right)$ and thus $B' \left( \frac{\cos(\beta)(b^2 - r^2)}{2b - 2r \cos(\beta)}, \frac{\sin(\beta)(b^2 - r^2)}{2b - 2r \cos(\beta)} \right)$.

We also verify that the line (CA') is:

$$y = \left( \frac{\sin(\alpha)(a^2 - r^2)}{(r^2 + a^2) \cos(\alpha) - 2ra} \right) x - \frac{r \sin(\alpha)(a^2 - r^2)}{(r^2 + a^2) \cos(\alpha) - 2ar}.$$

The points I and J are:

$$I \left\{ \begin{aligned} x_I &= \frac{r \cos(\beta) \sin(\alpha)(a^2 - r^2)}{\cos(\beta) \sin(\alpha)(a^2 - r^2) - \sin(\beta)((r^2 + a^2) \cos(\alpha) - 2ar)} \\ y_I &= \frac{r \sin(\beta) \sin(\alpha)(a^2 - r^2)}{\cos(\beta) \sin(\alpha)(a^2 - r^2) - \sin(\beta)((r^2 + a^2) \cos(\alpha) - 2ar)} \end{aligned} \right. $$

and by switching $\alpha \leftrightarrow \beta$, $a \leftrightarrow b$:

$$J \left\{ \begin{aligned} x_J &= \frac{r \cos(\alpha) \sin(\beta)(b^2 - r^2)}{\cos(\alpha) \sin(\beta)(b^2 - r^2) - \sin(\alpha)((r^2 + b^2) \cos(\beta) - 2br)} \\ y_J &= \frac{r \sin(\alpha) \sin(\beta)(b^2 - r^2)}{\cos(\alpha) \sin(\beta)(b^2 - r^2) - \sin(\alpha)((r^2 + b^2) \cos(\beta) - 2br)}. \end{aligned} \right. $$

Hereafter we assume that $\sin(\alpha)\sin(\beta) \neq 0$ since otherwise the configuration is trivial. The next lemma collects some identities that can be obtained at www.dcode.fr/math-expression-factor:

**Lemma 1.** Let $\alpha, \beta, r$ and $a$ be real numbers, set $s = \frac{\alpha+\beta}{2}$ and $d = \frac{\alpha-\beta}{2}$:

$$(2.1) \quad (\sin(\alpha)(a^2 - r^2))^2 + ((r^2 + a^2) \cos(\alpha) - 2ar)^2 = (a^2 + r^2 - 2ar \cos(\alpha))^2.$$
$$ (2.2) \qquad \begin{aligned} & r \sin(\alpha)(a^2 - r^2) + r \sin(\beta)(a^2 + r^2 - 2ar \cos(\alpha)) \\ &= 2r(a \sin(s) + r \sin(d))(a \cos(d) - r \cos(s)). \end{aligned} $$

$$ (2.3) \qquad \begin{aligned} & r \sin(\alpha)(a^2 - r^2) - r \sin(\beta)(a^2 + r^2 - 2ar \cos(\alpha)) \\ &= 2r(r \sin(s) + a \sin(d))(a \cos(s) - r \cos(d)). \end{aligned} $$

$$ (2.4) \qquad \begin{aligned} & \cos(\beta) \sin(\alpha)(a^2 - r^2) - \sin(\beta)((r^2 + a^2) \cos(\alpha) - 2ar) \\ &= 2(a \sin(d) + r \sin(s))(a \cos(d) - r \cos(s)). \end{aligned} $$

**Lemma 2.** With the given definitions we have:

$$ \bullet OA' = \frac{|a^2 - r^2|}{|2a - 2r \cos(\alpha)|}, \quad OB' = \frac{|b^2 - r^2|}{|2b - 2r \cos(\beta)|} $$

$$ \bullet CA' = \frac{a^2 + r^2 - 2ar \cos(\alpha)}{|2a - 2r \cos(\alpha)|} \quad \text{and} \quad CB' = \frac{b^2 + r^2 - 2br \cos(\beta)}{|2b - 2r \cos(\beta)|} $$

$$ \bullet OI = \frac{|r \sin(\alpha)(a^2 - r^2)|}{|\cos(\beta) \sin(\alpha)(a^2 - r^2) - \sin(\beta)((r^2 + a^2) \cos(\alpha) - 2ar)|} $$

$$ \bullet OJ = \frac{|r \sin(\beta)(b^2 - r^2)|}{|\cos(\alpha) \sin(\beta)(b^2 - r^2) - \sin(\alpha)((r^2 + b^2) \cos(\beta) - 2br)|} $$

$$ \bullet CI = \frac{r |\sin(\beta)| (a^2 + r^2 - 2ar \cos(\alpha))}{|\cos(\beta) \sin(\alpha)(a^2 - r^2) - \sin(\beta)((r^2 + a^2) \cos(\alpha) - 2ar)|} $$

$$ \bullet CJ = \frac{r |\sin(\alpha)| (b^2 + r^2 - 2br \cos(\beta))}{|\cos(\alpha) \sin(\beta)(b^2 - r^2) - \sin(\alpha)((r^2 + b^2) \cos(\beta) - 2br)|} $$

$$ \bullet \frac{CB'}{OB'} \cdot \frac{OI}{CI} = \frac{CJ}{OJ} \cdot \frac{OA'}{CA'} \quad (\text{see also } [5] \text{ page } 63). $$

*Proof.* A straight application of previous calculations. $\square$

<figure><img src="image_3.png" /></figure>

For any reals $\alpha, \beta, r$ and $a$ we define $F_{\alpha,\beta}^{\pm}(a)$ whenever the denominator is non zero by:

$$ (2.5) \qquad F_{\alpha,\beta}^{\pm}(a) = \frac{r \sin \alpha (a^2 - r^2) \pm r \sin \beta (a^2 + r^2 - 2ar \cos(\alpha))}{\cos(\beta) \sin(\alpha)(a^2 - r^2) - \sin(\beta)((r^2 + a^2) \cos(\alpha) - 2ar)}, $$

(with the same sign for $\pm$ on both sides). By Lemma 1 we know that

$$ F_{\alpha,\beta}^{+}(a) = \frac{r(a \sin(s) + r \sin(d))}{a \sin(d) + r \sin(s)} \quad \text{and} \quad F_{\alpha,\beta}^{-}(a) = \frac{r(a \cos(s) - r \cos(d))}{a \cos(d) - r \cos(s)}. $$

In order to get identities of the form (1.1) we need to solve in $b$, $F_{\alpha,\beta}^{\pm}(a) = \pm F_{\beta,\alpha}^{\pm}(b)$. This is summarized in the next theorems.

**Theorem 2.** Under previous notations assuming $\sin(\alpha) \sin(\beta) \neq 0$:
(1) $F_{\alpha, \beta}^{+}(a) = F_{\beta, \alpha}^{+}(b)$ if and only if

$$b = r \frac{a(1 - \cos(\alpha) \cos(\beta)) + r(\cos(\beta) - \cos(\alpha))}{r(1 - \cos(\alpha) \cos(\beta)) + a(\cos(\beta) - \cos(\alpha))}.$$

(2) $F_{\alpha, \beta}^{+}(a) = -F_{\beta, \alpha}^{+}(b)$ if and only if $b = -a$.

(3) $F_{\alpha, \beta}^{+}(a) = F_{\beta, \alpha}^{-}(b)$ if and only if $b = r \frac{a \cos(\alpha) - r}{a - r \cos(\alpha)}$.

(4) $F_{\alpha, \beta}^{+}(a) = -F_{\beta, \alpha}^{-}(b)$ if and only if $b = r \frac{a \cos(\beta) + r}{r \cos(\beta) + a}$.

(5) $F_{\alpha, \beta}^{-}(a) = F_{\beta, \alpha}^{-}(b)$ if and only if $b = a$.

(6) $F_{\alpha, \beta}^{-}(a) = -F_{\beta, \alpha}^{-}(b)$ if and only if

$$b = r \frac{a(1 + \cos(\alpha) \cos(\beta)) - r(\cos(\beta) + \cos(\alpha))}{a(\cos(\beta) + \cos(\alpha)) - r(1 + \cos(\alpha) \cos(\beta))}.$$

(7) $F_{\alpha, \beta}^{-}(a) = F_{\beta, \alpha}^{+}(b)$ if and only if $b = r \frac{r - a \cos(\beta)}{r \cos(\beta) - a}$.

(8) $F_{\alpha, \beta}^{-}(a) = -F_{\beta, \alpha}^{+}(b)$ if and only if $b = r \frac{r - a \cos(\alpha)}{a - r \cos(\alpha)}$.

**Theorem 3.** Take $\alpha$ and $\beta$ in $]0; \pi[$; $a \ge 0$ with $r > 0$:

1) For $b = r \frac{a(1 - \cos(\alpha) \cos(\beta)) + r(\cos(\beta) - \cos(\alpha))}{r(1 - \cos(\alpha) \cos(\beta)) + a(\cos(\beta) - \cos(\alpha))}$,

* if $\alpha \ge \beta$ then

- for $a \le r$; $b \le r$ and $|IO - IC| = |JO - JC|$.

- for $a \ge r$; $b \ge r$ and $IO + IC = JO + JC$,

* if $\alpha < \beta$ then

- for $\frac{r(\cos(\alpha) - \cos(\beta))}{1 - \cos(\alpha) \cos(\beta)} \le a \le r$; $0 \le b \le r$ and $|IO - IC| = |JO - JC|$.

- for $r \le a < \frac{r(1 - \cos(\alpha) \cos(\beta))}{(\cos(\alpha) - \cos(\beta))}$; $b \ge r$ and $IO + IC = JO + JC$.

2) For $b = \frac{r(a \cos(\alpha) - r)}{a - r \cos(\alpha)}$,

* if $0 < \alpha < \frac{\pi}{2}$ then

- for $0 \le a < r \cos(\alpha) \le r$; $b \ge r$ and $|IO - IC| = |JO - JC|$,

- for $a \ge \frac{r}{\cos(\alpha)} \ge r$; $b \le r$ and $IO + IC = JO + JC$.

3) For $b = \frac{r(a \cos(\beta) + r)}{a + r \cos(\beta)}$,

* if $0 < \beta \le \frac{\pi}{2}$ then

- for $a \ge r$; $b \le r$ and $IO + IC = JO + JC$,

- for $a \le r$; $b \ge r$ and $|IO - IC| = |JO - JC|$.

* if $\frac{\pi}{2} < \beta < \pi$ then

- for $-r \cos(\beta) < a \le r$; $b \ge r$ and $|IO - IC| = |JO - JC|$,

- for $r \le a \le \frac{-r}{\cos(\beta)}$; $b \le r$ and $IO + IC = JO + JC$.
4) For $b = a$,

* for $a \ge r$; $|IO - IC| = |JO - JC|$,

* for $a \le r$; $IO + IC = JO + JC$.

5) For $b = r \frac{a(1 + \cos(\alpha) \cos(\beta)) - r(\cos(\beta) + \cos(\alpha))}{a(\cos(\beta) + \cos(\alpha)) - r(1 + \cos(\alpha) \cos(\beta))}$,

* if $\alpha + \beta \le \pi$ then
  - for $0 \le a \le \frac{r(\cos(\alpha)+\cos(\beta))}{1+\cos(\alpha)\cos(\beta)} \le r$; $b \le r$ and $IO + IC = JO + JC$,
  - for $r \le \frac{r(1+\cos(\alpha)\cos(\beta))}{\cos(\alpha)+\cos(\beta)} < a$; $b \ge r$ and $|IO - IC| = |JO - JC|$.

6) For $b = \frac{r(r - a \cos(\beta))}{r \cos(\beta) - a}$,

* if $0 < \beta < \frac{\pi}{2}$ then
  - for $0 \le a < r \cos(\beta) \le r$; $b \ge r$ and $IO + IC = JO + JC$,
  - for $a \ge \frac{r}{\cos(\beta)} \ge r$; $b \le r$ and $|IO - IC| = |JO - JC|$.

7) For $b = \frac{r(r - a \cos(\alpha))}{a - r \cos(\alpha)}$,

* if $0 < \alpha < \frac{\pi}{2}$ then
  - for $r \cos(\alpha) < a \le r$; $b \ge r$ and $IO + IC = JO + JC$,
  - for $r \le a \le \frac{r}{\cos(\alpha)}$; $b \le r$ and $|IO - IC| = |JO - JC|$.

* if $\frac{\pi}{2} \le \alpha < \pi$ then
  - for $a \ge r$; $b \le r$ and $|IO - IC| = |JO - JC|$,
  - for $a \le r$; $b \ge r$ and $IO + IC = JO + JC$.

*Proof.* The expressions of $b$ are direct computations from Theorem 2. The bounds on $(\alpha, \beta, a)$ are to verify that $b$ is nonnegative and the bounds on $b$ can be deduced by taking the derivative of $b$ as a function in $a$. $\square$

The case $\alpha \in ]0; \pi[$ and $\beta \in [-\pi; 0[$ can be deduced from the previous theorem by replacing (verbatim) $\beta$ with $|\beta|$ and switching identities that is $IO + IC = JO + JC \leftrightarrow |IO - IC| = |JO - JC|$ and vice versa. This gives for example Theorem 1 as a consequence of case 4) which also appeared in [3].

**Example 1.** In this example we illustrate the proven identities for $a = 5$, $r = 3$, $\alpha = \frac{\pi}{3}$ and $\beta = -\frac{\pi}{6}$.
<figure><img src="image_4.png" /></figure>

$$1) b = \frac{288\sqrt{3}-21}{143}, |IO - IC| = |JO - JC|$$

<figure><img src="image_5.png" /></figure>

$$4) b = 5, IO + IC = JO + JC$$

<figure><img src="image_6.png" /></figure>

$$6) b = \frac{96\sqrt{3}-45}{73}, IO + IC = JO + JC$$

<figure><img src="image_7.png" /></figure>

$$3) b = \frac{96\sqrt{3}+45}{73}, |IO - IC| = |JO - JC|$$

<figure><img src="image_8.png" /></figure>

$$5) b = \frac{288\sqrt{3}+21}{143}, IO + IC = JO + JC$$

<figure><img src="image_9.png" /></figure>

$$7) b = \frac{3}{7}, IO + IC = JO + JC.$$

We point out at last that if the coordinates of $J$ are taken as real functions, the following couple values of $b$ give the same point $J$:

$$(a, r \frac{r - a \cos \beta}{r \cos(\beta) - a}), \left(-a, r \frac{r + a \cos \beta}{r \cos(\beta) + a}\right),$$

$$\left( r \frac{a \cos(\alpha) - r}{a - r \cos(\alpha)}, r \frac{a(1 - \cos(\alpha) \cos(\beta)) + r(\cos(\beta) - \cos(\alpha))}{r(1 - \cos(\alpha) \cos(\beta)) + a(\cos(\beta) - \cos(\alpha))} \right)$$
and

$$ \left( r \frac{r - a \cos(\alpha)}{a - r \cos(\alpha)}, r \frac{a(1 + \cos(\alpha) \cos(\beta)) - r(\cos(\beta) + \cos(\alpha))}{a(\cos(\beta) + \cos(\alpha)) - r(1 + \cos(\alpha) \cos(\beta))} \right). $$

ACKNOWLEDGMENTS

Computations are assisted through Maple and GeoGebra softwares. Some simplification formulas are obtained on https://www.dcode.fr.

REFERENCES

[1] H. Grossman, *Urquharts quadrilateral theorem*, The Mathematics Teacher, **66**, 1973, pp. 643-644.

[2] M. Hajja, *A Very Short and Simple Proof of The Most Elementary Theorem of Euclidean Geometry*, Forum Geometricorum, **Vol. 6**, 2006, pp. 167-169.

[3] R. T. Leslie, *A Theorem of M. L. Urquhart's and some Consequences*, The Mathematical Gazette, **Vol. 91**, (520), 2007, pp. 39-50.

[4] D. Pedoe, *The most elementary theorem of Euclidean geometry*, Math. Mag. **4**, 1976, pp. 40-42.

[5] G. Wanner, *The Cramer-Castillon problem and Urquhart's 'most elementary' theorem*, Elem. Math. **61**, 2006, pp. 58-64.

<footnote n="1">¹ KFARDEBIAN, LEBANON</footnote>

<footnote>E-mail address: tmhanat@yahoo.com</footnote>