## lunes, 6 de julio de 2020

### Generalization of two formulae and an alternative proof of Bretschneider's formula

"If we do not succeed in solving a mathematical problem, the reason frequently consists in our failure to recognize the more general standpoint from which the problem before us appears only as a single link in a chain of related problems. After finding this standpoint, not only is this problem frequently more accessible to our investigation, but at the same time we come into possession of a method which is applicable also to related problems."  David Hilbert

The following formulae generalize $(1)$ in my previous post Killing three birds with one stone. For implications in a triangle see also Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

Here, $a$, $b$, $c$, $d$ are the sides of a general convex quadrilateral, $s$ is the semiperimeter, and $\alpha$ and $\gamma$ are two opposite angles. Then

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)-bc\cos^2{\frac{\gamma}{2}}}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{1}$$

Proof. By the Law of Cosines,

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{\gamma}\tag{2}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2+2bc\cos{\gamma}}{2ad}$. Now, making use of the half angle formula for cosine,

\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{a^2+d^2+2ad-b^2-c^2+2bc\cos{\gamma}}{4ad}\tag{3}\\ &=\frac{a^2+d^2+2ad-b^2-c^2+2bc(1-2\sin^2{\frac{\gamma}{2}})}{4ad}\tag{4}\\&=\frac{(a+d)^2-(b-c)^2-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{5}\\&=\frac{(a+d+b-c)(a+d-b+c)-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{6}\\&=\frac{1}{ad}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)-\frac{bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{7}\\&=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{8}\end{align*}

$\square$

The other formula can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$ in $(3)$.

A proof of Bretschneider's formula
The formulae in $(1)$ can be rewritten as follows

$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\tag{9}$$

and

$$bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c)\tag{10}$$

Multiplying $(9)$ and $(10)$ we get

\begin{align*}\left(ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}\right)\left(bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}\right) &= (s-a)(s-b)(s-c)(s-d)\tag{11}\end{align*}

Expanding, factorizing, completing the squares and keeping in mind some well-known trigonometric identities,

\begin{align*}abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(ad\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+bc\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{12}\\abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{13} \end{align*}

Since the area of $ABCD$ can be expressed as the sum of the areas of $\triangle{ABD}$ and $\triangle{CBD}$, which in turn can be written as $\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}$, then we are done.
$\square$

An alternative form of Bretschneider's formula
We encourage readers to prove the following formula for themselves.

Prove that the area of a general convex quadrilateral is given by the following formula:

$$K=\sqrt{abcd\sin^2\left({\frac{\alpha+\gamma}{2}}\right)-s(s-c-d)(s-b-d)(s-b-c)},$$

where $a$, $b$, $c$ and $d$ are the sides lengths, $s$ is the semiperimeter, and  $\alpha$ and $\gamma$ are opposite angles.

A concept map of identities $(9)$ and $(10)$
Below you can find a concept map of the identities $(9)$ and $(10)$ so you can see clearly what's going on here (click on the image to have a better view).

It would be interesting to investigate whether identities $(9)$ and $(10)$ can be generalized to other geometries.

## sábado, 4 de julio de 2020

### Killing three birds with one stone

In this note we derive Heron's formula, Brahmagupta's formula and the bicentric quadrilateral's area formula, $\sqrt{abcd}$, from two formulae involving sine, cosine, semiperimeter and the side lenghts of a cyclic quadrilateral.

Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\frac{a+b+c+d}{2}$. If $\angle{BAD}=\alpha$, then

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc}\tag{1}$$

or

$$sin^2{\frac{\alpha}{2}}=\frac{(a+b+c-d)(-a+b+c+d)}{4(ad+bc)}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{2}$$

Proof. First we will find an expression for $\cos{\alpha}$ in terms of $a$, $b$, $c$ and $d$. Let $\angle{BCD}=\gamma$. By the Law of Cosines and keeping in mind that $\alpha$ and $\gamma$ are supplementary, we have

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{(180^\circ-\alpha)}\tag{3}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$. Now, making use of the half angle formula for cosine,

\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{2ad+2bc+a^2+d^2-b^2-c^2}{4(ad+bc)}\tag{4}\\ &=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\tag{5}\\&=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{6}\\&=\frac{1}{ad+bc}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)\tag{7}\\&=\frac{(s-b)(s-c)}{ad+bc}\end{align*}

$\square$

The other formulae can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$.

A generalization of $(1)$ together with a proof of Bretschneider's formula can be found here.

Remark$(1)$ appears as exercise 400 in V. Panagiotis' 1000 General Trigonometry Exercises, Volume B.

A proof of Heron's Formula
For a triangle, if in $(1)$ we assume $c=0$, then we have

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{s(s-b)}{ad}\tag{8}$$

Let $\Delta_0$ be the area of $\triangle{ABD}$. Making use of the double-angle identity for sine we have

$$\sin{\alpha}=2\sqrt{\frac{s(s-b)}{ad}}\sqrt{\frac{(s-a)(s-d)}{ad}}=2\frac{\sqrt{s(s-a)(s-b)(s-d)}}{ad}\tag{9}$$

Since $\Delta_0=\frac{ad\sin{\alpha}}{2}$, it follows

$$\Delta_0=\sqrt{s(s-a)(s-b)(s-d)}\tag{10}$$

$\square$

For more implications in a triangle see here.

A proof of the Brahmagupta's formula
Denote $\Delta_1$ the area of the cyclic quadrilateral, $ABCD$. Then

\begin{align*}\Delta_1&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{11}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{12}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{13}\\&=\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-b)(s-c)}{ad+bc}}(ad+bc)\tag{14}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\tag{15}\end{align*}

$\square$

A proof of the bicentric quadrilateral's area formula
Since $a+c=b+d$ in a bicentric quadrilateral, the formulae in $(2)$ reduce to

$$sin^2{\frac{\alpha}{2}}=\frac{bc}{ad+bc}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{ad}{ad+bc}\tag{16}$$

Assume $ABCD$ is a bicentric quadrilateral and let $\Delta_2$ be its area, then

\begin{align*}\Delta_2&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{17}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{18}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{19}\\&=\sqrt{\frac{bc}{ad+bc}}\sqrt{\frac{ad}{ad+bc}}(ad+bc)\tag{20}\\&=\sqrt{abcd}\tag{21}\end{align*}

$\square$