## sábado, 20 de junio de 2020

### Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

In Cut-the-knot's Relations between various elements of a triangle, the formulae (a generalization can be found here)

$$\sin^2{\frac{\gamma}{2}} = \frac{(s-a)(s-b)}{ab}\quad and\quad\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$

are derived using Heron's formulaHere we give an alternative proof without using Heron's Formula and we demonstrate several well-known theorems based on these formulae as a sample of its power. We will be using standard notation: $BC=a$, $AC=b$, $AB=c$, $\Delta$ for the area, $s$ for the semiperimeter, $R$ for the circumradius and $r$ for the inradius. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$.

Notice that $\frac{\cot{\frac{\gamma}{2}}}{s-c} = \frac{1}{r}$. Also, We know $Δ = rs$ and $Δ =\frac {ab\sin{\gamma}}{2}$, hence

$$\frac{\cot{\frac{\gamma}{2}}}{s-c}=\frac{1}{r}=\frac{s}{\Delta}=\frac{2s}{ab\sin{\gamma}}$$

But, $\sin{\gamma} = 2\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}$, so

$$\frac{\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}\cdot{\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}} = \frac{s(s-c)}{ab}$$

from which we get $\cos^2{\frac{\gamma}{2}} = \frac{s(s-c)}{ab}$.

The other formula can be obtained replacing $\cos^2{\frac{\gamma}{2}}$ by $1 - \sin^2{\frac{\gamma}{2}}$. Indeed,

\begin{aligned} 1-\sin^2{\frac{\gamma}{2}} &= \frac{s(s-c)}{ab} \\ \sin^2{\frac{\gamma}{2}} &= 1-\frac{s(s-c)}{ab} \\ &= \frac{\left(ab-s(s-c)\right)}{ab} \\ &= \frac{\left((y+z)(x+z)-(x+y+z)(z)\right)}{ab}\\ &= \frac{xy}{ab} \\ &=\frac{(s-a)(s-b)}{ab} \end{aligned}

$\square$

Purely geometrical proofs can be found at Trigonography.com.

1. A proof of Heron's Formula
Making use of the formulae proven above and the double angle identity for sine we have

$$\sin{\gamma}=2\sqrt{\frac{s(s-c)}{ab}}\sqrt{\frac{(s-a)(s-b)}{ab}}=2\frac{\sqrt{s(s-a)(s-b)(s-c)}}{ab}$$

Since $\Delta=\frac{ab\sin{\gamma}}{2}$, it follows

$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

$\square$

2. A proof of the Law of Cosines
Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=s(s-c)-(s-a)(s-b)$$
Substituting and multiplying by 4,
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Expanding and Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

3. Proofs for some trigonometric identities associated to a triangle
a) $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$.

As a consequence of the formulae proven at the beginning of the note,

$$\tan{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}, \quad\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\quad and \quad \tan{\frac{\gamma}{2}}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

So, by canceling and simplifying you get

\begin{aligned}\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}} &=\frac{s-c}{s}+\frac{s-b}{s}+\frac{s-a}{s}\\ &=\frac{z+y+x}{s}\\ &=\frac{s}{s}=1\end{aligned}

$\square$

b) $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$.

We make use of the well-known relationship $abc=4R\Delta$ (see here for a proof) and Heron's Formula.

\begin{aligned}r&=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\\&=4R\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{16R^2\Delta^2}}\\&=\frac{\Delta}{s}\\&=\frac{rs}{s}\\&=r\end{aligned}

$\square$

c) $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$.

\begin{aligned}s&=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-b)}{ac}}\sqrt{\frac{s(s-c)}{ab}}\\&=4R\sqrt{\frac{s^2\Delta^2}{a^2b^2c^2}}\\&=4R\frac{s\Delta}{abc}\\&=4R\frac{s\Delta}{4R\Delta}\\&=s\end{aligned}

$\square$

Consequently, the following relationship also holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{r}{s}$$

or

$$\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\frac{s}{r}$$

d) $\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}$.

To prove the above identity we will show that the right hand side equals $\frac{s}{r}$.

\begin{aligned}\frac{s}{r}&= \cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}\\&=\frac{\sqrt{s(s-a)}}{\sqrt{(s-b)(s-c)}}+\frac{\sqrt{s(s-b)}}{\sqrt{(s-a)(s-c)}}+\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\\&=\frac{\Delta(s-a)+\Delta(s-b)+\Delta(s-c)}{(s-a)(s-b)(s-c)}\\&=\frac{\Delta(x+y+z)}{xyz}\\&=\Delta\frac{s^2}{\Delta^2}\\&=\frac{s^2}{rs}\\&=\frac{s}{r}\end{aligned}

$\square$

We invite the reader to prove the following identity (possibly new) on their own.
$$\frac{\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}+\cot{\frac{\alpha}{2}}\cot{\frac{\gamma}{2}}+\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}}{(s-a)(s-b)+(s-a)(s-c)+(s-b)(s-c)}=\frac{1}{r^2}$$

4. The product $AI\cdot{BI}\cdot{CI}$
Consider a triangle $\triangle{ABC}$ and its Incenter, $I$. Denote $R$ and $r$ the circumradius and inradius, respectively. Also let $AI=k$; $BI=l$; $CI=m$. Then, the following identity holds

$$klm=4Rr^2$$

Proof. We make use of the half-angle formula,

$$\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$

Notice that $\cos{\frac{\gamma}{2}}=\frac{(s-c)}{m}$. Also, because of half-angle formula we have $\cos{\frac{\gamma}{2}}=\sqrt{\frac{s(s-a)}{ab}}$. Equating both expressions and solving for $m^2$,

$$m^2=\frac{ab(s-c)}{s}$$

Similarly you get $k^2=\frac{bc(s-a)}{s}$ and $l^2=\frac{ac(s-b)}{s}$. Hence,

$$(klm)^2=\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{s^3}=\frac{a^2b^2c^2s(s-a)(s-b)(s-c)}{s^4}$$

Substituting from Heron's formula,

$$(klm)^2=\frac{a^2b^2c^2\Delta^2}{s^4}$$

Simplifying and using the well-known formulas $abc=4R\Delta$ and $\Delta=rs$ you get the desired result.

$$klm=\frac{abc\Delta}{s^2}=\frac{4R\Delta^2}{s^2}=4Rr^2$$

## viernes, 5 de junio de 2020

### Yet Another Proof of the Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using standard notation, the law of cosines states

$$c^2=a^2+b^2-2ab\cos{\gamma},$$

where $\gamma$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$. For the same figure, the other two relations are analogous:

$$a^2=b^2+c^2-2ac\cos{\alpha},$$
$$b^2=a^2+c^2-2ac\cos{\beta}.$$

Proof. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$. We start from two well-known relationships of a triangle: $$\sin^2{\frac{\gamma}{2}}=\frac{(s-a)(s-b)}{ab} \qquad\text{and}\qquad \cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}$$
(See Cut-the-knot's Relations between various elements of a triangle for proofs), where $s$ denotes the semiperimeter of $\triangle{ABC}$. Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=sz-xy$$
Substituting and multiplying by 4,
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

AcknowledgementMy sincerest thanks to Angina Seng for giving helpful comments which allowed me to simplify the proof.

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