$\require{cancel}$
The sides of a triangle are in an arithmetic progression with $k>0$ and $a>b>c$. Prove that
$\tan\left(\frac{\hat{A}}{2}\right)\tan\left(\frac{\hat{C}}{2}\right)=\frac{1}{3}$
if $r$ is the radius of inscribed circle, then $r=\frac{2k}{3\left[\tan\left(\frac{\hat{A}}{2}\right)-\tan\left(\frac{\hat{C}}{2}\right)\right]}$.
My proof:
Let $\hat{A}=\alpha,\hat{B}=\beta,\hat{C}=\gamma,\overline{AB}=c,\overline{AC}=b,\overline{BC}=a$ and $r$ is the radius of the inscribed circle.
Let $s=\frac{a+b+c}{2}$ and $A$ the area of the circle. We know from some theorems that
$A=sr$
$A=\sqrt{s(s-a)(s-b)(s-c)}$.
From this we can conclude that
$$\begin{align} sr&=\sqrt{s(s-a)(s-b)(s-c)}\\ s^2r^2&=s(s-a)(s-b)(s-c)\\ r^2&=\frac{(s-a)(s-b)(s-c)}{s}\\ \frac{r}{(s-a)(s-c)}&=\frac{s-b}{sr} \end{align}$$
From some unknown theorem we know that
$r=\frac{s-a}{\cot\left(\frac{\alpha}{2}\right)}=\frac{s-b}{\cot\left(\frac{\beta}{2}\right)}=\frac{s-c}{\cot\left(\frac{\gamma}{2}\right)}$.
Since $\cot\theta=\frac{1}{\tan\theta}$, and since in an arithmetic progression with $a>b>c$ and $k>0$ we gets $a=b+k,c=b-k$
$$\begin{align} \tan\left(\frac{\alpha}{2}\right)&=\frac{r}{s-a}\\ \tan\left(\frac{\beta}{2}\right)&=\frac{r}{s-b}\\ \tan\left(\frac{\gamma}{2}\right)&=\frac{r}{s-c}. \end{align}$$
Then
$$\begin{align} \tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\gamma}{2}\right)&=\frac{r^2}{(s-a)(s-c)}\\ &=\frac{\cancel{(s-a)}(s-b)\cancel{(s-c)}}{s}\frac{1}{\cancel{(s-a)(s-c)}}\\ &=\frac{s-b}{s}\\ &=\frac{\frac{a+b+c}{2}-b}{\frac{a+b+c}{2}}\\ &=\frac{\frac{b+\cancel k+b+b-\cancel k}{2}-b}{\frac{b+\cancel k+b+b-\cancel k}{2}}\\ &=\frac{\frac{b}{2}}{\frac{3b}{2}}\\ &=\frac{1}{3}, \end{align}$$
which proves part 1. For 2 we have
$r=\frac{2k}{3\left[\tan\left(\frac{\alpha}{2}\right)-\tan\left(\frac{\gamma}{2}\right)\right]}\iff\tan\left(\frac{\alpha}{2}\right)-\tan\left(\frac{\gamma}{2}\right)=\frac{2k}{3r}$
we have that
$$\begin{align} \tan\left(\frac{\alpha}{2}\right)-\tan\left(\frac{\gamma}{2}\right)&=\frac{r}{s-a}-\frac{r}{s-c}\\ &=r\left(\frac{1}{s-a}-\frac{1}{s-c}\right)\\ &=\frac{r[(s-c)-(s-a)]}{(s-a)(s-c)}\\ &=\frac{r(a-c)}{(s-a)(s-c)}\\ &=\frac{(s-b)(a-c)}{sr}\\ &=\frac{a-c}{3r}\\ &=\frac{(b+k)-(b-k)}{3r}\\ &=\frac{2k}{3r}, \end{align}$$ which ends the second part.
- Is my proof correct?
- What can I improve?
- Is there another way to prove this?