<s>[INST] Find the area of a triangle given the radius of its incircle and a tangential point A friend gave me recently the following interesting problem and I would like to share a couple of solutions. Any additional contributions are welcome.
A triangle $\vartriangle ABC$ is given and we know the radius $r$ of its incircle $(O, r)$. Let $D$ be the tangential point of the incircle on $AC$ which partitions $AC$ into $AD=\alpha$ and $DC=\beta$. Determine the area of $\vartriangle ABC$ as a function of $\alpha$, $\beta$ and $r$.

 [/INST] Solution 1: Draw $AO$, $AO$ and $CO$. We know that these are the bisectors of $\angle BAC$, $\angle ABC$ and $\angle BCA$ respectively. 
Let $\angle OAC = \frac{1}{2} \angle BAC =: \phi_A$ and $\angle OBE = \frac{1}{2} \angle ABC =: \phi_B$ and $\angle OCA = \frac{1}{2}\angle BCA =: \phi_C$.
We have that $\angle ODA$ is a right angle (and so is $\angle ODC$). Thus $\tan \phi_A = \frac{r}{\alpha}$ and $\tan \phi_C = \frac{r}{\beta}$. Therefore $\angle AOD = \frac{\pi}{2} - \phi_A$. 
The two right-angled triangles $\vartriangle ADO$ and $\vartriangle AEO$ are equal (indeed, they share a common side $AO$ and $OE=OD=r$). Thus, $\angle EOA = \angle DOA = \frac{\pi}{2} - \phi_A$. 
Similarly, $\angle COD = \frac{\pi}{2}-\phi_C$ and since $\vartriangle COD = \vartriangle COF$ it is $\angle COF = \angle COD =  \frac{\pi}{2}-\phi_C$.
Using the fact that 
$$
\angle AOD + \angle DOC + \angle COF + \angle FOE + \angle EOA = 2\pi,\\
2(\frac{\pi}{2}-\phi_A) + 2(\frac{\pi}{2}-\phi_C) + \angle FOE = 2\pi,\\
\angle FOE = 2(\phi_A + \phi_C).
$$
Because of the equality of $\vartriangle BEO$ and $\vartriangle BFO$ we have $\angle EOB = \frac{\angle FOE}{2} = \phi_A + \phi_C$, so
$$
\tan \angle EOB = \frac{BE}{r},\\
BE = r \tan(\phi_A + \phi_C) = r\frac{\tan\phi_A + \tan\phi_C}{1-\tan\phi_A\tan\phi_C}=\\
=r\frac{\frac{r}{\alpha}+\frac{r}{\beta}}{1-\frac{r}{\alpha}\frac{r}{\beta}}=r\frac{\frac{r(\alpha+\beta)}{\alpha\beta}}{\frac{\alpha\beta-r^2}{\alpha\beta}}=r^2\frac{\alpha+\beta}{\alpha\beta-r^2}
$$
From the aforementioned triangle equalities we have $AE = \alpha$, $CF = \beta$ and $BF = BE$, therefore, the semiperimeter of $\vartriangle ABC$ is
$$
\tau = \alpha + \beta + BE = \alpha + \beta + r^2\frac{\alpha+\beta}{\alpha\beta-r^2}=\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2},
$$
and the area of the triangle is 
$$S = r\tau = r\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}$$
Solution 2: This is a solution without trigonometry and is less elegant. Since $\angle BOE = \phi_A + \phi_C$ there is a point $M$ on $BE$ so that $\angle MOE = \phi_A$ and $\angle MOB = \phi_C$ as in the following figure:

It is easy to see that the two triangles $\vartriangle MEO$ and $\vartriangle OEA$ are similar (they have all their angles equal to one another), so, 
$$
\frac{EM}{EO}=\frac{EO}{EA},\\
EM = \frac{r^2}{\alpha}.
$$
We now need to determine $BM$. What is the same, $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another). Note that (Pythagorean theorem on $\vartriangle BEO$)
$$BO=\sqrt{BE^2 + r^2} = \sqrt{(BM+ME)^2 + r^2},$$  
and (Pythagorean theorem on $\vartriangle OCD$)
$$OC = \sqrt{r^2 + \beta^2}.$$ 
We also have 
$$
MO = \sqrt{r^2 + ME^2} = \sqrt{r^2 + \frac{r^4}{\alpha^2}} = r\sqrt{1+(r/a)^2}
$$
Using the fact that $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another), we arrive at
$$
\frac{BM}{BO}=\frac{MO}{OC},\\
\frac{BM}{\sqrt{(BM+\frac{r^2}{\alpha^2})^2 + r^2}}=\frac{r\sqrt{1+\frac{r^2}{\alpha^2}}}{\sqrt{r^2 + \beta^2}},\\
\frac{BM^2}{(BM+\frac{r^2}{\alpha^2})^2 + r^2}=\frac{r^2 (1+\frac{r^2}{\alpha^2})}{r^2 + \beta^2},
$$
which is a quadratic equation one can solve to derive a formula for $BM$ (the non-positive solution should be discarded) in terms of $r$, $\alpha$ and $\beta$. Having determined $BM$ and $ME$ we can easily determine the semiperimeter of $\vartriangle ABC$, $\tau = \alpha + \beta + BM + ME$ and the area of $\vartriangle ABC$ is again $S = r\tau$.
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