Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side.

Question

In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$. Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having length $a$ and $b$, what is the area of $ABC$?

Answer 1

An alternative approach: let we just find the inradius $r$. Assuming that $\widehat{C}=60^\circ$, $I_C$ is the projection of the incenter on the $AB$-side, $AI_C=a,BI_C=b$, we have: $$\widehat{A}=2\arctan\frac{r}{a},\qquad \widehat{B}=2\arctan\frac{r}{b}\tag{1}$$ hence: $$ \frac{\pi}{3} = \arctan\frac{r}{a}+\arctan\frac{r}{b}\tag{2} $$ must hold, from which: $$ \sqrt{3} = \frac{\frac{r}{a}+\frac{r}{b}}{1-\frac{r^2}{ab}}\tag{3}$$ follows. Since: $$ \Delta = \frac{1}{2}(a+\sqrt{3}\,r)(b+\sqrt{3}\,r)\sin 60^\circ =\frac{\sqrt{3}}{4}ab\left(1+\sqrt{3}\left(\frac{r}{a}+\frac{r}{b}\right)+3\frac{r^2}{ab}\right)\tag{4}$$ by exploiting $(3)$ we have:

$$ \Delta = \color{red}{\sqrt{3}\,ab},\tag{5}$$

pretty nice, don't you think?

Answer 2

Let the triangle be $\triangle ABC$. Let $\angle A=60^{\circ}$. Let the incircle touch the triangle on the points $A_1,B_1,C_1$ (opposite to $A,B,C$, respectively). Let $A_1B=a,\,A_1C=b$. Then $C_1B=a,\, B_1C=b$. Let $AC_1=AB_1=x$. Then by the Law of cosines:

$$(a+b)^2=(x+a)^2+(x+b)^2-2(x+a)(x+b)\cos 60^{\circ}$$

Solve this quadratic equation. The area is $\frac{1}{2}(x+a)(x+b)\sin 60^{\circ}$.

Answer 3

Area $S_{ABC}$ in terms of $|AB|$ and altitude $h_C$ at the point $C$:

\begin{align} h_C&=(b+r\sqrt{3})\sin(60) =\frac{\sqrt{3}}{2}(b+r\sqrt{3}) \\ S&= \frac{\sqrt{3}}{4} (a+r\sqrt{3}) (b+r\sqrt{3}) \\ &= \frac{3}{4}\left( \sqrt{3}r^2+(a+b)r \right) +\frac{\sqrt{3}}{4}ab \quad(1) \end{align}

Also, $S_{ABC}=S_{OAB}+S_{OBC}+S_{OCA}$:

\begin{align} S_{ABC}&= \frac{r}{2} (2a+2b+2r\sqrt3) \\ &= \sqrt{3}r^2 + (a+b)r \quad(2) \end{align}

$\frac{4}{3} (1)-(2):$

\begin{align} \frac{1}{3}S_{ABC} &= \frac{\sqrt3}{3}ab, \end{align}

thus, $S_{ABC}=ab\sqrt{3}$.

Edit:

This approach also provides a generalization for $\angle BAC =\alpha$. By replacing $r\sqrt{3}$ with $r/\tan(\alpha/2)$, similar steps result in

\begin{align} S_{ABC}(\alpha)&=\frac{ab}{\tan{\alpha/2}} = \frac{\sin{\alpha}}{1-\cos{\alpha}}ab. \end{align}

Particularly, $S(90°)=ab$.