Let's calculate the area of the triangle if two sides $a$ and $b$ are given, as well as the length of the angle bisector of the angle formed by these two sides: $f_{c}$.
I. solution: Draw a line through vertex $A$ parallel to $f_{c}$, intersecting the extension of side $BC = a$ at a point $D$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_324add5b111524a1ee6fg-1.jpg?height=455&width=683&top_left_y=290&top_left_x=710)

From the angles, it is clear that $CD = CA = b$. The ratio of the areas of $\triangle ABC$ and $\triangle ACD$ is $a : b$, because the height from vertex $A$ is common to both triangles. If we denote the areas by $t$ and $t_{1}$, respectively, then $t : t_{1} = a : b$, so the area of $\triangle ABC$ we are looking for is

$$
t = \frac{a}{b} t_{1}
$$

However, $t_{1}$ can be easily expressed in terms of $f_{c}$. Let $d$ be the side $AD$ in $\triangle ACD$, and $m$ be the height perpendicular to $AD$, then

$$
d : f_{c} = (a + b) : a, \quad \text{or} \quad d = \frac{(a + b) f_{c}}{a}
$$

Furthermore, by the Pythagorean theorem,

$$
m = \sqrt{b^{2} - \left(\frac{d}{2}\right)^{2}} = \sqrt{b^{2} - \frac{(a + b)^{2} f_{c}^{2}}{4 a^{2}}} = \frac{1}{2 a} \sqrt{4 a^{2} b^{2} - (a + b)^{2} f_{c}^{2}}
$$

and thus

$$
\begin{aligned}
t = \frac{a}{b} \cdot t_{1} = & \frac{a}{b} \cdot \frac{d m}{2} = \frac{a}{b} \cdot \frac{(a + b) f_{c}}{4 a^{2}} \sqrt{4 a^{2} b^{2} - (a + b)^{2} f_{c}^{2}} = \\
& = \frac{(a + b) f_{c}}{4 a b} \sqrt{4 a^{2} b^{2} - (a + b)^{2} f_{c}^{2}}
\end{aligned}
$$

Gábor Halász (Bp. II., Rákóczi g. II. o. t.)

II. solution: The angle bisector $f_{c}$ divides $\triangle ABC$ into two triangles, whose area $t$ is equal to the sum of the areas of these two triangles, that is,

$$
\begin{gathered}
2 t = a b \sin \gamma = a f_{c} \sin \frac{\gamma}{2} + b f_{c} \sin \frac{\gamma}{2} = (a + b) f_{c} \sin \frac{\gamma}{2} \\
\text{Substituting } 2 \sin \frac{\gamma}{2} \cos \frac{\gamma}{2} \text{ for } \sin \gamma \text{ and dividing by } \sin \frac{\gamma}{2} \text{ (since } \sin \frac{\gamma}{2} \neq 0 \text{)}
\end{gathered}
$$

we get

$$
2 a b \cos \frac{\gamma}{2} = (a + b) f_{c}
$$

from which

$$
\cos \frac{\gamma}{2} = \frac{(a + b) f_{c}}{2 a b}
$$

Therefore, from (1) the desired area $t$ (considering (2)) is

$$
\begin{gathered}
t = \frac{(a + b) f_{c}}{2} \sqrt{1 - \cos^{2} \frac{\gamma}{2}} = \frac{(a + b) f_{c}}{2} \sqrt{1 - \frac{(a + b)^{2} f_{c}^{2}}{4 a^{2} b^{2}}} = \\
= \frac{(a + b) f_{c}}{4 a b} \sqrt{4 a^{2} b^{2} - (a + b)^{2} f_{c}^{2}}
\end{gathered}
$$

Iren Heinemann (Pécs, Leöwey K. lg. I. o. t.)