Given a triangle with three sides: $a, b$, and $c$. Determine the area of the triangle formed by the internal angle bisectors and their intersection points with the corresponding sides.
Let the intersection points of the angle bisectors and sides be $A_{1}, B_{1}, C_{1}$, and the area of triangle $ABC$ be $T$.

![](https://cdn.mathpix.com/cropped/2024_05_02_17883fd758989a31f520g-1.jpg?height=404&width=523&top_left_y=232&top_left_x=801)

From the diagram, it can be seen that the desired area is obtained by subtracting the areas of three smaller triangles from the area of the original triangle:

$$
t=T-t_{B_{1} A C_{1}}-t_{A_{1} B C_{1}}-t_{A_{1} C B_{1}}
$$

We determine the areas to be subtracted using two sides and the included angle:

$$
t=T-\frac{B_{1} A \cdot A C_{1} \sin \alpha}{2}-\frac{A_{1} B \cdot B C_{1} \sin \beta}{2}-\frac{A_{1} C \cdot C B_{1} \sin \gamma}{2}
$$

The distances and sines involved must be expressed in terms of the sides of the original triangle. According to the angle bisector division ratio:

$$
\begin{aligned}
B C_{1}=\frac{a c}{a+b}, & A C_{1}=\frac{b c}{a+b} \\
B A_{1}=\frac{a c}{b+c}, & C A_{1}=\frac{a b}{b+c} \\
A B_{1}=\frac{b c}{a+c}, & C B_{1}=\frac{a b}{a+c}
\end{aligned}
$$

From the previously used triangle area formula:

$$
\sin \alpha=\frac{2 T}{b c}, \quad \sin \beta=\frac{2 T}{a c}, \quad \sin \gamma=\frac{2 T}{a b}
$$

Substituting into equation (1), simplifying, and bringing to a common denominator:

$$
\begin{gathered}
t=T\left(1-\frac{b c}{(a+b)(a+c)}-\frac{a c}{(a+b)(b+c)}-\frac{a b}{(a+c)(b+c)}\right)= \\
=T \cdot \frac{2 a b c}{(a+b)(b+c)(c+a)}
\end{gathered}
$$

The area $T$ can be expressed using the sides with Heron's formula:

$$
t=\frac{2 a b c \sqrt{s(s-a)(s-b)(s-c)}}{(a+b)(b+c)(c+a)}
$$

Lajos Trón (Debrecen, Fazekas g. III. o. t.)