The area of a certain triangle is $T$. We divide its sides into $m>2, n>2$, and $p>2$ equal parts, respectively. By connecting the first and last division points on each side, we obtain a convex hexagon. What is the area $t$ of this hexagon?
I. Solution: The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_b5ca6f0e09b2fccc0a54g-1.jpg?height=361&width=531&top_left_y=210&top_left_x=797)

If the areas of the triangles $A B_{n-1} C_{1} \triangle, B C_{p-1} A_{1} \triangle$, and $C A_{m-1} B_{1} \triangle$ (shaded in the figure) are denoted by $t_{A}, t_{B}$, and $t_{C}$ respectively, then

$$
t_{A}=\frac{1}{2} \cdot \frac{b}{n} \cdot \frac{c}{p} \sin \alpha, \quad t_{B}=\frac{1}{2} \cdot \frac{a}{m} \cdot \frac{c}{p} \sin \beta, \quad \text { and } \quad t_{C}=\frac{1}{2} \cdot \frac{a}{m} \cdot \frac{b}{n} \sin \gamma
$$

Thus,

$$
t=T-\left(t_{A}+t_{B}+t_{C}\right)=T-\left(\frac{b c \sin \alpha}{2 n p}+\frac{a c \sin \beta}{2 m p}+\frac{a b \sin \gamma}{2 m n}\right)
$$

But

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

and thus

$$
t=T-\left(\frac{T}{n p}+\frac{T}{m p}+\frac{T}{m n}\right)=T\left(1-\frac{m+n+p}{m n p}\right)
$$

László Péntek (Kunszentmiklós, Damjanich g. IV. o. t.)

II. Solution: The area of $\triangle A B B_{n-1}$ is $\frac{T}{n}$, because the side $A B$ is common to $\triangle A B C$ and $\triangle A B B_{n-1}$, and the height corresponding to this common side is $\frac{1}{n}$ of the height of $\triangle A B C$. Similarly, the area of $\triangle A C_{1} B_{n-1}$ is $\frac{1}{p}$ of the area of $\triangle A B B_{n-1}$. Therefore, $t_{A}=\frac{T}{n p}$. It can be shown in the same way that $t_{B}=\frac{T}{m p}$, and $t_{C}=\frac{T}{m n}$, as in the first solution.

Gábor Forgó and Imre (Bp. V., Eötvös g. II. o. t.)

Note: Since we did not use the conditions $m>2, n>2, p>2$ anywhere, the formula for the area of the remaining figure is valid for $m \geq 2, n \geq 2, p \geq 2$ cases as well, although in this case the remaining figure is a hexagon, pentagon, quadrilateral, or triangle depending on whether the equality sign is valid for $0, 1, 2$, or $3$ cases.