A polygon $A_{1} A_{2} \ldots A_{n}$ is bounded by points that divide the sides $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$ in the same ratio $\lambda$. For which value of $\lambda$ is the area of the resulting polygon the smallest?
Let's determine the value of $\lambda$ at which the area of the triangle cut off by the line connecting two adjacent division points of sides $a$ and $b$ of a polygon is the largest!

![](https://cdn.mathpix.com/cropped/2024_05_02_f60a8259d33d10e0461eg-1.jpg?height=399&width=486&top_left_y=256&top_left_x=814)

The area of such a triangle is:

$$
t=\frac{(a-\lambda a) \cdot \lambda b \cdot \sin \gamma}{2}=(1-\lambda) \cdot \lambda \frac{a b \sin \gamma}{2}
$$

We need to examine the sum of such expressions.

Since the sum of the expressions $\frac{a b \cdot \sin \gamma}{2}$ is constant, it is sufficient to determine the value of $\lambda$ at which the expression $y=(1-\lambda) \lambda=-\lambda^{2}+\lambda$ is maximal. This function can be represented by a parabola with a vertical axis in the $\lambda, y$ coordinate system. This parabola intersects the $\lambda$ axis at $\lambda=1$ and $\lambda=0$, and its maximum value is at $\lambda=0.5$.

$$
\text { ifj. Csonka Pal (Budapest, III. o.) }
$$

Remark. $1^{\circ}$. The maximum value of $y=\frac{1}{4}-\left(1-\frac{1}{2}\right)^{2}$ is $\frac{1}{4}$, which is attained where the square term is 0, i.e., at $\lambda=\frac{1}{2}$.

$2^{\circ}$. The statement is not true for degenerate polygons, because in such cases, some of the triangles involved in the calculation represent a decrease, while others represent an increase in the area of the given polygon. Thus, with the increase of the product $\lambda(\lambda-1)$, the area of the new polygon may either increase or decrease. For example, if we place $45^{\circ}$ angles at the vertices of a square such that the diagonals are the angle bisectors, we get a star-shaped quadrilateral, for which the polygons created in the above manner all have the same area.