## Task 3 - 150923

Given the side length $a$ of a square $A B C D$ and a length $m$ such that $m \leq a$. Let $M$ be the point on the side $C D$ for which $M D=m$ holds.

We are looking for a point $N$ on the side $A D$ such that the area of triangle $N M D$ is to the area of square $A B C D$ as $1: 7$.

Determine all values of $m$ for which such a point $N$ exists on $A D$, and for each of these values, the length of the segment $D N$.
The area of the triangle $\triangle N M D$ is

$$
\frac{1}{2}|M D| \cdot|D N|=\frac{m}{2} \cdot|D N|
$$

For this to take the value of $\frac{1}{7} a^{2}$ as required by the problem, $|D N|=\frac{2 a}{7 m} \cdot a$ must hold. For the point $D$ to lie on the side $A D$ of the square (and not just on the line through this side), the factor $\frac{2 a}{7 m}$ must lie between 0 and 1.

Since $a>0$ and $m \geq 0$, it is always $\frac{2 a}{7 m}>0$. Finally, the inequality $\frac{2 a}{7 m} \leq 1$ is equivalent to $m \geq \frac{2}{7} a$, so that for exactly these values of $m$ there exists a corresponding point $N$ on the segment $A D$.