\section*{Problem \(2-341042=340943\)}

On the side \(AB\) of the square \(ABCD\), a point \(X \neq A\) is chosen. The square is then divided into four regions by the segments \(AC\) and \(X D\).

Determine all possibilities for choosing \(X\) such that there exist natural numbers \(p, q\), and \(r\) for which the areas of these regions are in the ratio \(1: p: q: r\) in some order!
}

We set up a coordinate system in the plane of the square such that \(A\) is at the origin and \(C\) has coordinates \((1,1)\). Then \(B\) is at \((1,0)\) and \(D\) is at \((0,1)\). Furthermore, let \(0 < a \leq 1\) be a real number and the point \(X\) have coordinates \((a, 0)\). Then \(X\) lies on the segment \(A B\), but is different from \(A\).

The line \(X D\) can be described by the function equation \(f(x) = \frac{0-1}{a-0} \cdot x + 1 = -\frac{1}{a} \cdot x + 1\) and the line \(A C\) by \(g(x) = x\). For the intersection point \(S\) of both lines and the coordinates \(\left(x_{S}, y_{S}\right)\), it holds that \(f\left(x_{S}\right) = g\left(x_{S}\right)\), i.e., \(-\frac{1}{a} x_{S} + 1 = x_{S}\) or \(1 = \left(1 + \frac{1}{a}\right) \cdot x_{S} = \frac{a+1}{a} \cdot x_{S}\), thus \(y_{S} = x_{S} = \frac{a}{a+1}\).

Therefore, the height \(h_{a}\) from \(S\) to \(A B\) has the length \(h_{a} = y_{s} = \frac{a}{a+1}\) and the height \(h_{d}\) from \(s\) to \(A D\) has the length \(h_{d} = x_{s} = \frac{a}{a+1}\). For the triangle \(\triangle A S X\), the area is

\[
F_{1} = \frac{1}{2} \cdot |A X| \cdot h_{a} = \frac{1}{2} \cdot a \cdot \frac{a}{a+1} = \frac{a^{2}}{2(a+1)}
\]

and analogously for the triangle \(\triangle A S D\), the area is

\[
F_{4} = \frac{1}{2} \cdot |A D| \cdot h_{d} = \frac{1}{2} \cdot 1 \cdot \frac{a}{a+1} = \frac{a}{2(a+1)}
\]

Since the triangle \(\triangle A C D\) has an area of \(\frac{1}{2}\) and is divided by the segment \(D S\) into the two triangles \(\triangle A S D\) and \(\triangle D C S\), the area of the triangle \(\triangle D C S\) is

\[
F_{3} = \frac{1}{2} - F_{4} = \frac{(a+1) - a}{2(a+1)} = \frac{1}{2(a+1)}
\]

Analogously, the area of the quadrilateral \(X B C S\) is

\[
F_{2} = \frac{1}{2} - F_{1} = \frac{(a+1) - a^{2}}{2(a+1)} = \frac{a+1-a^{2}}{2(a+1)}
\]

Thus, the areas are in the ratio \(F_{1} : F_{2} : F_{3} : F_{4} = a^{2} : a+1-a^{2} : 1 : a\).

Since \(0 < a \leq 1\), it follows that \(1 \geq a \geq a^{2}\), and thus \(1 - a^{2} \geq 0\) and \(a + 1 - a^{2} \geq a\). Therefore, \(a^{2}\) is the smallest of these values. Simplifying the ratio by \(a^{2}\), we get

\[
F_{1} : F_{2} : F_{3} : F_{4} = 1 : \frac{1}{a} + \frac{1}{a^{2}} - 1 : \frac{1}{a^{2}} : \frac{1}{a}
\]

The values \(\frac{1}{a} + \frac{1}{a^{2}} - 1\), \(\frac{1}{a^{2}}\), and \(\frac{1}{a}\) must correspond to the natural numbers \(p\), \(q\), and \(r\) in some order. In particular, \(0 < \frac{1}{a} = r\) must be a natural number, so \(a = \frac{1}{r}\). Then we have

\[
F_{1} : F_{2} : F_{3} : F_{4} = 1 : r + r^{2} - 1 : r^{2} : r = 1 : p : q : r
\]
with \(p := r + r^{2} - 1\) and \(q := r^{2}\), which is of the desired form.

In summary, we obtain a ratio of the areas of the four sub-regions as required by the problem statement if and only if the segment \(A B\) is a positive integer multiple of the segment \(A X\), i.e., \(|A X| = \frac{1}{r} \cdot |A B|\) with a positive integer \(r\).