\section*{Problem 3 - 340923}

Starting from a square \(A B C D\), for any two positive integers \(x\) and \(y\), the following constructions can be performed:

The side \(A B\) is extended beyond \(B\) by a length of \(x \cdot A B\) to the point \(S\), the side \(B C\) is extended beyond \(C\) by a length of \(y \cdot B C\) to the point \(T\), the side \(C D\) is extended beyond \(D\) by a length of \(x \cdot C D\) to the point \(U\), and the side \(D A\) is extended beyond \(A\) by a length of \(y \cdot D A\) to the point \(V\). Determine all pairs \((x ; y)\) of positive integers for which the resulting quadrilateral \(S T U V\) has exactly 11 times the area of the square \(A B C D\)!
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The quadrilateral STUV can be decomposed into the square \(A B C D\) and the four right-angled triangles \(\triangle V S A, \triangle S T B, \triangle T U C\), and \(\triangle U V D\).

Let the side length of the square \(A B C D\) be \(|A B|=|B C|=|C D|=|D A|=1\), so that \(|A V|=|C T|=y\), \(|B S|=|D U|=x\), and thus \(|A S|=|C U|=1+x\) and \(|B T|=|D V|=1+y\).

With this, we can calculate the areas of the four triangles, the square, and thus also the quadrilateral STUV, which is now

\(F=1 \cdot 1+\frac{1}{2} \cdot y \cdot(1+x)+\frac{1}{2} \cdot x \cdot(1+y)+\frac{1}{2} \cdot y \cdot(1+x)+\frac{1}{2} \cdot x \cdot(1+y)=1+y+x y+x+x y=1+x+y+2 x y\)

According to the problem, \(F=11\), so the equation \(11=1+x+y+2 x y\) or \(10=x+y+2 x y\) needs to be solved in positive integers \(x\) and \(y\). Since this equation is symmetric in \(x\) and \(y\), we can assume w.l.o.g. \(x \leq y\), so that \(10=x+y+2 x y \geq 2 x+2 x^{2}=2 x(1+x)\) holds, thus \(x=1\), as for all \(x \geq 2\) this inequality is not satisfied.

Then \(10=1+y+2 y\), so \(y=3\), which ultimately leads to the two solutions \((x ; y) \in\{(1 ; 3),(3 ; 1)\}\), for which the area of the quadrilateral \(S T U V\) is exactly 11 times the area of the square \(A B C D\).