\section*{Problem 1 - 211021}

In a right-angled triangle \(A B C\), the height \(D C\) perpendicular to the hypotenuse \(A B\) is exactly \(\frac{2}{5}\) times as long as the hypotenuse \(A B\). For the foot of the perpendicular \(D\), it holds that \(A D < D B\).

In what ratio \(A D: D B\) does it divide the hypotenuse?
}

For the lengths \(A B=c, D C=h, A D=q, D B=p\) it holds that

\[
h=\frac{2}{5} c \quad(1) \quad \text { and } \quad p+q=c \quad(2) \quad \text { and } \quad q<p
\]

Furthermore, according to the height theorem \(p g=h^{2} \quad\) (4). Substituting \(h\) from (1) into (4) yields

\[
p q=\frac{4}{25} c^{2}
\]

Substituting \(q\) from (2) into (5) yields

\[
\begin{aligned}
p(c-p) & =\frac{4}{25} c^{2} \\
p_{1,2} & =\frac{1}{2} c \pm \sqrt{\frac{1}{4} c^{2}-\frac{4}{25} c^{2}}=\frac{1}{2} c \pm \sqrt{\frac{1}{100} c^{2}(25-16)} \\
& =\frac{c}{10}(5 \pm 3)
\end{aligned}
\]

thus either \(p=\frac{4}{5} c\) and then further \(q=\frac{1}{5} c\) according to (2), or \(p=\frac{1}{5} c\) and then further \(q=\frac{4}{5} c\) according to (2). Of these two possibilities, only the first remains due to (3). Therefore, the sought ratio \(A D: D B=q: p=1: 4\).