2. In the plane, there is a right-angled triangle $ABC$, on whose hypotenuse $AB$ we consider an arbitrary point $K$. The circle constructed over the segment $CK$ as a diameter intersects the legs $BC$ and $CA$ at internal points, which we denote by $L$ and $M$ respectively. Determine for which point $K$ the quadrilateral $ABLM$ has the smallest possible area.
2. Since the angles $KLC$, $KMC$, and $LCM$ are right angles (Fig. 1), the quadrilateral $KLCM$ is a rectangle and triangles $AKM$ and $KBL$ are similar to triangle $ABC$. Let us denote

![](https://cdn.mathpix.com/cropped/2024_04_17_16872b628cb10877d512g-2.jpg?height=388&width=705&top_left_y=1459&top_left_x=710)

Fig. 1

as usual, $a = |BC|$, $b = |AC|$, $c = |AB|$, and let $|AK| = kc$, where $0 < k < 1$. If $|AK| > \frac{1}{2}|AB|$, we will use the similar trapezoid $BK'M'C$ (Fig. 2b) and conclude that

$$
S_{KLCM} = \frac{1}{2} S_{BK'M'C} < \frac{1}{2} S_{ABC}
$$

This proves the statement about the maximum area of $S_{KLCM}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_16872b628cb10877d512g-3.jpg?height=394&width=708&top_left_y=1091&top_left_x=297)
a)

![](https://cdn.mathpix.com/cropped/2024_04_17_16872b628cb10877d512g-3.jpg?height=405&width=697&top_left_y=1077&top_left_x=1045)
b)

Fig. 2

Answer: The quadrilateral $ABLM$ has the smallest possible area when point $K$ lies at the midpoint of the hypotenuse $AB$.

For a complete solution, award 6 points. If the solver only derives a functional expression for one of the areas $S_{ABLM}$, $S_{LMC}$, or $S_{KLCM}$, which would allow finding its extreme value, award 3 points. Award 1 point if the correct answer is only guessed or completely incorrectly justified.