2. On the legs $AC$ and $BC$ of a given right triangle $ABC$, determine points $K$ and $L$ respectively, such that the sum

$$
|AK|^{2}+|KL|^{2}+|LB|^{2}
$$

attains its minimum value, and express this value in terms of $c=|AB|$.
2. Let us denote $x=|C L|, y=|C K|$, then $|B L|=a-x$, and $|A K| = b-y$, where $a, b$ are the lengths of the legs $B C, A C$ respectively.

![](https://cdn.mathpix.com/cropped/2024_04_17_f5bf5e3935da2f7cce69g-2.jpg?height=509&width=1020&top_left_y=1553&top_left_x=518)

Fig. 1

By applying the Pythagorean theorem in the right triangle $K L C$, we get $|K L|^{2} = x^{2} + y^{2}$, so the expression we are examining can be modified as follows:

$$
\begin{aligned}
|A K|^{2} + |K L|^{2} + |L B|^{2} & = (b-y)^{2} + x^{2} + y^{2} + (a-x)^{2} = \\
& = 2 x^{2} + 2 y^{2} - 2 a x - 2 b y + a^{2} + b^{2} = \\
& = 2\left(x - \frac{a}{2}\right)^{2} + 2\left(y - \frac{b}{2}\right)^{2} + \frac{a^{2} + b^{2}}{2} = \\
& = 2\left(x - \frac{a}{2}\right)^{2} + 2\left(y - \frac{b}{2}\right)^{2} + \frac{c^{2}}{2}.
\end{aligned}
$$

Due to the non-negativity of the squares, we see that the expression we are examining attains its minimum value, which is $\frac{1}{2} c^{2}$, precisely when $x = \frac{1}{2} a$ and simultaneously $y = \frac{1}{2} b$, i.e., precisely when points $K, L$ are the midpoints of the legs $A C, B C$ of the given right triangle $A B C$.

Conclusion. The smallest possible value of the examined sum is $\frac{1}{2} c^{2}$. This value is obtained precisely when points $K, L$ are the midpoints of the legs $A C, B C$ of the given right triangle.

For a complete solution, award 6 points, of which 2 points for expressing the examined sum using two appropriately chosen parameters. Award another 3 points for the subsequent modification into a form with squares, from which the minimum value and the corresponding positions of points $K$ and $L$ are evident. Award 1 point for correctly stating the positions of both sought points $K, L$ and expressing the minimum value of the examined expression in terms of $c$.