5. (5 points) The sides of a parallelogram are 5 and 13, and the angle between them is $\arccos \frac{6}{13}$. Two mutually perpendicular lines divide this parallelogram into four equal-area quadrilaterals. Find the lengths of the segments into which these lines divide the sides of the parallelogram.

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# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_96b013e051a92576724eg-09.jpg?height=605&width=768&top_left_y=1645&top_left_x=358)

Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, $\angle BAC = \alpha = \arccos c$. Clearly, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.

Denote $|DN| = x$, $|LC| = y$.

By contradiction, it can be shown that the mutually perpendicular lines $MN$ and $KL$, specified in the problem, must pass through the intersection point of the diagonals of parallelogram $ABCD$. It is evident that $\square KON = \square LON$.

From this and the equality of areas, $S_{\text{KOND}} = S_{\text{LONC}}$, it follows that the areas coincide, $S_{\square KND} = S_{\square LNC}$. This fact implies the validity of the equality

$$
\frac{1}{2}(a-x) y \cdot \sin \alpha = \frac{1}{2}(b-y) x \cdot \sin (\pi - \alpha),
$$

from which the relation follows

$$
y = \frac{b}{a} x
$$

Based on the equality $|KN| = |LN|$ and the cosine theorem, we can write the equation

$$
(a-x)^2 + y^2 - 2(a-x) y \cdot \cos \alpha = (b-y)^2 + x^2 - 2(b-y) x \cdot \cos (\pi - \alpha)
$$

from which and from (1), in turn, follows the equation

$$
\frac{4 b c}{a} x^2 - \left(2a - \frac{2b^2}{a} + 4bc\right) x + a^2 - b^2 = 0
$$

For $a = 5$, $b = 13$, and $c = \frac{6}{13}$, the solutions to (2) are $x = -10$ and $x = 3$. The first is discarded for geometric reasons. The corresponding value of $y = \frac{39}{5}$.

Answer: 3 and $2, \frac{39}{5}$ and $\frac{26}{5}$.