5. (5 points) The sides of a parallelogram are 2 and 3, and the angle between them is $\arccos \frac{5}{16}$. 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-04.jpg?height=605&width=768&top_left_y=1588&top_left_x=356)

Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, and $\angle BAC = \alpha = \arccos c$.

Obviously, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.

Let $|DN| = x$ and $|LC| = y$.

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, $\quad S_{\square KND} = S_{\square LNC}$. 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

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

follows.

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

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

For $a = 2$, $b = 3$, and $c = \frac{5}{16}$, the solutions to (2) are $x = -2$ and $x = \frac{4}{3}$. The first is discarded for geometric reasons. The corresponding value of $y = 2$.

Answer: $\frac{4}{3}$ and $\frac{2}{3}, 2$ and 1.