A parallelogram's interior angle bisectors define a quadrilateral whose area is one-third of the parallelogram's area. Determine the ratio of the adjacent sides of the parallelogram.
Let the vertices of the parallelogram be denoted by $A, B, C, D$, the angles by $2 \alpha$ and $2 \beta$, the side lengths by $a$ and $b$, and the vertices of the quadrilateral formed by the angle bisectors by $E, F, G, H$. We can assume that $a > b$ (if two adjacent sides were equal, the angle bisectors would intersect at a single point rather than forming a quadrilateral).

We know that $2 \alpha + 2 \beta = 180^{\circ}$, because the sum of the adjacent angles in a parallelogram is $180^{\circ}$. In the triangle $A E D$,

$$
A E D \varangle = 180^{\circ} - (E A D \varangle + E D A \varangle) = 180^{\circ} - (\alpha + \beta) = 90^{\circ}
$$

Therefore, $H E F \varangle = A E D \varangle = 90^{\circ}$. Similarly, we can show that all the angles of the quadrilateral $EFGH$ are right angles, so the quadrilateral is a rectangle. $E$ is the intersection of the angle bisectors from $D$ and $A$, so it is equidistant from the lines $C D$, $D A$, and $A B$. Thus, $E$ lies on the midline of the parallelogram parallel to $A B$ and $C D$. Similarly, $G$ also lies on this midline, so $E G \parallel C D$; furthermore, $F H \parallel A D$.

Let $M$ be the intersection of the angle bisector from $A$ and the line $C D$. The triangle $A M D$ is isosceles because the angle bisector from $D$ is perpendicular to $A M$. Therefore, $D M = b$, and $C M = C D - M D = a - b$. Since $E M \parallel G C$ and $E G \parallel C M$, the quadrilateral $E G C M$ is a parallelogram, so $E G = C M = a - b$. We know that the diagonals of a rectangle are equal, so $F H = E G = a - b$.

The area of the rectangle can be calculated from the lengths of its diagonals and the angle between them:

$$
T_{E F G H} = \frac{1}{2} E G \cdot F H \cdot \sin 2 \alpha = \frac{1}{2} (a - b)^2 \sin 2 \alpha
$$

The area of the parallelogram is given by the known formula:

$$
T_{A B C D} = A B \cdot A D \cdot \sin 2 \alpha = a \cdot b \cdot \sin 2 \alpha
$$

According to our condition, $3 T_{E F G H} = T_{A B C D}$, so

$$
a \cdot b \cdot \sin 2 \alpha = 3 \cdot \frac{1}{2} (a - b)^2 \cdot \sin 2 \alpha
$$

Rearranging gives:

$$
2 a b = 3 a^2 - 6 a b + 3 b^2
$$

or

$$
3 \left(\frac{a}{b}\right)^2 - 8 \frac{a}{b} + 3 = 0
$$

Using the quadratic formula, we get:

$$
\frac{a}{b} = \frac{4 \pm \sqrt{7}}{3} \approx 2.215
$$

Since $a > b$, only $\frac{4 + \sqrt{7}}{3}$ is a valid solution (the other solution $\frac{4 - \sqrt{7}}{3}$ is its reciprocal).

Therefore, the longer side of the parallelogram is $\frac{4 + \sqrt{7}}{3}$ times the length of the shorter side.

Based on the work of Dávid Havas (Budapest, Móricz Zs. Gymnasium, 2nd year)

![](https://cdn.mathpix.com/cropped/2024_05_02_89b905a8f9b06eac906eg-1.jpg?height=431&width=745&top_left_y=1915&top_left_x=674)