A unit area parallelogram has side lengths $a$ and $b$, where $a<b<2a$. What is the area of the quadrilateral formed by the internal angle bisectors of the parallelogram?
Solution. Let the vertices of the parallelogram be denoted by $A, B, C, D$ - choosing the labeling such that $A B = C D = a$ and $A D = B C = b$ hold. Let the intersection points of the adjacent angle bisectors be $P, Q, R$, and $S$, and the intersection points of the angle bisectors with the boundary of the parallelogram, different from the vertices, be $K, L, M$, and $N$. From the condition $a < b < 2a$, it follows that the order of the points is as shown in the diagram.

Since the sum of the adjacent angles in a parallelogram is $180^{\circ}$, the angle bisectors from adjacent vertices are perpendicular to each other. Therefore, the quadrilateral $P Q R S$ is a rectangle because all its angles are right angles. From the perpendicularity of the angle bisectors $B M$ and $A L$, it follows that the triangle $A B L$ is isosceles, with $a = A B = B L$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b86ea2e293dcd0314019g-1.jpg?height=325&width=593&top_left_y=472&top_left_x=755)

Similarly, we find that the triangle $C D N$ is also isosceles. Therefore, the points $P$ and $R$ lie on the midline between the lines $A D$ and $B C$, and similarly, we can see that the line $Q S$ is the midline between the lines $A B$ and $C D$. If the intersection points of the line $P R$ with the other two sides of the parallelogram are denoted by $E$ and $F$ as shown in the diagram, then $E P$ is the midline of the triangle $A B L$, and $R F$ is the midline of the triangle $C D N$. Thus,

$$
E P = \frac{B L}{2} = \frac{a}{2} = \frac{N D}{2} = R F
$$

from which we obtain that the length of the diagonals of the rectangle $P Q R S$ is

$$
P R = E F - (E P + R F) = b - a
$$

The angle between the diagonals of the rectangle $P Q R S$ is the same as the angle $\alpha$ of the parallelogram $A B C \varangle = \alpha$. Therefore, using the known area formulas, the area of the parallelogram and the rectangle are:

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

and

$$
T_{P Q R S} = \frac{P R \cdot Q S \cdot \sin \alpha}{2} = \frac{(b-a)^{2} \sin \alpha}{2}
$$

The area of the parallelogram is one unit, so the desired area is

$$
T_{P Q R S} = \frac{T_{P Q R S}}{T_{A B C D}} = \frac{(b-a)^{2}}{2 a b}
$$