The lengths of the sides of a trapezoid are $5, 15, 15, 20 \mathrm{~cm}$ in some order. We drew the angle bisectors and marked the intersection points of adjacent angle bisector pairs. What quadrilateral is defined by these four points? What fraction of the trapezoid's area is the area of this quadrilateral?
The given segments can be assembled into a trapezoid in two different ways. The 5 cm segment will be the base in both cases, while the other base will be either 20 cm or $15 \mathrm{~cm}$. Indeed, the trapezoid can be divided into a parallelogram and a triangle, where the latter has two sides equal to the legs of the trapezoid, and the third side is the difference between the bases of the trapezoid. Therefore, the difference between the two bases must be greater than the difference between the two legs. This is not satisfied if the $5 \mathrm{~cm}$ segment is used as a leg.

![](https://cdn.mathpix.com/cropped/2024_05_02_2b2c8d6403460aa70c2fg-1.jpg?height=483&width=639&top_left_y=401&top_left_x=472)
B)

![](https://cdn.mathpix.com/cropped/2024_05_02_2b2c8d6403460aa70c2fg-1.jpg?height=477&width=480&top_left_y=396&top_left_x=1186)

In both cases, in the trapezoid $A B C D=T$, let $A B \| C D$, and the quadrilateral $P S R Q=N$ is such that the angle bisectors of $\angle D A B$ and $\angle A B C$ intersect at point $P$, and similarly for the other angles. The midpoints of the legs $A D$ and $B C$ are $E$ and $F$, respectively. The points $Q$ and $S$ lie on the midline $E F$, because by the property of the angle bisectors, for example, $Q$ is equidistant from $A B$ and $C D$, and thus half the distance of the trapezoid's height $m$. Furthermore, $\angle P Q R$ and $\angle P S R$ are right angles, because, for example, the angle bisector of $\angle A D C$ is parallel to the external angle bisector at $A$, and thus perpendicular to $A Q$. Therefore, the quadrilateral $N$ is a cyclic quadrilateral. - Moreover, $Q$ lies on the Thales circle above the diameter $A D$, $Q E = A D / 2$, and similarly $S F = B C / 2$. Based on this, $Q S$ can be calculated, equal to the absolute value of the difference between $(A D + B C) / 2$ and $E F = (A B + C D) / 2$.

If $Q$ and $S$ do not coincide, $P$ and $R$ lie on different sides of the line $Q S$, which is the midline $E F$. Therefore, $Q S$ divides $N$ into triangles $Q S P$ and $Q S R$, whose heights $m_{P}$ and $m_{R}$ perpendicular to $Q S$ can be expressed based on the fact that, for example, triangles $A B P$ and $Q S P$ are similar:

$$
m_{P} = \frac{m}{2} \cdot \frac{Q S}{A B \pm Q S}, \text{ similarly } m_{R} = \frac{m}{2} \cdot \frac{Q S}{Q S \pm C D}
$$

(the lower signs in the double signs are valid if the points appear in the order $E, Q, S, F$ on the midline). Thus, the area of $N$ is

$$
\frac{1}{2} \cdot Q S \left(m_{P} + m_{R}\right) = \frac{m \cdot Q S^{2}}{4} \left(\frac{1}{A B \pm Q S} + \frac{1}{C D \pm Q S}\right)
$$

On the other hand, the area of $T$ is $m \cdot E F = m (A B + C D) / 2$, so the desired ratio is

$$
k = \frac{Q S^{2}}{2 (A B + C D)} \left(\frac{1}{A B \pm Q S} + \frac{1}{C D \pm Q S}\right)
$$

In the case I. where $A B = 20, C D = 5, A D = B C = 15$ (part $\alpha$ of the figure), $E F = 12.5, E Q = F S = 7.5$, $Q S = E Q + S F - E F = 2.5$, and thus

$$
k_{\mathrm{I}} = \frac{2.5^{2}}{50} \left(\frac{1}{22.5} + \frac{1}{7.5}\right) = \frac{1}{45}
$$

In this case, $T$, and thus $N$, is symmetric with respect to the perpendicular bisector of $A B$, and $N$ is a kite.

In the case II. where $A B = A D = 15, C D = 5, B C = 20$ (part $\beta$ of the figure), $E F = 10 = F S$, $S$ coincides with $E$, so $Q S = E Q = 7.5$ and $k_{\mathrm{II}} = 7 / 40$.

Based on the work of Tamás Bulkai (Győr, Bencés g. II. o. t.)

Note. In the first case, a simpler calculation is possible by relying on the symmetry and the resulting $60^{\circ}$ angles, but the above solution handles both cases simultaneously.