4. The lateral sides $AB$ and $CD$ of trapezoid $ABCD$ are equal to 2 and 3, respectively, and angles $A$ and $D$ are acute. The bisectors of angles $A$ and $B$ of the trapezoid intersect at point $M$, and the bisectors of angles $C$ and $D$ intersect at point $N$. The length of segment $MN$ is 4. Find the radius of the circle inscribed in triangle $ABD$, if the area of trapezoid $ABCD$ is $26 \sqrt{2} / 3$.
Solution: Draw the angle bisectors of angles $A M$ and $B M$ of angles $A$ and $B$ respectively. Since $\angle A + \angle B = 180^{\circ}$, triangle $A B M$ is a right triangle, $\angle A M B = 90^{\circ}$. Let $L$ be the intersection of $B M$ with the base $A D$. Then triangle $A B L$ is isosceles, as $A M$ is both a bisector and an altitude, $A B = A L = 2$. Let $P$ be the intersection of the bisector $CN$ with the base $A D$. Then, by similar reasoning, triangle $D C P$ is isosceles, $C D = D P = 3$. Point $L$ lies between points $A$ and $P$, since otherwise $A L + D P \leq 5$, while $A L + D P \geq 2 M N = 8$ (angles $A$ and $D$ are acute). Since $B M = M L$ and $C N = N P$, $M N$ is the midline of trapezoid $B L P C$.

The midline of trapezoid $A B C D$ is $\frac{A L}{2} + M N + \frac{D P}{2} = \frac{13}{2}$. Let the height of trapezoid $A B C D$ be $h$. By the given condition, we have $S_{A B C D} = \frac{13}{2} h = \frac{26}{3} \sqrt{2}, h = \frac{4 \sqrt{2}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-15.jpg?height=362&width=985&top_left_y=453&top_left_x=1027)

Let $B K$ and $C S$ be the heights of trapezoid $A B C D$. Then $A K = \sqrt{4 - h^2} = \frac{2}{3}, D S = \sqrt{9 - h^2} = \frac{7}{3}$. Let $B C = x, A D = A K + x + D S = 3 + x$. On the other hand, $A D = A L + 2 M N - x + D P = 13 - x$. Thus, $3 + x = 13 - x \Rightarrow x = 5$. Therefore, $B C = 5, A D = 3 + x = 8$.

$S_{A B D} = A D \cdot h / 2 = 16 \sqrt{2} / 3, \quad B D^2 = h^2 + K D^2 = (2 \sqrt{129} / 3)^2, \quad P_{A B D} = (30 + 2 \sqrt{129}) / 3, \quad S_{A B D} = P_{A B D} \cdot r / 2$ $r = 16 \sqrt{2} / (15 + \sqrt{129})$.

Answer: $16 \sqrt{2} / (15 + \sqrt{129})$.