Given the sides of the convex quadrilateral $A B C D$ (in order), and the angle $\varphi$ between the lines containing the sides $A D$ and $C B$, within which the quadrilateral lies. We draw the (internal) angle bisectors, the bisectors at $A$ and $B$ intersect at $M$, and the bisectors at $C$ and $D$ intersect at $N$. What is the distance $M N$? Solve the problem generally, and then with the following data: $A B=70, B C=100, C D=35, D A=75$ units, and the projection of $B C$ onto the line $A D$ is 96 units.
1. Let the vertex of the angle $\varphi$ be denoted by $P$, and considering the relationship $AB = 2 \cdot CD$, we choose the labeling such that $P D P A$ is the starting point. According to this, the triangle $P D C$ is inside the triangle $P A B$. Considering the angle bisectors, $M$ is the center of the inscribed circle of the triangle $P A B$, and $N$ is the center of the circle externally tangent to the side $DC$ of the triangle $P D C$. Therefore, both are on the angle bisector of the common vertex $P$ of the triangles, and $M N = |P M - P N|$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e8a1dd5fdce754dcb8e9g-1.jpg?height=531&width=1303&top_left_y=388&top_left_x=411)

Let the projections of $M$ and $N$ onto the line $PA$ be denoted by $M'$ and $N'$, respectively, and the points of tangency of the mentioned circles by $M'$ and $N'$, respectively. Then,

$$
M N = \frac{M' N'}{\cos \varphi / 2} = \frac{|P M' - P N'|}{\cos \varphi / 2}
$$

and here, as is well-known for the segments extending to the points of tangency,

$$
\begin{gathered}
P M' = \frac{1}{2}(P A + P B - A B) = \frac{1}{2}(P D + D A + P C + C B - A B) \\
P N' = \frac{1}{2}(P D + P C + C D)
\end{gathered}
$$

therefore,

$$
M N = \frac{1}{2 \cos \frac{\varphi}{2}} \cdot |(D A + C B) - (A B - C D)|
$$

A special case of our result is when $M$ and $N$ coincide, $M N = 0$, which is precisely the known necessary condition for the existence of an inscribed circle in the quadrilateral $A B C D$ (which, as many know, is also a sufficient condition).

2. In the numerical example, first consider the case where $\varphi$ is an acute angle, i.e.,

$$
\cos \varphi = +0.96; \text{ then } \cos \frac{\varphi}{2} = +\sqrt{\frac{1 + \cos \varphi}{2}} = +\frac{1.4}{\sqrt{2}}
$$

and

$$
M N = \frac{\sqrt{2}}{2.8} |75 + 100 - 70 - 35| = 25 \sqrt{2} \approx 35 \text{ units. }
$$

If, however, $\varphi$ is taken as an obtuse angle:

$$
\cos \varphi = -0.96, \quad \text{ then } \quad \cos \frac{\varphi}{2} = \frac{0.2}{\sqrt{2}}
$$

from which $M N = 175 \sqrt{2} \approx 247.5$ results, and this is impossible, since the diagonal $A C$ of our quadrilateral is smaller than the smaller of the sums $A D + D C = 110$ and $A B + B C = 170$, i.e., smaller than 110, and the diagonal $B D$ is smaller than the smaller of the sums $D A + A B = 145$ and $B C + C D = 135$, i.e., smaller than 135, so our quadrilateral cannot contain a segment longer than the larger of 110 and 135.

Remark. With a slight additional calculation, it also follows that our quadrilateral is a trapezoid, and the side $C D$ is the midline of the triangle $P A B$ parallel to $B A$. We can arrive at this by following the procedure that would be used to construct the quadrilateral $A B C D$ from our data and the condition $0^{\circ} < \varphi < 90^{\circ}$ (by translating $D A$ to the position $C A'$, forming the triangle $B C A'$, forming the triangle $B A' A$ from its sides, and translating $A' C$ to $A D$; in this case, $B A' + A' A = B A$ results). The triangle $P A B$ is similar to the triangle examined in problem 2028 (KÖMAL 53. vol. (1976) p. 123).