Consider the trapezoid $ABCD$, where $BC \| AD$. Draw a line $XY$ in the trapezoid, which is parallel to $BC$ and $AD$ and has the property that the trapezoids $AXYD$ and $XBCY$ are isoperimetric (having equal perimeters). In what ratio does this line divide the area of the trapezoid?
According to the result of problem 109,

$$
\frac{p}{q}=\frac{A D^{2}-X Y^{2}}{X Y^{2}-B C^{2}}
$$

thus, only the length of the line $X Y$ needs to be determined.

Let $A B=a, \quad B C=b, \quad C D=c, \quad D A=d$ and $X Y=z$. Draw the line $B Y^{\prime} D^{\prime}$ from point $B$, parallel to $C D$, intersecting lines $X Y$ and $A D$ at points $Y^{\prime}$ and $D^{\prime}$, respectively.

According to the problem,

$$
X B+B Y^{\prime}=Y^{\prime} D^{\prime}+D^{\prime} A+A X
$$

Choose a point $B^{\prime}$ on the line $A D^{\prime}$ such that

$$
X B=B^{\prime} A+A X
$$

and

$$
Y^{\prime} B=B^{\prime} D^{\prime}+D^{\prime} Y^{\prime}
$$

But

$$
\begin{gathered}
X B+A X=a=B^{\prime} A+2 A X \\
Y^{\prime} B+D^{\prime} Y^{\prime}=c=B^{\prime} D^{\prime}+2 D^{\prime} Y^{\prime}
\end{gathered}
$$

and

$$
a: c=A X: D^{\prime} Y^{\prime}
$$

thus

$$
A B^{\prime}: B^{\prime} D^{\prime}=a: c
$$

which means $B$ is the intersection of the angle bisector of $\angle A B D^{\prime}$ with the line $A D^{\prime}$. Therefore,

$$
\begin{aligned}
A B^{\prime} & =\frac{a A D^{\prime}}{a+c} \\
B^{\prime} D^{\prime} & =\frac{c A D^{\prime}}{a+c}
\end{aligned}
$$

and

$$
\begin{gathered}
2 A X=a-\frac{a A D^{\prime}}{a+c}=\frac{a\left(a+c-A D^{\prime}\right)}{a+c} \\
2 D^{\prime} Y^{\prime}=c-\frac{c A D^{\prime}}{a+c}=\frac{c\left(a+c-A D^{\prime}\right)}{a+c} \\
B X=a-\frac{1}{2} \frac{a\left(a+c-A D^{\prime}\right)}{a+c}=\frac{a\left(a+c+A D^{\prime}\right)}{2(a+c)} \\
B Y^{\prime}=c-\frac{1}{2} \frac{c\left(a+c-A D^{\prime}\right)}{a+c}=\frac{c\left(a+c+A D^{\prime}\right)}{2(a+c)}
\end{gathered}
$$

Thus,

$$
B X=\frac{a(a+c+d-b)}{2(a+c)}
$$

But

$$
\begin{gathered}
X Y^{\prime}: A D^{\prime}=(z-b):(d-b)=\frac{a(a+c+d-b)}{2(a+c)}: a \\
(z-b):(d-b)=(a+c+d-b): 2(a+c) \\
z=\frac{d-b}{2(a+c)}(a+c+d-b)+b
\end{gathered}
$$

Finally, the sought value is:

$$
z=\frac{(d-b)^{2}+(a+c)(d+b)}{2(a+c)}=\frac{d+b}{2}+\frac{(d-b)^{2}}{2(a+c)}
$$

The problem was solved by: Friedmann Bernát, fg. VI. S.-A.-Ujhely; Meitner Elemér, fr. VIII. Budapest; Visnya Aladár, fr. VII. Pécs; Weisz Lipót, fr. VI. Pécs.