We divide the sides of a quadrilateral and a triangle in the same arbitrary ratio. We continue this procedure infinitely on the segments determined by the division points. How does the sum of the areas of the infinite number of quadrilaterals compare to the sum of the areas of the infinite number of triangles, if the area of the original triangle is $1 \frac{1}{2}$ times larger than the area of the original quadrilateral?
Let us mark those points which divide the sides of the triangle $ABC$ in the ratio $p: q$ as $M, N, P$; the area of triangle $APN$ as $T_{A}$, that of $BMP$ as $T_{B}$, that of $CMN$ as $T_{C}$, that of $ABC$ as $T$, and that of $MNP$ as $T_{1}$; then

$$
AP=AB \frac{p}{p+q}, AN=AC \frac{q}{p+q}
$$

thus

$$
T_{A}=\frac{AB \cdot AC \cdot \sin \alpha}{2} \cdot \frac{pq}{(p+q)^{2}}=T \cdot \frac{pq}{(p+q)^{2}}
$$

We see that the area of triangle $APN$ is independent of the sides of triangle $ABC$, and the same expression gives the areas of triangles $T_{B}$ and $T_{C}$. Therefore,

$$
T_{1}=T\left(1-\frac{3pq}{(p+q)^{2}}\right)
$$

Similarly,

$$
T_{2}=T_{1}\left(1-\frac{3pq}{(p+q)^{2}}\right)=T\left(1-\frac{3pq}{(p+q)^{2}}\right)^{2}, \text{ etc. }
$$

The sum of the areas of the triangles

The areas of the triangles thus form an infinite geometric series with a common ratio of $1-\frac{3pq}{(p+q)^{2}}$; thus, the sum of the areas is

$$
S=T \frac{(p+q)^{2}}{3pq}
$$

Let us mark those points which divide the sides of the quadrilateral $ABCD$ in the ratio $p: q$ as $M, N, P, Q$; the areas of the triangles at the vertices of the quadrilateral as $t_{a}, t_{b}, t_{c}, t_{d}$; the area of the original quadrilateral as $t$, and that of $MNPQ$ as $t_{1}$. Let us draw the diagonals of the quadrilateral and denote the area of triangle $ABC$ as $t_{A}$, that of $ABD$ as $t_{B}$, etc. Then, according to the previous,

$$
t_{a}=t_{A} \cdot \frac{pq}{(p+q)^{2}}, t_{b}=t_{B} \cdot \frac{pq}{(p+q)^{2}}, \text{ etc. }
$$

Thus,

$$
t_{1}=t-2t \frac{pq}{(p+q)^{2}}=t\left(1-\frac{2pq}{(p+q)^{2}}\right)
$$

The areas of the quadrilaterals again form an infinite geometric series with a common ratio of $1-\frac{2pq}{(p+q)^{2}}$, and the sum of the areas is:

$$
s=t \frac{(p+q)^{2}}{2pq}
$$

From (1) and (2),

$$
S: s=\frac{T}{3}: \frac{t}{2}
$$

But

$$
T: t=3: 2
$$

Thus,

$$
S: s=1: 1
$$

(Béla Krausz.)

The problem was also solved by: J. Filkorn, E. Freibauer, Ö. Kornis, Gy. Krisztián, Gy. Lukhaub, G. Sasvári, J. Weisz.