A plane intersects each side of the quadrilateral $ABCD$ at an internal point. Following a circumferential direction, we write down the ratio in which each intersection point divides the side it lies on. What will be the product of the four ratios?

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I. solution. Let the plane mentioned in the problem be denoted by $S$. The points of intersection of the sides of the skew quadrilateral with $S$ are $E, F, G, H$ as shown in the figure. Project the vertices of the skew quadrilateral perpendicularly onto the plane $S$, and let the projection points be $A^{\prime}, B^{\prime}$, $C^{\prime}, D^{\prime}$. According to the conditions of the problem, no point coincides with its projection, so for example, $A A^{\prime} \| B B^{\prime}$, and by the theorem of parallel segments, $\frac{A E}{E B}=\frac{A A^{\prime}}{B B^{\prime}}$. Similarly, we obtain $\frac{B F}{F C}=\frac{B B^{\prime}}{C C^{\prime}}, \frac{C G}{G D}=\frac{C C^{\prime}}{D D^{\prime}}, \frac{D H}{H A}=\frac{D D^{\prime}}{A A^{\prime}}$. The product of these four ratios is:

$$
\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=\frac{A A^{\prime}}{B B^{\prime}} \cdot \frac{B B^{\prime}}{C C^{\prime}} \cdot \frac{C C^{\prime}}{D D^{\prime}} \cdot \frac{D D^{\prime}}{A A^{\prime}}=1
$$

We get the same result with the other traversal direction.

# Formanek Csaba (Szeged, Radnóti M. Gymn., III. o.t.)

II. solution. Let us set up a spatial coordinate system such that the equation of the plane $S$ is $z=0$. The coordinates of the vertices of the skew quadrilateral are denoted as follows: $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right), C\left(x_{3}, y_{3}, z_{3}\right), D\left(x_{4}, y_{4}, z_{4}\right)$. Let the division ratio on the side $A B$ be $a_{1}$, i.e., $\frac{A E}{E B}=\frac{a_{1}}{1}$, and similarly for the other ratios $a_{2}, a_{3}, a_{4}$. Since, for example, the point $E$ lies on the plane $z=0$, its $z$-coordinate is 0, and thus by the formula for the coordinates of a division point, $0=\frac{z_{1}+a_{1} z_{2}}{1+a_{1}}$, from which $a_{1}=-\frac{z_{1}}{z_{2}}$. Similarly, we get $a_{2}=-\frac{z_{2}}{z_{3}}, a_{3}=-\frac{z_{3}}{z_{4}}, a_{4}=-\frac{z_{4}}{z_{1}}$. Clearly, none of the $z_{i}$ are zero, since the vertices of the skew quadrilateral do not lie on the plane $S$. The product of the four ratios is:

$$
a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4}=-\frac{z_{1}}{z_{2}} \cdot\left(-\frac{z_{2}}{z_{3}}\right) \cdot\left(-\frac{z_{3}}{z_{4}}\right) \cdot\left(-\frac{z_{4}}{z_{1}}\right)=1
$$

Szita István (Körmend, Kölcsei F. Gymn., III. o.t.)

III. solution. Consider the diagram prepared for the first solution. The points $E, H, B$, and $D$ lie in a plane, so the lines $E H$ and $B D$ either intersect or are parallel. Considering the first case, let the intersection point be $K$. The points $F$, $G$, and $K$ lie in both the plane $S$ and the plane determined by the points $B, C, D$. These two planes are different according to our conditions, so the points $F, G$, and $K$ lie on a line, which is the intersection of these two planes. We then apply Menelaus' theorem to the triangle $A B D$ and the line $E H$, and to the triangle $B C D$ and the line $F G$:

$$
\frac{A E}{E B} \cdot \frac{B K}{K D} \cdot \frac{D H}{H A}=-1, \quad \text { and } \quad \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D K}{K B}=-1
$$

From the product of these two equations:

$$
\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1, \quad \text { since } \quad \frac{B K}{K D} \cdot \frac{D K}{K B} \quad \text { is clearly } 1
$$

The product of the four ratios is therefore 1.

There is still the case where $E H \| B D$. In this case, $S \| B D$, which implies $B D \| G F$. Apply the theorem of parallel segments to the triangles $A B D$ and $B C D$:

$$
\frac{A E}{E B}=\frac{H A}{D H}, \quad \text { and } \quad \frac{B F}{F C}=\frac{G D}{C G}
$$

where the segments involved are not necessarily directed segments. From these two equalities:

$$
\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1
$$

Szabó Jácint (Győr, Révai M. Gymn., III. o.t.)

Note: It is easy to see that the essence of the first two solutions is the same, only the tools applied are different.

![](https://cdn.mathpix.com/cropped/2024_05_02_c7256e40e741423021f4g-1.jpg?height=561&width=610&top_left_y=2132&top_left_x=733)