## Task 6 - 110836

For a rectangle $A B C D$ with side lengths $\overline{A B}=a$ and $\overline{B C}=b, a>b$, a parallelogram $E F G H$ is inscribed such that the sides $D A$ and $B C$ of the rectangle are divided by the vertices of the parallelogram in the ratio $2: 3$ or $3: 2$, and the sides $A B$ and $C D$ in the ratio $3: 4$ or $4: 3$, with $E$ on $A B, F$ on $B C, G$ on $C D, H$ on $D A$.

Determine whether this is possible in one or more ways! In each of the possible cases, determine the ratio of the areas of the rectangle and the parallelogram to each other!
According to the problem, $E F G H$ is a parallelogram. This implies: $H E = G F, H G = E F, \angle H E G \cong \angle F G E$, and $\angle A E G \cong \angle C G E$ as alternate interior angles of intersected parallel lines.

Therefore, $\angle A E H \cong \angle C G F$. Thus, $\triangle A E H \cong \triangle C G F$ (ASA). Similarly, it can be shown that $\triangle B E F \cong \triangle D G H$.

Consequently, $A E = C G, B E = D G, A H = C F, D H = B F$, and there are exactly the following 4 possibilities to inscribe a parallelogram $E F G H$ in a rectangle $A B C D$ in the required manner (see figures).

Figure 1 can be obtained by reflecting Figure 3 and Figure 4 by reflecting Figure 2 across the perpendicular bisector of $A B$. Therefore, the following two cases need to be considered:

Figure 1

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0650.jpg?height=359&width=448&top_left_y=1057&top_left_x=541)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0650.jpg?height=355&width=465&top_left_y=1456&top_left_x=521)

Figure 2
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0650.jpg?height=748&width=450&top_left_y=1064&top_left_x=1202)

Case 1:

Let $D H: H A = B F: F C = 2: 3$ and $A E: E B = C G: G D = 3: 4$.

Then the area $A_{P}$ of the parallelogram $E F G H$ is equal to the difference between the area $A_{R}$ of the rectangle $A B C D$ and the sum of the areas of the triangles $\triangle A E H, \triangle E B F, \triangle F C G$, and $\triangle G D H$. Therefore,

\[
\begin{aligned}
A_{P} & = a b - \left( \frac{1}{2} \cdot \frac{3}{7} a \cdot \frac{3}{5} b + \frac{1}{2} \cdot \frac{4}{7} a \cdot \frac{2}{5} b + \frac{1}{2} \cdot \frac{3}{7} a \cdot \frac{3}{5} b + \frac{1}{2} \cdot \frac{4}{7} a \cdot \frac{2}{5} b \right) \\
& = a b - \left( \frac{9}{35} a b + \frac{8}{35} a b \right) = \frac{18}{35} a b
\end{aligned}
\]

This implies $A_{R}: A_{P} = 35: 18$.

Case 2:

Let $D H: H A = B F: F C = 2: 3$ and $A E: E B = C G: G D = 4: 3$.

Similarly to Case 1, we obtain

\[
A_{P} = a b - \left( \frac{1}{2} \cdot \frac{4}{7} a \cdot \frac{3}{5} b + \frac{1}{2} \cdot \frac{3}{7} a \cdot \frac{2}{5} b + \frac{1}{2} \cdot \frac{4}{7} a \cdot \frac{3}{5} b + \frac{1}{2} \cdot \frac{3}{7} a \cdot \frac{2}{5} b \right) = a b - \left( \frac{12}{35} a b + \frac{6}{35} a b \right) = \frac{17}{35} a b
\]

This implies $A_{R}: A_{P} = 35: 17$.

Solutions from the 3rd Round of 1971 adopted from [5]

### 5.14 12th Olympiad 1972

### 5.14.1 1st Round 1972, Class 8