3. Through a point lying inside a triangle, three lines parallel to its sides are drawn, which divide the triangle into six parts: three triangles and three quadrilaterals. The areas of all three inner triangles are equal. Determine the range in which the ratio of the area of each inner triangle to the area of the original triangle can lie.

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# Solution

Draw $K F\|A C, L Q\| B C, P E \| A B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_64e56d45f6f300cbe7d7g-2.jpg?height=506&width=721&top_left_y=674&top_left_x=645)

Triangles $K L O, O E F, P O Q$ are similar to each other and to triangle $A B C$ (their corresponding angles are equal as angles formed by parallel lines). Let their areas be $S_{1}, S_{2}, S_{3}$, and the area of the original triangle be $S$.

Since the areas of similar triangles are proportional to the squares of their (corresponding) sides, we have

$$
\frac{\sqrt{S_{1}}}{\sqrt{S}}=\frac{K O}{A C}, \quad \frac{\sqrt{S_{2}}}{\sqrt{S}}=\frac{O F}{A C}, \quad \frac{\sqrt{S_{3}}}{\sqrt{S}}=\frac{P Q}{A C}
$$

$A K O P$ and $Q O F C$ are parallelograms, therefore,

$$
K O+P Q+O F=A C .
$$

Thus,

$$
\frac{\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}}}{\sqrt{S}}=1
$$

According to the condition, $S_{1}=S_{2}=S_{3}$. Then

$$
1=\frac{\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}}}{\sqrt{S}}=\frac{3 \sqrt{S_{1}}}{\sqrt{S}}
$$

Therefore, $S_{1}=\frac{1}{9} S$.

Answer: The specified ratio can only be $\frac{1}{9}$.