4. 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 the three inner triangles are in the ratio $1: 4: 9$. Determine the range in which the ratio of the area of the largest of them 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_962c317ad84e313ab25eg-3.jpg?height=369&width=520&top_left_y=737&top_left_x=651)

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, hence,

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

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

Now, let $S_{1}: S_{2}: S_{3}=1: 4: 9$ (this does not match the diagram, but it is not important for further reasoning). Then

$$
S_{1}=\frac{1}{9} S_{3}, \quad S_{2}=\frac{4}{9} S_{3}
$$

and, further, $1=\frac{\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}}}{\sqrt{S}}=\frac{\frac{1}{3} \sqrt{S_{3}}+\frac{2}{3} \sqrt{S_{3}}+\sqrt{S_{3}}}{\sqrt{S}}=\frac{2 \sqrt{S_{3}}}{\sqrt{S}}$.

Therefore, $S_{3}=\frac{1}{4} S$.

Answer: The specified ratio can only be $\frac{1}{4}$.