4. Can the interior of a convex dodecagon be divided into ten triangles by drawing some of its diagonals, such that no two have a common interior, and four of these dividing triangles are enclosed triangles, and the following condition holds: If a real number different from zero is written in each triangle and the number in each enclosed triangle is the product of the numbers written in the triangles constructed over its three sides, then the product of all the numbers written in the dodecagon is constant?

![](https://cdn.mathpix.com/cropped/2024_06_05_edcf2076eef478aa4fe7g-09.jpg?height=394&width=586&top_left_y=1339&top_left_x=899)

(We say a triangle is enclosed if a new triangle is constructed over each of its sides)
Solution. The division of the dodecagon is unique and is given in the diagram. Let $a_{1}, a_{2}, \ldots, a_{10}$ be non-zero real numbers written in the divided triangles. From the condition of the problem, we have

$$
a_{7}=a_{1} \cdot a_{6} \cdot a_{10}, a_{8}=a_{2} \cdot a_{3} \cdot a_{10}, a_{9}=a_{4} \cdot a_{5} \cdot a_{10} \text { and } a_{10}=a_{7} \cdot a_{8} \cdot a_{9}
$$

From this, we get,

$$
\begin{aligned}
a_{10} & =a_{7} \cdot a_{8} \cdot a_{9} \\
& =\left(a_{1} \cdot a_{6} \cdot a_{10}\right) \cdot\left(a_{2} \cdot a_{3} \cdot a_{10}\right) \cdot\left(a_{4} \cdot a_{5} \cdot a_{10}\right) \\
& =a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} \cdot a_{5} \cdot a_{6} \cdot a_{10}^{3}
\end{aligned}
$$

i.e.

$$
a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} \cdot a_{5} \cdot a_{6} \cdot a_{10}^{2}=1
$$

Using (1) for the product of all the written numbers, we get,

$$
\begin{aligned}
& a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} \cdot a_{5} \cdot a_{6} \cdot a_{7} \cdot a_{8} \cdot a_{9} \cdot a_{10}= \\
& =a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} \cdot a_{5} \cdot a_{6} \cdot\left(a_{1} \cdot a_{6} \cdot a_{10}\right) \cdot\left(a_{2} \cdot a_{3} \cdot a_{10}\right) \cdot\left(a_{4} \cdot a_{5} \cdot a_{10}\right) \cdot a_{10} \\
& =a_{1}^{2} \cdot a_{2}^{2} \cdot a_{3}^{2} \cdot a_{4}^{2} \cdot a_{5}^{2} \cdot a_{6}^{2} \cdot a_{10}^{4} \\
& =\left(a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} \cdot a_{5} \cdot a_{6} \cdot a_{10}^{2}\right)^{2}=1
\end{aligned}
$$

Thus, the product of all the written numbers is constant and equals 1.