Let's consider the midpoints of the three diagonals: these are either different or two of them are the same. (All three cannot be the same, since in that case the three diagonals would intersect at one point and would not form a triangle.) Below, we will examine both cases.

a) If the midpoints of the diagonals are different, then the intersection of two diagonals bisects one diagonal and divides the other into a quarter. Let one vertex of the triangle determined by the diagonals be denoted by $A$, and let the endpoints of the diagonal passing through $A$ be $A_{1}$ and $B_{2}$, where $A$ divides $A_{1} B$ into a quarter, and the midpoint of this segment is denoted by $B$ ( $A$ bisects the segment $A_{1} B$, 1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_e156f260da1d5d429bf5g-1.jpg?height=534&width=743&top_left_y=432&top_left_x=672)

1st figure

The endpoints of the other diagonal passing through $A$ are denoted by $A_{2}$ and $C_{1}$, $A$ bisects the segment $A_{2} C_{1}$, and the midpoint of the segment $A C_{1}$ is the third vertex of the triangle, denoted by $C$. The midpoint of the third diagonal can only be $C$, and its quartering point is $B$, with the endpoints being $B_{1}$ and $C_{2}$ ( $B$ bisects the segment $B_{1} C$).

Let's choose the area unit to be the area of the triangle $A B C$. The triangles $A B C$ and $A A_{1} C$ have a common vertex at $C$, and the opposite sides lie on the same line and are equal, so the area of $A A_{1} C$ is also 1. Similarly, the areas of the triangles $B B_{1} A$ and $C C_{1} B$ are also 1. The triangles $A C A_{1}$ and $A_{2} A A_{1}$ have a common vertex at $A_{1}$, and the opposite sides lie on the same line, with $A_{2} A = 2 A C$, so the area of $A_{2} A A_{1}$ is 2. Similarly, the areas of the triangles $B_{2} B B_{1}$ and $C_{2} C C_{1}$ are also 2.

The triangles $A_{1} B C$ and $A_{1} C C_{2}$ have a common vertex at $A_{1}$, and the opposite sides lie on the same line, with $C C_{2} = 2 B C$. The area of the triangle $A_{1} B C$ is 2, because the median $A C$ divides it into two triangles of unit area, so the area of $A_{1} C C_{2}$ is 4. Similarly, the areas of the triangles $B_{1} A A_{2}$ and $C_{1} B B_{2}$ are also 4.

Thus, the area of the original hexagon in this case is $1 + 3(1 + 2 + 4) = 22$ times the area of the triangle $A B C$.

b) In the second case, let $C$ be the vertex of the triangle formed by the diagonals that bisects both diagonals passing through it, and let the midpoint of the third diagonal be $B$ (2nd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_e156f260da1d5d429bf5g-1.jpg?height=564&width=767&top_left_y=1648&top_left_x=663)

2nd figure

Let the sides $A B, B C, C A$ be the diagonals $A_{1} B_{2}, B_{1} C_{2}, C_{1} A_{2}$, respectively, and let the area of the triangle $A B C$ be 1 again.

Similarly to the above, we find that the areas of the triangles $A A_{1} C, B B_{1} A, C C_{1} B$ are 1, 1, and 2, respectively; the areas of the triangles $A A_{1} A_{2}, B B_{1} B_{2}, C C_{1} C_{2}$ are 1, 2, and 4, respectively; and the areas of the triangles $A_{1} C C_{2}, B_{1} A A_{2}, C_{1} B B_{2}$ are 4, 2, and 6, respectively. Thus, the area of the hexagon is 24 times the area of the triangle $A B C$.

Therefore, we have found that the answer to the given question is not unique; the area of the hexagon is either 22 times or 24 times the area of the triangle formed by the diagonals, depending on whether each vertex of the triangle is the midpoint of a diagonal or not.
We drew the diagonals connecting opposite vertices in a convex hexagon. At their intersection, a triangle was formed. On each diagonal, we find that one of the triangle's vertices on it bisects the diagonal, while the other bisects one half of the diagonal. How many times larger is the area of the hexagon compared to the area of the triangle?