In the triangle $A_{1} A_{2} A_{3}$, the point $B_{i}$ is the quarter point closest to $A_{i}$ on the side $A_{i} A_{i+1}$, and the point $C_{i}$ is the quarter point closest to $A_{i+1}$ on the same side $\left(i=1,2,3, A_{4} \equiv A_{1}\right)$. What fraction of the area of triangle $A_{1} A_{2} A_{3}$ is the area of the common part of triangles $B_{1} B_{2} B_{3}$ and $C_{1} C_{2} C_{3}$?
The common part of the triangles $B_{1} B_{2} B_{3}$ and $C_{1} C_{2} C_{3}$ is a hexagon. Let's denote its vertices as $X, Y, Z, P, Q, R$ as shown in the diagram. We will determine the area of the hexagon by subtracting the areas of the triangles $Y Z C_{1}$, $P Q C_{2}$, and $R X C_{3}$ from the area of the triangle $C_{1} C_{2} C_{3}$. In the calculation, we will use the known fact that if two triangles share an angle, then the ratio of their areas is equal to the ratio of the products of the sides enclosing the common angle (due to $T=\frac{a b \sin \gamma}{2}$).

From the definition of the points $C_{i}$, it follows that $A_{1} C_{1}=\frac{3}{4} A_{1} A_{2}$ and $A_{1} C_{3}=\frac{1}{4} A_{1} A_{3}$, therefore

$$
T_{A_{1} C_{1} C_{3}}=\frac{3}{4} \cdot \frac{1}{4} \cdot T_{A_{1} A_{2} A_{3}}
$$

Similarly, we get that

$$
T_{A_{2} C_{2} C_{1}}=T_{A_{3} C_{3} C_{2}}=\frac{3}{16} T_{A_{1} A_{2} A_{3}}
$$

Therefore, $T_{C_{1} C_{2} C_{3}}=T_{A_{1} A_{2} A_{3}}-\left(T_{A_{1} C_{1} C_{3}}+T_{A_{2} C_{2} C_{1}}+T_{A_{3} C_{3} C_{2}}\right)=\frac{7}{16} T_{A_{1} A_{2} A_{3}}$.

Since $B_{2}$ and $C_{3}$ are quarter points, $B_{2} C_{3}$ is parallel to $A_{1} A_{2}$, and $B_{2} C_{3}=\frac{3}{4} A_{1} A_{2}$. The triangles $B_{2} C_{3} Y$ and $B_{1} C_{1} Y$ are similar because their angles are equal. Therefore,

$$
\frac{C_{1} Y}{Y C_{3}}=\frac{C_{1} B_{1}}{C_{3} B_{2}}=\frac{\frac{1}{2} A_{1} A_{2}}{\frac{3}{4} A_{1} A_{2}}=\frac{2}{3}
$$

so $C_{1} Y=\frac{2}{5} C_{1} C_{3}$. The segments $B_{1} C_{2}$ and $B_{2} C_{1}$ are parallel to $A_{1} A_{3}$, and $B_{1} C_{2}=\frac{3}{4} A_{1} A_{3}$, $B_{2} C_{1}=\frac{1}{4} A_{1} A_{3}$. The angles of the triangles $B_{1} C_{2} Z$ and $B_{2} C_{1} Z$ are equal, so the ratio of the corresponding sides in the two triangles is also equal:

$$
\frac{C_{1} Z}{Z C_{2}}=\frac{C_{1} B_{2}}{C_{2} B_{1}}=\frac{\frac{1}{4} A_{1} A_{3}}{\frac{3}{4} A_{1} A_{3}}=\frac{1}{3}
$$

so $C_{1} Z=\frac{1}{4} C_{1} C_{2}$.

Thus, $T_{C_{1} Y Z}=\frac{2}{5} \cdot \frac{1}{4} T_{C_{1} C_{3} C_{2}}$. Similarly, we get that $T_{C_{2} Q P}=T_{C_{3} X R}=\frac{1}{10} T_{C_{1} C_{2} C_{3}}$.

Therefore,

$$
T_{P Q R X Y Z}=T_{C_{1} C_{2} C_{3}}-\left(T_{C_{1} Y Z}+T_{C_{2} Q P}+T_{C_{3} X R}\right)=\frac{7}{10} T_{C_{1} C_{2} C_{3}}=\frac{7}{10} \cdot \frac{7}{16} T_{A_{1} A_{2} A_{3}}
$$

Thus, the desired ratio is $\frac{49}{160}$.

Based on the work of Tamás Bosznay (Bp., ELTE Apáczai Csere J. Gymnasium, 2nd year).