XI OM - I - Task 3

Each side of a triangle with a given area $ S $ is divided into three equal parts, and the points of division are connected by segments, skipping one point to form two triangles. Calculate the area of the hexagon that is the common part of these triangles.
The solution to the problem can be easily read from Fig. 1. The area of the hexagon $123456$ equals $\frac{2}{3}S$, as this hexagon is obtained from the triangle $ABC$ by cutting off three equal corner triangles with an area of $\frac{1}{9}S$. The sides of triangles $135$ and $246$ bisect each other into equal segments because, for example, in the trapezoid $1234$, the diagonals $13$ and $24$ divide into parts in the same ratio as the bases $23$ and $14$, i.e., in the ratio $1 \colon 2$, so, for instance, segment $II\ 2$ equals one-third of segment $42$. The hexagon $123456$ consists of $18$ triangles of equal area, and the shaded hexagon $I\ II\ III\ IV\ V\ VI$ consists of $6$ triangles of the same area. Therefore, the area of the shaded hexagon is $\frac{2}{9} S$.