Khachaturyan A.V.

Side $AB$ of triangle $ABC$ is divided into $n$ equal parts (division points $B_{0}=A, B_{1}, B_{2}, B_{n}=B$), and side $AC$ of this triangle is divided into $n+1$ equal parts (division points $C_{0}=A, C_{1}, C_{2}, \ldots, C_{n+1}=C$). The triangles $C_{i} B_{i} C_{i+1}$ are shaded. What fraction of the area of the triangle is shaded?
Let's show that the shaded part constitutes exactly half the area of the entire triangle. For this, we drop perpendiculars from points $B_{1}, \ldots, B_{n}$ to the side $A C$. These perpendiculars are the heights of triangles $C_{i} B_{i} C_{i+1}$ with equal bases, and, as follows from similarity considerations, $h_{i}=i h_{1}$. From this, it follows that the areas of the shaded triangles will be related by the same ratio: $S_{i}=i S_{1}$. (The figure shows the case $n=4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-08.jpg?height=1009&width=1140&top_left_y=-1&top_left_x=459)

Dropping perpendiculars from points $C_{1}, \ldots, C_{n}$ to the side $A C$ and reasoning similarly, we obtain the same ratio for the areas of the unshaded triangles. It remains to note that the area of the first shaded triangle is equal to the area of the first unshaded triangle (their bases are equal, and the height $h_{1}$ is common).