Question 86
One chooses a point in the interior of $\triangle ABC$ with area $1$ and connects it with the vertices of the triangle. Then one chooses a point in the interior of one of the three new triangles and connects it with its vertices, etc. At any step one chooses a point in the interior of one of the triangles obtained before and connects it with the vertices of this triangle. Prove that after the $n$-th step:
a) $\triangle ABC$ is divided into $2n+1$ triangles;
б) there are two triangles with common side whose combined area is not less than $\frac{2}{2n+1}$.
a) $\triangle ABC$ is divided into $2n+1$ triangles;
б) there are two triangles with common side whose combined area is not less than $\frac{2}{2n+1}$.