Bulgarian Mathematical Olympiad

Solution:

a) At each step the triangle from which we choose a point is divided into three new triangles, i.e., the number of triangles increases by two. Hence at the $n$-th step we have $2n+1$ triangles.

b) We shall prove by induction on $n$ that removing any triangle, there is a pairing of the remaining triangles such that the triangles in any pair have a common side.

The statement is trivial for $n=1$. Assume that it is true for $n=k$. We shall prove it for $n=k+1$. Let a point $O$ in $\triangle MNP$ be added at the $(k+1)$-th step. Remove any $\triangle XYZ$. If it is some of $\triangle OMN$, $\triangle OMP$ or $\triangle ONP$ (we may assume $\triangle OMN$), we consider the configuration obtained by removing $\triangle MNP$ at the $k$-th step.

By the induction hypothesis, the remaining triangles can be paired in such a way that the triangles in any pair have a common side. Adding the pair $(\triangle OMP, \triangle ONP)$ we obtain the desired pairing.

If $\triangle XYZ$ does not coincide with $\triangle OMN$, $\triangle OMP$ and $\triangle ONP$, we consider the configuration obtained by removing $\triangle XYZ$ at the $k$-th step. Assume that $\triangle MNP$ is paired with $\triangle QMN$. Then replacing the pair $(\triangle MNP, \triangle QMN)$ with the pairs $(\triangle OMN, \triangle QMN)$ and $(\triangle OMP, \triangle OPN)$ completes the induction.

Since the area of $\triangle ABC$ equals $1$, then the minimal area of a triangle does not exceed $\frac{1}{2n+1}$. Remove a triangle of minimal area. Then as we proved above, the remaining triangles can be paired such that the triangles in any pair have a common side. Since the number of the pairs is equal to $n$, there is a pair of total area at least $\frac{1-\frac{1}{2n+1}}{n} = \frac{2}{2n+1}$.