6. Consider the plane tessellated with equilateral triangles, and let $F_{0}$ be any one of them. A sequence of increasingly larger figures is constructed as follows: $F_{1}$ is the polygon obtained by adding to $F_{0}$ the border formed by all the triangles of the tessellation that touch $F_{0}$ (on a side or at a vertex), $F_{2}$ is the polygon obtained by adding to $F_{1}$ the border formed by the triangles that touch $F_{1}$, and similarly, the subsequent ones are constructed up to $F_{10}$. How many triangles of the tessellation is the last polygon composed of?
(A) 541
(B) 661
(C) 691
(D) 721
(E) 841.
6. The answer is (B). For $n \geq 1$, the figure $F_{n}$ is a hexagon with three long sides and three short sides, alternating. Each of the long sides is composed of $n+1$ triangle sides, while each of the short sides is composed of $n$ triangle sides. Let $x_{n}$ be the number of triangles in the tiling that make up $F_{n}$. We calculate the analogous number $x_{n+1}$ by counting the triangles in the border that is added to $F_{n}$ to obtain $F_{n+1}$. This count provides $x_{n+1}-x_{n}$. We can build the border in three stages: in the first stage, we add one triangle for each side of the triangles that make up the sides of the hexagon $F_{n}$, so a total of $3(2 n+1)$ triangles; in the second stage, we add the triangles that are intercalated between those of the first stage, $n$ on each long side and $n-1$ on each short side, so a total of $3(2 n-1)$; finally, in the third stage, we add two triangles for each vertex of $F_{n}$, that is, 12 triangles. The total is

$$
x_{n+1}-x_{n}=3(2 n+1)+3(2 n-1)+12=12 n+12 \quad \text { i.e. } \quad x_{n+1}-x_{n}=12(n+1) \text {. }
$$

Now write $x_{1}=1+12$ (the result of a simple direct count) and the previous formula for $n=1, \ldots, 9$ (i.e., $x_{2}-x_{1}=12 \cdot 2, \ldots x_{10}-x_{9}=12 \cdot 10$), sum everything term by term, and simplify the first member. It remains written that $x_{10}=1+12+12 \cdot 2+12 \cdot 3+\cdots+12 \cdot 10$. Therefore,

$$
x_{10}=1+12(1+2+3+\cdots+10)=1+12 \cdot \frac{10 \cdot 11}{2}=661 .
$$

## Second Solution

The figure $F_{n}$ is an equilateral triangle of side $3 n+1$ from which 3 equilateral triangles of side $n$ have been removed, placed on the vertices. Therefore, $F_{n}$ is composed of $(3 n+1)^{2}-3 n^{2}=6 n^{2}+6 n+1$ small triangles of side 1.