, 2011

a.
Let $A_sB_sC_s$ be the given triangles, with $s \in \{1, 2, \dots, 2012\}$ and let $\{XYZ\}$ stand for the half-plane with the boundary line $XY$ and interior point $Z$. Each of the triangles is the intersection of the three half-planes $\{A_sB_sC_s\}$, $\{B_sC_sA_s\}$ and $\{C_sA_sB_s\}$, hence the intersection of all the triangles is the intersection of all such half-planes. Since the half-planes $\{A_sB_sC_s\}$, $1 \le s \le 2012$ differ only by translation, their intersection is the half-plane $\{A_iB_iC_i\}$ for a certain $i$. Similarly, the intersection of the half-planes $\{B_sC_sA_s\}$ is the half-plane $\{B_jC_jA_j\}$ for some $j$ and the intersection of the half-planes $\{C_sA_sB_s\}$ is the half-plane $\{C_kA_kB_k\}$ for some $k$. Hence the intersection of all the half-planes is the intersection of the three half-planes $\{A_iB_iC_i\}$, $\{B_jC_jA_j\}$ and $\{C_kA_kB_k\}$, which is a triangle $ABC$ which is similar to the original triangles, since its sides are parallel to the sides of the original triangles. Let the coefficient of similarity be $\lambda$. Clearly $0 < \lambda \le 1$; we wish to show that $\lambda \ge \frac{1}{3}$.

Let $v$ be the altitude from the vertex $C_i$ to the side $A_iB_i$ in triangle $A_iB_iC_i$. Since the line $A_iB_i$ coincides with the line $AB$, the distance of the centroid $G_i$ of triangle $A_iB_iC_i$ from the line $AB$ equals $\frac{1}{3}v$. Since every triangle contains the centroid of all the triangles, triangle $ABC$ contains all of them as well, and in particular, $ABC$ contains the centroid $G_i$, which means that the distance between $C$ and the line $AB$ is at least $\frac{1}{3}v$. Comparing the altitudes from the vertices $C_i$ and $C$ in the similar triangles $A_iB_iC_i$ and $ABC$, we thus obtain immediately that $\lambda \ge \frac{1}{3}$, and so the area of $ABC$ is at least $\frac{1}{9}$.

b.
Let $A_sB_sC_s$ be any one of the given triangles. The triangle $ABC$ is contained in it, and so the vertex $A_s$ lies in the half-plane $\{BCA\}$ at a distance at most $v$ from the line $BC$, where $v$ is the length of the altitude from the vertex $A_s$ in triangle $A_sB_sC_s$. On the other hand, the distance of the side $B_sC_s$ from $A$ is also at most $v$. Since the triangle $ABC$ contains centroids of all the given triangles, the distance between the side $B_sC_s$ and $BC$ cannot exceed $\frac{1}{3}v$. The distance of the vertex $A$ from the side $BC$ equals $\lambda v$, so altogether the distance between the parallel lines $BC$ and $B_sC_s$ is at most $\min\{1/3, 1-\lambda\} \cdot v$. All the given triangles thus lie in the belt bounded by the two parallel lines $a_0$ and $a_1$ parallel to $BC$ whose distance from $BC$ is $v$ and $\min\{1/3, 1-\lambda\} \cdot v$, respectively. A similar assertion clearly holds for the other two sides $AC$ and $AB$, so the union of the $2012$ triangles lie in the intersection of the three belts.

We now distinguish two cases, depending on the value of $\min\{1/3, 1-\lambda\}$.

If $\frac{1}{3} \le \lambda < \frac{2}{3}$, then the intersection of the belts is a hexagon which is obtained from the triangle $T$ determined by the triple of lines $(a_1, b_1, c_1)$ upon removing the three small triangles $T_a, T_b$ and $T_c$ determined by the triples of lines $(a_0, b_1, c_1)$, $(a_1, b_0, c_1)$ and $(a_1, b_1, c_0)$ respectively. From the way the lines $a_0, b_0, c_0, a_1, b_1, c_1$ are defined, it follows that the $T$ is similar to the given triangles with coefficient of similarity equal to $1 + \lambda$ and the triangles $T_a, T_b, T_c$ are also similar to the given triangles with coefficient of similarity $\lambda - \frac{1}{3}$.

The area of the hexagon is therefore given by (recalling that $\lambda < \frac{2}{3}$)
$$S = (1 + \lambda)^2 - 3 \left( \lambda - \frac{1}{3} \right)^2 = \frac{8}{3} - 2(\lambda - 1)^2 < \frac{8}{3} - \frac{2}{9} = \frac{22}{9}.$$

If $\frac{2}{3} \le \lambda \le 1$, then the intersection of the three belts is again a hexagon, and the corresponding triangle $T$ is similar to the given triangles with coefficient of similarity $3 - 2\lambda$, and the triangles $T_a, T_b, T_c$ have coefficients of similarity $1 - \lambda$. The area of the hexagon in this case is given by
$$(3 - 2\lambda)^2 - 3(1 - \lambda)^2 = (\lambda - 3)^2 - 3 \le \frac{49}{9} - 3 = \frac{22}{9},$$
with equality if $\lambda = \frac{2}{3}$.

However, since we only start with a finite number of triangles, the side $a_0$ of the hexagon can only contain a finite number of vertices of the given triangles (and cannot coincide with one of these triangles' sides), and so the entire area of the hexagon cannot be covered with only finitely many of these triangles, hence the inequality is strict.